Physics of the Universe

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Give the SI base unit of mass.
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   The kilogram

2. Express the newton in base units.
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   From \( F = ma \). \( \text{kg}\cdot\text{m}/\text{s}^2 \)

3. Convert 2.5 km to meters.
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   2500 m

4. How many significant figures are in 0.00340?
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   Leading zeros do not count; the trailing zero does, because it was written. Three

5. Convert 72 km/h to m/s.
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   Divide by 3.6. 20 m/s

6. A student writes \( a = \dfrac{v}{t^2} \) for acceleration. Check it by units and say whether it can be right.
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   Units of the right side: \( \dfrac{\text{m}/\text{s}}{\text{s}^2} = \dfrac{\text{m}}{\text{s}^3} \). Units of acceleration: \( \text{m}/\text{s}^2 \). These do not match, so the formula is wrong whatever the situation. The correct relation is \( a = \dfrac{\Delta v}{t} \), whose units are \( \dfrac{\text{m}/\text{s}}{\text{s}} = \text{m}/\text{s}^2 \) ✓ Cannot be right; the units give m/s³

7. Express the joule in base units, then use that to express the watt.
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   \( 1\ \text{J} = 1\ \text{N}\cdot\text{m} = \text{kg}\cdot\text{m}^2/\text{s}^2 \). A watt is a joule per second: \( 1\ \text{W} = \text{kg}\cdot\text{m}^2/\text{s}^3 \). Check the pattern. Each division by a second adds one to the power of s in the denominator, which is what "per second" means dimensionally. \( \text{J} = \text{kg}\cdot\text{m}^2/\text{s}^2 \), \( \text{W} = \text{kg}\cdot\text{m}^2/\text{s}^3 \)

8. A rectangular plate measures 12.4 cm by 3.0 cm. Give its area with the correct number of significant figures.
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   \( 12.4 \times 3.0 = 37.2 \) cm². The inputs have three and two significant figures, so the answer keeps two. \( 37.2 \to 37 \) cm². Why the rule exists. The 3.0 cm measurement means the true length is somewhere between about 2.95 and 3.05 cm, which puts the area between roughly 36.6 and 37.8 cm². Writing 37.2 claims a precision the second measurement never had. 37 cm²

9. Explain why dimensional analysis can prove a formula wrong but cannot prove it right.
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   Because it tests only the units, and units are blind to pure numbers. What it catches. If the two sides of an equation have different units, they cannot be equal for any values at all, so the formula is wrong universally rather than in some cases. That is a strong conclusion from a cheap check. What it misses. Any dimensionless factor. For a body falling from rest through height \( h \), the true speed is \( \sqrt{2gh} \), but \( \sqrt{gh} \), \( \sqrt{17gh} \) and \( 0.4\sqrt{gh} \) all pass the unit check identically, because 2, 17 and 0.4 carry no units. It also misses which of several correct-unit quantities you meant. Torque and energy both come out as \( \text{kg}\cdot\text{m}^2/\text{s}^2 \), so a unit check cannot tell them apart. How to use it anyway. Run the check first, because it is fast and rules things out; then get the numerical factor from a derivation. The two together are reliable, and neither is on its own. Matching units are necessary but not sufficient

10. A tank holds 4.0 cubic meters of water. Water has a density of 1000 kg/m³. Find the mass, then find the weight in newtons, and state which quantity a bathroom scale measures.
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    Find the mass. Density is mass per volume, so mass is density times volume: \( m = 1000\ \dfrac{\text{kg}}{\text{m}^3} \times 4.0\ \text{m}^3 = 4000\ \text{kg} \). The cubic meters cancel, leaving kilograms, which confirms the expression was set up the right way round. Find the weight. Weight is the gravitational force on the mass: \( W = mg = 4000\ \text{kg} \times 9.8\ \text{m}/\text{s}^2 = 39200\ \text{N} \). To two significant figures, matching the 4.0, that is \( 3.9 \times 10^4\ \text{N} \). Check the units. \( \text{kg} \times \text{m}/\text{s}^2 = \text{kg}\cdot\text{m}/\text{s}^2 = \text{N} \) ✓ Which one a scale measures. A bathroom scale measures force, because it works by compressing a spring or a strain gauge, and what compresses it is the weight. It then divides by a stored value of \( g \) and displays a mass, which is what you actually wanted. Why that distinction is not pedantry. The same scale on the Moon would read about one sixth as much, because the weight is smaller there, while the mass has not changed at all. The scale is not broken; it is measuring the quantity it always measured and converting with the wrong constant. Mass 4000 kg, weight about \( 3.9 \times 10^4 \) N; a scale measures force

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Is displacement a vector or a scalar?
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   A vector

2. Is distance a vector or a scalar?
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   A scalar

3. An object moves from \( x = 3\ \text{m} \) to \( x = 11\ \text{m} \). Find its displacement.
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   \( 11 - 3 \). \( +8\ \text{m} \)

4. An object moves from \( x = 5\ \text{m} \) to \( x = -2\ \text{m} \). Find its displacement.
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   \( -2 - 5 \). \( -7\ \text{m} \)

5. Can distance ever be negative?
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   It is a path length. No

6. A cyclist rides 3.0 km north, then 4.0 km south. Find the distance and the displacement.
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   Take north as positive. Distance: \( 3.0 + 4.0 = 7.0\ \text{km} \). Displacement: \( +3.0 + (-4.0) = -1.0\ \text{km} \), that is 1.0 km south. The sign is the answer's direction, so reporting "1.0 km" alone would be incomplete. Distance 7.0 km, displacement 1.0 km south

7. A ball is dropped from a height of 2.0 m, bounces, and is caught at 1.2 m on the way up. Find the distance traveled and the displacement.
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   Take upward positive with the floor at zero. Start at \( +2.0\ \text{m} \), reach \( 0 \), return to \( +1.2\ \text{m} \). Distance: \( 2.0 + 1.2 = 3.2\ \text{m} \). Displacement: \( 1.2 - 2.0 = -0.8\ \text{m} \), that is 0.8 m downward. The ball ends below where it began even though it was last moving upward. Displacement describes the net result, not the final direction of travel. Distance 3.2 m, displacement 0.8 m downward

8. A hiker walks 6.0 km east and then 8.0 km north. Find the distance and the magnitude of the displacement.
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   Distance is the path length: \( 6.0 + 8.0 = 14.0\ \text{km} \). The displacement is the straight line from start to finish, and the two legs are perpendicular, so use the Pythagorean theorem: \( \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = \sqrt{100} = 10.0\ \text{km} \). Direction: \( \tan^{-1}\left( \dfrac{8.0}{6.0} \right) = 53^\circ \) north of east. This is why displacement needs vector addition rather than ordinary addition once the motion leaves a single line. Lesson 1.7 builds on it. Distance 14.0 km, displacement 10.0 km at 53° north of east

9. Explain why the choice of positive direction cannot change the physical answer.
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   Because the choice is a description of reality, not a property of it, and every quantity in the problem is relabeled together. What changes. The sign attached to each displacement, velocity and acceleration flips. What does not. Which physical direction each sign now names. A displacement of \( +60\ \text{m} \) with east positive and one of \( -60\ \text{m} \) with west positive describe the identical journey, because in the second frame negative means east. Why the flip is consistent. Every vector in the problem is multiplied by \( -1 \) at once, and the equations relating them are unchanged by that, so the relationships survive intact. Where it goes wrong. Not in the choosing but in the forgetting. Taking down as positive for the fall and up as positive for the rebound, inside one problem, produces a calculation that describes no physical situation. Write the choice down at the top of the work and leave it there. A practical tip. Choose the direction of initial motion as positive. It tends to keep the most numbers positive and reduces the chance of a sign slip. The choice relabels every vector together, so the described physics is unchanged

10. A delivery van leaves the depot, drives 12 km north, 5.0 km east, then 12 km south, and stops. Find the distance, the displacement, and explain what the driver's odometer would read against what a straight-line tracker would report.
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    Set up a frame. Origin at the depot, north positive on one axis, east positive on the other. Track the two axes separately, which is the method lesson 1.7 formalizes. North-south axis. \( +12 \) then \( -12 \), so the net is \( 0\ \text{km} \). East-west axis. \( +5.0\ \text{km} \), unchanged by the other two legs. Displacement. The van ends 5.0 km east of the depot and level with it north to south, so the displacement is 5.0 km east. Distance. Add the three path lengths: \( 12 + 5.0 + 12 = 29\ \text{km} \). The odometer. It counts wheel rotations, so it measures path length and reads 29 km. It has no way to know the van turned around. A straight-line tracker. A device comparing the start and end positions reports 5.0 km east, and knows nothing about the route. Neither is wrong. They answer different questions, and both are needed for different purposes: the fuel used depends on the 29 km, while the time saved by a better route depends on how much of that 29 km was wasted relative to the 5.0 km that had to be covered. The general point. A quantity in physics is defined by the question it answers. Asking "which is the real distance?" is the wrong question; asking "which one does this situation depend on?" is the right one. Distance 29 km, displacement 5.0 km east

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write the definition of average velocity.
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   Displacement divided by time

2. What does the slope of a position-time graph give?
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   Velocity

3. An object moves \( +40\ \text{m} \) in 8.0 s. Find its average velocity.
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   \( +5.0\ \text{m/s} \)

4. What does a horizontal line on a position-time graph mean?
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   Zero slope. The object is at rest

5. A car covers 150 m in 10 s. Find its average speed.
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   15 m/s

6. A runner completes one lap of a 400 m track in 50 s. Find the average speed and the average velocity.
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   Average speed: \( \dfrac{400}{50} = 8.0\ \text{m/s} \). Average velocity: the start and finish are the same point, so the displacement is zero and \( \bar{v} = \dfrac{0}{50} = 0 \). Both are correct. She was certainly moving, but she got nowhere, and velocity measures getting somewhere. Speed 8.0 m/s, velocity zero

7. A car travels 60 km at 60 km/h, then 60 km at 30 km/h. Find the average speed for the whole trip.
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   Do not average the speeds. Find the times first. Leg 1: \( \dfrac{60\ \text{km}}{60\ \text{km/h}} = 1.0\ \text{h} \). Leg 2: \( \dfrac{60\ \text{km}}{30\ \text{km/h}} = 2.0\ \text{h} \). Total distance 120 km, total time 3.0 h. \( \dfrac{120}{3.0} = 40\ \text{km/h} \). Not 45. Twice as much time was spent at the slower speed, so the average is pulled toward it. 40 km/h

8. A position-time graph is a curve that gets steeper as time increases. Describe the motion.
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   Slope is velocity, and the slope is increasing, so the velocity is increasing. The object is moving in the positive direction and speeding up. Contrast the other curvatures. A curve flattening toward horizontal means slowing down. A curve bending below the horizontal means it reversed direction at the turning point. Moving in the positive direction and accelerating

9. Explain why average speed is not the magnitude of average velocity.
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   Because they are built from different numerators: one from distance, the other from displacement, and those agree only when the motion never reverses. The definitions. Average speed is \( \dfrac{\text{distance}}{t} \); average velocity is \( \dfrac{\text{displacement}}{t} \). The denominators match, so any difference comes entirely from the top. Why the tops differ. Lesson 1.2 established that distance is at least the magnitude of displacement, and strictly greater whenever the motion reverses. So average speed is at least the magnitude of the average velocity, and usually more. The extreme case. A closed loop has zero displacement, so the average velocity is zero while the average speed is not. The two cannot be the same quantity if one can be zero while the other is 8 m/s. When they do agree. Motion in a straight line without reversing. On a one-way trip the two are numerically equal, which is why the distinction is easy to miss until a problem involves a return leg. For instants the situation is different. Instantaneous speed *is* the magnitude of instantaneous velocity, because over a vanishingly small interval there is no opportunity to reverse. The mismatch is a property of averages only. Distance exceeds displacement whenever motion reverses, so the two averages come apart

10. A train accelerates from rest, travels at a steady speed, then brakes to a stop, covering 2400 m in 120 s altogether. Sketch the shape of its position-time graph, and explain why its average velocity of 20 m/s is not its speed during the middle phase.
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    Compute the average velocity. \( \bar{v} = \dfrac{2400\ \text{m}}{120\ \text{s}} = 20\ \text{m/s} \), in the direction of travel. Describe the graph in three parts. Phase 1, accelerating. The graph starts at the origin with zero slope, because the train starts at rest, and curves upward with increasing steepness. Phase 2, steady speed. The graph becomes a straight line, because constant velocity means constant slope. This is the steepest part. Phase 3, braking. The graph curves again, flattening until its slope reaches zero at the final position, where the train stops. The overall shape is a stretched letter S: shallow, steep, shallow, rising throughout because the train never reverses. Why 20 m/s is not the cruising speed. The average is the total displacement over the total time, and part of that time was spent moving more slowly than the cruise, during acceleration and braking. To average 20 m/s while spending time below it, the train must exceed 20 m/s during the middle phase. Put a number on it. Suppose the acceleration and braking take 20 s each and the cruise takes 80 s. Using the result from lesson 1.4 that a constant acceleration from rest covers half the distance a constant top speed would, accelerating to speed \( v \) over 20 s covers \( 10v \) meters, and the same for braking, while cruising covers \( 80v \). Total: \( 10v + 80v + 10v = 100v = 2400 \), so \( v = 24\ \text{m/s} \). Check. The cruise speed 24 m/s is above the 20 m/s average, as argued ✓ And the average sits between zero and 24, as any average must. The general reading. An average velocity is a single number standing in for a whole interval. It is exactly right about the net displacement and says nothing reliable about any particular instant. Whenever a problem gives you an average and asks about a moment, that gap is usually where the question lives. An S-shaped rising curve; the 20 m/s average is below the cruising speed of about 24 m/s

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Give the SI unit of acceleration.
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   \( \text{m}/\text{s}^2 \)

2. What does the slope of a velocity-time graph give?
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   Acceleration

3. What does the area under a velocity-time graph give?
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   Displacement

4. A car goes from 8.0 m/s to 20 m/s in 4.0 s. Find its acceleration.
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   \( \dfrac{20-8.0}{4.0} \). \( +3.0\ \text{m}/\text{s}^2 \)

5. An object has positive velocity and negative acceleration. Is it speeding up or slowing down?
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   Opposite signs. Slowing down

6. A velocity-time graph is a horizontal line at 15 m/s for 8.0 s. Find the acceleration and the displacement.
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   Slope is zero, so \( a = 0 \). Area is a rectangle: \( 15 \times 8.0 = 120\ \text{m} \). Constant velocity is a special case of the graph method, not a separate rule. \( a = 0 \), displacement 120 m

7. A ball rolls up a ramp, stops, and rolls back down. Describe the sign of its velocity and acceleration throughout, taking up the ramp as positive.
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   Going up: velocity positive, decreasing. Acceleration negative. At the top: velocity zero. Acceleration still negative. Coming down: velocity negative, growing more negative. Acceleration still negative. The acceleration is negative and constant for the whole motion, because the component of gravity along the ramp never changes direction. Only the velocity's sign changes. Acceleration negative throughout; velocity positive, then zero, then negative

8. A train brakes from 30 m/s at a steady \( -1.5\ \text{m}/\text{s}^2 \). Find how long it takes to stop and how far it travels.
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   Time: \( t = \dfrac{\Delta v}{a} = \dfrac{0 - 30}{-1.5} = 20\ \text{s} \). The negatives cancel, giving a positive time, which is the check that the setup was right. Distance: area of the triangle, \( \dfrac{30 + 0}{2} \times 20 = 300\ \text{m} \). Check by equation. \( v^2 = v_0^2 + 2a\Delta x \) gives \( 0 = 900 + 2(-1.5)\Delta x \), so \( \Delta x = \dfrac{900}{3.0} = 300\ \text{m} \) ✓ 20 s and 300 m

9. Explain how an object can be accelerating while its speed stays constant.
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   Because velocity is a vector, and a vector can change by turning without changing length. The definition is about velocity, not speed. Acceleration is \( \dfrac{\Delta \vec{v}}{\Delta t} \), and \( \vec{v} \) carries a direction as well as a magnitude. A change in either is a change in \( \vec{v} \). The standard case. A car going round a bend at a steady 20 m/s has constant speed and continuously changing direction, so its velocity is changing at every instant and it is accelerating the whole way round. Which way that acceleration points. Toward the center of the curve, which lesson 4.3 derives. It is perpendicular to the velocity, which is exactly why it turns the motion without changing its speed. The general rule worth carrying forward. An acceleration parallel to the velocity changes speed only; perpendicular to it changes direction only; at any other angle it does some of each. Projectile motion in lesson 1.7 is the mixed case. Why this matters for forces. Since force causes acceleration, a body moving in a circle at constant speed must have a net force on it, pointing inward. Anything moving in a curve is being pushed, and identifying what is doing the pushing is usually the whole problem. Velocity is a vector, so changing direction is a change in velocity

10. A motorcycle accelerates from rest at \( 2.5\ \text{m}/\text{s}^2 \) for 8.0 s, then holds that speed for 12 s. Find the final speed, the total displacement, and the average velocity for the whole 20 s.
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    Phase 1, the acceleration. Final speed: \( v = v_0 + at = 0 + 2.5(8.0) = 20\ \text{m/s} \). Displacement: \( \Delta x = \tfrac{1}{2}at^2 = \tfrac{1}{2}(2.5)(64) = 80\ \text{m} \). Check by area: a triangle of height 20 m/s and width 8.0 s has area \( \tfrac{1}{2}(20)(8.0) = 80\ \text{m} \) ✓ Phase 2, the steady speed. \( \Delta x = vt = 20 \times 12 = 240\ \text{m} \). Totals. Displacement: \( 80 + 240 = 320\ \text{m} \). Time: \( 8.0 + 12 = 20\ \text{s} \). Average velocity. \( \bar{v} = \dfrac{320}{20} = 16\ \text{m/s} \). Sanity check the average. It must lie between the slowest and fastest speeds reached, which are 0 and 20 m/s. It does, and it sits nearer 20 because most of the time was spent at the top speed ✓ A common wrong route. Averaging the two phase speeds, 10 and 20, to get 15 m/s. That treats the phases as equally long when the second lasts half again as long as the first. Average velocity is always total displacement over total time. Note the shape of the graph this describes. A velocity-time graph rising in a straight line from the origin to \( (8.0, 20) \), then horizontal to \( (20, 20) \). The total area is a triangle plus a rectangle, which is exactly the two phases computed separately. 20 m/s, 320 m, and an average of 16 m/s

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Take \( g = 9.8\ \text{m}/\text{s}^2 \) where needed.

1. What condition must hold for the kinematic equations to apply?
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   Constant acceleration

2. Which equation omits displacement?
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   \( v = v_0 + at \)

3. Which equation omits time?
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   \( v^2 = v_0^2 + 2a\Delta x \)

4. An object starts at rest with \( a = 3.0\ \text{m}/\text{s}^2 \). Find its speed after 5.0 s.
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   \( v = 0 + 3.0(5.0) \). 15 m/s

5. How many quantities appear across the four equations?
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   Five

6. A cyclist accelerates from 4.0 m/s at \( 1.2\ \text{m}/\text{s}^2 \) over 30 m. Find the final speed.
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   Time is neither given nor wanted, so use the fourth equation. \( v^2 = 4.0^2 + 2(1.2)(30) = 16 + 72 = 88 \). \( v = \sqrt{88} = 9.4\ \text{m/s} \). Check the size. Starting at 4.0 and accelerating over 30 m should give a moderate increase, and 9.4 is plausible ✓ 9.4 m/s

7. A stone is dropped from rest and falls for 2.0 s. Find its speed and the distance fallen.
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   Take down positive, so \( a = +9.8\ \text{m}/\text{s}^2 \) and \( v_0 = 0 \). Speed: \( v = 0 + 9.8(2.0) = 19.6\ \text{m/s} \). Distance: \( \Delta x = \tfrac{1}{2}(9.8)(2.0)^2 = \tfrac{1}{2}(9.8)(4.0) = 19.6\ \text{m} \). The two numbers matching at 19.6 is a coincidence of this particular time, not a rule. They have different units and different meanings. 19.6 m/s and 19.6 m

8. A car braking at \( -5.0\ \text{m}/\text{s}^2 \) stops in 10 m. Find its initial speed.
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   Time is not given, so use the fourth equation with \( v = 0 \). \( 0 = v_0^2 + 2(-5.0)(10) \), so \( v_0^2 = 100 \) and \( v_0 = 10\ \text{m/s} \). Watch the sign of \( a \). Using \( +5.0 \) would give \( v_0^2 = -100 \), which has no real solution, and that impossibility is itself the signal that the sign was wrong. 10 m/s

9. Explain why doubling a car's speed more than doubles its stopping distance.
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   Because the relationship is quadratic, not linear, and the fourth equation shows it directly. The derivation. Setting \( v = 0 \) in \( v^2 = v_0^2 + 2a\Delta x \) and solving: \( \Delta x = \dfrac{v_0^2}{2|a|} \). The stopping distance depends on the square of the initial speed. What that means numerically, at a braking deceleration of \( 5.0\ \text{m}/\text{s}^2 \): at 10 m/s, \( \dfrac{100}{10} = 10\ \text{m} \); at 20 m/s, \( \dfrac{400}{10} = 40\ \text{m} \); at 30 m/s, \( \dfrac{900}{10} = 90\ \text{m} \). Doubling the speed quadruples the distance, and tripling it multiplies the distance by nine. Why it is squared rather than linear. Two things worsen together at higher speed. The car takes longer to stop, and during that longer time it is covering ground faster. Each effect scales with the speed, so the product scales with the square. The energy reading, which unit 5 will formalize. Kinetic energy goes as \( v^2 \), and the brakes remove energy at a roughly constant rate per meter, so four times the energy needs four times the distance. Why this matters outside the exam. The difference between 30 and 35 mph in a residential street is a much larger difference in stopping distance than the speeds suggest, which is the reasoning behind speed limits near schools. The quadratic is doing the work that intuition does not. Stopping distance goes as the square of the speed

10. A rocket accelerates upward from rest at \( 4.0\ \text{m}/\text{s}^2 \) for 15 s, then its engine cuts out. Find its speed and height at cutout, and then the maximum height it reaches.
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    This needs two phases, because the acceleration changes when the engine stops. Applying one equation across both would be the error this lesson warns about. Phase 1, engine on. Take up positive, \( a = +4.0\ \text{m}/\text{s}^2 \), \( v_0 = 0 \), \( t = 15\ \text{s} \). Speed at cutout: \( v = 0 + 4.0(15) = 60\ \text{m/s} \). Height at cutout: \( \Delta x = \tfrac{1}{2}(4.0)(15)^2 = \tfrac{1}{2}(4.0)(225) = 450\ \text{m} \). Phase 2, engine off. The rocket is still moving up at 60 m/s, but now the only acceleration is gravity, \( a = -9.8\ \text{m}/\text{s}^2 \). It rises until its velocity reaches zero. Time is not wanted, so use the fourth equation: \( 0 = 60^2 + 2(-9.8)\Delta x \), giving \( \Delta x = \dfrac{3600}{19.6} = 183.7\ \text{m} \). Total height. \( 450 + 183.7 = 633.7\ \text{m} \), about \( 6.3 \times 10^2\ \text{m} \) to two significant figures. Check the coasting phase a second way. Time to stop: \( t = \dfrac{60}{9.8} = 6.12\ \text{s} \). Distance: \( \dfrac{60 + 0}{2}(6.12) = 30(6.12) = 183.7\ \text{m} \) ✓ Why the phases cannot be merged. The acceleration is \( +4.0 \) for 15 s and then \( -9.8 \) afterward. No single constant value describes the motion, so no single application of the equations can. A check on plausibility. The coasting phase adds about 40 percent to the powered height, which is reasonable: the rocket is moving fast at cutout, and gravity takes six seconds to stop it. 60 m/s and 450 m at cutout; maximum height about 634 m

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( g = 9.8\ \text{m}/\text{s}^2 \) and ignore air resistance unless told otherwise.

1. Give the acceleration of a freely falling object near Earth's surface.
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   \( 9.8\ \text{m}/\text{s}^2 \) downward

2. Does a heavier object fall faster in a vacuum?
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   No

3. What is the velocity of a ball at the highest point of its flight?
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   Zero

4. What is its acceleration at that point?
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   \( 9.8\ \text{m}/\text{s}^2 \) downward

5. An object is dropped from rest. Find its speed after 3.0 s.
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   \( 9.8 \times 3.0 \). 29.4 m/s

6. A ball is dropped from 20 m. Find the time to land and the impact speed.
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   Down positive, \( v_0 = 0 \). \( 20 = \tfrac{1}{2}(9.8)t^2 \), so \( t^2 = \dfrac{40}{9.8} = 4.082 \) and \( t = 2.02\ \text{s} \). \( v = \sqrt{2(9.8)(20)} = \sqrt{392} = 19.8\ \text{m/s} \). Check: \( v = 9.8(2.02) = 19.8\ \text{m/s} \) ✓ 2.02 s and 19.8 m/s

7. A ball is thrown up at 24.5 m/s. Find the maximum height and the total time in the air.
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   Up positive, \( a = -9.8\ \text{m}/\text{s}^2 \). Height: \( \Delta x = \dfrac{24.5^2}{2(9.8)} = \dfrac{600.25}{19.6} = 30.6\ \text{m} \). Time up: \( \dfrac{24.5}{9.8} = 2.5\ \text{s} \). Total time, by symmetry: \( 2 \times 2.5 = 5.0\ \text{s} \). The symmetry halves the work, and it is exact here because air resistance is being ignored. 30.6 m and 5.0 s

8. Two balls are released from the same height, one dropped and one thrown horizontally. Which lands first?
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   They land together. The horizontal throw gives the second ball a horizontal velocity, and horizontal velocity has no effect on vertical motion. Both start with zero vertical velocity and both have \( 9.8\ \text{m}/\text{s}^2 \) downward, so both satisfy the same vertical equation. This is the independence of components that lesson 1.7 develops, and it is one of the least intuitive results in the unit. They land at the same time

9. Explain why the mass of a falling object does not affect its acceleration, given that a heavier object is pulled harder by gravity.
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   Because the greater pull is exactly offset by the greater resistance to being accelerated, and the two effects cancel completely. Set it out with the second law, which unit 2 states formally as \( F = ma \). The gravitational force on a mass \( m \) is its weight, \( F = mg \). The acceleration that force produces is \( a = \dfrac{F}{m} = \dfrac{mg}{m} = g \). The mass cancels. Doubling the mass doubles the force, and doubling the force on a doubled mass gives the same acceleration. Why the cancellation is not trivial. The \( m \) in \( mg \) measures how strongly gravity pulls on the object. The \( m \) in \( F = ma \) measures how strongly the object resists being accelerated by any force at all. There is no obvious reason those two quantities should be the same number, and experiment says they are, to extraordinary precision. That equality has a name and consequences. It is the equivalence of gravitational and inertial mass, and its exactness was the observation Einstein built general relativity on. A student meeting it here is meeting the starting point of a much later theory. What would happen if they differed. Objects made of different materials would fall at measurably different rates, and every free-fall equation in this unit would need a material-dependent correction. Experiments looking for exactly that have found nothing to the limits of their precision. The extra pull is exactly canceled by the extra inertia

10. A ball is thrown straight up from the ground at 19.6 m/s from a 30 m building's roof. Find when it lands at the base of the building and its speed on landing.
    Show the full solution

    Set the frame carefully. Take up as positive with the origin at the roof. Then: \( v_0 = +19.6\ \text{m/s} \), \( a = -9.8\ \text{m}/\text{s}^2 \), and the ground is at \( \Delta x = -30\ \text{m} \), because it is below the origin. Use the equation with displacement and time. \( -30 = 19.6t + \tfrac{1}{2}(-9.8)t^2 \) \( -30 = 19.6t - 4.9t^2 \) Rearranged into standard form: \( 4.9t^2 - 19.6t - 30 = 0 \). Solve the quadratic. Discriminant: \( (-19.6)^2 - 4(4.9)(-30) = 384.16 + 588 = 972.16 \). \( \sqrt{972.16} = 31.18 \). \( t = \dfrac{19.6 \pm 31.18}{9.8} \). The two roots are \( t = \dfrac{50.78}{9.8} = 5.18\ \text{s} \) and \( t = \dfrac{-11.58}{9.8} = -1.18\ \text{s} \). Reject the negative root. It describes where the ball would have been before it was thrown, which the model does not cover. So the ball lands at 5.18 s. Find the landing speed. \( v = v_0 + at = 19.6 + (-9.8)(5.18) = 19.6 - 50.8 = -31.2\ \text{m/s} \). The negative sign means downward, so the speed is 31.2 m/s. Check with the equation that omits time. \( v^2 = 19.6^2 + 2(-9.8)(-30) = 384.16 + 588 = 972.16 \), so \( v = \sqrt{972.16} = 31.2\ \text{m/s} \) ✓ Notice that check is the discriminant. The same number appeared in the quadratic, which is not a coincidence: both routes solve the same physical relation. Sanity check the size. The ball goes up 19.6 m, comes back to the roof at 19.6 m/s downward after 4.0 s, then falls 30 m more. Falling 30 m from 19.6 m/s takes about 1.2 s and adds speed, so a total near 5.2 s and a landing speed above 19.6 m/s are both what should be expected ✓ A note on the sign of \( \Delta x \). Writing \( +30 \) for the building height is the commonest error here. The ball finishes below its starting point, and with up positive that displacement is negative. Getting this wrong gives a quadratic with no real roots, which at least announces the mistake. 5.18 s, landing at 31.2 m/s

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Ignore air resistance.

1. What is the horizontal acceleration of a projectile?
   Show the full solution

   Zero

2. What is its vertical acceleration?
   Show the full solution

   \( 9.8\ \text{m}/\text{s}^2 \) downward

3. Which quantity is shared between the horizontal and vertical calculations?
   Show the full solution

   Time

4. A projectile is launched at 20 m/s at \( 30^\circ \). Find its horizontal component.
   Show the full solution

   \( 20\cos 30^\circ = 20(0.866) \). 17.3 m/s

5. Find its vertical component.
   Show the full solution

   \( 20\sin 30^\circ = 20(0.5) \). 10.0 m/s

6. A stone is thrown horizontally at 25 m/s from a 45 m cliff. Find the time to land and the horizontal distance.
   Show the full solution

   Vertical: \( 45 = \tfrac{1}{2}(9.8)t^2 \), so \( t^2 = \dfrac{90}{9.8} = 9.184 \) and \( t = 3.03\ \text{s} \). Horizontal: \( x = 25 \times 3.03 = 75.8\ \text{m} \). The launch speed never entered the vertical calculation, which is the point of the method. 3.03 s and 75.8 m

7. For the projectile of questions 4 and 5, find the time of flight over level ground.
   Show the full solution

   It rises and falls symmetrically, so the total flight is twice the time to the top. Time to top: \( \dfrac{v_y}{g} = \dfrac{10.0}{9.8} = 1.02\ \text{s} \). Total: \( 2 \times 1.02 = 2.04\ \text{s} \). 2.04 s

8. Find the range and maximum height of that same projectile.
   Show the full solution

   Range: \( x = v_x t = 17.3 \times 2.04 = 35.3\ \text{m} \). Maximum height: \( \dfrac{v_y^2}{2g} = \dfrac{100}{19.6} = 5.1\ \text{m} \). Check the height a second way. Rising for 1.02 s from 10.0 m/s: \( \dfrac{10.0 + 0}{2}(1.02) = 5.1\ \text{m} \) ✓ Range 35.3 m, maximum height 5.1 m

9. Explain why the horizontal and vertical motions of a projectile are independent.
   Show the full solution

   Because the only force acting is vertical, so it can only change the vertical velocity, and perpendicular directions do not exchange information. Start from the cause. Acceleration is produced by force. For a projectile the only force is gravity, which points straight down. A vertical force changes only the vertical component. It has no horizontal part, so \( a_x = 0 \) and \( v_x \) is unchanged for the whole flight. And nothing acts horizontally, so the horizontal motion has no way to influence the vertical one either. The independence runs both ways. What makes this legitimate mathematically. Perpendicular components of a vector can be treated separately because neither contributes to the other. That is a property of perpendicularity, not a physical assumption, and it is why components are chosen at right angles in the first place. The consequence that surprises people. A bullet fired horizontally and a bullet dropped hit the ground together. The fired one travels hundreds of meters while doing so, and none of that horizontal travel delays its fall by any amount. Where the independence fails. Air resistance acts opposite to the total velocity, so it has both components, and each one then depends on the other. That couples the two motions, and projectile problems with drag cannot be separated this way. It is the idealization that makes the method work. Gravity is purely vertical, so it cannot alter the horizontal motion

10. A ball is kicked at 20 m/s at \( 30^\circ \) above the horizontal from the top of a 15 m wall, landing on the ground below. Find the time of flight and the horizontal distance from the wall.
    Show the full solution

    Resolve the launch velocity. \( v_x = 20\cos 30^\circ = 17.3\ \text{m/s} \) \( v_{0y} = 20\sin 30^\circ = +10.0\ \text{m/s} \) Set the vertical frame. Up positive, origin at the launch point, so the ground is at \( \Delta y = -15\ \text{m} \) and \( a_y = -9.8\ \text{m}/\text{s}^2 \). Write the vertical equation. \( -15 = 10.0t + \tfrac{1}{2}(-9.8)t^2 \) \( -15 = 10.0t - 4.9t^2 \) \( 4.9t^2 - 10.0t - 15 = 0 \) Solve it. Discriminant: \( 100 - 4(4.9)(-15) = 100 + 294 = 394 \). \( \sqrt{394} = 19.85 \). \( t = \dfrac{10.0 \pm 19.85}{9.8} \). Positive root: \( t = \dfrac{29.85}{9.8} = 3.05\ \text{s} \). The negative root, \( -1.01\ \text{s} \), is rejected as before. Find the horizontal distance. \( x = v_x t = 17.3 \times 3.05 = 52.8\ \text{m} \). Check the vertical result by stages. Rising: the ball takes \( \dfrac{10.0}{9.8} = 1.02\ \text{s} \) to reach the top, climbing \( \dfrac{100}{19.6} = 5.1\ \text{m} \) above the wall, so it is 20.1 m above the ground there. Falling: from rest through 20.1 m takes \( \sqrt{\dfrac{2(20.1)}{9.8}} = \sqrt{4.102} = 2.03\ \text{s} \). Total: \( 1.02 + 2.03 = 3.05\ \text{s} \) ✓ matching the quadratic. That two-stage check is worth the extra minute, because it tests the physics rather than repeating the algebra: if the quadratic had been set up with the wrong sign on the 15, the stages would disagree. Sanity check the distance. Flying for 3.05 s at 17.3 m/s gives about 53 m, and the ball spends longer aloft than it would on level ground because it lands lower than it started. Both are consistent ✓ 3.05 s and 52.8 m from the wall

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Give the SI unit of force.
   Show the full solution

   The newton

2. Is force a vector or a scalar?
   Show the full solution

   A vector

3. Name the two forces on a book resting on a table.
   Show the full solution

   Weight and the normal force

4. Forces of 12 N right and 5 N left act on a box. Find the net force.
   Show the full solution

   7 N to the right

5. Give one example of a field force.
   Show the full solution

   Gravity (electric or magnetic also acceptable)

6. Forces of 60 N east and 80 N north act on an object. Find the magnitude and direction of the net force.
   Show the full solution

   Perpendicular, so use the Pythagorean theorem. \( \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100\ \text{N} \). Direction: \( \tan^{-1}\left( \dfrac{80}{60} \right) = 53.1^\circ \) north of east. 100 N at 53.1° north of east

7. List the forces on a ball while it is in mid-flight after being thrown, ignoring air resistance.
   Show the full solution

   Weight only, the gravitational pull of Earth on the ball, directed downward. The hand is no longer touching it, so there is no throwing force once it has left. The force from the hand existed only during the throw. This is the commonest wrong diagram in the course: an arrow drawn forward along the ball's path labeled "force of the throw". Nothing is there to supply it. Weight alone

8. A lamp hangs from a single cord and is at rest. Draw its free-body diagram and state the relationship between the forces.
   Show the full solution

   Two forces: tension from the cord, upward, and weight from Earth, downward. The lamp is at rest, so its acceleration is zero and the net force must be zero: \( T = mg \). The two are equal in size and opposite in direction, but they are not a third-law pair, because they act on the same object. Lesson 2.4 returns to this. Tension up, weight down, equal in magnitude

9. Explain why a free-body diagram must show only the forces acting on one object.
   Show the full solution

   Because Newton's second law relates the net force on a single body to that body's acceleration, so mixing in forces on other bodies breaks the relationship. What the law says. \( F_{\text{net}} = ma \), where every force in the sum acts on the object whose mass and acceleration appear on the right. A force acting on something else has no place in that sum. What goes wrong if you include others. Consider a person pushing a box. If you draw both the push on the box and the box's push back on the person in one diagram, the two are equal and opposite, so they cancel and the net force comes out zero. That would predict the box never accelerates, which is false. The forces cancel only in a sum that has no physical meaning. The fix. One diagram per object. The push on the box goes in the box's diagram; the reaction goes in the person's. Each diagram then gives the correct acceleration for its own body. Why this is worth the discipline. Almost every wrong answer in mechanics traces back to a force in the wrong diagram, either a reaction force included with its partner or an invented force with no agent. Drawing the boundary around one object first, then asking what crosses it, prevents both. The law applies to one body, so the diagram must contain exactly that body's forces

10. A 5.0 kg box is pulled by a rope at 20 N and opposed by friction of 8.0 N, both horizontal. Find the net force and the acceleration, then state what happens to the acceleration if the rope tension is reduced to 8.0 N.
    Show the full solution

    Draw the diagram first. Four forces: tension right 20 N, friction left 8.0 N, weight down, normal up. The vertical pair cancels because the box does not accelerate vertically, so only the horizontal forces matter. Find the net force. Take right positive: \( F_{\text{net}} = 20 - 8.0 = 12\ \text{N} \) to the right. Find the acceleration, using the law that lesson 2.3 states formally: \( a = \dfrac{F_{\text{net}}}{m} = \dfrac{12\ \text{N}}{5.0\ \text{kg}} = 2.4\ \text{m}/\text{s}^2 \) to the right. Check the units. \( \dfrac{\text{N}}{\text{kg}} = \dfrac{\text{kg}\cdot\text{m}/\text{s}^2}{\text{kg}} = \text{m}/\text{s}^2 \) ✓ Now reduce the tension to 8.0 N. \( F_{\text{net}} = 8.0 - 8.0 = 0 \), so \( a = 0 \). What that means, and what it does not. Zero acceleration does not mean the box is at rest. It means the velocity is not changing. If the box was already sliding at 3 m/s, it continues at 3 m/s indefinitely. If it was at rest, it stays at rest. A subtlety worth flagging. Friction of 8.0 N here is the kinetic value, which applies while sliding. If the box were at rest, static friction would apply instead and would simply match whatever was applied, up to its maximum. Lesson 2.6 separates the two cases, and the distinction changes which of those two outcomes is right. 12 N and \( 2.4\ \text{m}/\text{s}^2 \); at 8.0 N the acceleration is zero and the velocity holds steady

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. State Newton's first law in your own words.
   Show the full solution

   An object keeps its velocity unless a net force acts on it

2. What quantity measures inertia?
   Show the full solution

   Mass

3. Does constant velocity require a net force?
   Show the full solution

   No

4. What force stops a sliding book?
   Show the full solution

   Kinetic friction from the surface

5. Two forces on an object cancel exactly. What is its acceleration?
   Show the full solution

   Zero net force. Zero

6. A car turns a corner at a steady 20 km/h. Is a net force acting on it?
   Show the full solution

   Yes. Constant velocity means constant speed and direction. The direction is changing, so the velocity is changing, so there is an acceleration and therefore a net force. What supplies it. Friction between the tires and the road, directed toward the center of the turn. On ice that friction is unavailable and the car continues straight, which is the first law being obeyed rather than violated. Yes, friction from the road

7. A spacecraft far from any star has its engines switched off. Describe its motion.
   Show the full solution

   Essentially no forces act, so it continues at constant velocity: the same speed, in the same straight line, indefinitely. No fuel is needed to keep going. Fuel is needed only to change the velocity, which is why course corrections are budgeted in spacecraft design and cruising is not. Constant velocity, indefinitely

8. A coin sits on a card on top of a glass. The card is flicked away sharply and the coin drops into the glass. Explain why.
   Show the full solution

   The coin's inertia resists the sudden change. The force on the coin is friction from the card, and it acts only while the card is under it. Flicking the card removes it in a very short time, so the sideways impulse delivered to the coin is tiny. With almost no sideways velocity, the coin stays essentially where it was, and once unsupported it falls straight into the glass. Why the flick must be fast. Sliding the card slowly gives friction a long time to act, and the coin travels with it. The trick works because a short contact time means a small change in motion, which lesson 3.2 quantifies as impulse. Inertia keeps the coin in place during the brief contact

9. Explain why the first law is not simply the second law with zero force.
   Show the full solution

   Because the first law says where the second law is allowed to be used, and that is a separate piece of information. The apparent redundancy. Setting \( F_{\text{net}} = 0 \) in \( F_{\text{net}} = ma \) gives \( a = 0 \), which looks like the first law. Why it is not. The second law is not true in every reference frame. Inside a braking bus, a bag on a smooth floor accelerates forward with nothing touching it. A passenger applying \( F = ma \) in the bus frame would have to invent a force that no object supplies. What the first law does. It identifies the frames where this does not happen. A frame in which an isolated object really does keep constant velocity is called inertial, and the second law holds in those frames and not in others. So the logical order is: the first law picks out the good frames, and the second law then applies within them. Stated as "\( F = 0 \) gives \( a = 0 \)", the first law would carry no such information and the braking-bus problem would have no resolution. How to spot a non-inertial frame. Something accelerates with no agent for the force. The frame is accelerating, and the answer is to solve the problem from the ground instead. It specifies the frames in which the second law is valid

10. A passenger in a car takes a sharp left turn and slides toward the right-hand door. Explain what happens from outside the car, and say why "centrifugal force" is not an explanation.
    Show the full solution

    Describe it from the ground first. Before the turn, the passenger is moving forward in a straight line at the car's speed. The first law says they will keep doing so unless something pushes them. What happens to the car. The road supplies friction to the tires, pushing the car leftward into the turn. The car changes direction. What happens to the passenger. Nothing is yet pushing them sideways, so they continue nearly straight ahead. The car curves left while they do not, so the right-hand door comes toward them. Then contact occurs. Once against the door, the door pushes them leftward, and that inward push is what finally turns them with the car. The seat and friction do the same job more gently in a mild turn. Why nothing was thrown outward. No object pushed the passenger to the right. They moved right relative to the car because the car moved left relative to them. A relative displacement is not evidence of a force. What centrifugal force actually is. Inside the turning car, which is an accelerating and therefore non-inertial frame, objects appear to accelerate outward with no agent. Bookkeeping in that frame requires adding a fictitious outward term so that \( F = ma \) still balances. It is a correction for using a rotating frame, not a push by any object. The test that shows it is fictitious. Name the body exerting it. For every real force you can: Earth for weight, the table for the normal force, the road for friction. For centrifugal force there is no answer, which is the diagnostic from lesson 2.1. What is real, and inward. The centripetal force, which is not a new kind of force but a name for whatever real force happens to point at the center. Here it is friction and the door. Lesson 4.3 develops this. The passenger goes straight while the car turns beneath them; no outward force acts

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write Newton's second law as an equation.
   Show the full solution

   \( F_{\text{net}} = ma \)

2. A 10 kg object experiences a net force of 50 N. Find its acceleration.
   Show the full solution

   \( 5.0\ \text{m}/\text{s}^2 \)

3. A 2000 kg truck accelerates at \( 1.5\ \text{m}/\text{s}^2 \). Find the net force.
   Show the full solution

   3000 N

4. If the mass doubles and the net force stays the same, what happens to the acceleration?
   Show the full solution

   It halves

5. In which direction does the acceleration point?
   Show the full solution

   Along the net force

6. A 1500 kg car goes from rest to 20 m/s in 8.0 s. Find the net force on it.
   Show the full solution

   \( a = \dfrac{20 - 0}{8.0} = 2.5\ \text{m}/\text{s}^2 \). \( F = 1500 \times 2.5 = 3750\ \text{N} \). 3750 N

7. A 0.50 kg ball is struck and leaves at 30 m/s after being in contact for 0.020 s. Find the average net force.
   Show the full solution

   \( a = \dfrac{30 - 0}{0.020} = 1500\ \text{m}/\text{s}^2 \). \( F = 0.50 \times 1500 = 750\ \text{N} \). The force is large because the time is short, which is the observation lesson 3.2 builds the impulse idea on. 750 N

8. A 3.0 kg object has forces of 12 N east and 16 N north acting on it. Find its acceleration.
   Show the full solution

   Net force: \( \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\ \text{N} \). Direction: \( \tan^{-1}\left( \dfrac{16}{12} \right) = 53.1^\circ \) north of east. \( a = \dfrac{20}{3.0} = 6.7\ \text{m}/\text{s}^2 \), in that same direction. Sum the forces before dividing, not after. Dividing each force separately and combining the results happens to work here because division by a scalar is linear, but summing first is the habit that survives harder problems. \( 6.7\ \text{m}/\text{s}^2 \) at 53.1° north of east

9. Explain why the same force produces different accelerations on different objects.
   Show the full solution

   Because acceleration is the force divided by the mass, and mass measures how strongly an object resists being accelerated. From the law. \( a = \dfrac{F}{m} \), so with \( F \) held fixed, \( a \) falls as \( m \) rises. It is an inverse proportion, not a subtraction: three times the mass gives a third of the acceleration. A concrete comparison. A 100 N push on a 10 kg cart gives \( 10\ \text{m}/\text{s}^2 \). The same push on a 500 kg cart gives \( 0.20\ \text{m}/\text{s}^2 \), fifty times less. What mass is doing here. It is not measuring size or weight. A large balloon has little mass and is easy to accelerate; a small lead block has a lot and is not. Mass is exactly the constant of proportionality between force and acceleration, and that is the most useful definition of it. Why the same quantity also appears in weight. The gravitational pull is \( mg \), so a heavier object is pulled harder and resists more, in the same proportion. Those two roles cancel in free fall, which is lesson 1.6's result and one of the more remarkable facts in the subject. Acceleration is force divided by mass, so larger mass means smaller acceleration

10. A 1200 kg car traveling at 25 m/s brakes to a stop in 40 m. Find the acceleration, the braking force, and the time taken, and state what the sign of each quantity means.
    Show the full solution

    Set a frame. Take the direction of travel as positive, so \( v_0 = +25\ \text{m/s} \), \( v = 0 \) and \( \Delta x = +40\ \text{m} \). Find the acceleration. Time is not given, so use the kinematic equation that omits it: \( v^2 = v_0^2 + 2a\Delta x \) \( 0 = 625 + 2a(40) \) \( a = \dfrac{-625}{80} = -7.81\ \text{m}/\text{s}^2 \). Find the braking force. \( F = ma = 1200 \times (-7.81) = -9375\ \text{N} \), about \( -9.4 \times 10^3\ \text{N} \). Find the time. \( t = \dfrac{v - v_0}{a} = \dfrac{0 - 25}{-7.81} = 3.20\ \text{s} \). Check with a second route. The average velocity while braking steadily is \( \dfrac{25 + 0}{2} = 12.5\ \text{m/s} \), so the time to cover 40 m is \( \dfrac{40}{12.5} = 3.20\ \text{s} \) ✓ Read the signs. The acceleration is negative, meaning it points backward, opposite the motion. That is what slowing down requires. The force is negative for the same reason: it is supplied by friction between the tires and the road, directed backward. The time is positive, as any duration must be. A negative time here would signal a sign error upstream. What the car is not doing. It is not moving backward at any point. The negative signs describe the direction of the change in velocity and of the force causing it, while the velocity itself stays positive until it reaches zero. Sanity check the magnitude. A deceleration of \( 7.81\ \text{m}/\text{s}^2 \) is about 0.8 g, which is near the limit of what tires on dry asphalt can supply. Stopping from 25 m/s, roughly 56 mph, in 40 m is therefore hard braking but physically possible ✓ A calculated value far above 1 g would have indicated an arithmetic error rather than a remarkable car. \( -7.81\ \text{m}/\text{s}^2 \), about \( 9.4 \times 10^3 \) N backward, and 3.20 s

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. State Newton's third law.
   Show the full solution

   If A exerts a force on B, B exerts an equal and opposite force on A

2. Do the two forces of a third-law pair act on the same object?
   Show the full solution

   No, on different objects

3. Can a third-law pair cancel each other out?
   Show the full solution

   They act on different bodies. No

4. Name the partner of "Earth pulls a ball down".
   Show the full solution

   The ball pulls Earth up

5. A 40 N force is exerted by a bat on a ball. What force does the ball exert on the bat?
   Show the full solution

   40 N, in the opposite direction

6. A swimmer pushes water backward and moves forward. Identify the pair and say which force moves the swimmer.
   Show the full solution

   Pair: the swimmer pushes the water backward, and the water pushes the swimmer forward. The force that moves the swimmer is the one acting on the swimmer: the water's forward push. This is why swimming is impossible without something to push against. The same reasoning explains walking, where the foot pushes the ground backward and the ground pushes the walker forward, and rockets, where exhaust is pushed one way and the rocket the other. The water's forward push on the swimmer

7. A truck collides head-on with a small car. Compare the forces on each and the accelerations of each.
   Show the full solution

   The forces are equal and opposite, by the third law, however different the vehicles are. The accelerations are not. With the same force \( F \), the car of mass \( m_c \) has \( a_c = \dfrac{F}{m_c} \) and the truck \( a_t = \dfrac{F}{m_t} \). A truck of ten times the mass has a tenth of the acceleration. Which is why the car is damaged more. Not because it was hit harder, but because the same force produces a far greater change in its motion, and injury follows from the acceleration of the occupants. Equal forces, very unequal accelerations

8. A book weighing 15 N rests on a table. Give the size and direction of the book's push on the table, and name the partner of the book's weight.
   Show the full solution

   The table pushes the book up with 15 N, since the book is in equilibrium. By the third law the book pushes the table down with 15 N. The partner of the book's weight is the book's gravitational pull on Earth, 15 N upward on Earth. Note the two 15 N values are not the same pair. One pair is book-and-table contact forces; the other is book-and-Earth gravitational forces. They happen to be equal in size here only because the book is in equilibrium. 15 N downward on the table; the weight's partner is the book's pull on Earth

9. Explain why a book's weight and the normal force on it are not a third-law pair, even though they are equal and opposite.
   Show the full solution

   Because they act on the same object, and a third-law pair by definition acts on two different ones. Check them against the criteria. Different objects? No. Both the weight and the normal force act on the book. That alone settles it. Same type of force? No. The weight is gravitational, exerted by Earth; the normal force is a contact force, exerted by the table. Pairs are always the same kind. Equal under all circumstances? No, and this is the decisive test. Break the equality to prove the point. Put the book in an elevator accelerating upward at \( 2.0\ \text{m}/\text{s}^2 \). Its weight is unchanged, since neither its mass nor \( g \) changed. But the normal force must now exceed the weight to produce the upward acceleration. The two are no longer equal. A third-law pair cannot be separated like that. Its two forces are equal by the law itself, in every situation, accelerating or not. So what makes them equal in the static case? The second law. The book is not accelerating, so the net force on it must be zero, which forces the two to balance. It is a consequence of equilibrium, not of the third law. The real partners. Weight pairs with the book's pull on Earth. Normal force pairs with the book's push on the table. Neither partner appears in the book's diagram. They act on the same body, so they are balanced forces rather than an interaction pair

10. Two ice skaters, 70 kg and 50 kg, push off from rest against each other. The push lasts 0.40 s and the 50 kg skater leaves at 1.4 m/s. Find the force, the other skater's speed, and explain what is conserved.
    Show the full solution

    Find the acceleration of the lighter skater. \( a = \dfrac{\Delta v}{t} = \dfrac{1.4 - 0}{0.40} = 3.5\ \text{m}/\text{s}^2 \). Find the force on her. \( F = ma = 50 \times 3.5 = 175\ \text{N} \). Apply the third law. The force on the 70 kg skater is also 175 N, in the opposite direction. Find his acceleration and speed. \( a = \dfrac{175}{70} = 2.5\ \text{m}/\text{s}^2 \). The push lasts the same 0.40 s, because the two are in contact for exactly the same interval. \( v = at = 2.5 \times 0.40 = 1.0\ \text{m/s} \), in the opposite direction. Check the ratio. The speeds should be in inverse proportion to the masses: \( \dfrac{1.4}{1.0} = 1.4 \) and \( \dfrac{70}{50} = 1.4 \) ✓ What is conserved. Take her direction as positive. Before: both at rest, so the total momentum is \( 50(0) + 70(0) = 0 \). After: \( 50(1.4) + 70(-1.0) = 70 - 70 = 0 \). The total momentum is zero before and zero after. Why it had to be. The forces are equal and opposite by the third law, and they act for the same length of time, so the two changes in momentum are equal and opposite and cancel exactly. That argument is the whole of lesson 3.3, arrived at here from the third law alone. What is not conserved. Kinetic energy. Before the push it was zero; afterward it is \( \tfrac{1}{2}(50)(1.4)^2 + \tfrac{1}{2}(70)(1.0)^2 = 49 + 35 = 84\ \text{J} \). That energy came from the skaters' muscles, so the system was not isolated with respect to energy even though it was with respect to momentum. Unit 5 takes this up. 175 N, the 70 kg skater at 1.0 m/s, and momentum conserved at zero

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( g = 9.8\ \text{m}/\text{s}^2 \).

1. Give the SI unit of weight.
   Show the full solution

   The newton

2. Find the weight of a 12 kg object.
   Show the full solution

   \( 12 \times 9.8 \). 117.6 N

3. Does mass change on the Moon?
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   No

4. What does a bathroom scale directly measure?
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   The normal force it exerts

5. What is the apparent weight of someone in free fall?
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   Zero

6. An 80 kg person is in an elevator accelerating upward at \( 3.0\ \text{m}/\text{s}^2 \). Find the scale reading.
   Show the full solution

   \( N = m(g + a) = 80(9.8 + 3.0) = 80(12.8) = 1024\ \text{N} \). Their true weight is \( 80 \times 9.8 = 784\ \text{N} \), so the scale reads about 31 percent high. 1024 N

7. The same person is now accelerating downward at \( 3.0\ \text{m}/\text{s}^2 \). Find the scale reading.
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   \( N = 80(9.8 - 3.0) = 80(6.8) = 544\ \text{N} \). Note the weight is still 784 N. Only the supporting force changed. 544 N

8. A scale reads 390 N for a 50 kg person. Find the elevator's acceleration and say which way it points.
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   Rearrange \( N = m(g + a) \): \( a = \dfrac{N}{m} - g = \dfrac{390}{50} - 9.8 = 7.8 - 9.8 = -2.0\ \text{m}/\text{s}^2 \). Negative with up positive, so the acceleration is downward at \( 2.0\ \text{m}/\text{s}^2 \). The reading is below the true weight of 490 N, which is the check: a low reading always means downward acceleration. Be careful what this does not say. The elevator could be moving downward and speeding up, or moving upward and slowing down. Both give a downward acceleration, and the scale cannot tell them apart. \( 2.0\ \text{m}/\text{s}^2 \) downward

9. Explain why astronauts on the space station float, given that gravity there is nearly as strong as on the ground.
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   Because they are in free fall, so nothing is pushing up on them, and floating is the absence of a supporting force rather than the absence of gravity. Establish that gravity is still there. At 400 km altitude the gravitational field is about \( 8.7\ \text{m}/\text{s}^2 \), roughly 89 percent of its surface value. If gravity really had switched off, the station would leave in a straight line and never return, which the first law requires. What floating really is. The sensation of weight is the normal force from a floor or seat. In free fall the station, the astronaut and everything loose inside all accelerate downward together at the same rate, so no surface has to push on any other. Every normal force is zero, and so is the apparent weight. Why they do not fall to Earth. They are falling, continuously. They also have a large sideways velocity, about 7.7 km/s, so as they fall the curved surface of Earth drops away beneath them at the same rate. The path is an orbit because the falling and the curving match. The equivalence with the elevator. Part (d) of the worked example is the same physics on a smaller scale. An astronaut in orbit is in a lift whose cable has been cut permanently, and the reason it never lands is the sideways motion. Why "microgravity" is the better word. It avoids claiming gravity is absent while acknowledging that the measurable effects of weight are. The tiny residual accelerations that give it the "micro" come from atmospheric drag and tidal differences across the station, not from any weakening of gravity. They are in continuous free fall, so no surface supports them

10. A 65 kg person stands in an elevator that starts from rest, accelerates upward at \( 1.5\ \text{m}/\text{s}^2 \) for 3.0 s, travels at constant speed for 5.0 s, then decelerates to rest over 2.0 s. Find the scale reading in each phase and the top speed.
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    True weight throughout. \( W = 65 \times 9.8 = 637\ \text{N} \). This never changes, in any phase. Phase 1, accelerating upward at \( +1.5\ \text{m}/\text{s}^2 \). \( N = m(g + a) = 65(9.8 + 1.5) = 65(11.3) = 734.5\ \text{N} \). Above the true weight, as an upward acceleration requires. Top speed at the end of phase 1. \( v = at = 1.5 \times 3.0 = 4.5\ \text{m/s} \) upward. Phase 2, constant speed. \( a = 0 \), so \( N = 65(9.8) = 637\ \text{N} \). Equal to the true weight. The elevator is moving, and quickly, but the scale cannot tell. Only acceleration affects the reading, which is the first law showing up in a measurement. Phase 3, decelerating to rest over 2.0 s. The elevator is moving up at 4.5 m/s and must reach zero, so \( a = \dfrac{0 - 4.5}{2.0} = -2.25\ \text{m}/\text{s}^2 \), pointing downward. \( N = 65(9.8 - 2.25) = 65(7.55) = 490.75\ \text{N} \), about 491 N. Collect the readings. 734.5 N, then 637 N, then 490.75 N. Check the pattern against experience. Heavy as the lift starts up, normal while it cruises, light as it slows near the top floor. That is exactly what a rising elevator feels like ✓ A trap worth naming. Phase 3 is deceleration while moving upward, and the acceleration is downward. Students often assign a positive acceleration because the elevator is still going up, which would predict a heavy feeling at the moment everyone reports a light one. Check with the total displacement, as a consistency test. Phase 1: \( \tfrac{1}{2}(1.5)(3.0)^2 = 6.75\ \text{m} \). Phase 2: \( 4.5 \times 5.0 = 22.5\ \text{m} \). Phase 3: \( \dfrac{4.5 + 0}{2}(2.0) = 4.5\ \text{m} \). Total 33.75 m, roughly eleven storeys, which is a plausible lift ride ✓ 734.5 N, 637 N, then 490.75 N; top speed 4.5 m/s

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( g = 9.8\ \text{m}/\text{s}^2 \).

1. In which direction does friction act?
   Show the full solution

   Opposite the relative sliding, along the surface

2. Write the formula for maximum static friction.
   Show the full solution

   \( f_{s,\max} = \mu_s N \)

3. Which is usually larger, \( \mu_s \) or \( \mu_k \)?
   Show the full solution

   \( \mu_s \)

4. What are the units of a coefficient of friction?
   Show the full solution

   A ratio of two forces. None; it is dimensionless

5. A 20 kg block rests on a level floor. Find the normal force.
   Show the full solution

   \( 20 \times 9.8 \). 196 N

6. For that 20 kg block, \( \mu_s = 0.50 \). Find the maximum static friction, and state what the friction is when an 80 N push is applied.
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   \( f_{s,\max} = 0.50 \times 196 = 98\ \text{N} \). An 80 N push is below 98 N, so the block does not move and the friction is 80 N, exactly matching the push. Not 98 N. Static friction supplies only what is needed. Maximum 98 N; the actual friction is 80 N

7. The same block has \( \mu_k = 0.35 \). Find its acceleration under a 120 N push.
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   \( 120 \gt 98 \), so it slides and kinetic friction applies. \( f_k = 0.35 \times 196 = 68.6\ \text{N} \). \( F_{\text{net}} = 120 - 68.6 = 51.4\ \text{N} \). \( a = \dfrac{51.4}{20} = 2.57\ \text{m}/\text{s}^2 \). \( 2.57\ \text{m}/\text{s}^2 \)

8. A box begins to slide when a ramp is tilted to \( 16.7^\circ \). Find \( \mu_s \).
   Show the full solution

   At the slipping angle, \( \tan\theta = \mu_s \). \( \mu_s = \tan 16.7^\circ = 0.30 \). No masses or forces were needed, because they cancel in the derivation. This is how coefficients of friction are measured in practice. \( \mu_s = 0.30 \)

9. Explain why static friction has no single value while kinetic friction does.
   Show the full solution

   Because static friction is whatever is needed to prevent motion, and what is needed depends on what is applied, while kinetic friction is a property of two surfaces sliding. The static case is constrained by the second law. If an object is not accelerating, the net force on it is zero. So whatever horizontal force is applied, friction must supply exactly that much in the opposite direction, or the object would accelerate. Push with 5 N and friction is 5 N. Push with 20 N and friction is 20 N. The value is set by the requirement of equilibrium, not by the surfaces. But it has a ceiling. The surfaces can only grip so hard, and that limit is \( \mu_s N \). Beyond it the object breaks free. The kinetic case is different. Once sliding, the object is not in equilibrium and there is no requirement for friction to match anything. It takes the value the surfaces produce, \( \mu_k N \), regardless of how hard you push. Why \( \mu_k \) is smaller. At rest the surfaces settle into contact at many points and can form momentary bonds. Sliding surfaces skim across high spots without the same time to settle, so the resistance is lower. The practical consequence. Getting something moving is harder than keeping it moving, and once it starts it tends to lurch. It is also why anti-lock brakes exist: a rolling tire grips by static friction, a skidding one by the smaller kinetic value, so a locked wheel stops the car less effectively and steers not at all. Static friction adjusts to match the applied force up to a limit; kinetic friction is fixed by the surfaces

10. A 15 kg crate is pushed across a floor by a 90 N horizontal force and accelerates at \( 2.0\ \text{m}/\text{s}^2 \). Find the friction force and \( \mu_k \), then find how far it slides if the push is removed while it is moving at 6.0 m/s.
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    Find the net force from the motion. \( F_{\text{net}} = ma = 15 \times 2.0 = 30\ \text{N} \). Find the friction. The push and friction are the only horizontal forces: \( f_k = 90 - 30 = 60\ \text{N} \). Find the coefficient. The normal force on a level floor is \( N = mg = 15 \times 9.8 = 147\ \text{N} \). \( \mu_k = \dfrac{f_k}{N} = \dfrac{60}{147} = 0.41 \). Sanity check. A coefficient of 0.41 is typical for something like wood on concrete, so the figure is plausible ✓ A value above 1 would have been possible but unusual, and one above about 1.5 would suggest an arithmetic error. Now remove the push. Friction is the only horizontal force left, and it is unchanged at 60 N because the normal force has not changed. \( a = \dfrac{-60}{15} = -4.0\ \text{m}/\text{s}^2 \), the sign showing it opposes the motion. Find the sliding distance. Time is not wanted, so use the kinematic equation that omits it: \( v^2 = v_0^2 + 2a\Delta x \) \( 0 = 6.0^2 + 2(-4.0)\Delta x \) \( \Delta x = \dfrac{36}{8.0} = 4.5\ \text{m} \). Check by a second route. Time to stop: \( t = \dfrac{6.0}{4.0} = 1.5\ \text{s} \). Distance at an average speed of 3.0 m/s for 1.5 s is \( 4.5\ \text{m} \) ✓ A detail worth noticing. The deceleration once the push is removed, \( 4.0\ \text{m}/\text{s}^2 \), is twice the acceleration while being pushed. That is not a coincidence of this problem but a consequence of the numbers: friction of 60 N alone is twice the 30 N net force available when the 90 N push was fighting it. A caution about the coefficient. The value 0.41 was derived assuming the push was horizontal. A push angled downward would increase the normal force and therefore the friction, and the same measurement would give a different and wrong coefficient. The direction of an applied force matters to friction through \( N \), not only through the horizontal balance. Friction 60 N, \( \mu_k = 0.41 \), and it slides 4.5 m

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Assume frictionless pulleys and massless ropes.

1. What do two connected objects share?
   Show the full solution

   The magnitude of their acceleration

2. In which direction does tension act on an object?
   Show the full solution

   Away from the object, along the rope

3. Is tension internal or external when the system is treated as one body?
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   Internal, so it cancels

4. Two blocks of 2.0 kg and 3.0 kg are pushed together by a 10 N force on a frictionless surface. Find the acceleration.
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   \( a = \dfrac{10}{5.0} \). \( 2.0\ \text{m}/\text{s}^2 \)

5. For those blocks, find the contact force on the 3.0 kg block.
   Show the full solution

   It is the only force on that block: \( F = 3.0 \times 2.0 \). 6.0 N

6. An Atwood machine has 2.0 kg and 6.0 kg. Find the acceleration and the tension.
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   \( a = \dfrac{(6.0 - 2.0)(9.8)}{8.0} = \dfrac{39.2}{8.0} = 4.9\ \text{m}/\text{s}^2 \). Tension from the rising mass: \( T = 2.0(9.8 + 4.9) = 2.0(14.7) = 29.4\ \text{N} \). Check on the falling mass: \( 6.0(9.8 - 4.9) = 6.0(4.9) = 29.4\ \text{N} \) ✓ Between the two weights of 19.6 N and 58.8 N, as it must be ✓ \( 4.9\ \text{m}/\text{s}^2 \) and 29.4 N

7. A 5.0 kg block on a frictionless table is connected to a 3.0 kg hanging mass. Find the acceleration and the tension.
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   \( a = \dfrac{3.0 \times 9.8}{8.0} = \dfrac{29.4}{8.0} = 3.675\ \text{m}/\text{s}^2 \). \( T = m_1a = 5.0 \times 3.675 = 18.4\ \text{N} \). Check: \( m_2g - T = 29.4 - 18.4 = 11.0\ \text{N} \), and \( m_2a = 3.0 \times 3.675 = 11.0\ \text{N} \) ✓ Less than the hanging weight of 29.4 N, as required for it to accelerate downward ✓ \( 3.675\ \text{m}/\text{s}^2 \) and 18.4 N

8. Three blocks of 1.0, 2.0 and 3.0 kg sit in a row on a frictionless surface and are pushed by a 12 N force applied to the 1.0 kg block. Find the acceleration and the force the 2.0 kg block exerts on the 3.0 kg block.
   Show the full solution

   System: \( a = \dfrac{12}{6.0} = 2.0\ \text{m}/\text{s}^2 \). Isolate the 3.0 kg block, which has only the contact force on it: \( F = 3.0 \times 2.0 = 6.0\ \text{N} \). Choosing the last block was deliberate, because it has the fewest forces. Isolating the front block would have needed both the 12 N push and the backward contact force. \( 2.0\ \text{m}/\text{s}^2 \) and 6.0 N

9. Explain why treating connected objects as a single system makes the tension disappear.
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   Because the tension acts twice, once on each object, in opposite directions, and summing over the whole system adds both so they cancel. Set it out. The rope pulls object A toward the pulley with force \( T \), and pulls object B toward the pulley with force \( T \). Those are two separate forces on two separate bodies. What the system view does. Drawing the boundary around both objects makes the rope internal. Newton's second law applied to the whole uses the sum of all forces crossing the boundary, and the tensions do not cross it: both endpoints are inside. Why they cancel exactly. They are equal in magnitude, because an ideal rope has one tension throughout, and opposite in the sense that one pulls the system one way round the pulley and the other pulls it back. In the coordinate scheme where each mass counts its own direction of motion as positive, one is \( +T \) and the other \( -T \). What survives. Only the external forces: the two weights, and the normal force if a surface is involved. That is why the acceleration formula contains only \( g \) and the masses. The general principle. Internal forces never change the motion of a system as a whole, because the third law guarantees they come in canceling pairs. This is the same argument that gives conservation of momentum in lesson 3.3, and it is why you cannot move a stationary car by pushing on its dashboard. How to get the tension back. Redraw the boundary around one object only. The rope now crosses it, the tension becomes external, and it appears in the equation. Internal forces occur in canceling pairs, so they vanish from the system equation

10. A 6.0 kg block on a table with \( \mu_k = 0.20 \) is connected over a pulley to a 4.0 kg hanging mass. Find the acceleration and the tension, and find the minimum \( \mu_s \) that would keep the system at rest.
    Show the full solution

    Find the friction on the table block. \( N = m_1g = 6.0 \times 9.8 = 58.8\ \text{N} \). \( f_k = \mu_k N = 0.20 \times 58.8 = 11.76\ \text{N} \), opposing the motion. System view. The driving force is the hanging weight; friction resists. Both are external to the system. \( F_{\text{net}} = m_2g - f_k = (4.0)(9.8) - 11.76 = 39.2 - 11.76 = 27.44\ \text{N} \). \( a = \dfrac{27.44}{10.0} = 2.744\ \text{m}/\text{s}^2 \), about \( 2.74\ \text{m}/\text{s}^2 \). Find the tension by isolating the table block. Its horizontal forces are tension forward and friction backward: \( T - f_k = m_1a \) \( T = m_1a + f_k = 6.0(2.744) + 11.76 = 16.46 + 11.76 = 28.2\ \text{N} \). Check on the hanging mass. \( m_2g - T = 39.2 - 28.2 = 11.0\ \text{N} \), and \( m_2a = 4.0 \times 2.744 = 10.98\ \text{N} \) ✓ agreeing to rounding. Now the static question. For the system to stay at rest, static friction must hold back the full pull of the hanging weight. With nothing accelerating, the tension equals the hanging weight: \( T = m_2g = 39.2\ \text{N} \). That tension must be balanced by static friction on the table block: \( f_s = 39.2\ \text{N} \). Find the coefficient that allows it. \( \mu_s \ge \dfrac{f_s}{N} = \dfrac{39.2}{58.8} = 0.667 \). Sanity check the two results against each other. The actual \( \mu_k = 0.20 \) is far below the 0.667 needed to hold, so the system should certainly move, and it does ✓ If the required coefficient had come out below 0.20, the acceleration calculation would have been self-contradictory. Why the tension differs between the two cases. Moving, it is 28.2 N; held at rest, it would be 39.2 N. A rope holding a stationary weight carries the full weight, while a rope accelerating one carries less, because the weight must exceed the tension to produce the downward acceleration. A note on the coefficient needed. A value of 0.667 is high but not impossible: rubber on dry concrete reaches about 1.0. So a rubber-footed block on concrete would hold, and a wooden one on a smooth table would not. \( 2.74\ \text{m}/\text{s}^2 \), tension 28.2 N, and \( \mu_s \ge 0.667 \) to hold

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write the formula for momentum.
   Show the full solution

   \( p = mv \)

2. Give the SI unit of momentum.
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   \( \text{kg}\cdot\text{m/s} \)

3. Is momentum a vector or a scalar?
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   A vector

4. Find the momentum of a 70 kg runner at 8.0 m/s.
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   \( 70 \times 8.0 \). \( 560\ \text{kg}\cdot\text{m/s} \)

5. What is the momentum of a stationary 5000 kg truck?
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   Zero

6. A 0.50 kg ball moves left at 12 m/s. Taking right as positive, give its momentum.
   Show the full solution

   \( p = 0.50 \times (-12) \). \( -6.0\ \text{kg}\cdot\text{m/s} \) The sign is part of the answer, and omitting it would make the collision arithmetic in lesson 3.4 come out wrong.

7. A 1500 kg car at 12 m/s and a 900 kg car at 20 m/s travel in the same direction. Find the total momentum.
   Show the full solution

   \( 1500 \times 12 = 18000\ \text{kg}\cdot\text{m/s} \) \( 900 \times 20 = 18000\ \text{kg}\cdot\text{m/s} \) Same direction, so they add: \( 36000\ \text{kg}\cdot\text{m/s} \). Note the two are equal individually, a lighter car going faster matching a heavier one going slower. \( 3.6 \times 10^4\ \text{kg}\cdot\text{m/s} \)

8. A 3.0 kg object moves east at 4.0 m/s and a 2.0 kg object moves west at 5.0 m/s. Find the total momentum.
   Show the full solution

   Take east positive. \( 3.0(+4.0) = +12\ \text{kg}\cdot\text{m/s} \) \( 2.0(-5.0) = -10\ \text{kg}\cdot\text{m/s} \) Total: \( +2.0\ \text{kg}\cdot\text{m/s} \), that is 2.0 units east. Adding the magnitudes would give 22, which describes no situation at all. \( 2.0\ \text{kg}\cdot\text{m/s} \) east

9. Explain why momentum is more useful than velocity for analyzing collisions.
   Show the full solution

   Because the total momentum of an isolated system is the same before and after, while the individual velocities are not predictable without knowing the details of the contact. What changes during a collision. Each object's velocity changes by an amount that depends on how stiff the objects are, how long they touch and how much they deform. None of that is usually known. What does not change. The vector sum \( m_1v_1 + m_2v_2 \) over the whole system. It has the same value after the collision as before, whatever happened in between. Why that is powerful. It gives one equation relating the unknown final velocities that is true regardless of the mechanism. A second piece of information, such as "they stick together" or "the collision is elastic", is then enough to solve completely. Why velocity alone fails. Velocity is not conserved and has no analogous rule. The sum of velocities before a collision is unrelated to the sum afterward, so it gives no equation to work with. Where the conservation comes from. Newton's third law. The two objects push on each other with equal and opposite forces for the same length of time, so their momentum changes are equal and opposite and cancel in the total. Lesson 3.3 sets that out properly. Total momentum is conserved while individual velocities are not predictable

10. A 0.045 kg golf ball leaves a club at 70 m/s and a 7.3 kg bowling ball rolls at 4.0 m/s. Compare their momenta and their kinetic energies, and say what the comparison shows.
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    Momenta. Golf ball: \( p = 0.045 \times 70 = 3.15\ \text{kg}\cdot\text{m/s} \). Bowling ball: \( p = 7.3 \times 4.0 = 29.2\ \text{kg}\cdot\text{m/s} \). The bowling ball has about 9.3 times the momentum. Kinetic energies, using \( \tfrac{1}{2}mv^2 \) from unit 5. Golf ball: \( \tfrac{1}{2}(0.045)(70)^2 = \tfrac{1}{2}(0.045)(4900) = 110.25\ \text{J} \). Bowling ball: \( \tfrac{1}{2}(7.3)(4.0)^2 = \tfrac{1}{2}(7.3)(16) = 58.4\ \text{J} \). The golf ball has about 1.9 times the kinetic energy. The comparison reverses. The bowling ball wins on momentum by a wide margin; the golf ball wins on energy. These cannot be two measures of the same thing. Why they disagree. Momentum is linear in speed and energy is quadratic. Doubling the speed doubles the momentum but quadruples the energy, so speed counts far more heavily in the energy comparison. The golf ball's speed advantage of 17.5 times is enough to overcome a mass disadvantage of 162 times in the energy accounting, but not in the momentum accounting. What each one predicts physically. Momentum governs what happens to velocities in a collision: the bowling ball would barely slow on striking something the golf ball would bounce off. Energy governs how much damage is done, how much heat is produced, how far something is pushed against resistance. The golf ball carries more capacity to do that. Why both are needed. Asking "which is moving more forcefully?" has no single answer. It depends on whether the question is about changing velocities, which is momentum, or about deforming, heating and breaking, which is energy. Unit 5 develops the second and this unit develops the first. Bowling ball has 9.3 times the momentum; golf ball has 1.9 times the energy, because energy depends on the square of speed

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write the impulse-momentum theorem.
   Show the full solution

   \( F\Delta t = \Delta p \)

2. Give the unit of impulse.
   Show the full solution

   \( \text{N}\cdot\text{s} \), equivalently \( \text{kg}\cdot\text{m/s} \)

3. What does the area under a force-time graph represent?
   Show the full solution

   Impulse

4. A 20 N force acts for 3.0 s. Find the impulse.
   Show the full solution

   \( 20 \times 3.0 \). \( 60\ \text{N}\cdot\text{s} \)

5. For a fixed change in momentum, what happens to the force if the contact time doubles?
   Show the full solution

   It halves

6. A 2.0 kg object speeds up from 3.0 m/s to 9.0 m/s in 4.0 s. Find the impulse and the average force.
   Show the full solution

   \( \Delta p = 2.0(9.0) - 2.0(3.0) = 18 - 6.0 = 12\ \text{kg}\cdot\text{m/s} \). That is the impulse, \( 12\ \text{N}\cdot\text{s} \). \( F = \dfrac{12}{4.0} = 3.0\ \text{N} \). Impulse \( 12\ \text{N}\cdot\text{s} \), force 3.0 N

7. A 0.060 kg tennis ball arrives at 25 m/s and leaves at 30 m/s in the opposite direction, with contact lasting 0.005 s. Find the average force.
   Show the full solution

   Take the incoming direction as positive. \( p_i = 0.060(+25) = +1.5\ \text{kg}\cdot\text{m/s} \) \( p_f = 0.060(-30) = -1.8\ \text{kg}\cdot\text{m/s} \) \( \Delta p = -1.8 - 1.5 = -3.3\ \text{kg}\cdot\text{m/s} \) \( F = \dfrac{-3.3}{0.005} = -660\ \text{N} \), so 660 N opposite the incoming motion. The magnitudes added rather than subtracted, because the direction reversed. 660 N

8. An 80 kg person jumps from a height and lands at 6.0 m/s. Find the force if they stop in 0.050 s, and if they stop in 0.40 s.
   Show the full solution

   \( \Delta p = 80 \times 6.0 = 480\ \text{kg}\cdot\text{m/s} \) in both cases. Stiff landing: \( F = \dfrac{480}{0.050} = 9600\ \text{N} \). Bent knees: \( F = \dfrac{480}{0.40} = 1200\ \text{N} \). An eightfold reduction from the same fall, purely from taking longer over the stop. 9600 N and 1200 N

9. Explain why a ball that bounces off a wall exerts a larger force on it than one that sticks.
   Show the full solution

   Because reversing the momentum requires a larger change than merely removing it, and a larger change in the same contact time means a larger force. Take the incoming direction as positive, with mass \( m \) and arrival speed \( v \), so \( p_i = +mv \). The sticking case. The ball ends at rest, so \( p_f = 0 \) and \( \Delta p = 0 - mv = -mv \). The magnitude is \( mv \). The bouncing case. Suppose it leaves at the same speed the other way, so \( p_f = -mv \) and \( \Delta p = -mv - mv = -2mv \). The magnitude is \( 2mv \), twice as much. Why the factor is two. The wall must first bring the ball to rest, which takes an impulse of \( mv \), and then push it back up to speed \( v \) the other way, which takes another \( mv \). Then the third law. Whatever force the wall exerts on the ball, the ball exerts an equal force back on the wall. A doubled impulse on the ball in the same contact time means a doubled force on the wall. A real bounce is less than double, because the rebound speed is usually lower than the arrival speed. But it always exceeds the sticking case, as long as the ball leaves at all. Where this shows up. Hail damages roofs more than rain of the same mass and speed, because raindrops splash and stay while hailstones bounce. And a bouncy rubber mallet transmits more force than a dead-blow hammer filled with shot, which is designed specifically not to rebound. Reversing momentum needs roughly twice the impulse of stopping it

10. A force on a 4.0 kg cart rises steadily from 0 to 20 N over 3.0 s, then stays at 20 N for 2.0 s. The cart starts from rest. Find the total impulse and the final speed.
    Show the full solution

    The force varies, so use the area under the force-time graph rather than \( F\Delta t \) with a single value. Splitting into a triangle and a rectangle handles it. Phase 1, the ramp. A triangle from \( (0, 0) \) to \( (3.0, 20) \): area \( = \tfrac{1}{2}(3.0)(20) = 30\ \text{N}\cdot\text{s} \). Phase 2, the constant force. A rectangle 20 N tall and 2.0 s wide: area \( = 20 \times 2.0 = 40\ \text{N}\cdot\text{s} \). Total impulse. \( 30 + 40 = 70\ \text{N}\cdot\text{s} \). Find the final speed. The impulse equals the change in momentum, and the cart began at rest: \( \Delta p = 70\ \text{kg}\cdot\text{m/s} \), so \( mv = 70 \) and \( v = \dfrac{70}{4.0} = 17.5\ \text{m/s} \). Check with an average force. The mean force over the 5.0 s is \( \dfrac{70}{5.0} = 14\ \text{N} \), giving \( a = \dfrac{14}{4.0} = 3.5\ \text{m}/\text{s}^2 \) and \( v = 3.5 \times 5.0 = 17.5\ \text{m/s} \) ✓ Why the graph method was necessary. Using the peak force of 20 N for the whole 5.0 s would give \( 100\ \text{N}\cdot\text{s} \) and a speed of 25 m/s, about 43 percent too high. The force was below 20 N for most of the first phase, and the area accounts for that automatically. The general point. \( F\Delta t \) with a single number is a shortcut valid only for a constant force. The area under the curve is the real definition, and it handles any force history, including the sharply peaked ones that occur in actual collisions. \( 70\ \text{N}\cdot\text{s} \) and 17.5 m/s

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. State the condition under which momentum is conserved.
   Show the full solution

   No net external force on the system

2. Which of Newton's laws does conservation of momentum follow from?
   Show the full solution

   The third

3. Is momentum conserved in an inelastic collision?
   Show the full solution

   Yes

4. Two carts at rest push apart. What is their total momentum afterward?
   Show the full solution

   It was zero before. Zero

5. Do internal forces change a system's total momentum?
   Show the full solution

   No

6. A 70 kg skater at rest throws a 3.0 kg ball at 6.0 m/s. Find the skater's recoil speed.
   Show the full solution

   Total momentum before is zero, so it is zero after. \( 0 = 3.0(6.0) + 70v \), so \( v = \dfrac{-18}{70} = -0.257\ \text{m/s} \). 0.26 m/s backward

7. A 1200 kg cannon fires a 8.0 kg shell at 300 m/s. Find the cannon's recoil speed.
   Show the full solution

   \( 0 = 8.0(300) + 1200v \) \( 1200v = -2400 \), so \( v = -2.0\ \text{m/s} \). 2.0 m/s backward. The shell's momentum, \( 2400\ \text{kg}\cdot\text{m/s} \), is matched exactly by the cannon's. Note the energies are not matched. The shell carries \( \tfrac{1}{2}(8.0)(300)^2 = 360000\ \text{J} \) and the cannon only \( \tfrac{1}{2}(1200)(2.0)^2 = 2400\ \text{J} \). Momentum splits evenly; energy does not. 2.0 m/s backward

8. A 2.0 kg cart at 4.0 m/s collides with a stationary 3.0 kg cart and they stick together. Find their common speed.
   Show the full solution

   Before: \( 2.0(4.0) + 3.0(0) = 8.0\ \text{kg}\cdot\text{m/s} \). After: the combined 5.0 kg moves at \( v \), so \( 5.0v = 8.0 \) and \( v = 1.6\ \text{m/s} \). Check the direction and size. Same direction as the original motion, and slower, because the same momentum is now spread over more mass ✓ 1.6 m/s

9. Explain why conservation of momentum does not require knowing anything about the forces involved.
   Show the full solution

   Because the forces cancel in pairs before their details ever matter, so the derivation goes through for any force at all. Trace the derivation. The third law guarantees that whatever force object 1 exerts on object 2, object 2 exerts exactly the negative of it on object 1. It says nothing about what that force is. Both act for the same duration. The objects are in contact over one shared interval, so the two impulses are \( F\Delta t \) and \( -F\Delta t \). The sum is zero regardless of \( F \). Whatever function of time \( F \) happens to be, its negative added to it gives zero. The force cancels out of the algebra entirely. What this buys practically. You can predict the outcome of a car crash without knowing how stiff the bumpers are, how long the contact lasted, or how the force varied during it. Those details determine the forces on the occupants, which lesson 3.7 needs, but they do not affect the final velocities. Contrast with energy. Predicting the energy lost does require knowing about the materials, because the loss depends on how much they deform and heat. That is why momentum gives a clean universal equation and energy does not. What is still required. That the system be isolated. The third law handles internal forces; external ones have no partner inside the system and do change the total. The internal forces cancel as a pair whatever their size or time dependence

10. A 5.0 kg object moving east at 6.0 m/s collides with a 3.0 kg object moving west at 4.0 m/s. After the collision the 5.0 kg object moves east at 1.5 m/s. Find the velocity of the 3.0 kg object and verify your answer.
    Show the full solution

    Set the convention. Take east as positive. Momentum before. \( p_i = 5.0(+6.0) + 3.0(-4.0) = 30 - 12 = +18\ \text{kg}\cdot\text{m/s} \). Note that the westward object contributes negatively. Adding 30 and 12 would be the commonest error here. Momentum after. \( p_f = 5.0(+1.5) + 3.0(v) = 7.5 + 3.0v \). Set them equal and solve. \( 7.5 + 3.0v = 18 \) \( 3.0v = 10.5 \) \( v = +3.5\ \text{m/s} \), that is 3.5 m/s east. Interpret it. The lighter object was moving west at 4.0 m/s and now moves east at 3.5 m/s. It reversed direction, which is what a head-on collision with a heavier object should do. Verify the momentum. \( 5.0(1.5) + 3.0(3.5) = 7.5 + 10.5 = 18\ \text{kg}\cdot\text{m/s} \) ✓ matching the initial total. Check the energy, as a plausibility test. Before: \( \tfrac{1}{2}(5.0)(36) + \tfrac{1}{2}(3.0)(16) = 90 + 24 = 114\ \text{J} \). After: \( \tfrac{1}{2}(5.0)(2.25) + \tfrac{1}{2}(3.0)(12.25) = 5.625 + 18.375 = 24\ \text{J} \). Energy fell from 114 J to 24 J, a loss of 90 J. That is allowed: the collision was inelastic and the missing energy went into deformation and heat. Why this check is worth doing. Had the final energy come out higher than the initial, the answer would be impossible, because a collision cannot create kinetic energy from nothing. An energy check cannot confirm a momentum answer, but it can rule one out, in the same way dimensional analysis works in lesson 1.1. 3.5 m/s east

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. What is conserved in every collision?
   Show the full solution

   Momentum

2. What extra quantity is conserved in an elastic collision?
   Show the full solution

   Kinetic energy

3. What defines a perfectly inelastic collision?
   Show the full solution

   The objects move off together

4. A 3.0 kg cart at 5.0 m/s hits a stationary 2.0 kg cart and they stick. Find their common speed.
   Show the full solution

   \( 5.0v = 15 \). 3.0 m/s

5. Where does the lost kinetic energy go?
   Show the full solution

   Into deformation, heat and sound

6. A 1.0 kg ball at 4.0 m/s strikes an identical stationary ball elastically. Find both final velocities.
   Show the full solution

   \( v_1 = \dfrac{1.0 - 1.0}{2.0}(4.0) = 0 \) \( v_2 = \dfrac{2(1.0)}{2.0}(4.0) = 4.0\ \text{m/s} \) The incoming ball stops dead and the target takes the whole velocity. This is the Newton's cradle result, and it happens only for equal masses. Check: momentum \( 4.0 \to 4.0 \) ✓, energy \( 8.0\ \text{J} \to 8.0\ \text{J} \) ✓ \( v_1 = 0 \), \( v_2 = 4.0\ \text{m/s} \)

7. A 4.0 kg ball at 5.0 m/s strikes a stationary 2.0 kg ball elastically. Find both final velocities.
   Show the full solution

   \( v_1 = \dfrac{4.0 - 2.0}{6.0}(5.0) = \dfrac{2.0}{6.0}(5.0) = 1.67\ \text{m/s} \) \( v_2 = \dfrac{2(4.0)}{6.0}(5.0) = \dfrac{8.0}{6.0}(5.0) = 6.67\ \text{m/s} \) The heavier ball continues forward, slowed, because \( m_1 \gt m_2 \) makes the first expression positive. Check momentum: \( 4.0(1.67) + 2.0(6.67) = 6.67 + 13.3 = 20 \), matching \( 4.0(5.0) \) ✓ 1.67 m/s and 6.67 m/s, both forward

8. Two 2.0 kg carts approach each other, each at 3.0 m/s, and stick together. Find the final velocity and the kinetic energy lost.
   Show the full solution

   Momentum before: \( 2.0(+3.0) + 2.0(-3.0) = 0 \). So the combined 4.0 kg mass has zero momentum and is at rest. Energy before: \( \tfrac{1}{2}(2.0)(9.0) \times 2 = 18\ \text{J} \). Energy after: zero. All 18 J is lost, which is the maximum possible loss. It happens whenever the total momentum is zero, because then the only way to conserve momentum is to end at rest. At rest; all 18 J lost

9. Explain why a perfectly inelastic collision loses the most kinetic energy possible.
   Show the full solution

   Because conserving momentum fixes the total, and of all the ways to share that total between the two objects, moving together carries the least kinetic energy. What is fixed. Momentum conservation requires \( m_1v_1 + m_2v_2 \) to equal the initial total, whatever the collision does. That constrains the final velocities but does not determine them. What varies. Among all the velocity pairs satisfying that constraint, the kinetic energy \( \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2 \) takes different values. The minimum occurs when the two velocities are equal, which is exactly the perfectly inelastic case. Why equal velocities minimize it. Any relative motion between the two objects represents kinetic energy that momentum conservation does not require them to have, because equal and opposite contributions to the relative motion cancel in the momentum sum while adding in the energy sum. Removing all relative motion removes all of that surplus. Check it on the head-on case. Two equal carts approaching at equal speeds have zero total momentum. Ending at rest conserves momentum and loses all the energy. Any other outcome, such as bouncing apart, also conserves momentum but keeps some energy. So sticking is the extreme. Why it cannot lose more. Kinetic energy cannot go negative, and the momentum requirement forces a nonzero energy whenever the total momentum is nonzero. A system with momentum must be moving. The engineering reading. A collision that ends with the objects moving together has converted the maximum available energy into deformation, which is precisely what a crumple zone is designed to do. Moving together is the lowest-energy way to conserve a given momentum

10. A 0.010 kg bullet travelling at 400 m/s embeds in a stationary 2.0 kg block on a frictionless surface. Find the final speed and the fraction of kinetic energy lost, and explain why the fraction is so large.
    Show the full solution

    Apply momentum conservation. \( p_i = 0.010(400) + 2.0(0) = 4.0\ \text{kg}\cdot\text{m/s} \). The combined mass is \( 0.010 + 2.0 = 2.01\ \text{kg} \): \( v = \dfrac{4.0}{2.01} = 1.99\ \text{m/s} \). Compute the energies. Before: \( \tfrac{1}{2}(0.010)(400)^2 = \tfrac{1}{2}(0.010)(160000) = 800\ \text{J} \). After: \( \tfrac{1}{2}(2.01)(1.99)^2 = \tfrac{1}{2}(2.01)(3.96) = 3.98\ \text{J} \). Fraction lost. \( \dfrac{800 - 3.98}{800} = 0.995 \), that is 99.5 percent. Why so much. The two quantities scale differently with the mass ratio. Momentum is linear in velocity, so the shared momentum forces the heavy block to a slow speed. Kinetic energy is quadratic, so that slow speed carries almost no energy even with 200 times the mass. Put it algebraically. For a light object of mass \( m \) embedding in a much heavier \( M \), the final speed is about \( \dfrac{mu}{M} \), and the final energy is about \( \dfrac{m^2u^2}{2M} \). Dividing by the initial \( \tfrac{1}{2}mu^2 \) leaves a fraction of roughly \( \dfrac{m}{M} \), here \( \dfrac{0.010}{2.0} = 0.005 \), or half a percent retained ✓ matching the computation. Where the 796 J went. Into tearing and heating the block along the bullet's path, deforming the bullet, and sound. A bullet embedding in wood raises its temperature measurably, and that heat is the missing energy. Why this is the basis of a real measurement. This arrangement is the ballistic pendulum, historically used to measure bullet speeds. The block is hung as a pendulum, its swing height gives the post-collision speed by energy conservation after the collision, and momentum conservation during the collision then gives the bullet's speed. The crucial subtlety. You must use momentum for the collision and energy for the swing, never the reverse. Using energy conservation through the collision would ignore the 99.5 percent that was lost and give a bullet speed wildly too low. 1.99 m/s, with 99.5 percent of the kinetic energy lost

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. In how many directions is momentum conserved in a two-dimensional collision?
   Show the full solution

   Both, independently

2. What is the first step in solving a two-dimensional collision?
   Show the full solution

   Resolve the velocities into components

3. A 2.0 kg object moves east at 5.0 m/s. Give its \( x \) and \( y \) momentum components.
   Show the full solution

   \( p_x = 10\ \text{kg}\cdot\text{m/s} \), \( p_y = 0 \)

4. How are momentum components recombined into a magnitude?
   Show the full solution

   \( p = \sqrt{p_x^2 + p_y^2} \)

5. Can \( x \)-momentum turn into \( y \)-momentum during a collision?
   Show the full solution

   No

6. A 3.0 kg object moving east at 8.0 m/s collides with a 5.0 kg object moving north at 3.0 m/s, and they stick. Find the velocity of the combined object.
   Show the full solution

   \( p_x = 3.0(8.0) = 24\ \text{kg}\cdot\text{m/s} \) \( p_y = 5.0(3.0) = 15\ \text{kg}\cdot\text{m/s} \) Combined mass 8.0 kg: \( v_x = \dfrac{24}{8.0} = 3.0\ \text{m/s} \), \( v_y = \dfrac{15}{8.0} = 1.875\ \text{m/s} \). \( v = \sqrt{9.0 + 3.52} = \sqrt{12.52} = 3.54\ \text{m/s} \) \( \theta = \tan^{-1}\left( \dfrac{1.875}{3.0} \right) = 32.0^\circ \) north of east. 3.54 m/s at 32.0° north of east

7. For that collision, find the kinetic energy lost.
   Show the full solution

   Before: \( \tfrac{1}{2}(3.0)(64) + \tfrac{1}{2}(5.0)(9.0) = 96 + 22.5 = 118.5\ \text{J} \). After: \( \tfrac{1}{2}(8.0)(12.52) = 50.1\ \text{J} \). Lost: \( 118.5 - 50.1 = 68.4\ \text{J} \), about 58 percent. Use the squared speed, not the components separately, and note that \( v^2 = v_x^2 + v_y^2 = 12.52 \) is already available from the previous question. About 68.4 J

8. An object at rest explodes into two pieces. One of mass 2.0 kg moves east at 6.0 m/s. Find the momentum of the other piece.
   Show the full solution

   Total momentum before is zero, and nothing external acted, so it is zero after. \( p_1 + p_2 = 0 \), so \( p_2 = -p_1 \). \( p_1 = 2.0(6.0) = 12\ \text{kg}\cdot\text{m/s} \) east. \( p_2 = 12\ \text{kg}\cdot\text{m/s} \) west. The mass of the second piece is not needed to answer the question as asked, and is not determinable from the information given. Its speed would be \( \dfrac{12}{m_2} \). \( 12\ \text{kg}\cdot\text{m/s} \) west

9. Explain why momentum conservation holds separately in each direction.
   Show the full solution

   Because momentum is a vector, and a vector equation is equivalent to one scalar equation per component. Start from the vector statement. Conservation says \( \vec{p}_{\text{before}} = \vec{p}_{\text{after}} \). Two vectors are equal only when all their components are equal, so that single statement contains one equation for \( x \) and one for \( y \). The physical reason behind it. The derivation in lesson 3.3 rested on the third law, which applies to force vectors. The force on object 1 is the exact negative of the force on object 2, component by component. So the \( x \) components cancel and the \( y \) components cancel, separately. Why the directions cannot mix. A force in the \( x \) direction produces an acceleration in the \( x \) direction only. There is no mechanism by which eastward momentum becomes northward momentum, in the same way that lesson 1.7's horizontal motion could not affect the vertical. What it buys. Two equations instead of one, so two unknowns can be found. That is exactly enough for a perfectly inelastic collision, where the two unknowns are \( v_x \) and \( v_y \) of the combined object. Where it stops being enough. A general elastic collision in two dimensions has four unknown final quantities and only three equations, two from momentum and one from energy. A fourth piece of information, such as one of the scattering angles, must be supplied. That is why elastic two-dimensional problems always give you an extra measurement. A vector equation is one equation per component, and the third law cancels each component separately

10. A 1500 kg car travelling north at 15 m/s is struck by a 1000 kg car travelling east. The wreckage moves off at \( 35^\circ \) east of north. Find the eastbound car's speed before impact.
    Show the full solution

    Set the axes. Take north as \( +y \) and east as \( +x \). Write the components before the collision. \( p_y = 1500 \times 15 = 22500\ \text{kg}\cdot\text{m/s} \), from the northbound car only. \( p_x = 1000 \times u \), from the eastbound car only, where \( u \) is what we want. Use the direction of the wreckage. The angle of a vector is determined by the ratio of its components, and momentum is conserved, so the wreckage's momentum has the same direction as the total momentum before. Measuring \( 35^\circ \) east of north means the angle is taken from the north axis toward the east axis, so \[ \tan 35^\circ = \frac{p_x}{p_y} \] Be careful which ratio this is. Because the angle is measured from north, the opposite side is the east component and the adjacent is the north one. Writing \( \tan 35^\circ = \dfrac{p_y}{p_x} \) would be the natural slip and would give a badly wrong speed. Solve for the east component. \( p_x = p_y \tan 35^\circ = 22500 \times 0.7002 = 15754\ \text{kg}\cdot\text{m/s} \). Find the speed. \( u = \dfrac{15754}{1000} = 15.8\ \text{m/s} \). Check the direction comes back out. \( \tan^{-1}\left( \dfrac{15754}{22500} \right) = \tan^{-1}(0.700) = 35.0^\circ \) ✓ Sanity check the size. The wreckage leans less than \( 45^\circ \) from north, so the northward momentum must exceed the eastward one. Indeed \( 22500 \gt 15754 \) ✓ And the lighter car needs a slightly higher speed than the heavier one to contribute comparably, which 15.8 against 15 reflects. Why this is the useful form of the problem. The angle of the wreckage is measurable from the scene long after the event, while neither pre-impact speed is. One measured angle plus the two masses recovers a speed that no witness could supply, which is exactly what makes momentum conservation valuable to an investigator. About 15.8 m/s

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write the formula for the center of mass of two objects.
   Show the full solution

   \( x_{\text{cm}} = \dfrac{m_1x_1 + m_2x_2}{m_1+m_2} \)

2. Two equal masses are 6.0 m apart. Where is the center of mass?
   Show the full solution

   Midway, 3.0 m from each

3. Is the center of mass closer to the heavier or lighter object?
   Show the full solution

   The heavier

4. Can internal forces move the center of mass of an isolated system?
   Show the full solution

   No

5. A 1.0 kg mass sits at \( x = 0 \) and a 3.0 kg mass at \( x = 2.0\ \text{m} \). Find the center of mass.
   Show the full solution

   \( \dfrac{(1.0)(0)+(3.0)(2.0)}{4.0} = \dfrac{6.0}{4.0} \). 1.5 m

6. A 1.0 kg and a 9.0 kg object are 10 m apart. Find the center of mass, measured from the light one.
   Show the full solution

   \( x_{\text{cm}} = \dfrac{(1.0)(0)+(9.0)(10)}{10} = \dfrac{90}{10} = 9.0\ \text{m} \). So it is 9.0 m from the light object and 1.0 m from the heavy one. A naive midpoint would give 5.0 m, wrong by four meters. 9.0 m from the light object

7. A hollow ring has all its mass in the rim. Where is its center of mass?
   Show the full solution

   At the geometric center, by symmetry. That point contains no material at all, which shows the center of mass need not lie inside the object. A boomerang, a doughnut and a horseshoe are all in the same situation. At the center of the hole

8. A firework shell rising through the air explodes into many fragments. Describe the subsequent path of the center of mass.
   Show the full solution

   It continues along the same parabola the unexploded shell was following. Why. The explosion supplies only internal forces, which cancel in pairs and cannot shift the center of mass. The only external force is gravity, so the center of mass keeps accelerating downward at \( g \), exactly as before. Ignoring air resistance, which does act externally on the spreading fragments and does eventually disturb the ideal path. The same parabola as before the burst

9. Explain why the center of mass of an isolated system moves at constant velocity.
   Show the full solution

   Because its velocity is the total momentum divided by the total mass, and both of those are constant for an isolated system. Write the relationship. \( v_{\text{cm}} = \dfrac{m_1v_1 + m_2v_2 + \cdots}{M} = \dfrac{p_{\text{total}}}{M} \). Apply conservation. An isolated system has no net external force, so lesson 3.3 gives \( p_{\text{total}} \) as constant. The total mass \( M \) is also constant. A constant divided by a constant is constant. What that rules out. No amount of internal activity can change it. Objects may collide, explode, stick, push apart or oscillate, and the center of mass proceeds unchanged. The intuitive version. You cannot move yourself by pushing on yourself. A person standing on frictionless ice cannot shift their center of mass by any motion of their arms and legs; they can only rotate about it or move parts of their body in compensating directions. Why an astronaut can still get somewhere. By throwing something. That makes the thrown object part of the system, and the center of mass of astronaut plus object still does not move. The astronaut drifts one way and the object the other, which is exactly the recoil calculation in lesson 3.3. What restores acceleration. An external force. A firework's fragments accelerate downward together because gravity acts from outside the system, and a car accelerates because the road pushes it. Its velocity is the conserved total momentum divided by the constant total mass

10. Find the center of mass of the Sun and Jupiter, and say whether it lies inside the Sun. Use \( M_{\text{Sun}} = 1.99 \times 10^{30}\ \text{kg} \), \( M_J = 1.898 \times 10^{27}\ \text{kg} \), separation \( 7.785 \times 10^{11}\ \text{m} \), and the Sun's radius \( 6.96 \times 10^{8}\ \text{m} \).
    Show the full solution

    Measure from the Sun's center, so \( x_{\text{Sun}} = 0 \) and \( x_J = 7.785 \times 10^{11}\ \text{m} \). Apply the definition. \[ x_{\text{cm}} = \frac{M_J\,d}{M_{\text{Sun}} + M_J} \] Compute the numerator. \( (1.898 \times 10^{27})(7.785 \times 10^{11}) = 1.478 \times 10^{39} \). Compute the denominator. \( 1.99 \times 10^{30} + 1.898 \times 10^{27} = 1.992 \times 10^{30} \). Jupiter adds only about a tenth of a percent, but keeping it matters for the ratio. Divide. \( x_{\text{cm}} = \dfrac{1.478 \times 10^{39}}{1.992 \times 10^{30}} = 7.42 \times 10^{8}\ \text{m} \). Compare with the Sun's radius. \( \dfrac{7.42 \times 10^{8}}{6.96 \times 10^{8}} = 1.066 \). The center of mass is outside the Sun, about 6.6 percent of a solar radius above its surface. What that means. The Sun does not sit still with Jupiter going round it. Both orbit a point just above the Sun's own surface, so the Sun traces a small circle with a period of about twelve years, Jupiter's orbital period. Contrast with the Earth-Moon case. There the shared point fell inside Earth, so Earth merely wobbles internally. Here it falls outside the Sun, so the Sun genuinely circles a point in empty space. Why this is not a curiosity. That wobble is detectable. A distant observer watching the Sun would see its light alternately blueshifted and redshifted as it moved toward and away from them, by the Doppler effect of lesson 9.5. Measuring such a wobble in another star is one of the main ways planets around other stars are found, and it works best for exactly this case: a massive planet far from a star. A check on the arithmetic. The ratio \( \dfrac{x_{\text{cm}}}{d} = \dfrac{7.42 \times 10^8}{7.785 \times 10^{11}} = 9.53 \times 10^{-4} \) should equal the mass fraction \( \dfrac{M_J}{M_{\text{Sun}}+M_J} = \dfrac{1.898 \times 10^{27}}{1.992 \times 10^{30}} = 9.53 \times 10^{-4} \) ✓ \( 7.42 \times 10^8 \) m from the Sun's center, just outside its surface

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. What quantity does a safety device actually reduce?
   Show the full solution

   The force

2. How does it reduce it?
   Show the full solution

   By extending the collision time

3. Can a crumple zone reduce the momentum change in a crash?
   Show the full solution

   No

4. Is a rigid car or a deformable one safer in a crash?
   Show the full solution

   Deformable, at the ends

5. Name the two separate collisions in a car crash.
   Show the full solution

   The car hitting the barrier, and the occupant hitting the interior

6. A 5.0 kg head moving at 6.0 m/s stops on impact. Find the force with no helmet (0.002 s) and with a helmet (0.020 s).
   Show the full solution

   \( \Delta p = 5.0 \times 6.0 = 30\ \text{kg}\cdot\text{m/s} \), the same either way. No helmet: \( F = \dfrac{30}{0.002} = 15000\ \text{N} \). Helmet: \( F = \dfrac{30}{0.020} = 1500\ \text{N} \). A tenfold reduction, and the whole function of the foam is to be crushed slowly rather than to be strong. 15000 N and 1500 N

7. Explain why a person landing on a mattress is less likely to be hurt than one landing on concrete, given that both stop from the same speed.
   Show the full solution

   Both undergo the same momentum change, since the mass and landing speed are the same. The mattress compresses, so the stop takes perhaps 0.5 s. Concrete does not, so the stop takes perhaps 0.01 s. With \( \Delta p \) fixed, the force differs by the ratio of those times, roughly fifty to one. The mattress is not absorbing momentum, which goes into Earth either way. It is extending the time. Same momentum change, far longer stopping time, far smaller force

8. An egg is dropped from 2.0 m onto a foam pad and survives. Find its landing speed, and explain what the foam must achieve for a 0.050 kg egg that breaks above 25 N.
   Show the full solution

   Landing speed, from lesson 1.6: \( v = \sqrt{2gh} = \sqrt{2(9.8)(2.0)} = \sqrt{39.2} = 6.26\ \text{m/s} \). \( \Delta p = 0.050 \times 6.26 = 0.313\ \text{kg}\cdot\text{m/s} \). For the force to stay below 25 N: \( \Delta t \ge \dfrac{\Delta p}{F} = \dfrac{0.313}{25} = 0.0125\ \text{s} \). The foam must stretch the stop to at least about 13 milliseconds. That is the design specification, stated as a number before any foam is chosen, which is what the standard asks for. 6.26 m/s; the stop must last at least 0.0125 s

9. Explain why making a car stronger and more rigid would make it more dangerous.
   Show the full solution

   Because rigidity shortens the collision time, and a shorter time means a larger force on the occupants for the same momentum change. The fixed quantity. In a crash from speed \( v \), the occupant's momentum must go from \( mv \) to zero. That change is set by the mass and the speed and is the same in any car. What rigidity does. A rigid structure resists deformation, so the car stops in a very short distance and therefore a very short time. With \( \Delta t \) small and \( \Delta p \) fixed, \( F = \dfrac{\Delta p}{\Delta t} \) is large. Put numbers on it. The worked example gave 420000 N for a 0.050 s rigid stop against 105000 N for a 0.20 s crumpling stop, from the identical crash. Why the intuition misleads. A rigid car looks undamaged afterward, which reads as "it survived better". But the damage to the car is the visible sign of energy being absorbed by the structure rather than by the people, so a badly crumpled front end is evidence the design worked. The necessary qualification. Not all rigidity is bad. The passenger compartment must stay rigid, because if it collapses the occupant's stopping distance is cut short by the structure arriving, and they may be struck directly. The design is soft at the ends and hard in the middle, and both halves follow from the same equation. A second reason rigidity hurts. A rigid car also rebounds more, and lesson 3.2 showed that a rebound requires a larger impulse than a dead stop. A car that bounces off a barrier subjects its occupants to a greater momentum change than one that stops against it. Rigidity shortens the stopping time, and force is momentum change divided by time

10. A designer must choose between two barrier designs for a 1200 kg vehicle striking at 20 m/s. Barrier A stops it in 0.15 s; barrier B stops it in 0.35 s but costs three times as much. Evaluate the options, state a criterion, and say what further information you would need.
    Show the full solution

    Compute the fixed momentum change. \( \Delta p = 1200 \times 20 = 24000\ \text{kg}\cdot\text{m/s} \), identical for both barriers. Barrier A. \( F = \dfrac{24000}{0.15} = 160000\ \text{N} \), that is \( 1.6 \times 10^5\ \text{N} \). Barrier B. \( F = \dfrac{24000}{0.35} = 68571\ \text{N} \), about \( 6.9 \times 10^4\ \text{N} \). Compare. Barrier B reduces the force to about 43 percent of barrier A's, a reduction of roughly 57 percent, for three times the cost. Convert to something meaningful for a person. The deceleration is what matters biologically. For barrier A: \( a = \dfrac{20}{0.15} = 133\ \text{m}/\text{s}^2 \), about 13.6 g. For barrier B: \( a = \dfrac{20}{0.35} = 57\ \text{m}/\text{s}^2 \), about 5.8 g. State a criterion before deciding. A defensible one: choose the cheapest barrier that keeps the occupant deceleration below the threshold associated with serious injury for a belted adult. Without that threshold stated, the comparison is just two numbers. What further information is needed. An injury threshold, from crash-test standards, so there is a line to be above or below. NHTSA publishes such criteria (US federal). The distances involved, since a 0.35 s stop at these speeds requires roughly 3.5 m of crush space, which the site may not have. The expected impact speeds, because a barrier optimized for 20 m/s may perform badly at 35 m/s. How often it is struck, to weigh the three times cost against the number of collisions it will mitigate. A provisional recommendation. If 13.6 g is below the injury threshold and the site cannot accommodate 3.5 m of crush space, barrier A is adequate and the extra cost is not justified. If the threshold sits between the two figures, barrier B is required regardless of cost, because the criterion is about injury rather than about relative improvement. The reasoning pattern worth taking away. Compute the fixed quantity first, then the quantity each design controls, then convert to something the criterion is actually written in terms of. A comparison of raw forces cannot decide anything until a threshold exists to compare them against. A 160000 N against 68571 N; the decision needs an injury threshold, the available crush distance and the expected impact speeds

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( G = 6.674 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2 \).

1. Write the law of universal gravitation.
   Show the full solution

   \( F = G\dfrac{m_1m_2}{r^2} \)

2. Give the value and units of \( G \).
   Show the full solution

   \( 6.674 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2 \)

3. From where is \( r \) measured?
   Show the full solution

   Center to center

4. If the distance doubles, what happens to the force?
   Show the full solution

   It falls to one quarter

5. Earth pulls the Moon with some force. How hard does the Moon pull Earth?
   Show the full solution

   Third-law pair. Exactly as hard

6. If the distance between two masses is tripled and one mass is doubled, what happens to the force?
   Show the full solution

   Doubling one mass doubles the force. Tripling the distance divides it by \( 3^2 = 9 \). Net: \( \dfrac{2}{9} \) of the original. \( \frac{2}{9} \) of the original

7. Find the gravitational force between a 70 kg person and a 90 kg person standing 0.50 m apart.
   Show the full solution

   \( F = \dfrac{(6.674 \times 10^{-11})(70)(90)}{0.50^2} = \dfrac{4.205 \times 10^{-7}}{0.25} = 1.68 \times 10^{-6}\ \text{N} \). About a millionth of a newton, roughly the weight of a large bacterium. Gravity between people is real and utterly negligible. \( 1.7 \times 10^{-6}\ \text{N} \)

8. A satellite orbits at an altitude equal to one Earth radius. By what factor is the gravitational force on it reduced compared with at the surface?
   Show the full solution

   At the surface, \( r = R_E \). At that altitude, \( r = R_E + R_E = 2R_E \). The force goes as \( \dfrac{1}{r^2} \), so doubling \( r \) divides the force by \( 4 \). The commonest error is treating "one Earth radius up" as \( r = R_E \), which forgets that the distance is measured from the center and gives no reduction at all. Reduced to one quarter

9. Explain why gravity shapes the universe despite being the weakest force.
   Show the full solution

   Because it never cancels, it accumulates with mass without limit, and its range is unlimited. It has only one sign. Every mass attracts every other mass. There is no negative mass to cancel the pull, so adding matter always adds attraction. Contrast with electricity. The electric force is about \( 10^{36} \) times stronger between two protons, but charge comes in two kinds. Ordinary matter contains almost exactly equal numbers of protons and electrons, so its net charge is nearly zero and its net electric force on distant objects is negligible. Contrast with the nuclear forces. The strong and weak forces are stronger still, but their range is about the size of a nucleus. They cannot reach from one atom to the next, let alone between stars. So gravity wins by default at large scales. It is the only force that both reaches far and fails to cancel. The scale at which it takes over. Roughly the size of an asteroid. Below that, chemical bonds hold an object's shape; above it, self-gravity does, which is why small bodies are irregular and large ones are round. Only attracts, accumulates without limit, and has unlimited range

10. Verify Newton's Moon test yourself: use \( g = 9.8\ \text{m}/\text{s}^2 \) at Earth's surface and the fact that the Moon is about 60 Earth radii away to predict its centripetal acceleration, then check it against its 27.3 day period at \( 3.84 \times 10^8\ \text{m} \).
    Show the full solution

    Prediction from the inverse square law. At 60 Earth radii the field is weaker by \( 60^2 = 3600 \): \( a = \dfrac{9.8}{3600} = 2.72 \times 10^{-3}\ \text{m}/\text{s}^2 \). Now compute what the orbit actually requires. Convert the period: \( T = 27.3 \times 24 \times 3600 = 2.359 \times 10^{6}\ \text{s} \). Centripetal acceleration for circular motion: \( a = \dfrac{4\pi^2 r}{T^2} \). Numerator: \( 4\pi^2(3.84 \times 10^8) = 39.478 \times 3.84 \times 10^8 = 1.516 \times 10^{10} \). Denominator: \( (2.359 \times 10^6)^2 = 5.565 \times 10^{12} \). \( a = \dfrac{1.516 \times 10^{10}}{5.565 \times 10^{12}} = 2.72 \times 10^{-3}\ \text{m}/\text{s}^2 \). The two agree to three significant figures. Why this is a genuine test and not a circular argument. The two calculations use entirely separate inputs. The first uses the acceleration of falling objects on Earth and a geometric ratio. The second uses the Moon's distance and its orbital period, both measured by astronomers. Nothing was fitted; no constant was adjusted. They agree because the same law governs both situations. What it would have meant if they disagreed. Either the force does not fall as \( \dfrac{1}{r^2} \), or the Moon is held by something other than what makes apples fall. Newton's check ruled out both at once. A note on the 60. \( \dfrac{3.84 \times 10^8}{6.37 \times 10^6} = 60.3 \), so the round number is a good approximation and the agreement survives using the exact value. Newton's own description of the test is that he compared the force required to keep the Moon in her orbit with the force of gravity at the surface of Earth, and found them to answer pretty nearly. The understatement is characteristic; the agreement is the foundation of the subject. Both give \( 2.72 \times 10^{-3}\ \text{m}/\text{s}^2 \)

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( M_E = 5.97 \times 10^{24}\ \text{kg} \) and \( R_E = 6.37 \times 10^{6}\ \text{m} \).

1. Write the formula for surface gravity.
   Show the full solution

   \( g = G\dfrac{M}{R^2} \)

2. Why does the falling object's mass not appear?
   Show the full solution

   It cancels between the weight and the gravitational force

3. Give an alternative unit for \( g \).
   Show the full solution

   N/kg

4. Is gravity significant at the altitude of the space station?
   Show the full solution

   About 89 percent of surface value. Yes

5. Surface gravity on the Moon is about what fraction of Earth's?
   Show the full solution

   About one sixth

6. Find \( g \) at a distance of two Earth radii from Earth's center.
   Show the full solution

   \( r = 2R_E = 1.274 \times 10^7\ \text{m} \). \( g = \dfrac{3.984 \times 10^{14}}{(1.274 \times 10^7)^2} = \dfrac{3.984 \times 10^{14}}{1.623 \times 10^{14}} = 2.46\ \text{m}/\text{s}^2 \). Check with the inverse square shortcut: \( \dfrac{9.82}{4} = 2.46 \) ✓ Doubling the distance quarters the field. \( 2.46\ \text{m}/\text{s}^2 \)

7. A planet has twice Earth's mass and twice its radius. Find its surface gravity.
   Show the full solution

   \( g = G\dfrac{2M}{(2R)^2} = G\dfrac{2M}{4R^2} = \dfrac{1}{2}G\dfrac{M}{R^2} \). So it is half Earth's, about \( 4.9\ \text{m}/\text{s}^2 \). Doubling the radius costs more than doubling the mass gains, because the radius is squared. This is why large low-density planets can have modest surface gravity. Half of Earth's

8. An 80 kg astronaut is on the space station. Find their weight there, and their apparent weight.
   Show the full solution

   Weight is the gravitational force, using \( g = 8.69\ \text{m}/\text{s}^2 \) at that altitude: \( W = 80 \times 8.69 = 695\ \text{N} \). On the ground it would be \( 80 \times 9.82 = 786\ \text{N} \), so their weight has fallen by only about 12 percent. Apparent weight is zero, because nothing supports them. They are in free fall, and lesson 2.5 showed the normal force is what a scale reads. Weight 695 N, apparent weight zero

9. Explain why \( g \) is a property of the planet rather than of the falling object.
   Show the full solution

   Because the falling object's mass cancels out of the derivation, leaving only quantities belonging to the planet. Follow the cancellation. The gravitational force on a mass \( m \) at the surface is \( G\dfrac{Mm}{R^2} \), and by the second law that produces an acceleration \( a = \dfrac{F}{m} \). Dividing removes \( m \) entirely: \( a = G\dfrac{M}{R^2} \). What is left. Only \( G \), a universal constant, and \( M \) and \( R \), which describe the planet. Nothing about what is falling survives. Why the cancellation happens. The same \( m \) appears in two places for two different reasons. It measures how strongly gravity pulls on the object, and it measures how strongly the object resists being accelerated. Those two roles are logically distinct, and it is an experimental fact that they are the same number. The consequence, stated plainly. A feather and an anvil dropped together in a vacuum land together, because \( g \) does not know anything about them. Lesson 1.6's Apollo 15 demonstration is this equation made visible. What \( g \) does depend on. Where you are. It falls with altitude as \( \dfrac{1}{r^2} \), it differs between planets, and it even varies slightly across Earth's surface because Earth is not a perfect uniform sphere. Every one of those is a change in the planet or the position, never in the object. Why this earns the name "field". Saying \( g \) belongs to the planet and to a location, rather than to the falling body, is exactly what a field is: a property assigned to every point in space, ready to act on whatever is placed there. The object's mass cancels, leaving only \( G \), the planet's mass and the distance

10. An astronaut can jump 0.50 m vertically on Earth. Estimate how high the same jump would take them on the Moon, and state what you assumed.
    Show the full solution

    Identify what stays the same. The astronaut's legs do the same work and give the same take-off speed, since that depends on muscle strength rather than on gravity. Call it \( v \). Relate jump height to take-off speed. From lesson 1.6, rising until the velocity reaches zero: \( v^2 = 2gh \), so \( h = \dfrac{v^2}{2g} \). Take the ratio between the two worlds. With \( v \) fixed, \( \dfrac{h_M}{h_E} = \dfrac{g_E}{g_M} \). Substitute the field strengths. \( \dfrac{9.82}{1.62} = 6.06 \). \( h_M = 0.50 \times 6.06 = 3.0\ \text{m} \). Check the take-off speed for reasonableness. \( v = \sqrt{2(9.82)(0.50)} = \sqrt{9.82} = 3.13\ \text{m/s} \), a plausible standing jump ✓ State the assumptions, because they matter. Same take-off speed. This is the key one and it is only roughly true. Legs also have to accelerate the astronaut's own mass, which is unchanged, so the muscles do behave similarly. But the push-off itself happens against a smaller weight, so the real take-off speed would be slightly higher. No spacesuit. An Apollo suit added over 80 kg of mass, which would cut the height substantially. The Apollo astronauts did not jump three meters, and this is the main reason. No air resistance, which is exact on the Moon and a small effect on Earth. Vertical jump only, measuring the rise of the center of mass rather than how high the feet get. Why the estimate is still worth making. It predicts the right order of magnitude and identifies the correct scaling, that jump height goes inversely with \( g \). Naming the assumptions is what turns a guess into an estimate, and the spacesuit assumption is the one that explains why the Apollo footage looks less dramatic than this number suggests. About 3.0 m, assuming the same take-off speed and no suit

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( g = 9.8\ \text{m}/\text{s}^2 \).

1. In which direction does centripetal acceleration point?
   Show the full solution

   Toward the center of the circle

2. Write the formula for centripetal acceleration in terms of speed and radius.
   Show the full solution

   \( a_c = \dfrac{v^2}{r} \)

3. Is centripetal force a new kind of force?
   Show the full solution

   No, it is a role filled by a real force

4. An object moves at 15 m/s in a circle of radius 25 m. Find its centripetal acceleration.
   Show the full solution

   \( \dfrac{225}{25} \). \( 9.0\ \text{m}/\text{s}^2 \)

5. What supplies the centripetal force for a car turning on a level road?
   Show the full solution

   Friction from the road

6. A 0.30 kg ball on a 1.2 m string is whirled in a horizontal circle at 4.0 m/s. Find the tension.
   Show the full solution

   The tension supplies the centripetal force. \( T = \dfrac{mv^2}{r} = \dfrac{(0.30)(16)}{1.2} = \dfrac{4.8}{1.2} = 4.0\ \text{N} \). 4.0 N

7. If the speed of a car on a curve doubles, what happens to the required friction force?
   Show the full solution

   \( F_c = \dfrac{mv^2}{r} \) depends on the square of the speed. It quadruples. Why this matters on the road. A curve safely taken at 40 km/h needs four times the grip at 80 km/h. Since the available friction does not change, the margin disappears far faster than the speed increases, which is the same quadratic trap as stopping distance in lesson 1.5. It quadruples

8. A satellite orbits at radius \( 7.0 \times 10^6\ \text{m} \) with a period of 5800 s. Find its centripetal acceleration.
   Show the full solution

   Use the period form. \( a_c = \dfrac{4\pi^2r}{T^2} = \dfrac{4\pi^2(7.0 \times 10^6)}{(5800)^2} \). Numerator: \( 39.478 \times 7.0 \times 10^6 = 2.763 \times 10^8 \). Denominator: \( 3.364 \times 10^7 \). \( a_c = 8.2\ \text{m}/\text{s}^2 \). Check against gravity at that radius: \( \dfrac{3.984 \times 10^{14}}{(7.0 \times 10^6)^2} = 8.13\ \text{m}/\text{s}^2 \) ✓ They agree because gravity is what supplies the centripetal force, which is lesson 4.4's whole point. About \( 8.2\ \text{m}/\text{s}^2 \)

9. Explain why an object in circular motion at constant speed is still accelerating.
   Show the full solution

   Because acceleration is the rate of change of velocity, and velocity is a vector that changes when its direction changes even if its length does not. The definitions are what settle it. Speed is a scalar, the magnitude alone. Velocity is a vector, carrying magnitude and direction. Acceleration is \( \dfrac{\Delta\vec{v}}{\Delta t} \), built from velocity rather than speed. So a turn is a change. An object moving north at 20 m/s that is later moving east at 20 m/s has a different velocity, and something must have changed it. How large the change is. Not zero, and not small. A car at 20 m/s on an 80 m curve has \( a_c = \dfrac{400}{80} = 5\ \text{m}/\text{s}^2 \), about half of \( g \), which is why a fast turn is so noticeable. Why it is perpendicular. Lesson 2.3 showed that the component of net force along the velocity changes the speed, and the perpendicular component changes the direction. Constant speed means the parallel component is zero, so the whole acceleration is perpendicular, pointing at the center. The consequence for forces. Since the acceleration is nonzero, the net force is nonzero, so something must be pushing inward. Anything moving in a curve is being pushed, and finding what is doing it is the first step of every problem in this lesson. What would happen if the force vanished. The object would leave along a tangent in a straight line, not fly outward radially. That is the first law, and it is what a stone released from a sling actually does. Velocity includes direction, so turning changes it

10. A car of mass 1000 kg rounds a curve of radius 60 m. The road is dry with \( \mu_s = 0.70 \). Find the maximum safe speed, and find how it changes if the road is wet with \( \mu_s = 0.35 \).
    Show the full solution

    Set up the condition. Friction must supply the centripetal force, and it cannot exceed its maximum: \[ \frac{mv^2}{r} \le \mu_s mg \] The mass cancels, so the answer is the same for every car on that surface: \[ v^2 \le \mu_s g r \] Dry road. \( v^2 \le (0.70)(9.8)(60) = 411.6 \) \( v \le \sqrt{411.6} = 20.3\ \text{m/s} \), about 73 km/h. Wet road. \( v^2 \le (0.35)(9.8)(60) = 205.8 \) \( v \le \sqrt{205.8} = 14.3\ \text{m/s} \), about 52 km/h. Note the relationship between the two. Halving the friction did not halve the safe speed. It reduced it by a factor of \( \sqrt{2} = 1.41 \), from 20.3 to 14.3, because the speed appears squared. Why that is dangerous rather than reassuring. The speed only needs to fall by 29 percent, which sounds modest, but a driver who keeps their dry-road speed on a wet curve is demanding twice the friction available, not 1.4 times. The quadratic hides how far past the limit they are. Check the mass really does not matter. A heavier car needs more centripetal force, in proportion to \( m \), and has more friction available, also in proportion to \( m \). The two scale identically and cancel. A loaded truck and an empty car slide at the same speed on the same curve, which is not what intuition suggests. What the model leaves out. Banking the road adds a component of the normal force pointing inward, which raises the safe speed considerably and is why highway curves are banked. Tire condition, temperature and load transfer during the turn all matter too. The calculation gives the flat-road limit, which is the conservative case. 20.3 m/s dry, 14.3 m/s wet

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( GM_E = 3.984 \times 10^{14}\ \text{m}^3/\text{s}^2 \) and \( R_E = 6.37 \times 10^{6}\ \text{m} \).

1. What force keeps a satellite in orbit?
   Show the full solution

   Gravity

2. Write the formula for circular orbital speed.
   Show the full solution

   \( v = \sqrt{\dfrac{GM}{r}} \)

3. Does a satellite's own mass affect its orbital speed?
   Show the full solution

   No

4. Is a higher orbit faster or slower?
   Show the full solution

   Slower

5. What is an orbiting astronaut's acceleration?
   Show the full solution

   Gravity is the only force acting. The local value of \( g \), directed toward Earth's center

6. Find the orbital speed at a radius of \( 1.0 \times 10^7\ \text{m} \).
   Show the full solution

   \( v = \sqrt{\dfrac{3.984 \times 10^{14}}{1.0 \times 10^{7}}} = \sqrt{3.984 \times 10^{7}} = 6.31 \times 10^{3}\ \text{m/s} \). About 6.31 km/s

7. Find the period of that orbit.
   Show the full solution

   \( T = \dfrac{2\pi r}{v} = \dfrac{2\pi(1.0 \times 10^7)}{6.31 \times 10^3} = \dfrac{6.283 \times 10^7}{6.31 \times 10^3} = 9.96 \times 10^{3}\ \text{s} \). That is about 9960 s, or 2.77 hours. Longer than the station's 92 minutes, as a higher orbit must be: further to go and moving more slowly ✓ About \( 1.0 \times 10^4 \) s, or 2.8 hours

8. A satellite orbits at four times Earth's radius from the center. By what factor is its speed smaller than one orbiting just above the surface?
   Show the full solution

   \( v \propto \dfrac{1}{\sqrt{r}} \), so quadrupling \( r \) divides the speed by \( \sqrt{4} = 2 \). Half the speed. Check with numbers. At the surface, \( v = \sqrt{\dfrac{3.984 \times 10^{14}}{6.37 \times 10^6}} = 7.91\ \text{km/s} \). At \( 4R_E = 2.548 \times 10^7 \), \( v = \sqrt{\dfrac{3.984 \times 10^{14}}{2.548 \times 10^7}} = 3.95\ \text{km/s} \) ✓ Half

9. Explain why an astronaut in orbit is weightless, using the ideas of this lesson.
   Show the full solution

   Because they are in free fall, so no surface pushes on them, and the apparent weight that lesson 2.5 identified as the normal force is zero. Gravity is present and strong. Lesson 4.2 computed \( 8.69\ \text{m}/\text{s}^2 \) at the station's altitude, about 89 percent of the surface value. The astronaut's weight is nearly what it is on the ground. What is absent is support. The station, the astronaut and everything loose inside are all accelerating toward Earth at the same rate, because that acceleration does not depend on mass. Nothing needs to push on anything else, so every normal force is zero. Why the fall never ends. The station has a sideways speed of 7.67 km/s. As it falls, Earth's surface curves away beneath it at the same rate, so it never gets closer. That is Newton's cannonball, and the orbit is the fall. The equivalence with a dropped elevator. Lesson 2.5 found a scale reads zero inside a freely falling lift. Orbit is that situation, made permanent by sideways motion. Why the word "weightless" is misleading. It suggests gravity has gone. The condition is free fall, and NASA's term "microgravity" acknowledges that the tiny residual accelerations come from atmospheric drag and from the station's size, not from any weakening of gravity. A test of the understanding. If gravity really were absent at that altitude, the station would travel in a straight line and leave, by the first law. That it curves around Earth is proof gravity is acting. They are in free fall, so nothing supports them

10. A spacecraft is in a circular orbit at \( 7.0 \times 10^6\ \text{m} \) and needs to catch another craft 100 km ahead in the same orbit. Explain why firing the engine forward makes it fall further behind, and describe what it should do instead.
    Show the full solution

    Compute the current orbit. \( v = \sqrt{\dfrac{3.984 \times 10^{14}}{7.0 \times 10^{6}}} = \sqrt{5.691 \times 10^{7}} = 7.54\ \text{km/s} \). \( T = \dfrac{2\pi(7.0 \times 10^6)}{7.54 \times 10^3} = \dfrac{4.398 \times 10^7}{7.54 \times 10^3} = 5833\ \text{s} \), about 97 minutes. What firing forward does. It adds kinetic energy. The craft is now moving faster than the circular speed for that radius, so gravity is insufficient to hold it in that circle and it climbs to a higher orbit. Why that is self-defeating. At the larger radius the required orbital speed is lower, and the circumference is longer. Both effects increase the period. The craft takes longer per lap and steadily falls behind the target it was trying to catch. Make it quantitative. Suppose the burn raises the orbit to \( 7.1 \times 10^6\ \text{m} \). \( v = \sqrt{\dfrac{3.984 \times 10^{14}}{7.1 \times 10^{6}}} = 7.49\ \text{km/s} \), slower than before. \( T = \dfrac{2\pi(7.1 \times 10^6)}{7.49 \times 10^3} = 5956\ \text{s} \). The period grew by about 123 s per orbit, so the craft loses roughly two minutes of angular position every lap. What it should do instead. Fire the engine backward, against its motion. That removes energy and drops it into a lower orbit, where the required speed is higher and the circumference shorter, so the period is shorter. It catches up from below. Then complete the maneuver. Once it has gained the required angular lead, fire forward to climb back to the original radius, arriving alongside the target. Check the direction of the lower orbit. At \( 6.9 \times 10^6\ \text{m} \): \( v = 7.60\ \text{km/s} \) and \( T = \dfrac{4.335 \times 10^7}{7.60 \times 10^3} = 5704\ \text{s} \), about 129 s shorter per orbit ✓ It gains rather than loses. Why this is genuinely counterintuitive. Every intuition about chasing comes from driving, where accelerating closes a gap. In orbit, speed and altitude are locked together by \( v = \sqrt{\dfrac{GM}{r}} \), and an engine burn changes the orbit rather than simply the speed. The rule of thumb pilots use is that to catch up you slow down, and the formula above is the reason. Slow down to drop lower and catch up, then climb back

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. What shape are planetary orbits?
   Show the full solution

   Ellipses, with the Sun at one focus

2. State Kepler's third law.
   Show the full solution

   \( T^2 \propto a^3 \)

3. Where in its orbit does a planet move fastest?
   Show the full solution

   Closest to the Sun

4. What is at the other focus of a planet's ellipse?
   Show the full solution

   Nothing

5. What quantity does Kepler's second law reflect?
   Show the full solution

   Angular momentum, conserved

6. Mars orbits at 1.52 times Earth's distance from the Sun. Find its period in years.
   Show the full solution

   Using \( T^2 \propto a^3 \) with Earth as the reference, so \( T \) in years and \( a \) in astronomical units: \( T^2 = a^3 = 1.52^3 = 3.51 \) \( T = \sqrt{3.51} = 1.87\ \text{years} \). The observed value is 1.88 years ✓ Using Earth units makes the constant equal to one, which removes \( G \) and the Sun's mass from the arithmetic entirely. About 1.87 years

7. A planet orbits a star with a period of 8.0 years at a distance of 4.0 AU. A second planet orbits the same star at 1.0 AU. Find its period.
   Show the full solution

   Take the ratio, which cancels the unknown stellar mass: \( \dfrac{T_1^2}{T_2^2} = \dfrac{a_1^3}{a_2^3} \) \( \dfrac{64}{T_2^2} = \dfrac{64}{1} \) \( T_2^2 = 1.0 \), so \( T_2 = 1.0\ \text{year} \). Using a ratio avoids needing the star's mass, which is usually the point of setting these problems up this way. 1.0 year

8. Io orbits Jupiter at \( 4.22 \times 10^8\ \text{m} \) with a period of \( 1.53 \times 10^5\ \text{s} \). Find Jupiter's mass.
   Show the full solution

   Rearrange the third law for the central mass: \( M = \dfrac{4\pi^2 a^3}{GT^2} \). Cube the radius: \( (4.22 \times 10^8)^3 = 7.515 \times 10^{25} \). Numerator: \( 39.478 \times 7.515 \times 10^{25} = 2.967 \times 10^{27} \). Denominator: \( (6.674 \times 10^{-11})(1.53 \times 10^5)^2 = (6.674 \times 10^{-11})(2.341 \times 10^{10}) = 1.562 \). \( M = \dfrac{2.967 \times 10^{27}}{1.562} = 1.90 \times 10^{27}\ \text{kg} \). The accepted value is \( 1.898 \times 10^{27}\ \text{kg} \) ✓ This is how the mass of any body with a moon is found, and it is the only practical way to weigh a planet. About \( 1.90 \times 10^{27} \) kg

9. Explain why Kepler's third law needs the mass of the body being orbited.
   Show the full solution

   Because that mass sets the strength of the gravitational field, and therefore how fast something must move at a given radius. Look at the constant. Newton's derivation gives \( T^2 = \dfrac{4\pi^2}{GM}a^3 \). The proportionality constant is \( \dfrac{4\pi^2}{GM} \), and \( M \) is the central mass. Why it enters physically. A more massive center pulls harder at the same distance, so a larger centripetal force is available, so an orbit at that radius is faster and its period shorter. Doubling the central mass shortens every period at a given radius by \( \sqrt{2} \). What does not enter. The orbiting body's own mass, which canceled in the derivation. A large moon and a small one at the same radius have the same period. The practical consequence. Kepler's original statement, that \( T^2 \propto a^3 \) with the same constant for all the planets, is correct only because they all orbit the same Sun. It does not extend to moons of Jupiter, which obey their own version with Jupiter's mass in the constant. Why this is a feature rather than a nuisance. Turning the equation around makes it a scale. Measure a satellite's orbital radius and period, and the central mass follows. That is how the masses of Earth, the Sun, Jupiter and distant stars with planets are all determined, and there is no other direct method. The constant is \( \dfrac{4\pi^2}{GM} \), so it depends on the central mass

10. Halley's Comet has a period of 76 years. Find its semi-major axis in astronomical units, and explain why it travels so much faster at perihelion than at aphelion.
    Show the full solution

    Apply the third law in Earth units, where \( T \) is in years and \( a \) in AU, so the constant is 1: \( T^2 = a^3 \) \( 76^2 = a^3 \) \( 5776 = a^3 \) \( a = 5776^{1/3} \). Take the cube root. \( 17^3 = 4913 \) and \( 18^3 = 5832 \), so the answer is just under 18. More precisely \( a = 17.9\ \text{AU} \). Interpret that. The semi-major axis is about 17.9 AU, roughly the distance of Uranus. But the orbit is extremely elongated: the comet comes inside Earth's orbit at perihelion, about 0.59 AU, and goes out beyond Neptune at aphelion, about 35 AU. Check the geometry. For an ellipse the two extremes average to the semi-major axis: \( \dfrac{0.59 + 35}{2} = 17.8 \) ✓ consistent with the computed 17.9. Now the speed question, via the second law. Equal areas are swept in equal times. Near perihelion the radius is about 0.59 AU; near aphelion it is about 35 AU, some 59 times larger. A thin swept triangle has area roughly \( \tfrac{1}{2}rv\Delta t \). Holding that constant while \( r \) grows by a factor of 59 requires \( v \) to fall by a factor of about 59. So the comet moves roughly sixty times faster at perihelion. It spends a few months racing through the inner solar system and decades crawling through the outer part. The angular momentum statement. Gravity points straight at the Sun, so it cannot change the comet's angular momentum about the Sun, and \( mvr \) is constant. That single conserved quantity is what the second law expresses geometrically. Why this matters for observing it. Halley is visible from Earth only during the brief perihelion passage. Its 76 year period is almost entirely spent far away and moving slowly, which is why a person is unlikely to see it twice. About 17.9 AU; it moves roughly sixty times faster at perihelion because angular momentum is conserved

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( GM_E = 3.984 \times 10^{14} \), \( R_E = 6.37 \times 10^{6}\ \text{m} \).

1. What is the period of a geostationary orbit?
   Show the full solution

   One day

2. Over what part of Earth must a geostationary satellite sit?
   Show the full solution

   The equator

3. Give the approximate altitude of a geostationary orbit.
   Show the full solution

   About 35800 km

4. Write the formula for escape speed.
   Show the full solution

   \( v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}} \)

5. Give Earth's escape speed.
   Show the full solution

   About 11.2 km/s

6. Find the escape speed from a planet with twice Earth's mass and twice its radius.
   Show the full solution

   \( v = \sqrt{\dfrac{2G(2M)}{2R}} = \sqrt{\dfrac{2GM}{R}} \). The factors of 2 cancel, so the escape speed is the same as Earth's, 11.2 km/s. A useful check on the structure of the formula, since mass and radius enter as a ratio. 11.2 km/s

7. Does a heavier spacecraft need a higher escape speed than a lighter one?
   Show the full solution

   No. The escaping mass cancels in the derivation, exactly as it did for \( g \) and for orbital speed. What does change with mass is the energy required, since \( \tfrac{1}{2}mv^2 \) grows with \( m \). A heavier craft needs more fuel to reach the same speed, but the speed itself is the same. No, the speed is identical

8. Find the escape speed from Mars, with \( M = 6.42 \times 10^{23}\ \text{kg} \) and \( R = 3.39 \times 10^{6}\ \text{m} \).
   Show the full solution

   \( GM = (6.674 \times 10^{-11})(6.42 \times 10^{23}) = 4.285 \times 10^{13} \). \( v = \sqrt{\dfrac{2(4.285 \times 10^{13})}{3.39 \times 10^{6}}} = \sqrt{2.528 \times 10^{7}} = 5.03 \times 10^{3}\ \text{m/s} \). About 5.0 km/s, roughly 45 percent of Earth's, which is part of why Mars has retained only a thin atmosphere. About 5.0 km/s

9. Explain why escape speed does not depend on the direction of launch.
   Show the full solution

   Because the derivation uses energy, and kinetic energy depends on speed alone, not on direction. Follow the argument. The condition for escape is that the total energy be at least zero: \( \tfrac{1}{2}mv^2 - \dfrac{GMm}{r} \ge 0 \). The first term contains \( v^2 \), a scalar. Nothing in the expression records which way the object is going. Why gravitational potential energy does not care either. It depends only on the distance \( r \) from the center, not on the direction of approach. Two points the same distance out have the same potential energy. So a projectile fired sideways escapes. Launched horizontally at 11.2 km/s from a mountain top, it follows a long curved path and never returns, provided it does not run into the planet on the way. That last clause is the practical catch. A shallow launch would intersect the ground or plough through kilometers of dense atmosphere, losing energy to drag. Real launches go up first for that reason, not because direction affects the speed requirement. Why the airless case is the clean one. On the Moon, a projectile fired in any direction at 2.37 km/s that clears the surface escapes. The direction is genuinely irrelevant. Contrast with orbital speed, which does depend on direction. To circle you must move perpendicular to the radius. Escaping has no such requirement, because it is a statement about energy rather than about a particular trajectory. Energy depends on speed, not direction

10. A satellite is to be placed in geostationary orbit. Find its orbital speed, compare it with the speed of a point on Earth's equator, and explain why launches head east from sites near the equator.
    Show the full solution

    Find the orbital speed. Use the radius from the worked example, \( r = 4.216 \times 10^7\ \text{m} \): \( v = \sqrt{\dfrac{3.984 \times 10^{14}}{4.216 \times 10^{7}}} = \sqrt{9.450 \times 10^{6}} = 3.07 \times 10^{3}\ \text{m/s} \), about 3.07 km/s. Check it a second way. \( v = \dfrac{2\pi r}{T} = \dfrac{2\pi(4.216 \times 10^7)}{86164} = \dfrac{2.649 \times 10^8}{86164} = 3.07 \times 10^3\ \text{m/s} \) ✓ Note it is much slower than the space station's 7.67 km/s, as a higher orbit must be. Now the speed of a point on the equator. It travels Earth's circumference in one sidereal day: \( v = \dfrac{2\pi R_E}{T} = \dfrac{2\pi(6.37 \times 10^6)}{86164} = \dfrac{4.002 \times 10^7}{86164} = 464\ \text{m/s} \). Why that matters for a launch. A rocket sitting on the pad at the equator is already moving east at 464 m/s relative to Earth's center. Launching eastward keeps that speed and adds to it, so the rocket needs 464 m/s less from its own engines. How much of a saving that is. Reaching low Earth orbit needs roughly 7.8 km/s, so 464 m/s is about 6 percent of the requirement. Because fuel requirements grow steeply with the speed needed, a 6 percent reduction translates into a considerably larger fraction of payload. Why the launch site's latitude matters. The eastward speed falls with the cosine of the latitude. At the equator it is the full 464 m/s; at 45 degrees it is about 328 m/s; at the poles it is zero. And there is a second reason for equatorial sites, specific to this orbit. A geostationary satellite must end up in the equatorial plane. A rocket launched from a higher latitude enters an inclined orbit and must spend extra fuel on a plane change, which is one of the most expensive maneuvers in spaceflight. Launching from near the equator avoids it. Which is why the sites are where they are. Kourou in French Guiana at 5 degrees north, Cape Canaveral at 28 degrees, and Sea Launch operating from a floating platform on the equator itself. Launching westward is possible and costs double. The rocket must first cancel its 464 m/s eastward motion and then build speed the other way, a penalty of 928 m/s. It is done only when the mission requires a retrograde orbit. 3.07 km/s orbital, against 464 m/s at the equator; eastward equatorial launches save both speed and a plane change

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. What causes tides?
   Show the full solution

   The difference in gravitational pull across Earth

2. How many tidal bulges are there?
   Show the full solution

   Two, on opposite sides

3. How does tidal force vary with distance?
   Show the full solution

   As \( \dfrac{1}{r^3} \)

4. When do spring tides occur?
   Show the full solution

   At new and full moon, when Sun and Moon align

5. At what rate is the Moon receding?
   Show the full solution

   About 3.8 cm per year

6. If the Moon were twice as far away, by what factor would its tidal effect change?
   Show the full solution

   Tidal effect goes as \( \dfrac{1}{r^3} \), so doubling \( r \) divides it by \( 2^3 = 8 \). Reduced to one eighth. Note that its ordinary gravitational pull would only fall to one quarter, so the tidal effect is far more sensitive to distance than the force itself. One eighth

7. Explain why the far side of Earth has a tidal bulge.
   Show the full solution

   Because it is pulled toward the Moon less strongly than Earth's center is, so relative to the center it falls behind, which shows up as a bulge pointing away from the Moon. The whole planet is falling toward the Moon. What matters for the shape is the difference between each part's acceleration and the center's. Near side: pulled harder than the center, so it pulls ahead, toward the Moon. Far side: pulled less than the center, so it lags, away from the Moon. Both are stretches in the same sense, which is why the two bulges are symmetric. Why the centrifugal explanation is best avoided. It requires an accelerating frame and a fictitious force, and it obscures the fact that a pure difference in gravity produces both bulges with no rotation needed at all. It is pulled less than the center, so it lags behind

8. Earth's day is lengthening by about 2 milliseconds per century. Estimate the change over 100 million years.
   Show the full solution

   \( 100 \) million years is \( 10^6 \) centuries. \( 2\ \text{ms} \times 10^6 = 2 \times 10^6\ \text{ms} = 2000\ \text{s} \). That is about 0.56 hours, so the day was roughly 23.4 hours long. Check against the fossil evidence. A shorter day means more days per year, and Devonian corals about 400 million years old show roughly 400 growth bands per year, implying a day near 22 hours ✓ consistent in direction and rough size. Why this is only an estimate. The rate is not constant. It depends on the arrangement of continents and ocean basins, which governs how much tidal friction occurs, and those have changed enormously. About 2000 s shorter, a day of roughly 23.4 hours

9. Explain why the Sun's tidal effect is weaker than the Moon's, and what the comparison would be if tides fell off as \( \dfrac{1}{r^2} \) instead.
   Show the full solution

   Because tides depend on \( \dfrac{M}{r^3} \), and the Sun's vastly greater distance, cubed, more than cancels its vastly greater mass. The two ratios. The Sun is about \( 2.7 \times 10^7 \) times more massive than the Moon and about 390 times further away. With the cube. \( 390^3 = 5.9 \times 10^7 \), so the ratio of solar to lunar tides is \( \dfrac{2.7 \times 10^7}{5.9 \times 10^7} = 0.46 \). The Sun contributes a little under half as much. Now suppose tides went as \( \dfrac{1}{r^2} \). \( 390^2 = 1.5 \times 10^5 \), giving a ratio of \( \dfrac{2.7 \times 10^7}{1.5 \times 10^5} = 180 \). The Sun would dominate tides by a factor of nearly two hundred. What we would observe in that case. Tides would follow the solar day exactly, high tide would occur at the same clock time every day, and the Moon would be irrelevant to them. That is emphatically not what happens: tides run on a 24 hour 50 minute cycle, matching the Moon's apparent motion rather than the Sun's. So the observation discriminates between the two models. The timing of tides is direct evidence that the dependence is \( \dfrac{1}{r^3} \) and not \( \dfrac{1}{r^2} \). And the spring-neap cycle confirms the ratio. If the Sun contributed nothing, tidal range would be constant; if it dominated, the Moon would not matter. That the range varies by roughly a factor of two over a fortnight is what a contribution of about 46 percent predicts. The reason for the extra power of \( r \). A tide is a difference between the field at two nearby points. Differencing an inverse square law over a small separation produces an inverse cube, which is the same mathematical step as the one that makes a rate of change one power steeper. Distance cubed beats mass; with \( \dfrac{1}{r^2} \) the Sun would dominate by about 180 times

10. The Moon is receding at 3.8 cm per year. Estimate how long until the day and the month are equal in length, and state why the simple extrapolation is unreliable.
    Show the full solution

    What the end state means. Tidal friction slows Earth's spin and lifts the Moon, lengthening both the day and the month. The process stops when they match, so Earth keeps one face permanently toward the Moon. That state is called mutual tidal locking, and the Moon has already reached its half of it, which is why it shows us one face. Set up a crude estimate. The day is lengthening at about 2 ms per century, or \( 2 \times 10^{-5}\ \text{s} \) per year. The day must grow from 24 hours to something like the eventual month, which would be perhaps 47 of today's days, so roughly \( 4 \times 10^6\ \text{s} \). Divide. \( \dfrac{4 \times 10^6\ \text{s}}{2 \times 10^{-5}\ \text{s/yr}} = 2 \times 10^{11}\ \text{years} \), about 200 billion years. Compare that with other timescales. The universe is about \( 1.4 \times 10^{10} \) years old, and the Sun will become a red giant and engulf or scorch Earth in roughly \( 5 \times 10^9 \) years. The locking timescale exceeds both by a wide margin, so it will not happen. Why the extrapolation is unreliable, beyond that. The rate is not constant. Tidal friction depends on how ocean basins are shaped, and continental drift rearranges them completely over a hundred million years. The present rate is unusually high because the current ocean basins happen to resonate well with the tidal period. The recession slows itself. As the Moon moves out, the tidal effect falls as \( \dfrac{1}{r^3} \), so the process weakens as it proceeds. A linear extrapolation from today's rate overestimates the speed of later stages. The Sun intervenes first. Long before any of this, the Sun's evolution changes the situation entirely. Running the extrapolation backward fails too. At the present rate the Moon would have been touching Earth about 1.5 billion years ago, which contradicts the evidence that it formed over 4 billion years ago. That contradiction is itself the proof that the rate has not been constant, and it is a genuine result rather than a flaw in the reasoning. The honest conclusion. The direction of the process is certain and measured. The timescale from a linear extrapolation is an order-of-magnitude figure at best, and in this case it exceeds the lifetime of the Earth, so the end state will never be reached. Order \( 10^{11} \) years, which is longer than Earth has left; the rate is not constant and the effect weakens as the Moon recedes

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( g = 9.8\ \text{m}/\text{s}^2 \).

1. Write the formula for work.
   Show the full solution

   \( W = Fd\cos\theta \)

2. Give the SI unit of work.
   Show the full solution

   The joule

3. How much work is done by a force if the object does not move?
   Show the full solution

   Zero

4. A 20 N force moves an object 5.0 m in its own direction. Find the work.
   Show the full solution

   100 J

5. What is the work done by a force perpendicular to the motion?
   Show the full solution

   Zero

6. A 120 N force acts at \( 30^\circ \) to the displacement of 8.0 m. Find the work.
   Show the full solution

   \( W = (120)(8.0)\cos 30^\circ = 960 \times 0.866 = 831\ \text{J} \). Only the component along the motion counts, \( 120\cos 30^\circ = 104\ \text{N} \), and \( 104 \times 8.0 = 831\ \text{J} \) ✓ About 831 J

7. A 3.0 kg object is lifted 2.5 m at constant speed. Find the work done by the lifting force.
   Show the full solution

   At constant speed the lifting force equals the weight: \( F = mg = 3.0 \times 9.8 = 29.4\ \text{N} \). Force and displacement are both upward, so \( \theta = 0 \): \( W = 29.4 \times 2.5 = 73.5\ \text{J} \). 73.5 J

8. Friction of 12 N acts on a box sliding 4.0 m. Find the work done by friction.
   Show the full solution

   Friction opposes the motion, so \( \theta = 180^\circ \) and \( \cos 180^\circ = -1 \): \( W = (12)(4.0)(-1) = -48\ \text{J} \). Negative work removes 48 J from the box, converting it into thermal energy in the surfaces. \( -48\ \text{J} \)

9. Explain why holding a heavy object still does no work, even though it is tiring.
   Show the full solution

   Because work requires displacement, and a stationary object has none, so no energy reaches it. Apply the definition. \( W = Fd\cos\theta \) with \( d = 0 \) gives \( W = 0 \) regardless of how large \( F \) is. Check it against energy. Work is a transfer of energy. A motionless object at a fixed height has unchanging kinetic and potential energy, so nothing was transferred to it. The definition and the energy bookkeeping agree. Where the effort goes. Muscle does not hold a load passively the way a shelf does. Individual fibers repeatedly contract and release, each cycle consuming chemical energy from ATP and releasing it as heat within the muscle. That is why you grow warm and tired holding something still. So energy is consumed but not transferred. The energy goes from chemical stores to thermal energy inside the person, never reaching the object. The contrast that makes it clear. A table supporting the same box exerts the same upward force indefinitely and consumes no energy at all. If holding required work in the physics sense, the table would need a power supply. Why the definition is worth defending. Defining work as effort would break the conservation laws in the rest of this unit, since the box's energy would have to increase without its speed or height changing. The narrow definition is what makes energy accounting possible. No displacement means no work, and no energy reaches the object

10. A 25 kg crate is dragged 12 m across a floor by a rope at \( 35^\circ \) above the horizontal with a tension of 90 N, against friction of 60 N. Find the work done by each force and the net work.
    Show the full solution

    List the four forces and their angles to the displacement. The displacement is horizontal, 12 m forward. Tension. At \( 35^\circ \) to the displacement: \( W_T = (90)(12)\cos 35^\circ = 1080 \times 0.8192 = 885\ \text{J} \). Friction. Directly opposed, \( \theta = 180^\circ \): \( W_f = (60)(12)(-1) = -720\ \text{J} \). Weight. Vertical, displacement horizontal, \( \theta = 90^\circ \): \( W_g = 0 \). Normal force. Also vertical, so \( W_N = 0 \). Net work. \( 885 - 720 + 0 + 0 = 165\ \text{J} \). Check by a second route. Find the net force along the motion first. The horizontal component of tension is \( 90\cos 35^\circ = 73.7\ \text{N} \), and friction is 60 N backward, so the net horizontal force is \( 13.7\ \text{N} \). \( W_{\text{net}} = 13.7 \times 12 = 165\ \text{J} \) ✓ What the positive net work means. The crate gains 165 J of kinetic energy, so it is speeding up. Lesson 5.2 turns that into a speed. Two things worth noticing. The vertical forces do no work even though the normal force is large, because nothing moves vertically. And the tension's vertical component, \( 90\sin 35^\circ = 51.6\ \text{N} \), reduces the normal force and therefore the friction, which is why dragging at an angle can be easier than pulling horizontally despite wasting part of the force. Tension 885 J, friction \( -720 \) J, weight and normal 0, net 165 J

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write the formula for kinetic energy.
   Show the full solution

   \( KE = \tfrac{1}{2}mv^2 \)

2. State the work-energy theorem.
   Show the full solution

   \( W_{\text{net}} = \Delta KE \)

3. Can kinetic energy be negative?
   Show the full solution

   The speed is squared. No

4. Find the kinetic energy of a 2.0 kg object at 3.0 m/s.
   Show the full solution

   \( \tfrac{1}{2}(2.0)(9.0) \). 9.0 J

5. If speed doubles, what happens to kinetic energy?
   Show the full solution

   It quadruples

6. Find the kinetic energy of a 0.145 kg baseball at 40 m/s.
   Show the full solution

   \( \tfrac{1}{2}(0.145)(1600) = 116\ \text{J} \). Compare with lesson 3.1, where its momentum was only \( 5.8\ \text{kg}\cdot\text{m/s} \) against a car's 30000. On energy the gap is far smaller, because the high speed is squared. 116 J

7. A net force of 200 N acts over 15 m on an object initially at rest. Find its final kinetic energy.
   Show the full solution

   \( W_{\text{net}} = 200 \times 15 = 3000\ \text{J} \). Starting from rest, \( \Delta KE = KE_f \), so \( KE_f = 3000\ \text{J} \). The mass was not needed for the energy, though it would be needed to find the speed. 3000 J

8. A 1500 kg car slows from 25 m/s to 15 m/s. Find the work done by the brakes.
   Show the full solution

   \( KE_i = \tfrac{1}{2}(1500)(625) = 468750\ \text{J} \) \( KE_f = \tfrac{1}{2}(1500)(225) = 168750\ \text{J} \) \( W = \Delta KE = 168750 - 468750 = -300000\ \text{J} \). Negative, because the brakes remove energy. The magnitude, \( 3.0 \times 10^5\ \text{J} \), is what ends up as heat in the brake discs. \( -3.0 \times 10^5\ \text{J} \)

9. Explain why the work-energy theorem is useful even though it follows from Newton's second law.
   Show the full solution

   Because it eliminates time and acceleration from the problem, and because it handles varying forces that the second law is awkward with. What it removes. The theorem relates force and distance directly to speed. Problems asking "how fast will it be going after this distance" need no acceleration and no duration. A concrete saving. To find the speed of a ball dropped 20 m using forces, you find the acceleration, then the time, then the speed. Using energy, you set \( mgh = \tfrac{1}{2}mv^2 \) and get \( v = \sqrt{2gh} \) in one line, with the mass canceling. Where it goes further than the second law conveniently can. The second law with a constant acceleration needs a constant force. A spring's force varies with extension, so the kinematic equations do not apply to it at all. Energy handles it, because the work is the area under the force-distance graph whatever shape it has. And it is a scalar equation. There are no components to resolve and no signs to track beyond positive and negative work. A curved path costs nothing extra, which is why the roller coaster in lesson 5.4 is tractable. The general pattern in this course. A derived result is worth having when it answers a different question more cheaply than its parent. Momentum and energy both come from the second law, and both are kept because each reorganizes it usefully. It eliminates time and acceleration, handles varying forces, and is a scalar

10. A 1100 kg car travelling at 22 m/s brakes to a stop in 38 m. Find the average braking force, then find the stopping distance from 33 m/s with the same brakes.
    Show the full solution

    Find the kinetic energy to be removed. \( KE = \tfrac{1}{2}(1100)(22)^2 = \tfrac{1}{2}(1100)(484) = 266200\ \text{J} \). Apply the work-energy theorem. The brakes do negative work equal to that energy: \( fd = 266200 \) \( f = \dfrac{266200}{38} = 7005\ \text{N} \), about \( 7.0 \times 10^3\ \text{N} \). Sanity check the force. The car's weight is \( 1100 \times 9.8 = 10780\ \text{N} \), so the braking force is about 0.65 of the weight, implying a friction coefficient near 0.65. That is realistic for tires on dry asphalt ✓ A value above about 1.0 would have signaled an error. Now the higher speed. \( KE = \tfrac{1}{2}(1100)(33)^2 = \tfrac{1}{2}(1100)(1089) = 598950\ \text{J} \). \( d = \dfrac{598950}{7005} = 85.5\ \text{m} \). Check the scaling. The speed rose by a factor of \( \dfrac{33}{22} = 1.5 \), so the distance should rise by \( 1.5^2 = 2.25 \): \( 38 \times 2.25 = 85.5\ \text{m} \) ✓ exactly. What this means on the road. Increasing speed by half adds 47.5 m to the stopping distance, more than doubling it. A hazard that was avoidable at 22 m/s is struck at speed from 33 m/s. And the impact is worse than the distance suggests. If the car does strike something after 38 m of braking from 33 m/s, it has shed only 266200 J of its 598950 J and arrives with 332750 J remaining, which is more energy than it had in total at 22 m/s. The collision is worse than if it had never braked from the lower speed. Why the energy framing is the honest one. Stopping distance is a consequence, not the fundamental quantity. What actually scales with the square is the energy the brakes must dissipate, and everything else follows from that. About \( 7.0 \times 10^3 \) N, and 85.5 m from 33 m/s

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( g = 9.8\ \text{m}/\text{s}^2 \).

1. Write the formula for gravitational potential energy near Earth.
   Show the full solution

   \( PE = mgh \)

2. Write Hooke's law.
   Show the full solution

   \( F = kx \)

3. Write the formula for elastic potential energy.
   Show the full solution

   \( PE = \tfrac{1}{2}kx^2 \)

4. Find the potential energy of a 4.0 kg mass 3.0 m above the floor.
   Show the full solution

   \( (4.0)(9.8)(3.0) \). 117.6 J

5. Can potential energy be negative?
   Show the full solution

   It depends on the chosen zero. Yes

6. A spring with \( k = 250\ \text{N/m} \) is stretched 0.20 m. Find the stored energy and the force required.
   Show the full solution

   Energy: \( \tfrac{1}{2}(250)(0.20)^2 = \tfrac{1}{2}(250)(0.040) = 5.0\ \text{J} \). Force at full stretch: \( F = kx = 250 \times 0.20 = 50\ \text{N} \). Note the energy is not \( 50 \times 0.20 = 10\ \text{J} \), because the force was 50 N only at the end. Averaged over the stretch it was 25 N, and \( 25 \times 0.20 = 5.0\ \text{J} \) ✓ 5.0 J and 50 N

7. A 60 kg person climbs 3.0 m of stairs. Find the gain in potential energy.
   Show the full solution

   \( (60)(9.8)(3.0) = 1764\ \text{J} \). About 1.8 kJ, which is roughly the energy in half a gram of sugar. The human body is inefficient enough that the metabolic cost is several times this. 1764 J

8. How much further must a spring of \( k = 300\ \text{N/m} \) be compressed to double the stored energy from a compression of 0.10 m?
   Show the full solution

   At 0.10 m: \( \tfrac{1}{2}(300)(0.010) = 1.5\ \text{J} \). For 3.0 J: \( 3.0 = \tfrac{1}{2}(300)x^2 \), so \( x^2 = \dfrac{6.0}{300} = 0.020 \) and \( x = 0.141\ \text{m} \). Additional compression: \( 0.141 - 0.10 = 0.041\ \text{m} \). Doubling the energy needs a factor of \( \sqrt{2} \) in the compression, not a factor of two, because of the square. About 4.1 cm further

9. Explain why the choice of reference level for gravitational potential energy does not affect any prediction.
   Show the full solution

   Because every physical prediction depends on a change in potential energy, and shifting the reference adds the same constant to every value, which cancels in any difference. Set it out. Suppose the reference is lowered by an amount \( c \). Every potential energy becomes \( mgh + mgc \). A difference between two heights is \( (mgh_2 + mgc) - (mgh_1 + mgc) = mg(h_2 - h_1) \), unchanged. What the physics actually uses. Conservation of energy relates \( \Delta KE \) to \( -\Delta PE \). Only the difference ever appears, so only the difference needs to be well defined. Why there is no natural zero. There is no special height in the universe. Sea level, the floor and the tabletop are all human choices, and nature provides no reason to prefer one. The practical version of this freedom. Choose the lowest point of the motion as zero and every potential energy in the problem is positive, which is the least error-prone arrangement. The one rule that is not negotiable. Use one reference throughout a single problem. Measuring the start from the floor and the end from the tabletop introduces a spurious energy difference and gives a wrong speed. The same freedom appears elsewhere. Electric potential in unit 7 has an arbitrary zero for exactly the same reason, and the convention there is usually to set it at infinity. Recognizing the pattern once makes the second instance unremarkable. Only differences enter any equation, and a shifted reference cancels in a difference

10. A 0.050 kg dart is pushed against a spring of \( k = 180\ \text{N/m} \), compressing it 0.12 m, then released horizontally. Find the launch speed, and find the compression needed to double it.
    Show the full solution

    Find the stored energy. \( PE = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(180)(0.12)^2 = \tfrac{1}{2}(180)(0.0144) = 1.296\ \text{J} \). Convert it to kinetic energy. Fired horizontally, the height does not change, so all the stored energy becomes kinetic: \( \tfrac{1}{2}mv^2 = 1.296 \) \( v^2 = \dfrac{2(1.296)}{0.050} = \dfrac{2.592}{0.050} = 51.84 \) \( v = 7.2\ \text{m/s} \). Check the units through the chain. \( \text{N/m} \times \text{m}^2 = \text{N}\cdot\text{m} = \text{J} \) ✓, and \( \sqrt{\dfrac{\text{J}}{\text{kg}}} = \sqrt{\dfrac{\text{kg}\cdot\text{m}^2/\text{s}^2}{\text{kg}}} = \text{m/s} \) ✓ Now double the speed to 14.4 m/s. Required kinetic energy: \( \tfrac{1}{2}(0.050)(14.4)^2 = \tfrac{1}{2}(0.050)(207.36) = 5.184\ \text{J} \), four times the original, as the square requires. Required compression: \( 5.184 = \tfrac{1}{2}(180)x^2 \), so \( x^2 = \dfrac{10.368}{180} = 0.0576 \) and \( x = 0.24\ \text{m} \). Exactly twice the compression. Why that came out so cleanly. Both energies go as a square, one in \( x \) and one in \( v \), so \( x \) and \( v \) are directly proportional for this device. Doubling one doubles the other. That is worth recognizing as a general result. For a spring launcher, launch speed scales linearly with compression, even though both energy expressions are quadratic. The two squares cancel. The practical caveat. Hooke's law holds only up to the spring's elastic limit. Compressing 0.24 m may exceed it, in which case the spring deforms permanently, \( F = kx \) fails, and the prediction is void. Any real design would check the manufacturer's maximum before assuming the calculation holds. 7.2 m/s, and 0.24 m to double it

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Ignore friction unless told otherwise, and use \( g = 9.8\ \text{m}/\text{s}^2 \).

1. What is mechanical energy?
   Show the full solution

   The sum of kinetic and potential energy

2. Under what condition is it conserved?
   Show the full solution

   When only conservative forces do work

3. Find the speed of an object dropped from 5.0 m.
   Show the full solution

   \( \sqrt{2(9.8)(5.0)} = \sqrt{98} \). 9.9 m/s

4. Does the mass of a falling object affect its landing speed?
   Show the full solution

   No

5. Is friction a conservative force?
   Show the full solution

   No

6. A 0.20 kg ball is thrown upward at 12 m/s. Find its maximum height using energy.
   Show the full solution

   At the top the speed is zero, so all the kinetic energy has become potential: \( \tfrac{1}{2}mv^2 = mgh \) The mass cancels: \( h = \dfrac{v^2}{2g} = \dfrac{144}{19.6} = 7.35\ \text{m} \). Same answer as the kinematic route in lesson 1.6, reached without finding the time. 7.35 m

7. A skier starts from rest at 25 m and reaches a point 10 m above the ground. Find the speed there.
   Show the full solution

   Only the height difference matters: \( 25 - 10 = 15\ \text{m} \). \( v = \sqrt{2(9.8)(15)} = \sqrt{294} = 17.1\ \text{m/s} \). The shape of the slope is irrelevant, which is what makes this solvable at all. 17.1 m/s

8. A 0.30 kg ball is launched by a spring storing 6.0 J, fired vertically. Find its maximum height.
   Show the full solution

   All the spring energy becomes gravitational potential energy at the top: \( 6.0 = mgh = (0.30)(9.8)h = 2.94h \) \( h = \dfrac{6.0}{2.94} = 2.04\ \text{m} \). Note the kinetic stage was skipped entirely. Energy conservation connects the first configuration to the last without needing the launch speed. About 2.0 m

9. Explain why a roller coaster can never have a later hill higher than its first, on a frictionless track.
   Show the full solution

   Because the car's total mechanical energy is fixed at the start, and reaching a greater height would require more potential energy than it has. Set up the accounting. Released from rest at height \( h_1 \), the car's total energy is \( mgh_1 \). With no friction, that total never changes. At any later point, \( E = \tfrac{1}{2}mv^2 + mgh \). Since kinetic energy cannot be negative, \( mgh \le mgh_1 \), so \( h \le h_1 \). What the equality means. The car could just reach \( h_1 \) again, but it would arrive with zero speed and stop there. In practice it must arrive with some speed to keep moving, so every later hill must be strictly lower. Why real coasters are lower still. Friction and air resistance remove energy continuously, so the usable height falls throughout the ride. That is why the profile of a coaster descends overall, and why the first hill is always the tallest. How the height is regained. Only by adding energy from outside. The chain lift at the start does exactly that, doing work on the car to raise its total energy before the ride begins. A loop is the interesting case. A car must not merely reach the top of a loop but arrive with enough speed for gravity alone to supply the centripetal force, or it leaves the track. So the usable height is lower than the energy limit suggests, and lesson 4.3's circular motion condition sets the real constraint. Total energy is fixed at the start and kinetic energy cannot be negative

10. A 1.5 kg block slides from rest down a frictionless ramp from 3.0 m, then compresses a spring of \( k = 800\ \text{N/m} \) at the bottom. Find the maximum compression, and find the block's speed when the spring is compressed half that far.
    Show the full solution

    Find the energy at the bottom of the ramp. All the potential energy becomes kinetic: \( E = mgh = (1.5)(9.8)(3.0) = 44.1\ \text{J} \). Find the maximum compression. At maximum compression the block is momentarily at rest, so all 44.1 J is stored in the spring: \( \tfrac{1}{2}kx^2 = 44.1 \) \( x^2 = \dfrac{2(44.1)}{800} = \dfrac{88.2}{800} = 0.11025 \) \( x = 0.332\ \text{m} \). Now at half that compression, \( x = 0.166\ \text{m} \). Energy stored in the spring: \( \tfrac{1}{2}(800)(0.166)^2 = \tfrac{1}{2}(800)(0.027556) = 11.02\ \text{J} \). The rest is still kinetic. \( KE = 44.1 - 11.02 = 33.08\ \text{J} \) \( v = \sqrt{\dfrac{2(33.08)}{1.5}} = \sqrt{44.1} = 6.64\ \text{m/s} \). Check the speed at the bottom before the spring. \( v = \sqrt{2gh} = \sqrt{2(9.8)(3.0)} = \sqrt{58.8} = 7.67\ \text{m/s} \). So at half compression the block has slowed from 7.67 to 6.64 m/s ✓ slower, as it must be. The point worth noticing. At half the maximum compression the spring holds only a quarter of the energy, not half, because the storage goes as \( x^2 \). So three quarters of the energy is still kinetic and the block is still moving at 87 percent of its original speed. Why that matters for a real stopper. A spring does most of its work in the final part of its travel. If the design requires a gentle deceleration, a spring is a poor choice, because it barely resists at first and then stops the object abruptly. A constant-force device, such as a crushable block, spreads the force evenly, which is the lesson 3.7 consideration applied to design rather than to safety. Three energy forms in one problem. Gravitational at the top, kinetic at the bottom, elastic at maximum compression, and a mixture in between. The total, 44.1 J, is the same at every stage, which is the whole content of the conservation law. 0.332 m maximum, and 6.64 m/s at half that

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write the formula for power.
   Show the full solution

   \( P = \dfrac{W}{t} \)

2. Give the SI unit of power.
   Show the full solution

   The watt

3. Is the kilowatt-hour a unit of power or energy?
   Show the full solution

   Energy

4. A machine does 600 J in 4.0 s. Find its power.
   Show the full solution

   150 W

5. Write power in terms of force and speed.
   Show the full solution

   \( P = Fv \)

6. A crane lifts a 500 kg load 12 m in 20 s. Find its power output.
   Show the full solution

   \( W = mgh = (500)(9.8)(12) = 58800\ \text{J} \). \( P = \dfrac{58800}{20} = 2940\ \text{W} \), about 2.9 kW. 2940 W

7. Convert 1 kWh to joules.
   Show the full solution

   \( 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^{6}\ \text{J} \). \( 3.6 \times 10^6 \) J

8. A 1500 W appliance runs for 4.0 hours. Find the energy used in kWh and the cost at 17 cents per kWh.
   Show the full solution

   \( 1500\ \text{W} = 1.5\ \text{kW} \). Energy: \( 1.5 \times 4.0 = 6.0\ \text{kWh} \). Cost: \( 6.0 \times 0.17 = \$1.02 \). 6.0 kWh, about $1.02

9. Explain why a car needs roughly eight times the power to travel twice as fast on a level road.
   Show the full solution

   Because drag grows as the square of the speed and power is drag times speed, so power grows as the cube. Start with the drag. At highway speeds air resistance is approximately proportional to \( v^2 \). Doubling the speed roughly quadruples the force opposing the car. Now the power. From \( P = Fv \), with \( F \) four times larger and \( v \) twice as large, the product is \( 4 \times 2 = 8 \) times larger. Algebraically, \( P \propto v^2 \cdot v = v^3 \). Put numbers on it. A car needing 16 kW at 20 m/s needs about 128 kW at 40 m/s, which is most of a family car's engine output devoted purely to pushing air aside. Why fuel economy collapses at speed. Fuel burned per unit distance is the power divided by the speed, which goes as \( v^2 \). Doubling the speed roughly quadruples the fuel used per kilometer. Why top speed is expensive. Adding 10 percent to a car's top speed requires about 33 percent more power, which is why high top speeds need disproportionately large engines. The qualification. This treats drag as the only resistance. At low speeds rolling resistance dominates and is roughly constant, so the cube law applies properly only above roughly 50 km/h. That is why there is a most-economical cruising speed rather than economy simply improving as you slow down. Drag goes as \( v^2 \) and power is drag times speed, giving \( v^3 \)

10. A 70 kg cyclist climbs a hill of 8.0 percent gradient at a steady 5.0 m/s, with the bicycle adding 10 kg. Find the power output against gravity, and compare it with the power a fit amateur can sustain.
    Show the full solution

    Interpret the gradient. An 8.0 percent gradient means 8.0 m of rise per 100 m travelled along the road, so the vertical speed is 8.0 percent of the speed along the slope. \( v_{\text{vertical}} = 0.080 \times 5.0 = 0.40\ \text{m/s} \). Find the total mass being lifted. \( 70 + 10 = 80\ \text{kg} \). The bicycle must be raised too. Compute the power against gravity. Use \( P = Fv \) with the force being the weight and the speed being the vertical component: \( P = mg\,v_{\text{vertical}} = (80)(9.8)(0.40) = 313.6\ \text{W} \). Check by the energy route. In one second the cyclist rises 0.40 m, gaining \( mgh = (80)(9.8)(0.40) = 313.6\ \text{J} \) of potential energy. Per second that is 313.6 W ✓ Compare with human capability. A fit amateur cyclist can sustain roughly 200 to 300 W for an extended climb; a professional can hold around 400 W for an hour. So 314 W against gravity alone is a hard effort, sustainable for a limited time by a strong rider. And this understates the real requirement. The figure covers only the work against gravity. The rider must also overcome rolling resistance and air drag, perhaps another 30 to 60 W at this speed, so the true output is closer to 350 to 375 W. A useful way to see why climbing is hard. On the flat at 5.0 m/s the same rider needs only the drag and rolling terms, maybe 50 W. The hill has multiplied the demand roughly sevenfold at the same speed, and all of that increase is the \( mgh \) term. Why weight matters so much on a climb and so little on the flat. The gravity term is proportional to total mass, so removing 1 kg cuts the climbing power by about 1.25 percent. On level ground, where drag dominates and depends on frontal area rather than mass, the same kilogram is almost irrelevant. This is precisely why competitive climbers care about bicycle mass and sprinters do not. About 314 W against gravity, which is a hard sustained effort for a fit amateur

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( g = 9.8\ \text{m}/\text{s}^2 \).

1. Is total energy conserved when friction acts?
   Show the full solution

   Yes

2. Is mechanical energy conserved when friction acts?
   Show the full solution

   No

3. What form does the mechanical energy become?
   Show the full solution

   Thermal energy

4. Write the work done by friction over a distance \( d \).
   Show the full solution

   \( -fd \)

5. Should friction use distance or displacement?
   Show the full solution

   Distance, the path length

6. A 4.0 kg block slides 5.0 m with a friction force of 8.0 N. Find the thermal energy produced.
   Show the full solution

   \( |W_f| = fd = 8.0 \times 5.0 = 40\ \text{J} \). The mass is not needed for this part, since the friction force was given directly. 40 J

7. A 3.0 kg block at 8.0 m/s slides to rest with \( \mu_k = 0.25 \). Find the distance.
   Show the full solution

   \( f = \mu_k mg = (0.25)(3.0)(9.8) = 7.35\ \text{N} \). \( KE = \tfrac{1}{2}(3.0)(64) = 96\ \text{J} \). \( d = \dfrac{96}{7.35} = 13.1\ \text{m} \). The mass cancels if done symbolically: \( d = \dfrac{v^2}{2\mu_k g} = \dfrac{64}{2(0.25)(9.8)} = 13.1\ \text{m} \) ✓ 13.1 m

8. A 2.0 kg block is pushed 4.0 m out and 4.0 m back to its starting point against a friction force of 6.0 N. Find the total thermal energy produced.
   Show the full solution

   Friction depends on path length, which is \( 4.0 + 4.0 = 8.0\ \text{m} \). \( |W_f| = (6.0)(8.0) = 48\ \text{J} \). Not zero. The displacement is zero, but friction opposed the motion on both legs and did negative work both times. This is exactly why friction is not conservative: a round trip does not return the energy. 48 J

9. Explain why thermal energy cannot be fully converted back into mechanical energy.
   Show the full solution

   Because mechanical energy is organized and thermal energy is not, and the disorganized arrangements vastly outnumber the organized ones. What the two forms are. A moving block has every particle sharing one common velocity. A warm block has the same total energy spread among particles moving in random directions at random speeds. The laws of motion do not forbid the reverse. Newton's laws are time-reversible: nothing in them says the random motions could not all align and send the block moving again. But the counting does. There are astronomically more ways for the energy to be distributed randomly than for it to be aligned. Left alone, a system moves toward the overwhelming majority of arrangements, which is the disorganized ones. Some conversion is possible. A heat engine converts thermal energy into work routinely. What it cannot do is convert all of it; lesson 6.6 shows the maximum fraction depends on the temperatures available. Why this makes energy quality a real idea. A joule of kinetic energy can be fully converted into a joule of potential energy or electrical energy. A joule of thermal energy at room temperature can be converted only partly into anything useful. The quantity is identical and the usefulness is not. The name for the difference. Entropy, which unit 6 introduces as a measure of how many arrangements a state has. Friction increases it, and that increase is why the process runs one way. Organized motion becomes disorganized, and the disorganized arrangements overwhelmingly outnumber the organized

10. A 25 kg sled starts from rest and slides down a 30 m slope that drops 12 m, arriving at the bottom at 12 m/s. Find the thermal energy produced and the average friction force, and state what fraction of the available energy was lost.
    Show the full solution

    Find the energy available at the top. \( PE = mgh = (25)(9.8)(12) = 2940\ \text{J} \). Starting from rest, that is the entire mechanical energy of the system. Find the energy that arrived as motion. \( KE = \tfrac{1}{2}(25)(12)^2 = \tfrac{1}{2}(25)(144) = 1800\ \text{J} \). The difference is the thermal energy. \( 2940 - 1800 = 1140\ \text{J} \). Find the average friction force. Friction acted along the 30 m of slope actually travelled, not the 12 m of drop: \( f = \dfrac{1140}{30} = 38\ \text{N} \). Check it for plausibility. The sled's weight is \( 25 \times 9.8 = 245\ \text{N} \). The normal force on a slope this steep is \( mg\cos\theta \), and with a 12 m rise over 30 m the angle is \( \sin^{-1}(0.40) = 23.6^\circ \), so \( N = 245\cos 23.6^\circ = 225\ \text{N} \). That gives \( \mu_k = \dfrac{38}{225} = 0.17 \), a reasonable value for a sled on packed snow ✓ Find the fraction lost. \( \dfrac{1140}{2940} = 0.388 \), about 39 percent. What a frictionless slope would have given. \( v = \sqrt{2gh} = \sqrt{2(9.8)(12)} = \sqrt{235.2} = 15.3\ \text{m/s} \), against the actual 12 m/s. Note how the percentages differ. The speed fell by 21 percent while the energy fell by 39 percent, because energy goes as the square of speed. Quoting one when the other is meant is a common source of confusion in these problems. The complete account. 2940 J of gravitational potential energy at the top, becoming 1800 J of kinetic energy at the bottom and 1140 J of thermal energy spread along 30 m of snow and the sled's runners. Total 2940 J at every stage. Why the path length mattered. Gravity cared only about the 12 m vertical drop, because it is conservative. Friction cared about all 30 m of contact, because it is not. Using 12 m for friction would have given a force of 95 N and a coefficient of 0.42, which is the single commonest error in this type of problem. 1140 J thermal, an average friction force of 38 N, and 39 percent of the energy converted

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( g = 9.8\ \text{m}/\text{s}^2 \).

1. Write the formula for efficiency.
   Show the full solution

   \( \eta = \dfrac{E_{\text{useful}}}{E_{\text{in}}} \)

2. Can a real device be 100 percent efficient?
   Show the full solution

   No

3. What form do most losses end up as?
   Show the full solution

   Thermal energy

4. A device takes in 200 J and usefully delivers 150 J. Find its efficiency.
   Show the full solution

   \( \dfrac{150}{200} \). 75 percent

5. Is energy destroyed in an inefficient device?
   Show the full solution

   No, it is converted into an unwanted form

6. A crane lifts a 200 kg load 15 m using 40 kJ of electrical energy. Find its efficiency.
   Show the full solution

   Useful output: \( mgh = (200)(9.8)(15) = 29400\ \text{J} \). \( \eta = \dfrac{29400}{40000} = 0.735 \), about 73.5 percent. Losses: \( 40000 - 29400 = 10600\ \text{J} \), into motor heating, gearbox friction and cable stretch. 73.5 percent

7. A 60 W bulb produces 3 W of light. Explain in what sense it is 5 percent efficient and in what sense it is 100 percent efficient.
   Show the full solution

   5 percent efficient at its purpose, which is producing light: \( \dfrac{3}{60} = 0.05 \). 100 percent efficient at converting energy, since all 60 W leaves as light plus heat and none is destroyed. Efficiency is defined against the intended output, so the same device has different efficiencies for different purposes. Used as a small heater in winter it is nearly 100 percent efficient, which is why the "wasted" heat is not entirely wasted in a cold room. 5 percent at lighting, 100 percent at converting

8. A hydroelectric plant releases water falling 60 m at 500 kg/s and generates 250 kW. Find its efficiency.
   Show the full solution

   Input power: the rate at which potential energy arrives. \( P_{\text{in}} = \dot{m}gh = (500)(9.8)(60) = 294000\ \text{W} = 294\ \text{kW} \). \( \eta = \dfrac{250}{294} = 0.850 \), about 85 percent. Hydroelectric plants really are this efficient, far better than thermal power stations, because they convert mechanical energy directly and are not bound by the heat-engine limit of lesson 6.6. About 85 percent

9. Explain why a heat engine has a lower maximum efficiency than an electric motor.
   Show the full solution

   Because a heat engine must convert disorganized thermal energy into organized work, and lesson 6.6's second law places a hard ceiling on that conversion, while a motor starts from already-organized electrical energy. What a motor does. It converts electrical energy, which is organized motion of charge, into mechanical energy, which is organized motion of matter. Both are ordered forms, so the conversion is limited only by practical losses: resistance, friction and magnetic effects. Good motors reach 90 to 95 percent. What a heat engine does. It takes thermal energy, which is disorganized particle motion, and extracts organized work from it. That is the conversion lesson 5.6 showed runs against the statistical grain. The ceiling. The maximum possible efficiency depends only on the temperatures between which the engine works, and lesson 6.6 derives it as \( 1 - \dfrac{T_{\text{cold}}}{T_{\text{hot}}} \). For a car engine that is roughly 60 percent, and real engines achieve 25 to 35 percent. The crucial difference. The motor's losses are engineering problems, which better materials and tighter tolerances can reduce. The engine's ceiling is a law of physics, and no improvement in manufacturing can pass it. Why this drives technology. It is one reason electric vehicles convert stored energy to motion more effectively than combustion vehicles, though a fair comparison must include how the electricity was generated, and if that was a thermal power station the heat-engine limit has simply been moved rather than removed. A motor converts ordered energy to ordered energy; an engine must convert disordered to ordered, which the second law caps

10. Design a device to convert the kinetic energy of a falling 2.0 kg mass into stored elastic energy, state how you would measure its efficiency, and predict a realistic value.
    Show the full solution

    State the conversion required. Gravitational potential energy to kinetic energy to elastic potential energy. The design. A 2.0 kg mass is released from a measured height \( h \) above a vertical spring of known stiffness \( k \), mounted on a rigid base with a guide rod to keep the fall vertical. The mass compresses the spring, and the maximum compression is recorded by a light sliding marker on the rod that stays at the lowest point reached. State the measurements. Input: \( E_{\text{in}} = mg(h + x) \), where \( x \) is the compression, because the mass falls that extra distance while compressing the spring. Omitting the \( x \) is the commonest error in this design. Output: \( E_{\text{out}} = \tfrac{1}{2}kx^2 \), from the measured compression. Efficiency: the ratio of the two. Work a numerical prediction. Take \( h = 0.50\ \text{m} \) and \( k = 2000\ \text{N/m} \). If the conversion were perfect, \( mg(h + x) = \tfrac{1}{2}kx^2 \): \( (2.0)(9.8)(0.50 + x) = 1000x^2 \) \( 9.8 + 19.6x = 1000x^2 \) \( 1000x^2 - 19.6x - 9.8 = 0 \). Discriminant: \( 384.16 + 39200 = 39584 \), root \( 198.96 \). \( x = \dfrac{19.6 + 198.96}{2000} = 0.109\ \text{m} \). So a perfect device gives about 10.9 cm of compression, storing \( \tfrac{1}{2}(2000)(0.109)^2 = 11.9\ \text{J} \), which matches \( mg(h + x) = (2.0)(9.8)(0.609) = 11.9\ \text{J} \) ✓ Predict a realistic value. Expect 70 to 85 percent. A compression of about 0.095 m would indicate \( \tfrac{1}{2}(2000)(0.095)^2 = 9.0\ \text{J} \) stored against \( (2.0)(9.8)(0.595) = 11.7\ \text{J} \) in, an efficiency of 77 percent. Name the losses, as the criterion requires. Internal friction in the spring, which warms it slightly on compression. Friction on the guide rod, reduced by keeping the rod smooth and vertical. Sound, audible at impact and therefore a real loss. Vibration of the base, which carries energy away from the spring. The marker itself, which requires a small force to move and so removes a little energy. State the evaluation criterion before testing. The device succeeds if it converts at least 70 percent of the input into measured elastic energy across five trials, with the spread between trials under 10 percent. Why the repeats matter. A single trial cannot distinguish a genuine efficiency from a mis-read marker. Reporting a mean and a spread is what makes the number a measurement rather than an anecdote, which is the standard lesson 11.1 will set out for data generally. A design improvement worth naming. Replacing the sliding marker with a high-speed video recording removes the marker's own energy cost and gives the whole compression history rather than just the maximum, which would let the losses be separated rather than lumped together. A spring-and-guide-rod design measuring \( mg(h+x) \) against \( \tfrac{1}{2}kx^2 \), with 70 to 85 percent expected

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. What does temperature measure?
   Show the full solution

   The average kinetic energy per particle

2. What is heat?
   Show the full solution

   Energy transferred because of a temperature difference

3. Convert \( 100^\circ\text{C} \) to kelvin.
   Show the full solution

   373 K

4. What is absolute zero in Celsius?
   Show the full solution

   \( -273.15^\circ\text{C} \)

5. In which direction does heat flow spontaneously?
   Show the full solution

   From hot to cold

6. A temperature rises from \( 15^\circ\text{C} \) to \( 55^\circ\text{C} \). Give the change in Celsius degrees and in kelvin.
   Show the full solution

   Celsius: \( 55 - 15 = 40^\circ\text{C} \). Kelvin: \( 328.15 - 288.15 = 40\ \text{K} \). The same number, because the two scales have identical degree sizes and differ only in their zero. 40 in both

7. Two objects of different mass are at the same temperature. Compare their average particle kinetic energies and their total thermal energies.
   Show the full solution

   Average kinetic energy per particle: the same, because that is what equal temperature means. Total thermal energy: greater in the more massive object, assuming the same material, because it has more particles each carrying that average. If the materials differ the comparison is less direct, since the energy is also shared among rotational and vibrational modes in ways that vary by substance. That is exactly what specific heat in lesson 6.2 quantifies. Same average, greater total for the larger mass

8. A gas at \( 27^\circ\text{C} \) is heated until its absolute temperature doubles. Find the new temperature in Celsius.
   Show the full solution

   Convert first: \( 27 + 273 = 300\ \text{K} \). Double it: \( 600\ \text{K} \). Convert back: \( 600 - 273 = 327^\circ\text{C} \). Not \( 54^\circ\text{C} \). Doubling in Celsius is meaningless because the Celsius zero is arbitrary, and this is precisely the error the warning names. \( 327^\circ\text{C} \)

9. Explain why an object can be said to contain thermal energy but not heat.
   Show the full solution

   Because thermal energy is a property of the object's current state, while heat is a process by which energy crosses a boundary. Thermal energy is a state quantity. Given an object's temperature, mass and material, its thermal energy is determined. It does not matter how it got that way. Heat is a process quantity. It describes energy in transit, and it only exists while the transfer is happening. Once the energy has arrived it is simply part of the object's thermal energy and is no longer identifiable as having been heat. The decisive demonstration. Two identical blocks are raised to the same temperature, one by placing it near a flame and one by rubbing it. Their final states are indistinguishable, yet one received energy as heat and the other as work. If either contained a measurable amount of heat, the two would differ, and they do not. Why the distinction is not pedantry. Lesson 6.5's first law, \( \Delta U = Q - W \), depends on it. The change in internal energy is a state change; \( Q \) and \( W \) are the two different routes by which it happened. Merging them would make the equation meaningless. The same structure elsewhere. Work has exactly the same status: nothing contains work either. Both are transfers, and the object has only its energy. Thermal energy is a state; heat is a transfer

10. A 0.25 kg cup of water at \( 100^\circ\text{C} \) and a 50000 kg pool at \( 21^\circ\text{C} \) are compared, both measured against \( 20^\circ\text{C} \). Compute the thermal energy above that reference for each, and explain what the comparison shows.
    Show the full solution

    Use \( Q = mc\Delta T \) with \( c = 4180\ \text{J/(kg}\cdot\text{K)} \) for water, taking \( 20^\circ\text{C} \) as the reference. The cup. \( \Delta T = 100 - 20 = 80\ \text{K} \) \( Q = (0.25)(4180)(80) = 83600\ \text{J} \), about \( 8.4 \times 10^4\ \text{J} \). The pool. \( \Delta T = 21 - 20 = 1\ \text{K} \) \( Q = (50000)(4180)(1) = 2.09 \times 10^{8}\ \text{J} \). The ratio. \( \dfrac{2.09 \times 10^8}{8.36 \times 10^4} = 2500 \). The pool holds about 2500 times the thermal energy above the reference, while being \( 79^\circ\text{C} \) colder. What the comparison shows. Temperature and thermal energy are independent. Temperature is intensive, meaning it does not depend on how much substance there is; thermal energy is extensive, meaning it scales with the amount. Knowing one tells you nothing about the other without the mass. Why it matters practically. Which one hurts depends on the situation. Splashing the cup on your hand transfers energy quickly from a high temperature and burns you. Swimming in the pool transfers far more energy over time and does nothing, because the transfer rate depends on the temperature difference, which is tiny. The general principle. Temperature drives the flow; thermal energy is the reservoir. A large cold reservoir can hold enormous energy and still be unable to heat anything warmer than itself, which is the second law of lesson 6.6 in a single sentence. And it explains a real engineering problem. The oceans hold an immense quantity of thermal energy that is almost entirely unusable, because there is nothing colder nearby to run an engine against. Energy and available energy are different things. \( 8.4 \times 10^4 \) J against \( 2.1 \times 10^8 \) J; the colder body holds 2500 times more

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( c_{\text{water}} = 4180 \), \( c_{\text{Al}} = 900 \), \( c_{\text{Cu}} = 385 \) in \( \text{J/(kg}\cdot\text{K)} \).

1. Write the specific heat equation.
   Show the full solution

   \( Q = mc\Delta T \)

2. Give the units of specific heat.
   Show the full solution

   \( \text{J/(kg}\cdot\text{K)} \)

3. Give the specific heat of water.
   Show the full solution

   4180 J/(kg·K)

4. Find the energy to heat 1.0 kg of copper by 50 K.
   Show the full solution

   \( (1.0)(385)(50) \). 19250 J

5. In calorimetry, what does the heat lost equal?
   Show the full solution

   The heat gained

6. Find the energy to heat 0.50 kg of aluminum from \( 20^\circ\text{C} \) to \( 100^\circ\text{C} \).
   Show the full solution

   \( \Delta T = 80\ \text{K} \). \( Q = (0.50)(900)(80) = 36000\ \text{J} \). 36000 J

7. A 3.0 kg block absorbs 54000 J and rises by 40 K. Find its specific heat and suggest the material.
   Show the full solution

   \( c = \dfrac{Q}{m\Delta T} = \dfrac{54000}{(3.0)(40)} = \dfrac{54000}{120} = 450\ \text{J/(kg}\cdot\text{K)} \). That is close to iron or steel, whose specific heat is about 450. This is how specific heats are measured, and identifying an unknown metal by its specific heat is a standard laboratory exercise. 450 J/(kg·K), probably iron

8. How much water can be heated from \( 15^\circ\text{C} \) to \( 100^\circ\text{C} \) by \( 1.0 \times 10^6\ \text{J} \)?
   Show the full solution

   \( \Delta T = 85\ \text{K} \). \( m = \dfrac{Q}{c\Delta T} = \dfrac{1.0 \times 10^6}{(4180)(85)} = \dfrac{1.0 \times 10^6}{355300} = 2.81\ \text{kg} \). About 2.8 kg

9. Explain why coastal areas have smaller annual temperature ranges than inland areas at the same latitude.
   Show the full solution

   Because water's specific heat is about five times that of soil and rock, so the sea changes temperature far less than land for the same energy input, and it moderates the air that passes over it. Compare the numbers. Water is 4180 J/(kg·K); dry soil and rock are around 800. The same energy per kilogram raises land roughly five times as much. And water mixes. Solar energy absorbed by the sea is carried down and spread through a deep layer by currents and waves, while land heats only its top few centimeters. The effective mass being warmed is vastly larger for water. Evaporation adds to it, since some of the incoming energy goes into evaporating water rather than warming it, which lesson 6.3's latent heat explains. The result. A coastal town's air is repeatedly brought toward the sea's slow-changing temperature. Summers are cooler and winters milder than inland, and the seasonal extremes are compressed. The observed size of the effect. San Francisco's annual range is about \( 9^\circ\text{C} \); Sacramento, 120 km inland at the same latitude, is roughly double that (NOAA, US federal). There is also a lag. The sea's slowness shifts its warmest and coldest months later than the land's, which is why coastal waters are often coldest in late spring and warmest in early autumn. Water's high specific heat and its mixing make the sea a slow-changing reservoir that moderates nearby air

10. A 0.15 kg copper block at \( 200^\circ\text{C} \) is dropped into 0.40 kg of water at \( 18^\circ\text{C} \) in an insulated container. Find the final temperature, and state what assumption the calculation makes.
    Show the full solution

    Set up the energy balance. The container is insulated, so all the energy leaving the copper enters the water. Call the final temperature \( T \). Copper cools. \( Q_{\text{lost}} = m_{Cu}c_{Cu}(200 - T) = (0.15)(385)(200 - T) = 57.75(200 - T) \). Water warms. \( Q_{\text{gained}} = (0.40)(4180)(T - 18) = 1672(T - 18) \). Set them equal. \( 57.75(200 - T) = 1672(T - 18) \) \( 11550 - 57.75T = 1672T - 30096 \) \( 41646 = 1729.75T \) \( T = 24.1^\circ\text{C} \). Check it is between the starting values. 18 and 200, and 24.1 lies between them ✓ Check it is close to the water's start, as it should be: the water has nearly three times the mass and about eleven times the specific heat, so it dominates overwhelmingly ✓ Verify the energies balance. \( 57.75(200 - 24.1) = 57.75 \times 175.9 = 10158\ \text{J} \) \( 1672(24.1 - 18) = 1672 \times 6.1 = 10199\ \text{J} \) ✓ agreeing to rounding. Note how little the water warmed. A block at \( 200^\circ\text{C} \) raised 0.40 kg of water by only \( 6.1^\circ\text{C} \). That is the practical meaning of water's high specific heat. State the assumptions. Perfect insulation, so no energy escapes to the surroundings. Real containers leak, which would give a slightly lower final temperature. The container itself absorbs nothing. A real calorimeter has its own mass and specific heat, and a careful experiment includes a "water equivalent" term for it. No water is vaporized. At \( 200^\circ\text{C} \) the copper is well above boiling, so some water in immediate contact may flash to steam, which would carry away energy through the latent heat of lesson 6.3 and lower the result. Specific heats are constant over the range, which is a good approximation here but not exact. Which assumption matters most. The vaporization one, because latent heat is large. If even 5 g of water boiled off, that alone would absorb \( 0.005 \times 2.26 \times 10^6 = 11300\ \text{J} \), more than the entire energy the copper released. The calculation would then be badly wrong, and a careful experiment would use a cooler block to avoid it. About \( 24.1^\circ\text{C} \), assuming perfect insulation, a negligible container and no boiling

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( L_f = 3.34 \times 10^5 \), \( L_v = 2.26 \times 10^6\ \text{J/kg} \), \( c_{\text{water}} = 4180 \), \( c_{\text{ice}} = 2100\ \text{J/(kg}\cdot\text{K)} \).

1. What happens to temperature during a phase change?
   Show the full solution

   It stays constant

2. Write the equation for energy during a phase change.
   Show the full solution

   \( Q = mL \)

3. Give the latent heat of fusion of water.
   Show the full solution

   \( 3.34 \times 10^5\ \text{J/kg} \)

4. Which takes more energy per kilogram, melting or boiling water?
   Show the full solution

   Boiling

5. Where does the energy go during melting?
   Show the full solution

   Into breaking intermolecular bonds

6. Find the energy to melt 2.5 kg of ice at \( 0^\circ\text{C} \).
   Show the full solution

   \( Q = mL_f = (2.5)(3.34 \times 10^5) = 8.35 \times 10^5\ \text{J} \). \( 8.35 \times 10^5 \) J

7. Find the energy to take 1.0 kg of ice at \( 0^\circ\text{C} \) to water at \( 50^\circ\text{C} \).
   Show the full solution

   Two stages. Melt: \( (1.0)(3.34 \times 10^5) = 334000\ \text{J} \). Warm: \( (1.0)(4180)(50) = 209000\ \text{J} \). Total: \( 543000\ \text{J} \). Melting took more energy than heating the resulting water by \( 50^\circ\text{C} \), which is worth noticing. \( 5.43 \times 10^5 \) J

8. How much ice at \( 0^\circ\text{C} \) can be melted by 1.0 kg of water cooling from \( 80^\circ\text{C} \) to \( 0^\circ\text{C} \)?
   Show the full solution

   Energy released by the water: \( Q = (1.0)(4180)(80) = 334400\ \text{J} \). Ice melted: \( m = \dfrac{Q}{L_f} = \dfrac{334400}{3.34 \times 10^5} = 1.00\ \text{kg} \). Almost exactly one kilogram, a coincidence of the numbers worth remembering as a rough check: melting ice takes about as much energy as heating the same mass of water by \( 80^\circ\text{C} \). About 1.0 kg

9. Explain why sweating cools the body.
   Show the full solution

   Because evaporating sweat absorbs a large latent heat, and that energy comes from the skin. The mechanism. Liquid water on the skin evaporates. To change from liquid to vapor, each molecule must break free of the others, which requires energy. That energy is drawn from the warmest thing in contact, which is the skin. Why the effect is large. The latent heat of vaporization is \( 2.26 \times 10^6\ \text{J/kg} \), so evaporating just 100 g of sweat removes about \( 2.26 \times 10^5\ \text{J} \). Cooling the same 100 g of water by \( 10^\circ\text{C} \) would remove only 4180 J, about fifty times less. A subtlety worth stating. Evaporation at skin temperature is not boiling, and the latent heat is slightly larger at \( 35^\circ\text{C} \) than at \( 100^\circ\text{C} \), around \( 2.4 \times 10^6\ \text{J/kg} \). The effect is even stronger than the boiling figure suggests. Why sweat that drips off does nothing. Only evaporation removes the latent heat. Sweat that runs off the body carries away a negligible amount, which is why fanning helps and why heavy sweating in still air is inefficient. Why humidity matters. In humid air the rate of evaporation falls because the air is already near saturation, so the cooling mechanism largely stops. That is why humid heat is dangerous in a way dry heat of the same temperature is not, and it is the reasoning behind wet-bulb temperature as a measure of heat stress (NOAA, US federal). The same physics elsewhere. A dog panting, water evaporating from a porous clay pot, and alcohol feeling cold on skin are all the same latent heat being taken from the surroundings. Evaporation absorbs latent heat, and it is taken from the skin

10. How much steam at \( 100^\circ\text{C} \) must be condensed into 2.0 kg of water at \( 20^\circ\text{C} \) to raise it to \( 60^\circ\text{C} \)? Assume the container absorbs nothing.
    Show the full solution

    Find the energy the water needs. \( Q_{\text{needed}} = mc\Delta T = (2.0)(4180)(60 - 20) = (2.0)(4180)(40) = 334400\ \text{J} \). Identify what the steam does. Two stages, and both release energy. Condensing at \( 100^\circ\text{C} \): releases \( mL_v \). Cooling from \( 100^\circ\text{C} \) to the final \( 60^\circ\text{C} \): releases \( mc(100 - 60) = m(4180)(40) = 167200m \). Set the released energy equal to the required energy. \( m(2.26 \times 10^6) + m(167200) = 334400 \) \( m(2.26 \times 10^6 + 1.672 \times 10^5) = 334400 \) \( m(2.4272 \times 10^6) = 334400 \) \( m = \dfrac{334400}{2.4272 \times 10^6} = 0.1378\ \text{kg} \). About 138 g of steam. Check the result. Condensation: \( (0.1378)(2.26 \times 10^6) = 311428\ \text{J} \). Cooling: \( (0.1378)(4180)(40) = 23040\ \text{J} \). Total released: \( 334468\ \text{J} \) ✓ matching the 334400 J required, to rounding. Note where the energy came from. Condensation supplied 93 percent of it, and cooling the condensed water only 7 percent. The phase change dominates completely. The commonest error here. Forgetting the second stage and using only \( mL_v \), which gives \( m = 0.148\ \text{kg} \), about 7 percent too high. The condensed steam does not vanish; it becomes water at \( 100^\circ\text{C} \) that must cool with the rest. A second error worth naming. Writing the steam's cooling range as \( 100 - 20 \) instead of \( 100 - 60 \). Everything ends at the final temperature of \( 60^\circ\text{C} \), not at the water's starting temperature. Sanity check the size. 138 g of steam heats 2.0 kg of water by \( 40^\circ\text{C} \), so a small mass of steam does a great deal of heating. That is why steam heating systems work and why steam is used to move energy around a power station. About 0.138 kg, or 138 g

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( \sigma = 5.67 \times 10^{-8}\ \text{W/(m}^2\cdot\text{K}^4) \).

1. Name the three mechanisms of heat transfer.
   Show the full solution

   Conduction, convection and radiation

2. Which mechanism works in a vacuum?
   Show the full solution

   Radiation

3. Write the conduction equation.
   Show the full solution

   \( P = \dfrac{kA\Delta T}{L} \)

4. What drives convection?
   Show the full solution

   Density differences in a heated fluid

5. Radiated power depends on which power of absolute temperature?
   Show the full solution

   The fourth

6. Find the conduction rate through \( 10\ \text{m}^2 \) of loft insulation 150 mm thick with \( k = 0.040\ \text{W/(m}\cdot\text{K)} \) and \( \Delta T = 20\ \text{K} \).
   Show the full solution

   \( P = \dfrac{(0.040)(10)(20)}{0.150} = \dfrac{8.0}{0.150} = 53.3\ \text{W} \). About 53 W. The same thickness of masonry, with \( k = 0.60 \), would pass 800 W, fifteen times more. 53 W

7. An object's absolute temperature is doubled. By what factor does its radiated power increase?
   Show the full solution

   \( P \propto T^4 \), so doubling \( T \) multiplies \( P \) by \( 2^4 = 16 \). This is why hot things lose energy so fast and why a star's luminosity is dominated by its surface temperature rather than its size, a point unit 11 returns to. 16 times

8. Explain why metals feel colder than wood at the same temperature.
   Show the full solution

   Because metal conducts energy away from the skin far faster, and the sensation reports the rate of energy loss rather than the temperature. Both objects are at room temperature, so there is no temperature difference between them to feel. Metals have high \( k \), around 200 for aluminum against about 0.15 for wood, because free electrons carry energy through the lattice rather than relying on vibrations passing from atom to atom. The nerve endings measure rate, not temperature. The metal draws energy out of the skin quickly, the skin surface cools, and the nerves report cold. The wood draws energy slowly, the skin surface stays near its usual temperature, and the wood reports as neutral. The prediction that confirms it. Warm both to \( 45^\circ\text{C} \) and the metal will now feel far hotter than the wood, because the same high conductivity now delivers energy quickly. If the sensation reported temperature, the two would feel identical in both cases, and they do not. High conductivity means a fast transfer rate, and touch measures rate

9. A radiator is placed low on a wall and an air-conditioning vent high on the same wall. Explain why each is positioned as it is.
   Show the full solution

   Both positions exploit convection, and each is placed so its circulation sweeps the whole room. The radiator, low down. Air heated at the radiator expands, becomes less dense, and rises. It spreads across the ceiling, cools against the walls and windows, and sinks back to floor level where it returns to the radiator. Starting the warm air at the bottom means the circulation covers the full height of the room. Put the radiator high and the warm air would rise the short remaining distance to the ceiling and stay there. The floor would stay cold, because nothing would drive cold air upward. The vent, high up. Cooled air is denser, so it sinks. Releasing it at the ceiling lets it fall through the whole room, displacing warm air upward to the vent to be cooled in turn. Again the circulation spans the room. Put the cold vent low and the cold air would pool on the floor, and the warm air above it would never be reached. The common principle. Introduce the buoyant fluid at the end of the room it will travel away from. Warm air rises, so introduce it low; cold air sinks, so introduce it high. Both are placed so buoyancy carries the conditioned air across the full room

10. A radiator surface of \( 1.2\ \text{m}^2 \) at \( 60^\circ\text{C} \) with \( e = 0.90 \) sits in a room at \( 20^\circ\text{C} \). Find the net radiated power, and comment on whether radiation is the main way it heats the room.
    Show the full solution

    Convert to kelvin first. \( T = 60 + 273 = 333\ \text{K} \); \( T_s = 20 + 273 = 293\ \text{K} \). Use the net form. \( P_{\text{net}} = \sigma A e (T^4 - T_s^4) \). Compute the fourth powers. \( 333^4 = 1.2293 \times 10^{10} \) \( 293^4 = 7.3701 \times 10^{9} \) Difference: \( 4.923 \times 10^{9} \). Compute the prefactor. \( (5.67 \times 10^{-8})(1.2)(0.90) = 6.124 \times 10^{-8} \). Multiply. \( P_{\text{net}} = (6.124 \times 10^{-8})(4.923 \times 10^{9}) = 302\ \text{W} \). About 300 W. Now judge whether radiation dominates. A domestic radiator of this size typically delivers 1000 to 1500 W in total. So radiation supplies roughly a quarter of the output, and convection supplies most of the rest. Why convection wins here. The temperature difference is only 40 K, which is modest, and the \( T^4 \) law only becomes dominant at much higher temperatures. Meanwhile the radiator is continuously replacing the air at its surface, so convection runs at full rate indefinitely. The design evidence. Radiators are built with fins and folded panels that greatly increase contact with air while adding little to the outward-facing area that can radiate. If radiation were the point, the shape would be a flat black plate instead. Where radiation does dominate. An open fire or a glowing electric bar at 1000 K radiates thousands of watts by the fourth-power law, and its warmth reaches you across the room at the speed of light, which is why you feel it instantly on the side facing it and not on the other. The name is wrong. A household radiator is mostly a convector, and the two calculations together show it rather than just asserting it. About 300 W, roughly a quarter of the output; convection dominates

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Take \( W \) as work done by the system throughout.

1. State the first law of thermodynamics.
   Show the full solution

   \( \Delta U = Q - W \)

2. What does \( U \) represent?
   Show the full solution

   The internal energy of the system

3. What is \( Q \) in an adiabatic process?
   Show the full solution

   Zero

4. What is \( \Delta U \) in an isothermal process?
   Show the full solution

   Zero

5. A gas is compressed. Is \( W \) positive or negative in this convention?
   Show the full solution

   Negative, because work is done on the gas

6. A system absorbs 800 J of heat and does 300 J of work. Find \( \Delta U \).
   Show the full solution

   \( \Delta U = Q - W = 800 - 300 = +500\ \text{J} \). +500 J

7. A system releases 400 J of heat while 100 J of work is done on it. Find \( \Delta U \).
   Show the full solution

   Heat leaves, so \( Q = -400\ \text{J} \). Work is done on the system, so \( W = -100\ \text{J} \). \( \Delta U = -400 - (-100) = -400 + 100 = -300\ \text{J} \). The internal energy falls by 300 J, and the system cools. Both signs had to be set deliberately, which is the whole difficulty of this equation. \( -300 \) J

8. A gas expands isothermally and does 250 J of work. Find the heat absorbed, and explain why it cannot be zero.
   Show the full solution

   Isothermal means constant temperature, so \( \Delta U = 0 \) for an ideal gas. \( 0 = Q - W \), therefore \( Q = W = 250\ \text{J} \). It cannot be zero because the gas is doing work on its surroundings, which drains energy from it. If no heat were supplied, that energy could come only from the internal energy, and the gas would cool. Holding the temperature constant requires heat to be supplied at exactly the rate work is done. 250 J absorbed

9. Explain why a gas cools when it expands rapidly, and name a place this is observed.
   Show the full solution

   Because rapid expansion is close to adiabatic, so the work the gas does comes out of its own internal energy. Apply the first law. Rapid means there is no time for heat to flow in from the surroundings, so \( Q \approx 0 \) and \( \Delta U = -W \). The gas is expanding, so it pushes the surrounding air back and \( W \) is positive. Therefore \( \Delta U \) is negative and the internal energy falls. Internal energy falling means temperature falling, since for a gas the internal energy is the molecular kinetic energy. The microscopic picture. Molecules bouncing off a receding wall rebound more slowly than they arrived, the exact reverse of the compression case. Their average speed drops, so the temperature drops. Where it is observed. A can of compressed air becomes cold when sprayed continuously. A carbon dioxide extinguisher can form dry ice at its nozzle. Air rising over a mountain expands as the pressure falls, cools, and forms the cloud that sits on the summit. And every refrigerator and air conditioner runs on this step: the working fluid is forced through an expansion valve and emerges cold enough to absorb heat from the compartment. The counterpart. The bicycle pump warming on compression is the same equation with both signs reversed, which is a useful check that the reasoning is consistent. The work of expansion is paid for out of internal energy when no heat can flow in

10. Explain why a perpetual motion machine that produces more energy than it consumes is impossible, and explain why the first law alone does not rule out every proposed perpetual motion machine.
    Show the full solution

    The first kind of perpetual motion machine. This is a device claiming to deliver net work with no energy input at all, or more work out than energy in. Why the first law forbids it. A machine that runs in cycles returns to its starting state each cycle, so \( \Delta U = 0 \) over the cycle. The first law then gives \( Q = W \) exactly: the work delivered must equal the heat absorbed, no more. Producing work with \( Q = 0 \) would require \( W \gt 0 \) with \( \Delta U = 0 \), which contradicts the equation. Where such machines actually get their energy. Every proposed example has turned out to be drawing on something unnoticed: a wound spring, a temperature difference, a chemical reaction, a slow leak of stored potential energy. When the accounting is complete, the balance closes. How strong the evidence is. No violation of energy conservation has ever been observed, in any experiment, at any scale from subatomic to astronomical. The United States Patent and Trademark Office requires a working model for perpetual motion applications, essentially the only field where it does so, and none has been supplied (USPTO, US federal). Now the second part, and it is the more interesting one. The first law is only a bookkeeping rule. It says the totals must balance; it says nothing about which direction things go. The machine the first law permits. Consider a ship that draws thermal energy from the ocean, converts it entirely to work to drive its propellers, and leaves slightly cooler water behind. Energy in equals energy out, so the first law is perfectly satisfied. The ocean holds an enormous quantity of thermal energy, so the fuel would be free and effectively limitless. Why it still cannot work. This is a perpetual motion machine of the second kind, and it is forbidden by the second law, not the first. Heat will not flow out of a single reservoir and become work with nothing else changing. An engine needs a cold reservoir to reject heat into, and the ship has none. What this shows about the two laws. The first law says you cannot win; the second says you cannot break even. Energy conservation alone is not enough to describe how the world behaves, which is precisely why a second law is needed, and lesson 6.6 states it. The first law forbids creating energy; it does not forbid converting thermal energy completely into work, and the second law does

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use kelvin in every efficiency and entropy calculation.

1. State the second law in terms of heat flow.
   Show the full solution

   Heat flows spontaneously from hot to cold, never the reverse

2. What does entropy measure?
   Show the full solution

   The number of microscopic arrangements matching a state, that is, its disorder

3. Write the Carnot efficiency formula.
   Show the full solution

   \( \eta = 1 - \dfrac{T_c}{T_h} \)

4. Write the entropy change for a transfer at constant temperature.
   Show the full solution

   \( \Delta S = \dfrac{Q}{T} \)

5. What happens to the total entropy of an isolated system?
   Show the full solution

   It never decreases

6. Find the maximum efficiency of an engine with \( T_h = 1200\ \text{K} \) and \( T_c = 350\ \text{K} \).
   Show the full solution

   \( \eta = 1 - \dfrac{350}{1200} = 1 - 0.2917 = 0.708 \). 70.8 percent. A real gasoline engine manages about 25 percent, barely a third of this, because the ideal cycle is never achieved and much energy leaves through the exhaust and cooling system. 70.8 percent

7. Find the entropy change when 2000 J is transferred into a large reservoir at 400 K.
   Show the full solution

   \( \Delta S = \dfrac{Q}{T} = \dfrac{2000}{400} = +5.0\ \text{J/K} \). The reservoir is specified as large so that its temperature does not change measurably during the transfer, which is what lets a single value of \( T \) be used. +5.0 J/K

8. An engine claims 65 percent efficiency operating between 600 K and 300 K. Evaluate the claim.
   Show the full solution

   Find the Carnot limit for those reservoirs: \( \eta_{\max} = 1 - \dfrac{300}{600} = 0.500 \), that is 50 percent. The claim of 65 percent exceeds the limit, so it is impossible. No further information is needed. The Carnot limit depends only on the two temperatures, so no detail of the design, fuel or workmanship could rescue the claim. It would require a net entropy decrease. What is probably wrong. Either the temperatures are misreported, or the efficiency is being measured against something other than the total heat input, or the engine is drawing on a hotter reservoir than stated. Impossible; the limit is 50 percent

9. Explain why a refrigerator does not violate the second law even though it moves heat from cold to hot.
   Show the full solution

   Because the law forbids that transfer only when it happens spontaneously, and a refrigerator does work to make it happen. Read the law precisely. Heat does not flow from cold to hot of its own accord. The Clausius statement includes that qualifier, and a refrigerator does not meet it. What the refrigerator supplies. A compressor driven by electrical work forces the refrigerant around the cycle. Without that work the machine does nothing and the food warms up, which is exactly what the law predicts. Do the entropy accounting. Heat \( Q_c \) is removed from the cold compartment, lowering its entropy by \( Q_c/T_c \). Heat \( Q_h = Q_c + W \) is released at the back, raising the kitchen's entropy by \( Q_h/T_h \). Why the total still rises. The released quantity is larger, because it includes the work, and although the kitchen's temperature \( T_h \) is higher, the added \( W \) is more than enough to make \( Q_h/T_h \) exceed \( Q_c/T_c \) in any real machine. A worked case. Removing 300 J from a compartment at 270 K lowers its entropy by \( 300/270 = 1.11\ \text{J/K} \). With 100 J of work, 400 J is released into a 300 K kitchen, raising entropy by \( 400/300 = 1.33\ \text{J/K} \). The net change is \( +0.22\ \text{J/K} \), positive, so the process is allowed. The limiting case. Reducing the work input drives the total toward zero, and at zero it would be the reversible ideal, which no real machine reaches. Below that no amount of engineering helps, and that limit sets the maximum coefficient of performance of any refrigerator or heat pump. The system that matters. The compartment alone is not isolated, so its entropy decrease means nothing on its own. The isolated system is the compartment plus the kitchen plus the power station, and its entropy rises. Work is done, so the transfer is not spontaneous and the total entropy still rises

10. Explain why the second law gives time a direction while the other laws of physics do not.
    Show the full solution

    Start with what is strange about the other laws. Newton's laws, the conservation of momentum and energy, and the laws of electromagnetism are all time-symmetric. Reverse every velocity in a system and the reversed motion is an equally valid solution. Nothing in those equations distinguishes forward from backward. The film test. A film of two billiard balls colliding looks entirely normal run backward. A film of a cup shattering does not, and neither does a film of a drop of ink un-mixing itself from water or of a warm room spontaneously heating a cold cup of coffee. The puzzle this creates. The shattering cup is made of atoms obeying time-symmetric laws. Every individual collision in the reversed film is legal. So where does the obvious wrongness come from? The answer is statistical. The reversed process is not forbidden by any law of mechanics; it is overwhelmingly improbable. There are vastly more microscopic arrangements corresponding to a shattered cup than to an intact one, and vastly more corresponding to mixed ink than to a concentrated drop. Why that is decisive. The numbers involved are not merely large. For a system of \( 10^{23} \) particles the ratio of arrangements is such that waiting for a spontaneous reversal would take far longer than the age of the universe. Improbable at that scale is indistinguishable from impossible. So entropy defines the arrow. The direction in which total entropy increases is the direction we call the future. This is the only place in physics where a direction of time appears, and it appears not as a new law of mechanics but as a consequence of counting arrangements. Note what is not being claimed. The second law does not say a reversal is impossible in principle. It says it is so improbable for any macroscopic system that it has never been observed and never will be. For a handful of particles fluctuations against the law are routinely seen, which is itself good evidence that the statistical account is correct. The deeper question it raises. If entropy increases toward the future, it must have been lower in the past, and lower still further back. Following that chain leads to the early universe having had remarkably low entropy, a fact no one has fully explained and which unit 11 returns to when it reaches cosmology. Why this belongs in a physics course rather than philosophy. The arrow of time is measurable. Entropy has units, it can be computed, and the direction of every spontaneous process can be predicted from its sign. It is a quantitative claim, not a metaphor. Entropy increase is the only physical process that distinguishes past from future, and it does so statistically rather than by a new law

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Use \( k = 8.99 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2 \) and \( e = 1.60 \times 10^{-19}\ \text{C} \).

1. Write Coulomb's law.
   Show the full solution

   \( F = \dfrac{kq_1q_2}{r^2} \)

2. Give the magnitude of the charge on an electron.
   Show the full solution

   \( 1.60 \times 10^{-19}\ \text{C} \)

3. What happens between two negative charges?
   Show the full solution

   They repel

4. What does it mean that charge is conserved?
   Show the full solution

   It is transferred, never created or destroyed

5. Name the property that distinguishes a conductor from an insulator.
   Show the full solution

   A conductor has charges free to move through it

6. Find the force between \( +4.0\ \mu\text{C} \) and \( -2.0\ \mu\text{C} \) separated by 0.20 m.
   Show the full solution

   Use magnitudes: \( F = \dfrac{(8.99 \times 10^9)(4.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.20)^2} = \dfrac{0.07192}{0.040} = 1.80\ \text{N} \). The signs are opposite, so the force is attractive. 1.80 N, attractive

7. How many electrons make up a charge of \( -1.0\ \mu\text{C} \)?
   Show the full solution

   \( n = \dfrac{1.0 \times 10^{-6}}{1.60 \times 10^{-19}} = 6.25 \times 10^{12} \). Over six trillion electrons for one microcoulomb, which shows how enormous a coulomb is as a unit and why everyday static charges are measured in nanocoulombs and microcoulombs. \( 6.25 \times 10^{12} \) electrons

8. Two charges attract with 12 N. The distance between them is tripled and one charge is doubled. Find the new force.
   Show the full solution

   Work with the proportionality \( F \propto \dfrac{q_1q_2}{r^2} \). Doubling one charge multiplies \( F \) by 2. Tripling \( r \) divides \( F \) by \( 3^2 = 9 \). New force: \( 12 \times \dfrac{2}{9} = 2.67\ \text{N} \). Still attractive, since neither sign changed. 2.67 N

9. Explain how a charged balloon sticks to a neutral wall.
   Show the full solution

   By inducing a separation of charge in the wall, so that the nearer surface carries the opposite charge and attraction wins. Start with the puzzle. The wall is neutral overall, and a neutral object should feel no net force from a charged one. So the explanation must involve the charges inside the wall moving. What the balloon does. Rubbing it on hair transfers electrons to it, leaving it negatively charged. What happens in the wall. The balloon's negative charge repels electrons in the wall's surface molecules, pushing them slightly away, and leaves the nearer side of each molecule slightly positive. This is polarization, and it happens even in an insulator where no charge can travel far, because the molecules themselves distort. Why attraction wins. The wall now has a slightly positive layer near the balloon and a slightly negative layer further away. Coulomb's law is an inverse square, so the nearer positive layer exerts a larger attraction than the more distant negative layer exerts repulsion. The forces do not cancel, and the balloon is pulled in. Why the distance dependence is essential. If the force fell off linearly, or not at all, the two layers would cancel exactly and the balloon would fall. The effect exists only because closer charges count for more. The prediction that tests it. The attraction should also work on any neutral insulator, and it does: a charged rod attracts small pieces of paper, a thin stream of water bends toward it, and none of those objects carries a net charge. Why it eventually falls. Charge slowly leaks from the balloon to the air, especially in humid conditions where water molecules carry it away, which is why the trick works poorly on a damp day. Polarization creates a nearer opposite charge, and the inverse square makes its attraction dominate

10. Two identical conducting spheres carry charges of \( +8.0\ \mu\text{C} \) and \( -2.0\ \mu\text{C} \). They are touched together and separated to 0.30 m. Find the force before contact and after, and explain the change.
    Show the full solution

    Before contact, at 0.30 m. \( F = \dfrac{(8.99 \times 10^9)(8.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.30)^2} = \dfrac{0.14384}{0.090} = 1.60\ \text{N} \), attractive, since the signs are opposite. What contact does. The spheres are conductors and identical, so charge is free to move between them and distributes equally. Charge is conserved, so the total is unchanged. Total charge. \( +8.0 + (-2.0) = +6.0\ \mu\text{C} \). Each sphere afterward. \( 6.0/2 = +3.0\ \mu\text{C} \). After separation, at 0.30 m. \( F = \dfrac{(8.99 \times 10^9)(3.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.30)^2} = \dfrac{0.080910}{0.090} = 0.899\ \text{N} \), repulsive, since both are now positive. Explain the change in direction. This is the striking part. The force reverses from attraction to repulsion without either sphere being touched by anything external. Contact allowed the excess electrons on the negative sphere to move onto the positive one, neutralizing part of it, and what remained was a net positive charge shared between them. Explain the change in size. The magnitude fell from 1.60 N to 0.899 N, a factor of 0.562. That is exactly \( \dfrac{3 \times 3}{8 \times 2} = \dfrac{9}{16} \) ✓ which confirms the arithmetic. Why the product matters, not the total. Sharing equally maximizes the product of the two charges for a fixed total, so for a given total charge, the equal split gives the strongest possible repulsion. Here the total also fell from an effective \( 8 \times 2 = 16 \) to \( 3 \times 3 = 9 \) because neutralization destroyed part of the imbalance. The assumption being made. That the spheres are identical, which is what justifies the equal split. If one were larger it would take a larger share, and the problem would need their relative sizes. A second assumption. That 0.30 m is large compared with the spheres, so they can be treated as point charges. If they nearly touched, the charge on each would redistribute toward the far side under mutual repulsion and Coulomb's law in this simple form would not apply. 1.60 N attractive before; 0.899 N repulsive after

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Electron mass \( 9.11 \times 10^{-31}\ \text{kg} \).

1. Define electric field strength.
   Show the full solution

   Force per unit positive charge, \( E = F/q \)

2. Give the units of electric field.
   Show the full solution

   N/C, equivalently V/m

3. Which way do field lines point near a negative charge?
   Show the full solution

   Inward, toward the charge

4. Write the field between parallel plates.
   Show the full solution

   \( E = \dfrac{V}{d} \)

5. What is the electric field inside a conductor in equilibrium?
   Show the full solution

   Zero

6. Find the field 0.50 m from a \( -8.0\ \mu\text{C} \) charge.
   Show the full solution

   \( E = \dfrac{(8.99 \times 10^9)(8.0 \times 10^{-6})}{(0.50)^2} = \dfrac{7.192 \times 10^4}{0.25} = 2.88 \times 10^5\ \text{N/C} \). Directed toward the charge, since it is negative. \( 2.88 \times 10^5 \) N/C, inward

7. A field of 2500 N/C exerts a force of 0.015 N on a charge. Find the charge.
   Show the full solution

   \( q = \dfrac{F}{E} = \dfrac{0.015}{2500} = 6.0 \times 10^{-6}\ \text{C} \). \( 6.0\ \mu\text{C} \)

8. Two plates 5.0 mm apart are connected to 250 V. Find the field and the force on a proton in it.
   Show the full solution

   \( E = \dfrac{V}{d} = \dfrac{250}{0.0050} = 5.0 \times 10^4\ \text{V/m} \). \( F = qE = (1.60 \times 10^{-19})(5.0 \times 10^4) = 8.0 \times 10^{-15}\ \text{N} \). Directed from the positive plate toward the negative one, since the proton's charge is positive and it follows the field. \( 5.0 \times 10^4 \) V/m and \( 8.0 \times 10^{-15} \) N

9. Explain why the electric field is zero everywhere inside a hollow charged metal sphere.
   Show the full solution

   Because any field inside the metal would drive the free charges until it was canceled, and equilibrium is reached only when it is zero. The dynamic argument, which is the essential one. Suppose a field existed somewhere inside the conducting material. The metal contains free electrons, so they would feel a force and move. Moving charge is a current, and a current means the situation is changing, so it is not equilibrium. Charges keep moving until they have rearranged into a configuration where the field they produce cancels whatever field was there. At that point nothing moves, and the field inside the metal is exactly zero. Extending it to the hollow cavity. The same result holds in the empty space inside the shell. Any field there would have to begin and end on charges, and there are none in the cavity. A field line entering the cavity would have to leave again, which would require work around a closed loop, and the electric force is conservative, so that cannot happen. Where the charge sits. All the excess charge ends up on the outer surface, driven there by mutual repulsion, spreading as far apart as the geometry allows. The historical test. This prediction is an exact consequence of the inverse square law, and would fail if the exponent were not precisely 2. Cavendish tested it in the 1770s by looking for charge on an inner sphere and finding none, and modern versions of the same experiment confirm the exponent to within about one part in \( 10^{16} \). It remains one of the most precisely verified statements in physics. Why the shape does not matter. The argument nowhere assumed a sphere. Any closed conducting shell works, which is why a car, an aircraft and a wire mesh cage all shield their interiors. What it does not protect against. Fields from charges placed inside the cavity are not shielded; the shell blocks external fields, not internal ones. Free charges rearrange until no field remains, and that is the equilibrium condition

10. An electron enters a uniform field of \( 3.0 \times 10^4\ \text{V/m} \) moving perpendicular to it at \( 2.0 \times 10^6\ \text{m/s} \), and the field region is 4.0 cm long. Find its sideways deflection on exit, and explain what this arrangement is used for.
    Show the full solution

    Recognize the structure. The electron has a constant velocity along the field region and a constant acceleration perpendicular to it. That is projectile motion from lesson 1.6 with the electric force replacing gravity, and the two directions are independent. Find the time in the field. \( t = \dfrac{L}{v} = \dfrac{0.040}{2.0 \times 10^6} = 2.0 \times 10^{-8}\ \text{s} \). Find the acceleration. \( F = qE = (1.60 \times 10^{-19})(3.0 \times 10^4) = 4.8 \times 10^{-15}\ \text{N} \) \( a = \dfrac{F}{m} = \dfrac{4.8 \times 10^{-15}}{9.11 \times 10^{-31}} = 5.27 \times 10^{15}\ \text{m/s}^2 \). Find the sideways displacement. It starts with no sideways velocity, so \( y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(5.27 \times 10^{15}) (2.0 \times 10^{-8})^2 \) \( y = \tfrac{1}{2}(5.27 \times 10^{15})(4.0 \times 10^{-16}) = 1.05\ \text{m} \). Stop and check that. A deflection of over a meter inside a 4 cm region is impossible, so something is wrong with the setup rather than the arithmetic. The electron would strike a plate long before exiting. What the check reveals. The numbers chosen describe a field far too strong, or a speed far too low, for the electron to cross. This is exactly the kind of sanity check the course keeps insisting on: an answer must be compared against the geometry it came from. Find what would work. For a deflection of, say, 5.0 mm across the same 4.0 cm at the same field, the required speed follows from \( y = \tfrac{1}{2}a\left(\dfrac{L}{v}\right)^2 \), so \( v = L\sqrt{\dfrac{a}{2y}} = 0.040\sqrt{\dfrac{5.27 \times 10^{15}}{0.010}} = 0.040 \times 7.26 \times 10^{8} = 2.9 \times 10^{7}\ \text{m/s} \). About \( 3 \times 10^7\ \text{m/s} \), roughly ten percent of the speed of light, which is a realistic figure for an electron beam and the reason real instruments accelerate electrons through thousands of volts first. What the arrangement is for. Deflecting a charged beam by a controlled amount is how the cathode ray oscilloscope works: the voltage to be measured is applied to the plates, and the beam's deflection on the screen is proportional to it, giving a direct visual reading of a rapidly changing voltage. The same principle elsewhere. Inkjet printers steer charged droplets onto paper this way, and mass spectrometers separate ions by how much a field deflects them, which depends on their charge-to-mass ratio. The relationship worth extracting. \( y = \dfrac{qEL^2}{2mv^2} \), so the deflection is proportional to the field, and therefore to the applied voltage, which is what makes the instrument linear and readable. The given numbers give an impossible 1.05 m, showing the beam would strike a plate; about \( 3 \times 10^7 \) m/s is needed for a workable 5 mm deflection

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Define potential difference.
   Show the full solution

   Energy transferred per unit charge, \( V = W/q \)

2. One volt equals how many joules per coulomb?
   Show the full solution

   One

3. Write the potential near a point charge.
   Show the full solution

   \( V = \dfrac{kQ}{r} \)

4. Give the value of one electronvolt in joules.
   Show the full solution

   \( 1.60 \times 10^{-19}\ \text{J} \)

5. Is potential a vector or a scalar?
   Show the full solution

   A scalar

6. Find the energy delivered when a 12 V battery moves 500 C.
   Show the full solution

   \( W = qV = (500)(12) = 6000\ \text{J} \). 6000 J

7. A charge of 0.25 C gains 4.5 J crossing a component. Find the potential difference.
   Show the full solution

   \( V = \dfrac{W}{q} = \dfrac{4.5}{0.25} = 18\ \text{V} \). 18 V

8. Find the potential at a point 0.40 m from \( +3.0\ \mu\text{C} \) and 0.60 m from \( -4.0\ \mu\text{C} \).
   Show the full solution

   Potential is a scalar, so the contributions add with their signs and no vector work is needed. This is the main practical advantage of potential over field. From the positive charge: \( \dfrac{(8.99 \times 10^9)(3.0 \times 10^{-6})}{0.40} = +6.74 \times 10^4\ \text{V} \). From the negative charge: \( \dfrac{(8.99 \times 10^9)(-4.0 \times 10^{-6})}{0.60} = -5.99 \times 10^4\ \text{V} \). Sum: \( 6.74 \times 10^4 - 5.99 \times 10^4 = +7.5 \times 10^3\ \text{V} \). Note that the geometry was never needed. The angle between the two charges does not appear, whereas finding the field at that point would have required resolving two vectors. \( +7.5 \times 10^3 \) V

9. Explain why potential falls off as \( 1/r \) while field falls off as \( 1/r^2 \).
   Show the full solution

   Because potential is the accumulated effect of the field over distance, and accumulating an inverse square over distance gives an inverse first power. State the relationship. The potential difference between two points is the work per unit charge to move between them, and work is force times distance. So potential is field multiplied by distance, accumulated along the path. Apply it dimensionally. Field has units N/C; multiplying by meters gives N·m/C, which is J/C, which is volts ✓ So potential must carry one more power of distance than field, which is exactly the difference between \( 1/r^2 \) and \( 1/r \). See it in the formulas. \( E = \dfrac{kQ}{r^2} \) and \( V = \dfrac{kQ}{r} \), and \( V = Er \) for a point charge, which is the same statement. The physical reading. The field weakens quickly with distance, but moving from \( r \) to infinity covers an unlimited distance, and those two effects combine so that the total work is finite and proportional to \( 1/r \). Why this matters practically. Potential is far easier to work with: it is a scalar, so contributions from many charges add as numbers, while fields must be added as vectors. Most real calculations find the potential first and get the field from how fast it changes with position. The parallel with gravity. Gravitational field goes as \( 1/r^2 \) and gravitational potential energy as \( 1/r \), for exactly the same reason, which is why the escape speed derivation in unit 4 had the form it did. Potential accumulates the field over distance, adding one power of \( r \)

10. A proton is accelerated from rest through 2000 V. Find its final speed, its kinetic energy in joules and in electronvolts, and compare it with the electron in the worked example.
    Show the full solution

    Find the energy first. \( W = qV = (1.60 \times 10^{-19})(2000) = 3.20 \times 10^{-16}\ \text{J} \). In electronvolts. By definition, a charge of one elementary unit crossing 2000 V gains 2000 eV, so the answer is 2000 eV, or 2.0 keV, with no arithmetic required. This is why the unit exists. Find the speed. All the work becomes kinetic energy, since it starts at rest. \( \tfrac{1}{2}mv^2 = 3.20 \times 10^{-16} \) \( v = \sqrt{\dfrac{2(3.20 \times 10^{-16})}{1.673 \times 10^{-27}}} = \sqrt{3.825 \times 10^{11}} = 6.18 \times 10^{5}\ \text{m/s} \). About \( 6.2 \times 10^5\ \text{m/s} \). Now compare with the electron. The electron through 100 V reached \( 5.9 \times 10^6\ \text{m/s} \), nearly ten times faster, despite receiving twenty times less energy. Explain why. The proton is about 1836 times more massive. For the same energy, \( v \propto 1/\sqrt{m} \), so the proton would be \( \sqrt{1836} = 42.8 \) times slower. Check that the two effects combine correctly. The proton got 20 times more energy, which alone would make it \( \sqrt{20} = 4.47 \) times faster. Combined with being 42.8 times slower from mass: \( 4.47/42.8 = 0.104 \), so the proton should be about a tenth of the electron's speed. Check: \( \dfrac{6.18 \times 10^5}{5.93 \times 10^6} = 0.104 \) ✓ exactly as predicted. Why this matters for machine design. Accelerating heavy particles to a given speed takes far more energy than light ones, which is why electron accelerators are compact and proton accelerators are enormous. The Large Hadron Collider's 27 km ring exists because protons are hard to accelerate. The relativistic check. The proton at \( 6.2 \times 10^5 \) m/s is at 0.2 percent of \( c \), so the non-relativistic formula is very safe here, more so than for the electron in the worked example. The design consequence worth keeping. Because the energy in electronvolts depends only on the charge and the voltage, and not on the mass, quoting accelerator energies in eV lets machines using different particles be compared directly, while quoting speeds would not. \( 6.2 \times 10^5 \) m/s, \( 3.2 \times 10^{-16} \) J, 2.0 keV; about a tenth the electron's speed despite twenty times the energy

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Define electric current.
   Show the full solution

   The rate of flow of charge, \( I = Q/t \)

2. One ampere equals how many coulombs per second?
   Show the full solution

   One

3. Write the relationship defining resistance.
   Show the full solution

   \( R = \dfrac{V}{I} \)

4. In which direction does conventional current flow?
   Show the full solution

   From positive to negative, opposite to the electron flow

5. What shape is the current-voltage graph of an ohmic conductor?
   Show the full solution

   A straight line through the origin

6. Find the resistance when 9.0 V drives 0.30 A.
   Show the full solution

   \( R = \dfrac{V}{I} = \dfrac{9.0}{0.30} = 30\ \Omega \). \( 30\ \Omega \)

7. Find the charge passing in 2.0 minutes at 0.75 A, and the energy it delivers across 6.0 V.
   Show the full solution

   \( t = 120\ \text{s} \). \( Q = It = (0.75)(120) = 90\ \text{C} \). \( W = QV = (90)(6.0) = 540\ \text{J} \). 90 C and 540 J

8. A wire is replaced by one of the same material, twice the length and half the diameter. Find the factor by which its resistance changes.
   Show the full solution

   Use \( R = \dfrac{\rho L}{A} \). Doubling the length doubles \( R \). Halving the diameter quarters the area, since \( A = \pi r^2 \), and quartering the area multiplies \( R \) by 4. Combined: \( 2 \times 4 = 8 \). The diameter is the more powerful lever, because it enters squared, which is why cables for heavy loads are made thick rather than short. 8 times greater

9. Explain why Ohm's law is better described as a useful approximation than as a law of nature.
   Show the full solution

   Because it is an empirical description of how certain materials behave under certain conditions, not a universal statement, and many common components disobey it. Separate the two statements first. \( V = IR \) is the definition of resistance and is always true, because \( R \) is defined as the ratio. Ohm's law is the stronger claim that this ratio stays constant as the voltage changes, and that claim can fail. Where it holds. Metals at constant temperature, over a wide range of currents. This covers most wiring and most resistors, which is why the approximation is so useful. Where it fails: temperature. A filament lamp's resistance rises roughly fifteenfold between room temperature and operating temperature, so its graph curves badly. Even an ordinary resistor drifts if it warms. Where it fails: direction. A diode conducts readily one way and barely at all the other. Its resistance is not a number at all, but depends on both the size and the sign of the applied voltage. Every rectifier and every LED relies on this failure. Where it fails: the opposite temperature sign. A thermistor's resistance falls as it warms, because heating frees more charge carriers rather than just increasing lattice vibration. Semiconductors generally behave this way, opposite to metals, and this is the basis of temperature sensors. Where it fails completely. A superconductor below its critical temperature has exactly zero resistance, and a gas discharge tube has a region where increasing the current lowers the voltage across it, a negative resistance that no constant \( R \) can describe. Compare with a genuine law. Newton's second law has no known exceptions within its domain, and where it fails, at relativistic speeds, a deeper theory replaces it and explains why. Ohm's law has ordinary exceptions sitting in every household appliance, and the exceptions are the basis of most modern electronics. Why the distinction is worth making. Treating it as universal leads students to apply \( V = IR \) with a fixed \( R \) to a lamp or diode and get answers that are simply wrong. Knowing it is conditional prompts the question "is this component ohmic?", which is the right question. It describes a class of materials under restricted conditions, and its exceptions are the foundation of electronics

10. A filament lamp draws 0.50 A at 12 V when running. Its resistance at room temperature is \( 1.6\ \Omega \). Find its operating resistance, the switch-on current, and explain what this means for the circuit design.
    Show the full solution

    Find the operating resistance. \( R = \dfrac{V}{I} = \dfrac{12}{0.50} = 24\ \Omega \). Compare with the cold value. \( \dfrac{24}{1.6} = 15 \), so the filament's resistance is fifteen times greater when hot, which matches the known behavior of tungsten. Find the switch-on current. At the instant of switching, the filament is still cold and its resistance is \( 1.6\ \Omega \): \( I = \dfrac{12}{1.6} = 7.5\ \text{A} \). Fifteen times the running current, which follows directly since the voltage is fixed and the current is inversely proportional to the resistance. How long the surge lasts. Only milliseconds. The filament has a very small mass and the power dissipated in the first instant is \( VI = 12 \times 7.5 = 90\ \text{W} \), fifteen times the running 6 W, so it heats to operating temperature almost immediately and the current falls back. Consequence one: this is when lamps fail. The surge stresses the filament mechanically and thermally, and the thinnest point, where the filament has already been eroded by evaporation, heats fastest and breaks. Almost every filament lamp fails at switch-on rather than during steady operation, which is a prediction this calculation makes and everyday experience confirms. Consequence two: fuse and breaker ratings. A fuse rated just above the running current would blow every time the lamp was switched on. Protective devices must either be rated well above the steady load or, better, be of a slow-blow type that tolerates brief surges but responds to sustained overcurrent. Consequence three: scaling to installations. A lighting circuit with twenty such lamps draws 10 A running but 150 A for a few milliseconds at switch-on. This is why large installations stagger their switching across several contactors rather than energizing everything at once, and why theatrical dimmers often keep filaments slightly warm between cues. Consequence four: why LEDs changed this. An LED's current is set by its driver circuit, which limits it regardless of the LED's own characteristics, so there is no inrush of this kind. That removed a whole category of design constraint along with the energy saving. What the lamp is not. This whole analysis exists because the lamp is not ohmic. If it were, the cold and hot resistances would be equal and there would be no surge at all. \( 24\ \Omega \) hot, 7.5 A at switch-on, fifteen times the running current

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. In a series circuit, what is the same at every point?
   Show the full solution

   The current

2. In a parallel circuit, what is the same across every branch?
   Show the full solution

   The voltage

3. Write the rule for resistances in series.
   Show the full solution

   \( R_{\text{total}} = R_1 + R_2 + \dots \)

4. Write the rule for resistances in parallel.
   Show the full solution

   \( \dfrac{1}{R_{\text{total}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots \)

5. State Kirchhoff's current rule.
   Show the full solution

   Current into a junction equals current out of it

6. Find the total resistance of \( 4.0\ \Omega \) and \( 6.0\ \Omega \) in parallel.
   Show the full solution

   \( \dfrac{1}{R} = \dfrac{1}{4.0} + \dfrac{1}{6.0} = \dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12} \). \( R = \dfrac{12}{5} = 2.4\ \Omega \). Smaller than both branches ✓ \( 2.4\ \Omega \)

7. A 12 V supply drives \( 4.0\ \Omega \) and \( 6.0\ \Omega \) in series. Find the current and the voltage across each.
   Show the full solution

   \( R = 4.0 + 6.0 = 10.0\ \Omega \). \( I = \dfrac{12}{10.0} = 1.2\ \text{A} \), the same through both. \( V_4 = (1.2)(4.0) = 4.8\ \text{V} \) \( V_6 = (1.2)(6.0) = 7.2\ \text{V} \). Check: \( 4.8 + 7.2 = 12\ \text{V} \) ✓ The larger resistance takes the larger share of the voltage, in the same ratio as the resistances, which is the principle behind a potential divider. 1.2 A, with 4.8 V and 7.2 V

8. The same \( 4.0\ \Omega \) and \( 6.0\ \Omega \) are connected in parallel across 12 V. Find each branch current and the total.
   Show the full solution

   Both have the full 12 V across them. \( I_4 = \dfrac{12}{4.0} = 3.0\ \text{A} \) \( I_6 = \dfrac{12}{6.0} = 2.0\ \text{A} \) Total: \( 5.0\ \text{A} \). Check against the combined resistance: \( \dfrac{12}{2.4} = 5.0\ \text{A} \) ✓ Compare with the series case above: 5.0 A instead of 1.2 A from the same supply, so the parallel arrangement draws more than four times the current and dissipates more than four times the power. 3.0 A and 2.0 A, totaling 5.0 A

9. Explain why adding a resistor in parallel lowers the total resistance, when adding one in series raises it.
   Show the full solution

   Because a parallel resistor adds a new route for charge rather than lengthening the existing one, and more routes mean more current for the same voltage. Define resistance in terms of what it measures. Resistance is voltage divided by current, so a lower resistance means more current flows for the same voltage. The question becomes whether the addition increases or decreases the total current. The series case. The added resistor sits in the only path, so every coulomb must pass through it as well. The obstruction is longer, the current falls, and the resistance rises. The current is forced through both. The parallel case. The original branch is untouched: it still has the same voltage across it and still carries exactly the same current as before. The new branch carries additional current. So the total current has increased while the voltage is unchanged, which means the total resistance has fallen. The key realization. Adding a parallel branch takes nothing away from the existing one. This is why the result is not paradoxical: nothing got easier for the original path, but more paths exist. The physical analogy. Opening a second checkout lane does not make the first lane faster, but it doubles the rate people leave the store. The store's resistance to customer flow has halved. Why the reciprocals appear. Since the currents add and each current is \( V/R \), the total is \( V/R_1 + V/R_2 \). Dividing through by \( V \) gives \( 1/R_{\text{total}} = 1/R_1 + 1/R_2 \). The reciprocal form is not a convention; it falls out of adding currents. The check it provides. Since every added branch only increases the total current, the parallel resistance must always be below the smallest branch. Any answer failing that test is wrong, and this catches the commonest arithmetic error. The limiting cases. Two equal resistances in parallel give exactly half the value. And a branch of zero resistance, a short circuit, drives the total to zero and carries all the current, which is precisely why a short circuit is dangerous. A parallel branch adds current without reducing the existing current, so the total resistance falls

10. Three resistors of \( 10\ \Omega \), \( 20\ \Omega \) and \( 30\ \Omega \) are available with a 24 V supply. Find the arrangement giving the largest current and the one giving the smallest, and calculate both.
    Show the full solution

    Identify what determines the current. \( I = V/R \) with \( V \) fixed at 24 V, so the largest current comes from the smallest total resistance and the smallest current from the largest. The smallest possible resistance: all three in parallel. \( \dfrac{1}{R} = \dfrac{1}{10} + \dfrac{1}{20} + \dfrac{1}{30} \) Common denominator 60: \( \dfrac{6}{60} + \dfrac{3}{60} + \dfrac{2}{60} = \dfrac{11}{60} \) \( R = \dfrac{60}{11} = 5.45\ \Omega \). Check: below 10, the smallest branch ✓ \( I = \dfrac{24}{5.45} = 4.4\ \text{A} \). The largest possible resistance: all three in series. \( R = 10 + 20 + 30 = 60\ \Omega \) \( I = \dfrac{24}{60} = 0.40\ \text{A} \). The ratio. \( \dfrac{4.4}{0.40} = 11 \), so the parallel arrangement draws eleven times the current. That ratio is exactly \( 60/5.45 \) ✓ as it must be. Confirm no other arrangement beats either extreme. Consider the intermediate cases. \( 10 \) in series with \( (20 \parallel 30) \): \( 20 \parallel 30 = \dfrac{600}{50} = 12\ \Omega \), total \( 22\ \Omega \), giving 1.09 A. \( 30 \) in series with \( (10 \parallel 20) \): \( 10 \parallel 20 = \dfrac{200}{30} = 6.67\ \Omega \), total \( 36.7\ \Omega \), giving 0.65 A. \( (10 + 20) \) in parallel with \( 30 \): \( 30 \parallel 30 = 15\ \Omega \), giving 1.6 A. All intermediate values fall between 0.40 A and 4.4 A ✓ confirming the extremes. Why the extremes are guaranteed. Series always adds, so putting everything in series can only maximize the total. Parallel always reduces below the smallest branch, so putting everything in parallel minimizes it. No mixed arrangement can do better in either direction. The power consequence. Power is \( VI \), so the parallel arrangement dissipates \( 24 \times 4.4 = 106\ \text{W} \) while the series one dissipates \( 24 \times 0.40 = 9.6\ \text{W} \). Eleven times the heat from the same components and the same supply, purely from how they are connected. The practical warning this contains. Adding devices in parallel is what household wiring does, and every one raises the total current. There is no limit built into the physics, only into the fuse. This is the mechanism behind overloaded circuits and the reason extension leads carry a maximum rating. All parallel gives 4.4 A; all series gives 0.40 A

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Take electricity at 15 cents per kilowatt-hour.

1. Write electrical power in terms of voltage and current.
   Show the full solution

   \( P = VI \)

2. Write power in terms of current and resistance.
   Show the full solution

   \( P = I^2R \)

3. How many joules are in a kilowatt-hour?
   Show the full solution

   \( 3.6 \times 10^6\ \text{J} \)

4. Find the power of a device drawing 2.0 A at 120 V.
   Show the full solution

   \( P = VI = (2.0)(120) = 240\ \text{W} \). 240 W

5. Which formula should be used for losses in a transmission cable?
   Show the full solution

   \( P = I^2R \), using the cable's own resistance

6. Find the power dissipated when 3.0 A flows through \( 8.0\ \Omega \).
   Show the full solution

   \( P = I^2R = (3.0)^2(8.0) = (9.0)(8.0) = 72\ \text{W} \). 72 W

7. Find the cost of running a 900 W microwave for 20 minutes a day for 30 days.
   Show the full solution

   Daily energy: \( 0.900\ \text{kW} \times \dfrac{20}{60}\ \text{h} = 0.30\ \text{kWh} \). Monthly: \( 0.30 \times 30 = 9.0\ \text{kWh} \). Cost: \( 9.0 \times 0.15 = \$1.35 \). A high-power appliance used briefly costs little, which is the opposite of the intuition that wattage alone determines cost. Energy is power times time, and the time matters as much. \$1.35

8. A heater is designed for 240 V but connected to 120 V. Find the fraction of its rated power it delivers.
   Show the full solution

   The resistance of the element is fixed, so use \( P = \dfrac{V^2}{R} \). Halving \( V \) quarters \( P \), since the voltage is squared. It delivers one quarter of its rated power. The common wrong answer is one half, from assuming power is proportional to voltage. It is not, because the current also halves, and power is the product of both. Check by following both quantities. At half the voltage the current is \( I = V/R \), also halved. Power \( VI \) is then \( \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4} \) ✓ One quarter

9. Explain why a thin extension cord can overheat while the appliance it feeds runs normally.
   Show the full solution

   Because the cord and the appliance carry the same current, and the cord dissipates \( I^2R \) in its own resistance, which is concentrated in a small volume with poor cooling. Establish that the current is the same. The cord is in series with the appliance, so every coulomb passes through both. There is no sense in which the appliance uses up the current. The appliance runs normally because the cord's resistance is small compared with the appliance's, so most of the supply voltage still appears across the appliance and it receives nearly its rated power. But the cord dissipates power too. \( P_{\text{cord}} = I^2R_{\text{cord}} \), and although \( R_{\text{cord}} \) is small, a large current makes this significant because the current is squared. Put numbers on it. A 1500 W heater on 120 V draws 12.5 A. A thin cord of \( 1.0\ \Omega \) dissipates \( (12.5)^2(1.0) = 156\ \text{W} \) along its own length. That is more than a soldering iron, spread over a few meters of plastic-covered wire. Why the geometry makes it dangerous. The appliance is designed to dissipate its heat, with vents, metal casing and often a fan. The cord is a thin plastic-insulated conductor, often coiled on a reel or run under a rug, with almost no cooling. The same power in a worse thermal environment is what causes the failure. Why coiling makes it worse. A coiled cord on a reel has its inner turns insulated by its outer ones, so the heat cannot escape. Cable reel ratings are always given twice, fully unwound and coiled, with the coiled rating far lower, precisely for this reason. Why thin cords are the problem. \( R = \dfrac{\rho L}{A} \), so halving the cross-sectional area doubles the resistance and doubles the heat for the same current. A long thin cord is the worst case on both counts. Why the fuse does not protect against it. The circuit fuse is sized for the household wiring, typically 15 or 20 A. A cord rated for 5 A carrying 12.5 A is badly overloaded and the fuse never notices, because the current is within the circuit's limit. The protection has to come from choosing the right cord. The design rule this leads to. Match the cord's rating to the appliance's current, use the shortest cord that reaches, uncoil it fully, and never cover it. Series means equal current, and \( I^2R \) in a poorly cooled thin conductor is what overheats

10. A power station delivers 5.0 MW along a line of resistance \( 2.0\ \Omega \). Compare the losses at 25 kV and at 275 kV, and explain what limits how high the voltage can go.
    Show the full solution

    At 25 kV, find the current. \( I = \dfrac{P}{V} = \dfrac{5.0 \times 10^6}{2.5 \times 10^4} = 200\ \text{A} \). Find the loss. \( P_{\text{loss}} = I^2R = (200)^2(2.0) = 80000\ \text{W} = 80\ \text{kW} \). As a fraction: \( \dfrac{80}{5000} = 1.6 \) percent. At 275 kV, find the current. \( I = \dfrac{5.0 \times 10^6}{2.75 \times 10^5} = 18.2\ \text{A} \). Find the loss. \( P_{\text{loss}} = (18.2)^2(2.0) = (331)(2.0) = 662\ \text{W} \). As a fraction: \( \dfrac{0.662}{5000} = 0.013 \) percent. Compare. The loss fell from 80 kW to 662 W, a factor of 121. Check: the voltage ratio is \( 275/25 = 11 \), and \( 11^2 = 121 \) ✓ exactly as the square law demands. Put the 80 kW in context. That is enough to run about thirty homes, lost as heat in the cables, on a single line. Across a national grid the difference between these two choices is a whole power station's worth of generation. Now the second part: what limits the voltage. Insulation breakdown. Air ionizes at roughly \( 3 \times 10^6\ \text{V/m} \), so higher voltages require greater clearances between conductors and between conductors and ground. This sets pylon height, arm length and the size of the insulator strings, and all of it costs money. Corona discharge. Near a high-voltage conductor the field can exceed the breakdown value at the surface even without a full arc, ionizing the surrounding air. This wastes power continuously, generates audible buzzing and radio interference, and produces ozone. It is countered by using bundled conductors and larger effective diameters to reduce the surface field, which adds weight and cost. Transformer cost and size. Every voltage step needs a transformer rated for it, and insulation requirements make high-voltage transformers enormous and expensive. The saving must exceed this capital cost. Safety and land use. Higher voltages demand wider rights of way, which means more land acquisition and greater public objection. The resulting engineering compromise. Voltage is chosen by distance: long trunk routes use 275 kV or 400 kV and above, regional distribution steps down to tens of kilovolts, and local delivery to 120 or 240 V. Each step trades transformer cost against transmission loss over the distance involved. Why it is never taken to extremes. The loss is already below 0.02 percent at 275 kV, so doubling again would save almost nothing measurable while multiplying the insulation problem. Beyond a point the square law has already won and there is nothing left to gain. 80 kW against 662 W, a factor of 121; insulation breakdown, corona loss and transformer cost set the ceiling

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write the definition of capacitance.
   Show the full solution

   \( C = \dfrac{Q}{V} \)

2. One farad equals how many coulombs per volt?
   Show the full solution

   One

3. Write the energy stored in a capacitor.
   Show the full solution

   \( E = \tfrac{1}{2}CV^2 \)

4. Does charge cross the gap between a capacitor's plates?
   Show the full solution

   No

5. Find the charge on a \( 50\ \mu\text{F} \) capacitor at 20 V.
   Show the full solution

   \( Q = CV = (50 \times 10^{-6})(20) = 1.0 \times 10^{-3}\ \text{C} \). 1.0 mC

6. Find the energy stored on a \( 220\ \mu\text{F} \) capacitor at 9.0 V.
   Show the full solution

   \( E = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(220 \times 10^{-6})(81) \) \( = (1.1 \times 10^{-4})(81) = 8.9 \times 10^{-3}\ \text{J} \). 8.9 mJ

7. A capacitor holds 4.4 mC at 11 V. Find its capacitance.
   Show the full solution

   \( C = \dfrac{Q}{V} = \dfrac{4.4 \times 10^{-3}}{11} = 4.0 \times 10^{-4}\ \text{F} \). \( 400\ \mu\text{F} \)

8. The voltage across a capacitor is tripled. Find the factors by which the charge and the stored energy change.
   Show the full solution

   \( Q = CV \) is linear in \( V \), so the charge triples. \( E = \tfrac{1}{2}CV^2 \) is quadratic, so the energy rises by \( 3^2 = 9 \). The two scale differently, and confusing them is the most common error with capacitors. Charge \( \times 3 \), energy \( \times 9 \)

9. Explain why a capacitor blocks direct current but passes alternating current.
   Show the full solution

   Because charge cannot cross the insulating gap, so a steady flow is impossible, but a changing charge on the plates drives a corresponding current in the connecting wires. The direct current case. When a capacitor is first connected to a steady voltage, charge flows onto one plate and off the other, and a current is measured in the circuit. But as charge accumulates, the voltage across the capacitor rises and opposes the supply. When it equals the supply voltage there is no net driving voltage left, the flow stops, and the current is zero from then on. So direct current is blocked, but only after a transient. The blocking is not instantaneous, and the time it takes depends on the circuit resistance. That delay is the basis of timing circuits. The alternating current case. With an alternating supply the voltage is continuously reversing. The capacitor charges one way, then discharges, then charges the opposite way, and never reaches a steady state. Why this counts as current. Charge is constantly flowing onto and off the plates through the wires, so an ammeter in the circuit reads a current. No charge ever crosses the gap, and none needs to: the current in the wires is real and measurable. Why frequency matters. The faster the reversals, the more charge moves per second, so the current rises with frequency. The opposition a capacitor presents, its reactance, is \( 1/(2\pi fC) \), falling as frequency rises. At high frequency a capacitor behaves almost like a wire; at zero frequency, like a break in the circuit. What this is used for: coupling. An audio signal riding on a steady supply voltage can be passed to the next stage through a capacitor, which transmits the varying signal and blocks the steady bias. Almost every amplifier stage does this. And filtering. A capacitor across a supply line gives high-frequency noise an easy route to ground while leaving the steady supply voltage untouched, which is why circuit boards are covered in small capacitors beside every integrated circuit. And tuning. Because the reactance depends on frequency, a capacitor with an inductor selects one frequency from many, which is how a radio is tuned and what unit 9 returns to with resonance. Steady voltage produces no ongoing charge movement; a changing voltage produces continuous charge movement in the wires

10. A camera flash uses a \( 1200\ \mu\text{F} \) capacitor charged to 300 V, discharging in 1.5 ms. Find the stored energy, the peak power, and explain why the capacitor must be charged slowly.
    Show the full solution

    Find the stored energy. \( E = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(1200 \times 10^{-6})(300)^2 \) \( = \tfrac{1}{2}(1.2 \times 10^{-3})(9.0 \times 10^4) = 54\ \text{J} \). Find the average power during the flash. \( P = \dfrac{E}{t} = \dfrac{54}{1.5 \times 10^{-3}} = 36000\ \text{W} = 36\ \text{kW} \). Find the charge and average current. \( Q = CV = (1.2 \times 10^{-3})(300) = 0.36\ \text{C} \) \( I = \dfrac{Q}{t} = \dfrac{0.36}{1.5 \times 10^{-3}} = 240\ \text{A} \). Two hundred and forty amperes from a camera, which is more than an arc welder draws, for a millisecond and a half. Now the third part: why the charging must be slow. The battery cannot supply that power. A small lithium cell might deliver 10 W comfortably. Charging the capacitor in the same 1.5 ms would require the same 36 kW, roughly three thousand times more than the cell can give. What the cell can do. At 10 W, delivering 54 J takes \( t = \dfrac{54}{10} = 5.4\ \text{s} \), and in practice a few seconds is exactly what a flash takes to recycle, which is a prediction the calculation makes and everyday experience confirms. Why the inefficiency makes it longer. Charging a capacitor through a resistance dissipates as much energy as it stores, so the cell must actually supply about 108 J, roughly doubling the time. Real flash units use a switching converter to beat this, which is why they recycle faster than a simple resistor circuit would allow. What limits the discharge instead. Once charged, the discharge rate is set by the resistance of the flash tube and the circuit, not by the battery at all. The battery is out of the loop entirely during the flash, which is the whole point of the arrangement. Why the flash must be brief rather than merely bright. A short flash freezes motion, and the exposure is determined by the flash duration rather than the shutter. Spreading the same 54 J over a second would give the same total light and a blurred photograph, so brevity is the requirement and the capacitor is the only way to meet it. The general structure, which is worth extracting. Energy is gathered slowly from a low-power source and released quickly from a store. The same pattern appears in a defibrillator, a railgun, a laser pulse and a lightning strike, where the atmosphere itself acts as the capacitor. The safety note. 54 J at 300 V is dangerous, and the capacitor stays charged after the camera is switched off. Disposable cameras have caused serious shocks to people opening them, and this is the mechanism. 54 J, about 36 kW, and 240 A; the cell can supply only about 10 W so charging takes seconds

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Give the unit of magnetic field strength.
   Show the full solution

   The tesla

2. What happens if a bar magnet is cut in half?
   Show the full solution

   Two complete magnets are produced, each with both poles

3. State the right hand grip rule.
   Show the full solution

   Thumb along the current, fingers curl in the direction of the field

4. What produces every magnetic field?
   Show the full solution

   Moving charge

5. Give the approximate strength of Earth's magnetic field.
   Show the full solution

   About \( 5 \times 10^{-5}\ \text{T} \)

6. Describe the shape of the field around a long straight current-carrying wire.
   Show the full solution

   Concentric circles centered on the wire, lying in planes perpendicular to it, with the direction given by the right hand grip rule. The strength falls as \( 1/r \), more slowly than a point charge's field, because the source is a line rather than a point. Circles around the wire

7. Give two ways to increase the field strength of an electromagnet.
   Show the full solution

   Any two of: Increase the current, since the field is proportional to it. Increase the number of turns per unit length, since each turn contributes. Insert a soft iron core, whose domains align and multiply the field by hundreds or thousands. This is by far the largest effect of the three. More current, more turns, or an iron core

8. Explain why magnetic field lines form closed loops while electric field lines do not.
   Show the full solution

   Because field lines must begin and end on the sources of the field, and magnetic charge does not exist. The electric case. Electric field lines start on positive charge and end on negative charge. Isolated charges exist, so lines can have genuine ends. The magnetic case. No isolated magnetic pole has ever been found, so there is nothing for a line to begin or end on. A line that cannot end must close on itself. Where the lines go inside a magnet. Outside they run north to south; inside the magnet they continue from south back to north, completing the loop. Diagrams that stop at the magnet's surface are incomplete. The experimental basis. Cutting a magnet never isolates a pole, and this holds down to the atomic level. Closed loops, because there is no magnetic charge to terminate on

9. Explain why heating a permanent magnet above a certain temperature destroys its magnetism, and why this supports the domain model.
   Show the full solution

   Because thermal motion eventually overcomes the alignment between neighboring atomic magnetic moments, and once alignment is lost the fields cancel. Recall what makes iron magnetic. Unpaired electron spins in neighboring atoms align through exchange coupling, forming domains of aligned atoms. A magnetized sample has its domains pointing predominantly one way. What heating does. Raising the temperature raises the average kinetic energy per atom, as lesson 6.1 established. Atoms vibrate more violently, and the vibration disrupts the alignment. The competition. The exchange interaction favors alignment; thermal energy favors randomness. Below a critical temperature alignment wins; above it, randomness does. The Curie temperature. That critical value is 1043 K for iron, 1394 K for cobalt and 631 K for nickel. Above it the material becomes paramagnetic, retaining no permanent magnetization at all. Why this supports the domain model specifically. The transition is sharp, occurring over a narrow temperature range rather than fading gradually. A sharp transition is what a collective alignment predicts, because once neighbors begin disagreeing the whole cooperative arrangement collapses. If magnetism were a substance stored in the metal, there would be no reason for a critical temperature at all, and certainly no reason for a sharp one. A second piece of support. Hammering or dropping a magnet also weakens it, by mechanically jolting domains out of alignment. This is a different mechanism reaching the same conclusion. A third. Cooling the iron back down does not restore the magnetism unless an external field is present to re-align the domains. The material has no memory of its former orientation, which is exactly what a randomized-domain picture predicts. The practical consequence. Magnets in motors and speakers must be rated for their operating temperature, and a magnet that has been overheated cannot be repaired, only remagnetized. Thermal energy overcomes exchange alignment at the Curie temperature, and the sharpness of the transition is the evidence for domains

10. Explain how Earth's magnetic field is generated, what evidence shows it changes, and why it matters for life.
    Show the full solution

    Rule out the simple explanation first. Earth cannot contain a permanent magnet. The core's temperature is far above iron's Curie temperature of 1043 K, reaching perhaps 5000 K, so no permanent magnetization could survive there. Whatever produces the field must be active rather than stored. The accepted mechanism: the geodynamo. The outer core is liquid iron and nickel, electrically conducting, and in constant motion driven by heat escaping from the inner core and by Earth's rotation. Moving conducting fluid carries moving charge, and moving charge produces a magnetic field. Why it sustains itself. The motion of the conductor through an existing field induces currents, by the induction of lesson 8.4, and those currents produce a field that reinforces the original. It is a self-sustaining loop, which is why the arrangement is called a dynamo. What it needs to keep running. An energy source, which is the heat from the core, and rotation to organize the flow. Both are present, and the field has persisted for billions of years. The evidence that it changes: magnetic reversals. Lava cooling below the Curie temperature locks in the direction of the field at that moment. Volcanic rock of different ages therefore records the field's past orientation, and it shows that north and south have swapped many times. The decisive evidence: sea floor striping. New crust forms at mid-ocean ridges and spreads outward. Magnetic surveys show symmetric bands of alternating polarity on both sides of every ridge, a matching record read outward in both directions. This was one of the strongest confirmations of plate tectonics, and it dates the most recent reversal to about 780000 years ago (USGS, US federal). What changes now. The magnetic north pole has moved hundreds of kilometers over the past century, at times exceeding 50 km per year, fast enough that navigation charts and aviation systems are revised on a regular schedule (NOAA, US federal). Why it matters for life: the solar wind. The Sun emits a stream of charged particles. Lesson 8.2 shows that a magnetic field exerts a force on moving charge perpendicular to its motion, so Earth's field deflects these particles around the planet rather than letting them strike the atmosphere. The visible consequence. Particles that do penetrate are funneled along field lines toward the poles, where they excite atmospheric atoms and produce the aurora. The aurora is therefore direct visual evidence of the field's shape. The consequence for the atmosphere. Without deflection, the solar wind would gradually strip lighter atmospheric gases away. Mars, which lost its global field billions of years ago, has an atmosphere under one percent of Earth's density, and measurements from orbiting spacecraft show it still losing gas to the solar wind today (NASA, US federal). The consequence for technology. Geomagnetic storms induce currents in long conductors such as power lines and pipelines, by exactly the induction of lesson 8.4. A storm in 1989 collapsed the Quebec power grid in under two minutes, and protecting against this is now a standard part of grid design. What is not known. Whether a reversal poses a serious threat. During one the field weakens and becomes complex rather than vanishing, and the fossil record shows no mass extinction coinciding with past reversals, so the honest answer is that the risk is to technology rather than to life. A self-sustaining dynamo in the liquid outer core; rock and sea floor magnetism record its reversals; it shields the atmosphere from the solar wind

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Proton mass \( 1.673 \times 10^{-27}\ \text{kg} \).

1. Write the magnetic force on a moving charge.
   Show the full solution

   \( F = qvB\sin\theta \)

2. When is the magnetic force on a moving charge zero?
   Show the full solution

   When it moves parallel to the field, or when it is stationary

3. How much work does the magnetic force do on a moving charge?
   Show the full solution

   None

4. Write the radius of the circular path.
   Show the full solution

   \( r = \dfrac{mv}{qB} \)

5. Does the period of circular motion in a magnetic field depend on speed?
   Show the full solution

   No

6. Find the force on a proton moving at \( 5.0 \times 10^5\ \text{m/s} \) perpendicular to a 0.80 T field.
   Show the full solution

   \( F = qvB = (1.60 \times 10^{-19})(5.0 \times 10^5)(0.80) = 6.4 \times 10^{-14}\ \text{N} \). \( 6.4 \times 10^{-14} \) N

7. Find the radius of a proton's path at \( 1.0 \times 10^7\ \text{m/s} \) in a 1.5 T field.
   Show the full solution

   \( r = \dfrac{mv}{qB} = \dfrac{(1.673 \times 10^{-27})(1.0 \times 10^7)} {(1.60 \times 10^{-19})(1.5)} = \dfrac{1.673 \times 10^{-20}}{2.4 \times 10^{-19}} = 0.070\ \text{m} \). About 7.0 cm, far larger than an electron's path at the same speed and field, by the mass ratio of 1836. 0.070 m

8. Explain why the magnetic force does no work on a moving charge, and what this implies about its speed.
   Show the full solution

   Because work requires a force component along the displacement, and the magnetic force is always exactly perpendicular to the velocity. Recall the definition of work from lesson 5.1: \( W = Fd\cos\theta \), where \( \theta \) is the angle between force and displacement. When \( \theta = 90^\circ \), \( \cos\theta = 0 \) and the work is zero regardless of how large the force is. The magnetic force satisfies this at every instant. Its direction is perpendicular to both the velocity and the field by definition, and the displacement is along the velocity, so the angle is always exactly 90 degrees. Therefore no energy is transferred, so the kinetic energy is constant, so the speed is constant. But the velocity is not constant. Velocity is a vector, and its direction changes continuously. The force changes where the particle is going without changing how fast it goes. Why this makes the motion circular. A constant-magnitude force always perpendicular to a constant-speed velocity is exactly the definition of centripetal force from lesson 4.2, and circular motion follows necessarily. The contrast with the electric force. An electric force acts along the field regardless of the velocity, so it generally does work and changes the speed. This is why accelerators use electric fields to speed particles up and magnetic fields to steer them, with each doing the job the other cannot. Zero work, constant speed, changing direction

9. A charged particle enters a magnetic field at an angle between 0 and 90 degrees. Describe its path and explain why.
   Show the full solution

   It follows a helix, a spiral that advances steadily along the field direction. Resolve the velocity into two components, exactly as lesson 1.6 did with projectiles: one parallel to the field and one perpendicular to it. The parallel component. With \( \theta = 0 \) for this component, \( \sin\theta = 0 \) and it experiences no magnetic force at all. It therefore continues unchanged, at constant velocity, forever. The perpendicular component. This experiences the full \( qv_\perp B \) and, being perpendicular to the force, moves in a circle of radius \( r = \dfrac{mv_\perp}{qB} \). Combining them. Uniform circular motion in one plane plus uniform straight-line motion perpendicular to that plane is a helix. The particle winds around a field line while drifting steadily along it. The pitch of the helix, meaning the distance advanced per turn, is \( v_\parallel T = \dfrac{2\pi m v_\parallel}{qB} \), and since the period is independent of speed, the pitch depends only on the parallel component. Where this is seen: the aurora. Solar wind particles reaching Earth spiral along its field lines toward the poles. As the field strengthens near a pole the helix tightens, and the particles are funneled into the polar atmosphere, which is why the aurora appears in rings around the magnetic poles rather than uniformly. The magnetic mirror effect. As the field converges, more of the particle's energy is transferred into the circular component until the parallel motion reverses, sending the particle back. Particles bounce between the poles this way, trapped, forming the Van Allen radiation belts. Where this is used: fusion research. A tokamak confines plasma at millions of kelvin using the same principle, since no material container could touch it. The particles spiral along closed field lines within a toroidal chamber. The special cases check out. At 90 degrees the parallel component is zero and the helix collapses to a circle ✓ At 0 degrees the perpendicular component is zero and the helix becomes a straight line ✓ A helix, because the parallel component is unaffected while the perpendicular component circles

10. A velocity selector uses perpendicular electric and magnetic fields so that only particles of one speed pass through undeflected. Derive the condition, and explain why this is needed before a mass spectrometer.
    Show the full solution

    Set up the arrangement. A charged particle travels through a region where an electric field \( E \) and a magnetic field \( B \) are perpendicular to each other and both perpendicular to the particle's velocity. Identify the two forces. Electric: \( F_E = qE \), along the electric field, independent of speed. Magnetic: \( F_B = qvB \), perpendicular to both \( v \) and \( B \), proportional to speed. Arrange them to oppose. The geometry is chosen so the two forces point in opposite directions. Set them equal for undeflected passage. \( qE = qvB \) The charge cancels: \( E = vB \), so \( v = \dfrac{E}{B} \). Note what the condition does not contain. Neither the charge nor the mass appears. The selected speed depends only on the two field strengths, so the device selects purely on speed and treats every particle alike. What happens to the others. A particle moving faster than \( E/B \) has a larger magnetic force and is deflected toward the magnetic force's direction. A slower one has a smaller magnetic force and is deflected the other way. Only the exact speed passes straight through to the slit. Now the second part: why the spectrometer needs this. The spectrometer sorts by radius, using \( r = \dfrac{mv}{qB} \). The radius depends on both mass and speed. That is the problem. If the incoming ions have a spread of speeds, a light fast ion and a heavy slow one can produce the same radius and land at the same place. The measurement becomes ambiguous and the peaks blur into each other. The selector removes the ambiguity. Fixing \( v \) for every entering ion means the radius depends only on \( m/q \), and the position on the detector maps directly to mass. Why ions arrive with a spread of speeds in the first place. The ion source is hot, so the ions emerge with a thermal distribution of speeds superimposed on whatever acceleration they received. Even careful acceleration through a fixed voltage leaves a spread. The alternative approach. Some instruments instead accelerate through a precisely known voltage so that \( qV = \tfrac{1}{2}mv^2 \) fixes the energy rather than the speed, giving \( r \propto \sqrt{m/q} \). Both designs work; both exist to remove the speed as an unknown. How sharp the resolution can be. Modern instruments distinguish masses differing by less than one part in \( 10^5 \), enough to tell apart molecules with the same nominal mass but different atomic composition, which is how unknown compounds are identified rather than merely weighed. The historical note. J. J. Thomson used crossed fields in exactly this way in 1897 to measure the charge-to-mass ratio of the electron, the experiment that established the electron as a particle. The technique is older than the spectrometer it now feeds. \( v = \dfrac{E}{B} \), independent of charge and mass, needed so that the spectrometer's radius depends on mass alone

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write the force on a current-carrying wire in a magnetic field.
   Show the full solution

   \( F = BIL\sin\theta \)

2. State the left hand rule for a motor.
   Show the full solution

   First finger field, second finger current, thumb force

3. What is the net force on a current loop in a uniform field?
   Show the full solution

   Zero

4. Write the maximum torque on a coil.
   Show the full solution

   \( \tau = NBIA \)

5. What does a commutator do?
   Show the full solution

   Reverses the current in the coil every half turn

6. Find the force on 0.60 m of wire carrying 8.0 A perpendicular to a 0.20 T field.
   Show the full solution

   \( F = BIL = (0.20)(8.0)(0.60) = 0.96\ \text{N} \). 0.96 N

7. A wire experiences 0.45 N in a 0.30 T field over 0.50 m. Find the current.
   Show the full solution

   \( I = \dfrac{F}{BL} = \dfrac{0.45}{(0.30)(0.50)} = \dfrac{0.45}{0.15} = 3.0\ \text{A} \). Assuming the wire is perpendicular to the field, since no angle was given. At any other angle the current would need to be larger. 3.0 A

8. Give three ways to increase the torque of a motor, and say which is most limited in practice.
   Show the full solution

   From \( \tau = NBIA \): More turns \( N \), which is cheap and is why motor coils carry hundreds of windings. Stronger field \( B \), achieved with better permanent magnets or a stronger electromagnet and an iron core. More current \( I \), which works directly but is the most limited in practice, because \( I^2R \) heating rises as the square and the windings have a temperature limit set by their insulation. Doubling the current quadruples the heat. Larger area \( A \) also works but means a physically bigger motor. Any three; current is the most limited, by heating

9. Explain why a simple current loop in a magnetic field would stop rotating without a commutator.
   Show the full solution

   Because the torque falls to zero at one orientation and then acts to push the coil back, so the motion reverses rather than continuing. Follow the coil through a half turn. Starting with the coil's plane parallel to the field, the forces on the two sides are opposite and act at maximum perpendicular distance from the axis, giving maximum torque. As it rotates, the perpendicular distance from the axis to each force shrinks, so the torque falls, even though the forces themselves are unchanged in size. At 90 degrees of rotation, the coil's plane is perpendicular to the field. Both forces now act along lines passing through the axis, so neither has any turning effect. The torque is exactly zero. This is the dead point. What happens next is the crucial part. The coil has angular momentum and carries past the dead point. Now the same force directions produce a torque in the opposite rotational sense, because the geometry has flipped relative to the axis. So the coil decelerates, stops, and swings back. It oscillates about the dead point, losing energy to friction and resistance, and settles there. This is a useful device, just not a motor. A coil that settles at an angle depending on the current, opposed by a spring, is a moving coil galvanometer, the basis of every analog ammeter and voltmeter. What the commutator changes. Reversing the current at the dead point reverses both forces, so the torque after the dead point acts in the same rotational sense as before it. The coil continues turning instead of swinging back. Why the timing must be exact. Switching too early or too late produces a period of opposing torque, which wastes energy and causes sparking at the brushes. Real motors adjust the brush position slightly to compensate for this. The torque reverses past the dead point, so without switching the coil oscillates instead of rotating

10. A wire of mass 0.015 kg and length 0.20 m rests on horizontal rails in a vertical magnetic field of 0.50 T. Find the current needed to make it lift off, and explain what changes once it is moving.
    Show the full solution

    Set up the condition for lift-off. The magnetic force must equal the weight. \( BIL = mg \). Solve for the current. \( I = \dfrac{mg}{BL} = \dfrac{(0.015)(9.8)}{(0.50)(0.20)} = \dfrac{0.147}{0.10} = 1.47\ \text{A} \). About 1.5 A. Check the geometry is consistent. The field is vertical and the current is horizontal along the wire, so they are perpendicular and \( \sin\theta = 1 \) ✓ But the force \( BIL \) is perpendicular to both, which would be horizontal, not vertical. So the stated geometry does not work, and this is worth catching. To lift the wire vertically with a horizontal current, the field must be horizontal and perpendicular to the wire, not vertical. The arithmetic stands; the description of the field direction does not. Restate it correctly. With a horizontal 0.50 T field perpendicular to a horizontal wire, the force is vertical and 1.5 A lifts it. This is exactly the sort of check the left hand rule exists to provide, and it must be done before trusting any answer here. Now the second part: what changes once the wire moves. It becomes a generator as well as a motor. A conductor moving through a magnetic field has an emf induced across it, \( \varepsilon = BLv \), which is lesson 8.4's result. The induced emf opposes the supply. By Lenz's law in lesson 8.5, it is directed against the current that produced the motion. This is called back emf. The consequence for the current. The net driving voltage becomes \( V_{\text{supply}} - BLv \), so as the wire speeds up the current falls. Which means the force falls too, since \( F = BIL \). The acceleration decreases as the wire rises. Why this matters far beyond this problem. Every motor works this way. At rest a motor has no back emf and draws its maximum current, which is why starting currents are large and why a stalled motor can burn out: with no rotation there is no back emf, nothing limits the current but the winding resistance, and the windings overheat. And it explains load response. Loading a motor slows it, which reduces the back emf, which raises the current, which raises the torque. A motor automatically draws more power when worked harder, with no control system required. That self-regulation is a direct consequence of induction opposing the change that caused it. About 1.5 A, with the field horizontal and perpendicular to the wire; once moving, back emf reduces the current and the force

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. Write the definition of magnetic flux.
   Show the full solution

   \( \Phi = BA\cos\theta \)

2. Give the unit of magnetic flux.
   Show the full solution

   The weber

3. State Faraday's law.
   Show the full solution

   \( \varepsilon = -N\dfrac{\Delta\Phi}{\Delta t} \)

4. What is the induced emf when the flux is large but constant?
   Show the full solution

   Zero

5. Write the emf induced in a rod moving across a field.
   Show the full solution

   \( \varepsilon = BLv \)

6. Find the flux through a \( 0.020\ \text{m}^2 \) coil in a 0.45 T field perpendicular to it.
   Show the full solution

   Perpendicular to the coil means parallel to the normal, so \( \theta = 0 \) and \( \cos\theta = 1 \). \( \Phi = BA = (0.45)(0.020) = 9.0 \times 10^{-3}\ \text{Wb} \). 9.0 mWb

7. A 500-turn coil has its flux change by \( 2.0 \times 10^{-4}\ \text{Wb} \) in 0.010 s. Find the emf.
   Show the full solution

   \( |\varepsilon| = N\dfrac{\Delta\Phi}{\Delta t} = \dfrac{(500)(2.0 \times 10^{-4})}{0.010} = \dfrac{0.10}{0.010} = 10\ \text{V} \). 10 V

8. Name the three ways the flux through a coil can be changed, and give a device using each.
   Show the full solution

   Changing the field strength \( B \). A transformer does this: an alternating current in the primary produces a continuously changing field through the secondary, with nothing moving at all. Changing the area \( A \). A sliding rod on rails does this, and so does any circuit whose enclosed area changes. Changing the angle \( \theta \). A generator does this: the coil rotates in a fixed field, so \( \cos\theta \) varies sinusoidally and the flux with it. All three are the same law, since Faraday's law responds to \( \Delta\Phi \) regardless of which factor caused it. Field, area, angle: transformer, sliding rod, generator

9. Explain why a magnet dropped through a copper tube falls far more slowly than through a plastic one.
   Show the full solution

   Because the changing flux induces currents in the copper, and those currents produce a field opposing the magnet's motion. Follow the flux. As the magnet falls, the flux through each ring-like cross-section of the copper tube first increases as the magnet approaches and then decreases as it recedes. Faraday's law gives an emf around each such loop of copper, and since copper is an excellent conductor and the loops are complete, large currents flow. These are called eddy currents. Those currents produce their own fields. Below the magnet the induced current opposes the approaching flux and repels it; above the magnet the induced current opposes the departing flux and attracts it. Both effects retard the fall, which is Lenz's law in lesson 8.5. Why the motion reaches a steady speed. The faster the magnet falls, the faster the flux changes, the larger the induced current, and the larger the retarding force. It accelerates until that force equals its weight, then falls at constant terminal velocity, exactly as lesson 2.6's drag analysis described. Why plastic does nothing. Plastic is an insulator. An emf is still induced, but no current can flow, so no opposing field is produced and there is no force. The magnet falls freely. The decisive test. Cutting a vertical slit along the length of the copper tube breaks the circular current paths, and the magnet then falls almost normally, even though the tube is still copper and still the same shape. The effect depends on complete conducting loops, not on the material being near the magnet. A second test. Dropping an unmagnetized steel slug of the same mass and shape through the copper tube produces no slowing at all, confirming that it is the magnet's field rather than the object's presence that matters. Where this is used. Eddy current braking in trains and roller coasters, with no contact and no wear. Metal detectors, which sense the eddy currents induced in buried conductors. Induction cooktops, which drive large eddy currents in the pan itself. And where it is a nuisance. Eddy currents in a transformer's iron core waste energy as heat, which is why cores are built from thin laminations separated by insulating varnish rather than solid iron, to break up the current paths. Induced eddy currents oppose the changing flux, producing a retarding force

10. A square coil of side 0.20 m with 100 turns is pulled completely out of a 0.40 T field in 0.25 s. Find the average emf, the flux change, and explain what determines whether it matters how the coil is pulled.
    Show the full solution

    Find the area. \( A = (0.20)^2 = 0.040\ \text{m}^2 \). Find the initial flux per turn. \( \Phi_i = BA = (0.40)(0.040) = 0.016\ \text{Wb} \). Find the final flux. Completely out of the field, so \( \Phi_f = 0 \). The flux change per turn. \( \Delta\Phi = 0.016\ \text{Wb} \). Apply Faraday's law. \( |\varepsilon| = N\dfrac{\Delta\Phi}{\Delta t} = \dfrac{(100)(0.016)}{0.25} = \dfrac{1.6}{0.25} = 6.4\ \text{V} \). Total flux linkage change: \( N\Delta\Phi = (100)(0.016) = 1.6\ \text{Wb} \). Now the third part, which is the interesting one. For the average emf, the path does not matter. Faraday's law depends only on the total flux change and the total time. Whether the coil was pulled steadily, jerked out at the end, or moved in fits and starts, the average emf over the 0.25 s is 6.4 V, because the endpoints and the duration are the same. For the instantaneous emf, the path matters completely. A steady pull gives a constant 6.4 V throughout. A pull that removes most of the coil in the first 0.05 s and then dawdles would give a large spike followed by nearly nothing, with the same average. Why this distinction is practical. Insulation breakdown depends on the peak voltage, not the average. A device designed for 6.4 V could be destroyed by a 50 V spike even though the average was unchanged. This is why switching an inductive load produces damaging voltage spikes and why protective diodes are fitted across relay coils. A related consequence: the charge is path independent. The total charge that flows is \( q = \displaystyle\int I\,dt = \dfrac{N\Delta\Phi}{R} \), which contains no time at all. So however fast or slow the coil is removed, the same total charge circulates. A fast removal gives a large brief current; a slow one gives a small prolonged current; the integral is identical. Where that is used. A ballistic galvanometer measures total charge and therefore measures flux change directly, independent of how the change was made. This is how magnetic field strengths were measured before electronic instruments. A qualification on the geometry. The 6.4 V assumes the coil leaves the field region entirely and that the field is uniform within it and zero outside. A real field has fringing at its edges, so the flux tails off gradually rather than stopping sharply, and the actual emf profile is smoother than the idealization suggests. The average over the full removal is unaffected. 6.4 V average, 1.6 Wb of flux linkage; the average depends only on the endpoints and the time, but the peak depends entirely on how it is pulled

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.

1. State Lenz's law.
   Show the full solution

   An induced current opposes the change that produced it

2. Which conservation law does Lenz's law express?
   Show the full solution

   Conservation of energy

3. A north pole approaches a coil. Which pole is induced facing it?
   Show the full solution

   North, so that it repels

4. What shape is a simple generator's output?
   Show the full solution

   Sinusoidal

5. When in the rotation is a generator's emf greatest?
   Show the full solution

   When the coil's plane is parallel to the field, where the flux changes fastest

6. A south pole is pulled away from a coil. Determine the induced pole facing it.
   Show the full solution

   The flux through the coil is decreasing, so the induced current acts to maintain it. That means the coil's near face must become a north pole, which attracts the departing south pole and opposes its removal. The check: in every case the coil resists what is being done, so approaching poles are repelled and departing poles attracted. North

7. A generator produces a peak emf of 170 V. Give two ways to double it.
   Show the full solution

   From \( \varepsilon_0 = NBA\omega \), any two of: Double the rotation speed \( \omega \), which also doubles the frequency, so this is not free if the frequency must stay at 60 Hz. Double the number of turns \( N \). Double the field strength \( B \). Double the coil area \( A \). In a grid generator only the last three are available, since the frequency is fixed by the grid, which is why generator design concentrates on field strength and winding count. Any two of speed, turns, field or area

8. Explain why an aluminum ring jumps off an electromagnet when it is switched on, even though aluminum is not magnetic.
   Show the full solution

   Because switching on creates a rapidly changing flux through the ring, inducing a large current whose field opposes the electromagnet's, producing repulsion. Note what makes this possible. Aluminum is not ferromagnetic, so it is not attracted to a magnet at all. But it is an excellent conductor, and that is the only property the effect requires. The sequence. Switching on the electromagnet makes its field rise from zero very rapidly. The flux through the ring rises with it. Faraday's law gives a large emf, because the change is fast. The ring is a complete loop of low resistance, so a very large current flows, often thousands of amperes for a few milliseconds. Lenz's law gives the direction. The induced current opposes the increasing flux, so the ring's field points opposite to the electromagnet's. Two opposing fields repel. Why it jumps rather than merely being pushed. The force is enormous and brief, delivering a large impulse in a short time, which is exactly the situation lesson 3.3 analyzed. The ring leaves with substantial momentum. Why it happens only at switch-on. Once the current is steady the flux stops changing, the induced current dies, and the ring feels nothing. Switching off produces a jump too, but in the opposite direction, since the falling flux induces a current that attracts. The decisive test. A ring with a slit cut in it does not jump at all. Nothing about its material or position changed, only the completeness of the current path. This proves the effect depends on induced current, not on any magnetic property of aluminum. A second test. Cooling the ring in liquid nitrogen lowers its resistance and makes it jump noticeably higher, which is what a larger induced current predicts. Where it is used. This is the principle of a coilgun and of electromagnetic forming, where metal sheets are shaped by magnetic pulses with no die touching them. Induced current in a closed conducting loop opposes the rising flux and repels

9. Explain why the minus sign in Faraday's law cannot be a matter of convention.
   Show the full solution

   Because reversing it would permit a machine that creates energy, and no choice of coordinate convention can change whether energy is conserved. Distinguish two kinds of sign. Some signs in physics genuinely are conventions: whether work done on or by a gas is called positive, or which direction is called positive in a coordinate system. Choosing the other convention changes the equations' appearance but no prediction. Test whether this sign is of that kind. The test is whether flipping it changes any physical prediction. It does, immediately and dramatically. Construct the machine. With the sign reversed, a magnet approaching a coil would be attracted rather than repelled. Release a magnet near a coil and it accelerates in, inducing a current. The faster it goes, the stronger the current and the stronger the attraction. Count the energy. The magnet gains kinetic energy. The coil dissipates heat through \( I^2R \). Nothing supplies either. Both increase without limit, and the machine would run forever while delivering power. Why no bookkeeping can save it. Kinetic energy and dissipated heat are measurable quantities, not conventions. A coordinate choice cannot make a magnet accelerate or a resistor warm. The prediction is physical and it is wrong. What this makes the sign. It encodes a fact about nature: induced effects oppose their causes. Lenz's law is that fact stated separately, and the minus sign is the same fact written into the equation. The deeper point about how physics works. Faraday's law could have been discovered experimentally with either sign. It was not necessary to know energy conservation to measure it. But once measured, the sign turns out to be exactly the one conservation demands, and that agreement between an independently measured detail and a general principle is strong evidence that both are right. Where else this reasoning appears. The same argument fixes the sign in Le Chatelier's principle in chemistry and in the damping of any feedback system. Systems that reinforce their own disturbances do not survive to be observed. A convention changes appearance, not predictions; reversing this sign predicts an energy-creating machine

10. A 400-turn coil of area \( 0.025\ \text{m}^2 \) rotates at 60 revolutions per second in a 0.20 T field. Find the peak emf and the frequency, and explain why real power stations must hold the frequency to within a fraction of a hertz.
    Show the full solution

    Find the angular speed. \( \omega = 2\pi f = 2\pi(60) = 377\ \text{rad/s} \). Find the peak emf. \( \varepsilon_0 = NBA\omega = (400)(0.20)(0.025)(377) \) \( = (400)(0.20)(0.025) = 2.0 \), then \( (2.0)(377) = 754\ \text{V} \). About 750 V peak. Find the frequency. One revolution produces one complete cycle, so \( f = 60\ \text{Hz} \), the United States grid frequency. Find the root mean square value, which is what a voltmeter reads: \( V_{rms} = \dfrac{754}{\sqrt{2}} = 533\ \text{V} \). Now the second part: why the frequency must be held so tightly. Reason one: generators must stay synchronized. Every generator on the grid is electrically connected to every other. They must turn in step, at the same frequency and nearly the same phase, or a generator running ahead will be violently pulled back by the others. The mechanical stresses of a loss of synchronization can destroy a turbine shaft. Reason two: frequency is the demand signal. Generation and consumption must balance instant by instant, because the grid stores almost no energy. If demand exceeds generation, the extra energy comes from the rotating machines' kinetic energy, and they slow down. Frequency falling below 60 Hz is therefore a direct measurement that demand exceeds supply, and it is the signal operators act on continuously. Reason three: equipment depends on it. Synchronous motors and clocks run at a speed set by the frequency. Transformers designed for 60 Hz saturate and overheat at lower frequencies, because the flux swing per cycle is larger when each cycle lasts longer. Reason four: protection cascades. Generators disconnect automatically if the frequency strays too far, to protect themselves. If the frequency is falling because of a shortfall, each disconnection worsens the shortfall, and the result is a cascading blackout. This is the mechanism behind most large-scale grid failures. The tolerances actually used. North American grids hold 60 Hz within roughly 0.02 Hz under normal conditions, with load shedding triggered automatically below about 59.5 Hz. Operators also manage the long-term average so that frequency-driven clocks stay accurate over days. Why this connects back to Lenz's law. The whole mechanism rests on the opposing torque of lesson 8.5. Demand produces current, current produces retarding torque, retarding torque slows the machine, and the slowing is the signal. Without the minus sign there would be no feedback path at all, and no way for a kettle in a kitchen to reach a turbine a hundred kilometers away. About 754 V peak, 533 V rms, at 60 Hz; the frequency must be held because it is simultaneously the synchronization reference and the real-time measure of supply against demand

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Treat transformers as ideal unless stated otherwise.

1. Write the transformer voltage equation.
   Show the full solution

   \( \dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} \)

2. Why can a transformer not work on direct current?
   Show the full solution

   A steady current gives no changing flux, so no emf is induced

3. What happens to the current when a transformer steps the voltage up?
   Show the full solution

   It steps down in the same proportion

4. Why is a transformer core laminated?
   Show the full solution

   To break up eddy current paths and reduce heating losses

5. Write the power relationship for an ideal transformer.
   Show the full solution

   \( V_pI_p = V_sI_s \)

6. A transformer has 1500 primary turns and 75 secondary turns on a 240 V supply. Find the output voltage.
   Show the full solution

   \( V_s = V_p \cdot \dfrac{N_s}{N_p} = 240 \times \dfrac{75}{1500} = 240 \times 0.050 = 12\ \text{V} \). 12 V

7. An ideal transformer delivers 3.0 A at 12 V from a 120 V supply. Find the primary current.
   Show the full solution

   Output power: \( P = V_sI_s = (12)(3.0) = 36\ \text{W} \). Ideal, so the input power is also 36 W. \( I_p = \dfrac{P}{V_p} = \dfrac{36}{120} = 0.30\ \text{A} \). The voltage stepped down by 10 and the current stepped up by 10 ✓ 0.30 A

8. A laptop adapter takes 0.50 A at 120 V and outputs 19 V at 85 percent efficiency. Find the output current and the power wasted.
   Show the full solution

   Input power: \( P_{in} = (120)(0.50) = 60\ \text{W} \). Output power: \( P_{out} = (0.85)(60) = 51\ \text{W} \). Output current: \( I_s = \dfrac{51}{19} = 2.7\ \text{A} \). Power wasted: \( 60 - 51 = 9.0\ \text{W} \). Nine watts of heat is why adapters are warm to the touch, and the heat is entirely the copper, core and switching losses. 2.7 A, with 9.0 W wasted

9. Explain why the transformer is the reason alternating current was adopted for power distribution.
   Show the full solution

   Because transmission requires high voltage while generation and use require low voltage, and only alternating current could be converted between them efficiently. Establish the transmission requirement. Line losses are \( I^2R \), so a given power must be sent at the lowest possible current, which means the highest possible voltage. Raising the voltage forty times cuts the loss by a factor of 1600. Establish the conflicting requirement. Generators cannot practically produce hundreds of kilovolts, because of insulation limits in the windings. And appliances cannot use it, for obvious safety reasons. So the voltage must be high in between and low at both ends. Identify what that demands. An efficient, cheap, reliable way to change voltage in both directions, at very high power levels. What the transformer provides. Exactly that: over 99 percent efficiency, no moving parts, nothing to wear out, and scalable to hundreds of megawatts. A large unit can run for decades with only oil maintenance. Why it requires alternating current. It works entirely through Faraday's law, which needs a changing flux. Direct current produces a constant flux and therefore no output at all. What direct current could do at the time. Nothing comparable. Converting DC voltage required rotating machinery, a motor driving a generator, which was inefficient, expensive, noisy and needed constant maintenance. It could not be justified at every distribution point. The practical consequence in the 1880s. A DC station had to distribute at roughly the voltage customers used, so currents were large, losses were severe, and the economical service radius was about a mile. Cities would have needed a generating station every few blocks. How the argument was settled. Alternating current systems demonstrated transmission over tens of miles at acceptable loss, and the 1893 Chicago exposition and the 1895 Niagara Falls project settled the matter commercially. The physics had already settled it. The modern qualification, which matters. High-voltage direct current is now used for the longest links and all undersea cables, because solid-state converters can change DC voltage efficiently, and because DC avoids the capacitive losses that plague long AC cables. The original decision was correct for its technology, and the technology changed. What did not change. The \( I^2R \) argument for high-voltage transmission is as valid as ever. Only the means of changing voltage has an alternative. Transmission needs high voltage, use needs low voltage, and only AC could be converted efficiently between them

10. Trace the voltage from a 25 kV generator to a 120 V outlet, explaining what happens at each stage and why the system is built this way rather than with one conversion.
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    Stage one: generation at about 25 kV. The generator's own windings limit the voltage, since insulating a rotating machine for hundreds of kilovolts is impractical. This is a compromise between insulation difficulty and the current the windings must carry. Stage two: step up to 400 kV at the station. A large transformer raises the voltage by a factor of 16, reducing the current by 16 and the \( I^2R \) loss by \( 16^2 = 256 \). This single transformer is what makes long distance transmission viable. Stage three: transmission. Power travels tens or hundreds of kilometers on overhead lines at 400 kV. Losses across a national grid run at a few percent total. Stage four: step down to about 33 kV at a regional substation. The extreme voltage is no longer needed once the distance is shorter, and lower voltages need far less clearance, allowing lines to run closer to towns. Stage five: step down to about 4 kV for local distribution. These are the lines on residential streets, at a voltage low enough to be carried on ordinary poles. Stage six: step down to 240 V at the pole or pad transformer, supplying a handful of homes, and split into two 120 V legs for ordinary outlets with 240 V available across both for large appliances. Now the second part: why not one conversion? Reason one: insulation cost scales badly. Equipment rated for 400 kV requires large clearances, tall structures, long insulator strings and wide rights of way. Using it for the final hundred meters to a house would be absurdly expensive. Reason two: the loss argument weakens with distance. Line loss is proportional to the line's resistance, which is proportional to its length. A hundred-meter run has almost no resistance, so there is nothing to save by raising its voltage. High voltage pays only where the distance is long. Reason three: safety. Every additional meter of high-voltage conductor near people is a hazard. Stepping down progressively keeps the highest voltages on remote transmission corridors and away from streets. Reason four: the network branches. One transmission line feeds many regional substations, each feeding many local ones, each feeding many homes. The stepping-down mirrors the branching, and each stage serves a smaller area at a lower voltage, matching the voltage to the distance still to travel. Reason five: fault isolation. Each transformer stage is a natural place to install protection. A fault in one neighborhood is isolated at its local transformer without affecting the region, and a fault on a transmission line does not reach individual homes. The cost of the extra stages. Every transformer loses perhaps one percent, so six stages cost several percent overall. That is far less than the transmission losses avoided, and far less than the cost of insulating a whole distribution network at 400 kV. The underlying principle. Each stage's voltage is chosen so that the transmission loss over the distance that stage covers is comparable to the cost of the equipment needed at that voltage. It is an economic optimization built on one physical relationship, \( P_{\text{loss}} = I^2R \). Progressive stepping, because high voltage pays only over long distances while insulation cost is paid everywhere

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Take the speed of sound as 343 m/s and \( c = 3.00 \times 10^8\ \text{m/s} \).

1. What does a wave transfer?
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   Energy, without transferring matter

2. Write the wave equation.
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   \( v = f\lambda \)

3. Give the relationship between frequency and period.
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   \( f = \dfrac{1}{T} \)

4. Energy carried by a wave is proportional to which power of the amplitude?
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   The second

5. In a transverse wave, how does the oscillation relate to the direction of travel?
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   Perpendicular to it

6. Find the frequency of a wave of wavelength 2.5 m traveling at 12 m/s.
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   \( f = \dfrac{v}{\lambda} = \dfrac{12}{2.5} = 4.8\ \text{Hz} \). 4.8 Hz

7. Find the wavelength of an FM radio station broadcasting at 100 MHz.
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   \( \lambda = \dfrac{c}{f} = \dfrac{3.00 \times 10^8}{1.00 \times 10^8} = 3.0\ \text{m} \). This sets the antenna length. An efficient antenna is a substantial fraction of a wavelength, which is why FM car antennas are roughly 75 cm, a quarter of 3.0 m, and why cell phone antennas at 2 GHz can be a few centimeters and fit inside the case. 3.0 m

8. A wave's amplitude is tripled. Find the factor by which its energy changes, and explain why the frequency is irrelevant here.
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   Energy is proportional to \( A^2 \), so tripling the amplitude multiplies the energy by \( 3^2 = 9 \). The frequency does not appear because it describes how often the oscillation repeats, not how far each particle moves. A high-frequency wave of tiny amplitude can carry far less energy than a low-frequency wave of large amplitude. The exception worth flagging. For a mechanical wave the energy per particle also depends on frequency, since faster oscillation means larger particle speeds. The clean statement is that intensity is proportional to \( A^2f^2 \) for a mechanical wave, and the amplitude dependence is what this question asks about. 9 times

9. Explain why the speed of a wave is determined by the medium rather than the source.
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   Because the wave propagates by each part of the medium pushing on the next, and how fast that happens is a property of the medium's stiffness and inertia. The mechanism of propagation. A disturbance displaces one region of the medium. That region exerts a restoring force on its neighbor, which accelerates, displaces, and passes the disturbance on. What sets the rate. Two competing properties. A stiffer medium transmits the push more quickly, raising the speed. A denser medium has more inertia per unit volume, so each part accelerates more slowly, lowering the speed. In general \( v = \sqrt{\dfrac{\text{stiffness}}{\text{density}}} \) for a mechanical wave. Why the source cannot affect it. The source determines how often and how strongly the medium is disturbed, which sets the frequency and amplitude. But once the disturbance exists, its onward propagation is handled entirely by the medium, which has no information about what created it. The consequence for wavelength. With \( v \) fixed by the medium and \( f \) fixed by the source, \( \lambda = v/f \) is forced. Raising the frequency shortens the wavelength; it does not speed the wave up. The observational confirmation. A high note and a low note from a single instrument reach a listener simultaneously. If speed depended on frequency, music would arrive scrambled, with chords separating into a sequence, and this is obviously not what happens. The exception, which proves the point. In a dispersive medium the speed does depend slightly on frequency, and the consequence is exactly what the argument predicts: white light entering glass separates into colors, because each travels at a slightly different speed. Dispersion is observable precisely because non-dispersive propagation is the norm. Propagation is a handover between neighboring parts of the medium, governed by its stiffness and density

10. A sound wave of frequency 500 Hz travels from air into water, where the speed is 1480 m/s. Find the wavelength in each, explain what stays the same at the boundary, and say why.
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    In air. \( \lambda_{\text{air}} = \dfrac{v}{f} = \dfrac{343}{500} = 0.686\ \text{m} \). In water. \( \lambda_{\text{water}} = \dfrac{1480}{500} = 2.96\ \text{m} \). The ratio. \( \dfrac{2.96}{0.686} = 4.31 \), exactly the ratio of the speeds \( \dfrac{1480}{343} = 4.31 \) ✓ Now the crucial question: what stays the same? The frequency stays the same. It is 500 Hz in both media, and the wavelength changes to accommodate the new speed. Why the frequency must be conserved. Consider the boundary itself. The air molecules at the surface push on the water molecules, and the two are in contact. Whatever rate the air oscillates at is the rate it drives the water at. If the frequencies differed, the two sides of the boundary would separate or overlap, which is physically impossible for a continuous medium. A counting argument for the same point. Five hundred wave crests arrive at the boundary each second. Five hundred must leave each second, or crests would accumulate at the surface without limit. Conservation of crests forces conservation of frequency. Why the wavelength must therefore change. With \( f \) fixed and \( v \) changing, \( \lambda = v/f \) has no choice. The wave spreads out in the faster medium. The same result for light. Light entering glass slows down, keeps its frequency and shortens its wavelength. This is why color, which is determined by frequency, does not change underwater, even though the wavelength does. Defining color by wavelength is a convenient shorthand that fails at a boundary. What this implies for refraction. Since the frequency is fixed and the speed changes, a wave crossing a boundary at an angle must change direction, because the parts of the wavefront that enter first begin traveling at the new speed while the rest are still in the old medium. That is lesson 9.6's refraction, derived here from nothing but the constancy of frequency. A practical consequence. Sound crosses from air into water very poorly, with over 99 percent reflected, because of the enormous mismatch in the two media's acoustic properties. This is why shouting at a swimmer underwater does not work, and why sonar transmitters must be immersed rather than aimed from above. 0.686 m and 2.96 m; the frequency is unchanged because the boundary drives both media at the same rate

Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Take the speed of sound in air as 343 m/s and in water as 1480 m/s.

1. What type of wave is sound?
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   Longitudinal

2. Name the two regions in a sound wave.
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   Compressions and rarefactions

3. Can sound travel through a vacuum?
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   No

4. Which wave property determines pitch?
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   Frequency

5. Give the approximate range of human hearing.
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   20 Hz to 20 kHz

6. A ship's sonar pulse returns after 0.40 s. Find the depth.
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   The pulse travels down and back, so the depth is half the total path. \( d = \dfrac{vt}{2} = \dfrac{(1480)(0.40)}{2} = \dfrac{592}{2} = 296\ \text{m} \). 296 m

7. Find the speed of sound in air at \( 30^\circ\text{C} \) using \( v = 331 + 0.6T \), and the fractional change from \( 0^\circ\text{C} \).
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   \( v = 331 + (0.6)(30) = 331 + 18 = 349\ \text{m/s} \). At \( 0^\circ\text{C} \): 331 m/s. Fractional change: \( \dfrac{18}{331} = 0.054 \), about 5.4 percent. Why temperature matters. Warmer air has faster-moving molecules, so collisions transmit the disturbance more quickly. The effect is small but enough that wind instruments go noticeably sharp as they warm up, which is why orchestras tune after the players have been blowing for a while. 349 m/s, a 5.4 percent rise

8. A hammer strikes one end of a long steel rail. A listener at the other end, 1200 m away, hears two distinct sounds. Explain and find the time between them. Take the speed in steel as 5000 m/s.
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   Why there are two sounds. The impact sends a wave along the rail and also through the air. They travel at very different speeds and arrive at different times. Through the steel. \( t = \dfrac{1200}{5000} = 0.24\ \text{s} \). Through the air. \( t = \dfrac{1200}{343} = 3.50\ \text{s} \). The gap. \( 3.50 - 0.24 = 3.26\ \text{s} \). Which arrives first. The one through the rail, by over three seconds, which is easily distinguishable by ear. The historical use. This is the classic method for measuring the speed of sound in a solid, and putting an ear to a rail to detect a distant train works for exactly this reason. About 3.3 s apart, steel first

9. Explain why sound travels faster in solids than in gases, and why citing density as the reason is wrong.
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   Because the speed depends on the ratio of stiffness to density, and solids gain far more from stiffness than they lose to density. Write the relationship. For a mechanical wave, \( v = \sqrt{\dfrac{\text{stiffness}}{\text{density}}} \). Two properties appear, and they work in opposite directions. What density does. It appears in the denominator, so greater density means slower sound, not faster. More massive particles accelerate less for a given force, so the handover takes longer. Why the naive claim is backward. Saying solids carry sound faster because they are denser names the one factor that works against the conclusion. Among gases, sound is faster in helium than in denser air, which confirms the direction directly: helium at around 965 m/s against air's 343. What actually wins. Stiffness, meaning resistance to compression. Steel is about a million times stiffer than air while being only a few thousand times denser. The square root of that ratio gives roughly a factor of 15, and the observed speeds confirm it. Why solids are so stiff. Their atoms are bonded to their neighbors, so displacing one immediately pulls on the next. In a gas the molecules are unattached and interact only during collisions, so the disturbance must be carried across empty space between encounters. The ordering this predicts. Solids fastest, then liquids, then gases, which is what is observed: about 5000 m/s in steel, 1480 in water, 343 in air. The counterexample that tests the rule. Lead is much denser than aluminum but not proportionally stiffer, and sound travels at about 1200 m/s in lead against 6400 m/s in aluminum. Density dominates in that comparison, and it slows the wave, exactly as the formula says it should. Stiffness raises the speed and density lowers it; solids win on stiffness by a far larger margin than they lose on density

10. A person claps once in a large empty hall and hears a reverberation lasting about 2 s rather than a single clean echo. Explain the difference, and work out roughly how many reflections are involved in a hall 30 m across.
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    What an echo is. A single distinct reflection from one large distant surface, arriving late enough to be heard as a separate sound. The ear resolves two sounds as separate when they are about 0.1 s apart, which at 343 m/s means the reflecting surface must be at least \( \dfrac{(343)(0.1)}{2} = 17\ \text{m} \) away. What reverberation is. A dense overlapping succession of reflections from many surfaces at many distances, arriving too close together to be separated, and decaying gradually as energy is absorbed at each bounce. Estimate the time between reflections. In a hall 30 m across, a typical path between surfaces is of that order, so \( t = \dfrac{30}{343} = 0.087\ \text{s} \) per bounce. Estimate the number of reflections. \( n = \dfrac{2.0}{0.087} = 23 \), so roughly twenty or more reflections contribute before the sound becomes inaudible. Why they blur together. At 0.087 s apart they are below the ear's resolution threshold of about 0.1 s, and in a real hall the paths have many different lengths, so the arrivals are spread continuously rather than evenly spaced. The result is a smooth decay rather than a series of claps. Why the sound dies away at all. Each reflection absorbs some energy into the wall material, and the air itself absorbs high frequencies over distance. If the surfaces absorbed nothing, the sound would persist far longer. Why high notes fade first. Absorption rises with frequency, so the tail of a reverberation sounds duller than the original. This is audible in any large stone building. Why this matters for building design. Reverberation time is the central parameter in acoustic design. A concert hall wants roughly 1.8 to 2.2 s, which blends the orchestra and gives the sound warmth. A lecture room wants under 1 s, because long reverberation smears consonants together and destroys speech intelligibility. A recording studio wants a very short time so that the reverberation can be added artificially and controlled. How it is adjusted. Absorptive materials such as carpet, curtains, upholstered seats and perforated panels shorten it. Hard parallel surfaces lengthen it and also risk flutter echo, a rapid repeating reflection between two parallel walls, which is why studio walls are deliberately non-parallel. The audience effect. People absorb sound, so a full hall reverberates less than an empty one. Modern concert halls use heavily upholstered seats chosen to absorb about as much as a seated person, so the acoustics do not change with attendance. An echo is one resolvable reflection; reverberation is roughly twenty or more overlapping ones decaying together