AP Physics 1

Lesson 1.1 · Unit 1 · CED topics 1.1–1.2

Scalars, vectors, and the language of motion

Physics starts by being picky about words. "How far" and "how far from where you started" are different questions, and so are "how fast" and "how fast in which direction." Every unit in this course leans on the distinction, so it pays to get it exactly right now.

Exam tip: when a question says "velocity," it wants a direction with the number. A runner who completes one 400 m lap in 50 s has an average speed of 8.0 m/s but an average velocity of zero: the displacement is zero, and graders check that you know why.

Lesson 1.2 · Unit 1 · CED topic 1.3

Representing motion with graphs

A motion graph is a story told in slopes and areas. The exam will hand you one graph and ask about a quantity that lives on a different graph, so you need the translation rules cold, and you need to be able to say them in words, not just use them.

Exam tip: a curved \(x\)–\(t\) graph means changing velocity: concave up means positive acceleration, concave down means negative. A falling \(x\)–\(t\) graph means negative velocity, not "slowing down."

Lesson 1.3 · Unit 1 · CED topic 1.3

The constant-acceleration equations

When acceleration is constant, the graphs of Lesson 1.2 are straight lines and parabolas, and three algebraic equations replace the picture. These are on the AP equation sheet; the skill is choosing the right one from what's given and what's asked.

Lesson 1.4 · Unit 1 · CED topic 1.3

Free fall

Drop a bowling ball and a golf ball together and they land together. Near Earth's surface, with air resistance ignored, every object accelerates downward at the same rate regardless of its mass. That makes free fall the cleanest possible application of Lesson 1.3.

Common error: writing \(a = 0\) at the top because "the ball stops." If the acceleration were zero there, the ball would hover. Velocity and acceleration are independent quantities.

Lesson 1.5 · Unit 1 · CED topic 1.4

Reference frames and relative motion

Velocity is always measured relative to something. A passenger walking forward at 1 m/s inside a train doing 30 m/s is moving at 31 m/s relative to the ground. When the two velocities aren't along the same line, they still add, as vectors, component by component.

Lesson 1.6 · Unit 1 · CED topic 1.5

Projectile motion

A projectile is anything in free fall that also has horizontal velocity. The key insight: gravity pulls only downward, so it changes only the vertical velocity. The horizontal motion is constant-velocity; the vertical motion is Lesson 1.4. The two share only one thing: time.

Exam tip: for a horizontal launch, the fall time is set entirely by the height: a ball fired at 50 m/s and one dropped from the same height hit the ground at the same instant. Say this in words when asked to compare.

Unit 1 practice · 10 problems

Unit 1 practice: Kinematics

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer: the reveal shows the governing equation, the substitution, and the answer with units. Use \(g = 9.8\ \text{m/s}^2\). A calculator is fine for the arithmetic.

1. A hiker walks 12 m north, then 5.0 m east, taking 10 s. Find the distance traveled, the displacement (magnitude and direction), the average speed, and the average velocity.

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   Distance is path length: \(12 + 5.0 = 17\) m. Displacement is the straight arrow from start to finish: \(|\Delta\vec{x}| = \sqrt{12^2 + 5.0^2} = 13\) m at \(\tan^{-1}(5.0/12) = 23^\circ\) east of north. Average speed \(= 17\ \text{m}/10\ \text{s} = 1.7\) m/s. Average velocity \(= \Delta\vec{x}/\Delta t = 13\ \text{m}/10\ \text{s} = 1.3\) m/s, 23° east of north.

2. A position–time graph rises in a straight line from \(x = 0\) at \(t = 0\) to \(x = 20\) m at \(t = 5.0\) s, is flat until \(t = 8.0\) s, then falls in a straight line to \(x = -10\) m at \(t = 13\) s. Find the velocity in each segment, the total distance traveled, and the displacement.

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   Velocity is the slope of \(x\)–\(t\). Segment 1: \(v = \dfrac{20\ \text{m}}{5.0\ \text{s}} = +4.0\) m/s. Segment 2: \(v = 0\) (at rest). Segment 3: \(v = \dfrac{-10 - 20\ \text{m}}{5.0\ \text{s}} = -6.0\) m/s. Distance \(= 20 + 0 + 30 = 50\) m; displacement \(= x_f - x_0 = -10\) m (10 m in the negative direction). The \(v\)–\(t\) graph is three horizontal steps at \(+4.0\), \(0\), and \(-6.0\) m/s.

3. A velocity–time graph rises in a straight line from 0 to 12 m/s during the first 4.0 s, stays at 12 m/s until \(t = 10\) s, then falls in a straight line to 0 at \(t = 12\) s. Find the acceleration in each part and the total displacement.

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   Acceleration is the slope of \(v\)–\(t\): \(a_1 = \dfrac{12\ \text{m/s}}{4.0\ \text{s}} = 3.0\) m/s², \(a_2 = 0\), \(a_3 = \dfrac{0 - 12\ \text{m/s}}{2.0\ \text{s}} = -6.0\) m/s². Displacement is the area: triangle \(\tfrac{1}{2}(4.0)(12) = 24\) m, rectangle \((6.0)(12) = 72\) m, triangle \(\tfrac{1}{2}(2.0)(12) = 12\) m. Total \(\Delta x = 108\) m.

4. A car traveling at 30 m/s brakes with constant acceleration and stops in 75 m. Find the acceleration and the time to stop.

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   No time given, so use \(v_x^2 = v_{x0}^2 + 2a_x(x - x_0)\): \(0 = (30\ \text{m/s})^2 + 2a_x(75\ \text{m})\), so \(a_x = -\dfrac{900}{150} = -6.0\) m/s² (opposite the velocity). Then \(v_x = v_{x0} + a_x t\): \(t = \dfrac{0 - 30\ \text{m/s}}{-6.0\ \text{m/s}^2} = 5.0\) s.

5. A ball is thrown straight up at 24.5 m/s from ground level. Find its maximum height, its total time in the air, and its velocity just before it lands.

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   Up positive, \(a_y = -9.8\) m/s². At the top \(v_y = 0\): \(y_{\max} = \dfrac{v_{y0}^2}{2g} = \dfrac{(24.5\ \text{m/s})^2}{2(9.8\ \text{m/s}^2)} = 31\) m (30.6 m). Time up: \(t = \dfrac{24.5}{9.8} = 2.5\) s, so total flight \(= 5.0\) s by symmetry. It lands at \(24.5\) m/s downward: same speed as launch, opposite direction.

6. At the highest point of the ball's flight in the previous problem, which statement is correct?

   velocity is zero and acceleration is zero velocity is zero and acceleration is 9.8 m/s² downward velocity is nonzero and acceleration is zero velocity and acceleration are both directed upward

   The velocity is indeed zero for an instant, but acceleration is the rate of change of velocity, and the velocity is changing from upward to downward right through that instant. If \(a\) were zero at the top, the ball would hover there.

   The velocity passes through zero as it reverses, but gravity acts the whole time, so \(a = -g\), or 9.8 m/s² downward, at the top just as everywhere else in the flight.

   For a ball thrown straight up, the velocity at the peak is exactly zero: that is what defines the peak. And free-fall acceleration is never zero while gravity acts, so both halves of this statement fail.

   Nothing about the motion points upward at the peak: the velocity is momentarily zero, and the acceleration due to gravity is downward throughout. Confusing “was moving up” with “is accelerating up” is the misconception here.

7. A boat moves at 5.0 m/s relative to the water and points straight across a 200 m wide river whose current flows at 2.0 m/s. Find the crossing time, the downstream drift, and the boat's velocity relative to the shore.

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   \(\vec{v}_{\text{boat,shore}} = \vec{v}_{\text{boat,water}} + \vec{v}_{\text{water,shore}}\): 5.0 m/s across and 2.0 m/s downstream. Crossing time uses only the across component: \(t = \dfrac{200\ \text{m}}{5.0\ \text{m/s}} = 40\) s. Drift \(= (2.0\ \text{m/s})(40\ \text{s}) = 80\) m downstream. Speed relative to shore \(= \sqrt{5.0^2 + 2.0^2} = 5.4\) m/s at \(\tan^{-1}(2.0/5.0) = 22^\circ\) downstream of straight across. The current does not change the crossing time.

8. A ball is kicked from level ground at 25 m/s, 30° above the horizontal. Find the time of flight, the range, and the maximum height.

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   Components: \(v_{x0} = 25\cos 30^\circ = 21.7\) m/s, \(v_{y0} = 25\sin 30^\circ = 12.5\) m/s. Lands when \(y = 0\): \(0 = v_{y0}t - \tfrac{1}{2}gt^2 \Rightarrow t = \dfrac{2v_{y0}}{g} = \dfrac{2(12.5)}{9.8} = 2.6\) s. Range \(x = v_{x0}t = (21.65)(2.55) = 55\) m. Max height, where \(v_y = 0\): \(y_{\max} = \dfrac{v_{y0}^2}{2g} = \dfrac{(12.5)^2}{19.6} = 8.0\) m.

9. Three identical balls are thrown from the edge of a 30 m cliff, all at 20 m/s: ball X horizontally, ball Y at 30° above the horizontal, ball Z at 30° below the horizontal. Ignore air resistance. Rank the balls by time in the air, then by speed just before landing. Explain each ranking in words.

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   Time: \(Y \gt X \gt Z\). Only the vertical motion sets the flight time. Y starts with upward velocity and must rise and fall back before descending the cliff; X starts with zero vertical velocity; Z starts already moving downward and reaches the ground soonest. Landing speed: \(X = Y = Z\). Gravity is the only force, and \(v^2 = v_0^2 + 2g\,\Delta h\) holds for the full velocity regardless of launch angle: the same 20 m/s launch and the same 30 m drop give the same landing speed, \(\sqrt{20^2 + 2(9.8)(30)} = 31\) m/s. (Unit 3 says the same thing with energy.) Their landing directions differ.

10. Experimental design. You have a steel ball, a meterstick, and a motion sensor or high-speed camera that records position versus time. Describe a procedure to measure \(g\), say what graph you would make and how you would get \(g\) from it, and name one way to reduce the uncertainty.

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    Drop the ball from rest from several measured heights \(h\) (e.g. 0.5 m to 2.0 m) and record the fall time \(t\) for each; repeat each height at least three times and average. Since \(h = \tfrac{1}{2}gt^2\) for release from rest, plot \(h\) on the vertical axis against \(t^2\) on the horizontal axis: the data should be a straight line through the origin with slope \(g/2\), so \(g = 2 \times \text{slope}\) (expected slope 4.9 m/s²). Reduce uncertainty by using the sensor or frame-by-frame video rather than a hand-held stopwatch (reaction time is comparable to the fall time), using the larger heights, and fitting a best-fit line rather than using one point.

Lesson 2.1 · Unit 2 · CED topics 2.1–2.3

Systems, center of mass, and free-body diagrams

Before you can apply Newton's laws you have to decide what "the object" is and what is pushing or pulling on it. Those two decisions, choosing a system and drawing its free-body diagram, are worth points on their own, and most Unit 2 errors trace back to one of them.

Common FBD mistakes: drawing a "force of motion" or velocity arrow (velocity is not a force); labeling a force by its value instead of its type; including forces the object exerts on other things; and drawing the normal force equal to \(mg\) on a tilted surface. Only forces acting on the system belong.

Lesson 2.2 · Unit 2 · CED topics 2.4–2.5

Newton's first and third laws

The first law says what happens when nothing is pushing: velocity stays constant. The third law says forces never come alone. Both are simple to state and easy to misapply, which is why the exam keeps asking about them.

Exam tip: to name a third-law partner, swap the two objects and keep the type: the partner of "normal force of table on book" is "normal force of book on table," never gravity.

Lesson 2.3 · Unit 2 · CED topic 2.6

Newton's second law

When the forces don't balance, the object accelerates, in the direction of the net force, by an amount that depends on its mass. This one equation, applied component by component, solves most of Unit 2.

Lesson 2.4 · Unit 2 · CED topic 2.7

Gravitational force and weight

Weight is a force: the gravitational pull of a planet on an object. Mass is the amount of stuff, the same everywhere. Confusing them costs points; so does assuming a bathroom scale always reads your weight.

Exam tip: at a height of two Earth radii above the surface, \(r = 3R_E\), so \(g\) drops to \(9.8/9 = 1.1\) m/s². Inverse-square reasoning like this, "triple the distance, one-ninth the force", is a common qualitative question.

Lesson 2.5 · Unit 2 · CED topic 2.8

Friction

Friction is the contact force parallel to a surface. It comes in two kinds with two different rules, and the single most important habit in this lesson is computing the normal force honestly rather than assuming it equals \(mg\).

Lesson 2.6 · Unit 2 · CED topic 2.9

Spring forces

A spring pushes back harder the more you deform it. That simple proportionality, Hooke's law, makes springs the standard force sensor in the lab and the starting point for oscillations in Unit 7.

Exam tip: the force in Hooke's law is the force the spring exerts. The force you exert to hold it stretched is \(+k\,\Delta x\), the third-law partner. Also, \(\Delta x\) is measured from the natural length, not from wherever the mass happens to hang.

Lesson 2.7 · Unit 2 · CED topic 2.10

Circular motion

An object moving in a circle at constant speed is accelerating, because its velocity keeps changing direction. That acceleration points toward the center, and by Newton's second law some real force must point there too. Finding that force is the whole game.

Why "centrifugal force" isn't on the FBD: the outward pull you feel on a merry-go-round is your inertia: your body trying to continue in a straight line while the seat pushes you inward. No object exerts an outward force on you, so nothing outward goes on the diagram.

Unit 2 practice · 10 problems

Unit 2 practice: Force and Translational Dynamics

Ten problems covering the whole unit: one of the two heaviest units on the exam. Draw the free-body diagram before you write any equation. Use \(g = 9.8\ \text{m/s}^2\).

1. (a) A 3.0 kg block sits at \(x = 0\) and a 5.0 kg block at \(x = 2.0\) m. Find the center of mass. (b) A block slides down a rough incline. List every force on its free-body diagram with its direction, and state which arrow must be shorter than the weight arrow and why.

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   (a) \(x_{\text{cm}} = \dfrac{\sum m_i x_i}{\sum m_i} = \dfrac{(3.0)(0) + (5.0)(2.0)}{8.0} = 1.25\) m, closer to the heavier block. (b) Gravity \(mg\) straight down; normal force perpendicular to the incline surface, away from it; kinetic friction parallel to the surface, up the slope (opposite the sliding). The normal arrow is shorter than the weight arrow because it balances only the perpendicular component \(mg\cos\theta\). No "force of motion" arrow: velocity is not a force.

2. An 8.0 kg sign hangs at rest from two identical cables, each making 40° with the horizontal. Find the tension in each cable.

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   FBD: weight down, two equal tensions \(T\) at 40° above horizontal leaning opposite ways; horizontal components cancel. \(\sum F_y = 2T\sin 40^\circ - mg = 0\), so \(T = \dfrac{mg}{2\sin 40^\circ} = \dfrac{(8.0)(9.8)}{2(0.643)} = \dfrac{78.4\ \text{N}}{1.29} = 61\) N. Each cable carries more than half the weight because only part of its pull is vertical.

3. A book rests on a table. Which force is the Newton's third-law partner of Earth's gravitational pull on the book?

   the normal force of the table on the book the gravitational pull of the book on Earth the normal force of the book on the table the weight of the table

   The normal force happens to be equal and opposite to the book's weight here, but it is a different type of force (contact, not gravitational) acting on the same object; it balances gravity by Newton's first law, not the third. Third-law pairs never act on the same object.

   Third-law pairs swap the two objects and keep the type: Earth pulls the book gravitationally, so the book pulls Earth gravitationally, with equal magnitude and opposite direction.

   The normal force of the book on the table is the third-law partner of the normal force of the table on the book: a contact-force pair. It is not paired with the gravitational force at all.

   The table's weight is Earth's pull on a different object entirely and is paired with nothing acting on the book. Its own third-law partner would be the table's gravitational pull on Earth.

4. Box A (3.0 kg) and box B (5.0 kg) sit on a frictionless floor, connected by a light string. A horizontal 24 N force pulls B, dragging A behind it. Find the acceleration and the tension in the string.

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   Whole system (8.0 kg): the string is internal, so \(a = \dfrac{\sum F_x}{m_{\text{tot}}} = \dfrac{24\ \text{N}}{8.0\ \text{kg}} = 3.0\) m/s². Box A alone: the only horizontal force is \(T\), so \(T = m_A a = (3.0)(3.0) = 9.0\) N. Check on B: \(24 - T = m_B a \Rightarrow 24 - 9.0 = (5.0)(3.0) = 15\) N ✓.

5. (a) An 80 kg person stands on a scale in an elevator accelerating downward at 3.0 m/s². What does the scale read, and does the person feel heavier or lighter? (b) A planet has three times Earth's mass and twice Earth's radius. Find \(g\) at its surface.

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   (a) Up positive: \(\sum F_y = N - mg = m(-3.0)\), so \(N = m(g - a) = (80)(9.8 - 3.0) = 544\) N ≈ 540 N, less than the 784 N weight: lighter. (b) \(g = \dfrac{GM}{r^2}\), so \(g_{\text{planet}} = g_E \cdot \dfrac{3}{2^2} = \dfrac{3}{4}(9.8) = 7.4\) m/s². Tripling the mass triples \(g\); doubling the radius cuts it by four.

6. A 4.0 kg block slides down a 25° incline with \(\mu_k = 0.15\). Find the normal force, the friction force, and the acceleration.

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   Axes along and perpendicular to the slope. Perpendicular: \(N = mg\cos\theta = (4.0)(9.8)\cos 25^\circ = 35.5\) N. Friction: \(f_k = \mu_k N = (0.15)(35.5) = 5.3\) N, up the slope. Along the slope: \(mg\sin\theta - f_k = ma\), so \(a = \dfrac{(4.0)(9.8)(0.423) - 5.33}{4.0} = \dfrac{16.6 - 5.33}{4.0} = 2.8\) m/s². Equivalently \(a = g(\sin\theta - \mu_k\cos\theta)\); the mass cancels.

7. A student hangs masses from a spring and records spring force versus stretch. Two points on the best-fit line are (0.10 m, 12 N) and (0.25 m, 30 N). Find \(k\), then find how far the spring stretches when a 0.60 kg mass hangs at rest from it.

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   \(k = \text{slope} = \dfrac{30 - 12\ \text{N}}{0.25 - 0.10\ \text{m}} = \dfrac{18}{0.15} = 120\) N/m. Hanging mass at rest: \(\sum F_y = k\,\Delta x - mg = 0\), so \(\Delta x = \dfrac{mg}{k} = \dfrac{(0.60)(9.8)}{120} = 0.049\) m ≈ 4.9 cm. The line passes through the origin because an unstretched spring exerts no force.

8. (a) A 0.50 kg ball on a 1.2 m string moves in a horizontal circle at 6.0 m/s on a frictionless table. Find the tension. (b) What is the fastest a car can round a flat curve of radius 60 m if \(\mu_s = 0.70\)?

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   (a) Tension is the only inward force: \(T = m\dfrac{v^2}{r} = (0.50)\dfrac{(6.0)^2}{1.2} = 15\) N. (b) Static friction supplies the inward force, at most \(\mu_s N = \mu_s mg\): \(\mu_s mg = m\dfrac{v^2}{r} \Rightarrow v = \sqrt{\mu_s g r} = \sqrt{(0.70)(9.8)(60)} = 20\) m/s. The mass cancels: a truck and a bike have the same limit on the same road.

9. A passenger in a car turning sharply left says, "I was thrown outward against the door." In a short paragraph, explain what actually happened using Newton's laws, and state whether an outward force belongs on the passenger's free-body diagram.

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   By Newton's first law the passenger tends to keep moving in a straight line at constant velocity. The car turns left beneath her, so relative to the car she slides toward the right-hand door, but nothing pushed her outward. When she reaches the door, the door pushes inward on her (a normal force), and that inward force is what makes her follow the curve: \(\sum F_{\text{in}} = mv^2/r\). Her free-body diagram shows gravity, the seat's normal force, and the door's inward push. No outward "centrifugal" force appears, because no object exerts one.

10. Experimental design. You have a wooden block, a set of masses, a spring scale, and a horizontal table. Design a procedure to measure the coefficient of kinetic friction between block and table. State what you measure, what you vary, what graph you plot, and how \(\mu_k\) comes from the graph.

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    Pull the block horizontally with the spring scale at constant speed; then \(\sum F_x = 0\), so the scale reading equals \(f_k\). Vary the normal force by stacking known masses on the block; on a horizontal table \(N = (m_{\text{block}} + m_{\text{added}})g\). Repeat each load several times and average. Plot \(f_k\) (vertical) against \(N\) (horizontal): \(f_k = \mu_k N\) predicts a straight line through the origin with slope \(\mu_k\). Using the slope of a best-fit line rather than one ratio reduces the effect of random error; pulling with the scale parallel to the table keeps \(N\) equal to the weight.

Lesson 3.1 · Unit 3 · CED topic 3.1

Kinetic energy

Forces tell you how motion changes moment by moment. Energy gives you a shortcut: a single number that summarizes how much "motion" an object has, no matter which direction it's going. That number is kinetic energy, and it's the first piece of the bookkeeping system this unit builds.

Exam tip: when two objects have the same kinetic energy but different masses, the lighter one is moving faster by a factor of \(\sqrt{m_{\text{heavy}}/m_{\text{light}}}\), not by the mass ratio itself. Ratio questions like this are common multiple-choice items; set up \(K_A = K_B\) symbolically before touching numbers.

Lesson 3.2 · Unit 3 · CED topic 3.2

Work

Work is how a force moves energy into or out of an object. Push something along its motion and you give it energy; push against its motion and you take energy away; push sideways and you change nothing. This lesson connects the force picture of Unit 2 to the energy picture.

Lesson 3.3 · Unit 3 · CED topic 3.3

Potential energy

Lift a book and let go: it speeds up without anything pushing it. The kinetic energy came from somewhere: energy that was stored in the book–Earth system by virtue of their separation. Stored energy of this kind is potential energy, and it belongs to a system, not to a single object.

Lesson 3.4 · Unit 3 · CED topic 3.4

Conservation of energy

Energy is never created or destroyed; it only moves between forms and between systems. Name every form present at the start and the end, and a single equation replaces pages of force analysis.

Lesson 3.5 · Unit 3 · CED topic 3.5

Power

Two motors can lift the same load to the same height, do the same work, yet one finishes in seconds while the other takes a minute. Power is the quantity that tells them apart: how fast energy is transferred.

Unit 3 practice · 10 problems

Unit 3 practice: Work, Energy, and Power

Ten problems covering the whole unit. For each energy problem, name the system and the forms of energy present before and after: that sentence earns points on the exam. Use \(g = 9.8\ \text{m/s}^2\).

1. Find the kinetic energy of a 1200 kg car at 15 m/s and at 30 m/s. By what factor does it change, and why?

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   \(K = \tfrac{1}{2}mv^2\): \(K_1 = \tfrac{1}{2}(1200)(15)^2 = 1.35\times10^{5}\) J; \(K_2 = \tfrac{1}{2}(1200)(30)^2 = 5.4\times10^{5}\) J. Doubling the speed quadruples \(K\), because speed enters squared.

2. Objects A and B have the same kinetic energy, and A has four times the mass of B. The ratio of B's speed to A's speed is

   4 2 1/2 16

   A ratio of 4 assumes speed scales inversely with mass, as it would for equal momentum (\(p = mv\)). Kinetic energy depends on \(v^2\), so the speed ratio is the square root of the mass ratio.

   Set \(\tfrac{1}{2}m_A v_A^2 = \tfrac{1}{2}m_B v_B^2\) with \(m_A = 4m_B\): \(4v_A^2 = v_B^2\), so \(v_B = 2v_A\). The lighter object is faster by the square root of the mass ratio, not by the mass ratio itself.

   1/2 has the right size but the wrong direction: it is A's speed over B's. The lighter object B must be the faster one to carry the same kinetic energy with a quarter of the mass.

   16 comes from squaring the mass ratio instead of taking its square root. Since \(K \propto v^2\), a factor of 4 in mass corresponds to a factor of \(\sqrt{4} = 2\) in speed.

3. A rope pulls a 15 kg crate 8.0 m across a floor with a 40 N force directed 25° above the horizontal, while friction opposes with 10 N. The crate starts from rest. Find the work done by each force, the net work, and the final speed.

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   \(W = Fd\cos\theta\). Rope: \((40)(8.0)\cos 25^\circ = 290\) J. Friction: \(-(10)(8.0) = -80\) J. Gravity and normal force: 0 J (perpendicular to the motion). \(W_{\text{net}} = 290 - 80 = 210\) J. Work–energy theorem: \(W_{\text{net}} = \Delta K = \tfrac{1}{2}mv^2\), so \(v = \sqrt{\dfrac{2(210)}{15}} = 5.3\) m/s.

4. A force–position graph for a 4.0 kg cart on a frictionless track rises in a straight line from 0 to 30 N between \(x = 0\) and \(x = 2.0\) m, then stays at 30 N from 2.0 m to 5.0 m. The cart starts from rest at \(x = 0\). Find the work done and the cart's speed at \(x = 5.0\) m.

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   Work is the area under \(F\)–\(x\): triangle \(\tfrac{1}{2}(2.0)(30) = 30\) J plus rectangle \((3.0)(30) = 90\) J, total \(W = 120\) J. Then \(W = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{\dfrac{2(120)}{4.0}} = 7.7\) m/s. The acceleration was changing during the first 2.0 m; energy never needed it.

5. A spring with \(k = 500\) N/m is compressed 0.20 m and used to launch a 0.25 kg ball straight up. Ignoring air resistance and the small initial compression distance, how high above the launch point does the ball rise? State the energy forms before and after.

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   Before: elastic potential energy only, \(U_s = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(500)(0.20)^2 = 10\) J. At the top: gravitational potential energy only (\(K = 0\)). \(U_s = mg\,\Delta h \Rightarrow \Delta h = \dfrac{10\ \text{J}}{(0.25)(9.8)} = 4.1\) m. Halving the compression would cut the height to one quarter.

6. A 3.0 kg block starts from rest at the top of a frictionless ramp 4.0 m high, slides down, then moves across a horizontal floor with \(\mu_k = 0.25\). Find its speed at the bottom of the ramp and how far it slides on the floor. Describe the energy bar chart at the start, the bottom, and the end.

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   Ramp: \(mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2(9.8)(4.0)} = 8.9\) m/s (\(K = 118\) J). Floor: friction \(f_k = \mu_k mg = (0.25)(3.0)(9.8) = 7.35\) N converts all of \(K\) to thermal energy: \(f_k d = K \Rightarrow d = \dfrac{117.6}{7.35} = 16\) m (or \(d = h/\mu_k\)). Bars: start, one tall \(U_g\) bar (118 J); bottom, the same height, now all \(K\); end, the same height, now all thermal energy. The total never changes.

7. A roller-coaster car is at rest at a height of 30 m and rolls down a frictionless track to a point 10 m above the ground. Find its speed there, and say what fraction of the original potential energy (measured from the ground) is still potential.

   Show answer

   System: car + Earth, no friction, so \(K_i + U_i = K_f + U_f\): \(0 + mg(30) = \tfrac{1}{2}mv^2 + mg(10)\). Mass cancels: \(v = \sqrt{2g(20)} = \sqrt{2(9.8)(20)} = 20\) m/s (19.8 m/s). Potential energy remaining: \(\dfrac{mg(10)}{mg(30)} = \dfrac{1}{3}\); two-thirds has become kinetic.

8. Two identical blocks are released from rest at the same height on two frictionless ramps, one steep and one shallow, both ending at the same floor. Compare the blocks' speeds at the bottom and their times to reach the bottom. Explain both comparisons in words.

   Show answer

   Same speed at the bottom: on a frictionless ramp the normal force does no work (it is perpendicular to the motion), so only gravity transfers energy, and gravity's work depends only on the height dropped, \(mgh\). Equal \(mgh\) means equal \(\tfrac{1}{2}mv^2\). Different times: the steep ramp gives a larger acceleration \(g\sin\theta\) along a shorter path, so the block on it arrives first. The shallow ramp's block travels farther at lower average speed. Energy tells you the final speed; only forces and kinematics tell you the time.

9. (a) A hoist lifts a 120 kg load through 15 m at constant speed in 12 s. What average power does it deliver? (b) A car cruises at a constant 20 m/s against 500 N of total resistance. What power does the engine deliver to the wheels?

   Show answer

   (a) At constant speed the work equals the gain in \(U_g\): \(P = \dfrac{mgh}{\Delta t} = \dfrac{(120)(9.8)(15)}{12} = \dfrac{17\,640\ \text{J}}{12\ \text{s}} = 1470\) W ≈ 1.5 kW. (b) Constant velocity means the forward force equals the 500 N resistance, so \(P = Fv = (500)(20) = 1.0\times10^{4}\) W = 10 kW, all of it going to thermal energy of air and tires.

10. Experimental design. A cart on a frictionless track is launched by a spring compressed a distance \(x\). You have a motion sensor, a meterstick, and a balance. Design an experiment to test whether the cart's kinetic energy is proportional to \(x^2\), and say how you would use a graph to find \(k\).

    Show answer

    Measure the cart's mass \(m\). Compress the spring by several measured distances \(x\), release, and use the motion sensor to record the cart's speed \(v\) after it leaves the spring; repeat each \(x\) and average. If \(\tfrac{1}{2}kx^2 = \tfrac{1}{2}mv^2\), then \(v^2 = \dfrac{k}{m}x^2\). Plot \(v^2\) (vertical) against \(x^2\) (horizontal): a straight line through the origin confirms the proportionality, and \(k = m \times \text{slope}\). Plotting \(v\) against \(x\) also works (a line with slope \(\sqrt{k/m}\)), but the squared plot tests the \(x^2\) dependence directly.

Lesson 4.1 · Unit 4 · CED topic 4.1

Linear momentum

Kinetic energy measures how much motion an object has but throws away the direction. Momentum keeps it. That one difference makes momentum the right tool for collisions, where objects push on each other in specific directions and energy alone can't tell you where they end up.

Exam tip: "Which object has more momentum?" and "which has more kinetic energy?" can have different answers for the same pair of objects. Two objects with equal momentum but different mass have unequal kinetic energies: the lighter one has more, since \(K = p^2/2m\).

Lesson 4.2 · Unit 4 · CED topic 4.2

Impulse

Newton's second law can be rewritten so that it talks about momentum instead of acceleration: a force acting for a time changes momentum. The product of force and time is called impulse, and it explains why catching a ball with "soft hands" hurts less.

Lesson 4.3 · Unit 4 · CED topic 4.3

Conservation of linear momentum

When two objects collide, each gives the other an impulse. By Newton's third law those impulses are equal and opposite, so whatever momentum one object gains the other loses. The total never changes, and that single fact lets you predict the outcome of a collision without knowing anything about the forces inside it.

Exam tip: before writing the equation, say why momentum is conserved: "no net external force acts on the two-cart system during the collision." Graders award a point for that justification and deduct when the system is chosen so that an external force sneaks in (a cart hitting a wall bolted to Earth, for instance).

Lesson 4.4 · Unit 4 · CED topic 4.4

Elastic and inelastic collisions

Momentum is conserved in every collision. Kinetic energy is not: it can become thermal energy, sound, and permanent deformation. How much survives sorts collisions into types, and it supplies the extra equation you need when both final velocities are unknown.

Unit 4 practice · 10 problems

Unit 4 practice: Linear Momentum

Ten problems covering the whole unit. Choose a positive direction and carry signs on every momentum. Before you conserve momentum, say why it is conserved. Use \(g = 9.8\ \text{m/s}^2\).

1. Compare the momentum and the kinetic energy of a 0.045 kg golf ball at 70 m/s with those of an 80 kg jogger at 3.0 m/s.

   Show answer

   Momentum \(p = mv\): golf ball \((0.045)(70) = 3.2\) kg·m/s; jogger \((80)(3.0) = 240\) kg·m/s: the jogger has about 75 times more. Kinetic energy \(K = \tfrac{1}{2}mv^2\): ball \(\tfrac{1}{2}(0.045)(70)^2 = 110\) J; jogger \(\tfrac{1}{2}(80)(3.0)^2 = 360\) J. Momentum is linear in \(v\), energy is quadratic, so the two comparisons can come out differently.

2. Cart A (4.0 kg) rolls right at 2.0 m/s while cart B (6.0 kg) rolls left at 1.5 m/s. Find the total momentum of the two-cart system, magnitude and direction.

   Show answer

   Right positive. \(p_A = (4.0)(+2.0) = +8.0\) kg·m/s; \(p_B = (6.0)(-1.5) = -9.0\) kg·m/s. \(p_{\text{sys}} = +8.0 - 9.0 = -1.0\) kg·m/s, i.e. 1.0 kg·m/s to the left. Momenta add with signs; kinetic energies (8.0 J and 6.75 J) would simply add.

3. The force on a 0.50 kg ball, initially at rest, is graphed against time: it rises in a straight line from 0 to 400 N in 0.010 s, holds at 400 N until \(t = 0.030\) s, then falls in a straight line to 0 at \(t = 0.050\) s. Find the impulse, the ball's final speed, and the average force.

   Show answer

   Impulse is the area under \(F\)–\(t\): \(\tfrac{1}{2}(0.010)(400) + (0.020)(400) + \tfrac{1}{2}(0.020)(400) = 2.0 + 8.0 + 4.0 = 14\) N·s. \(J = \Delta p = m v_f - 0 \Rightarrow v_f = \dfrac{14}{0.50} = 28\) m/s. \(F_{\text{avg}} = \dfrac{J}{\Delta t} = \dfrac{14}{0.050} = 280\) N, the height of the rectangle with the same area.

4. A 0.25 kg ball hits the floor moving downward at 8.0 m/s and rebounds upward at 6.0 m/s. Contact lasts 0.015 s. Find the impulse on the ball and the average force from the floor. Is the ball's weight important during the contact?

   Show answer

   Up positive: \(v_i = -8.0\) m/s, \(v_f = +6.0\) m/s. \(J = m(v_f - v_i) = (0.25)\big(6.0 - (-8.0)\big) = +3.5\) kg·m/s, upward. \(F_{\text{avg}} = \dfrac{J}{\Delta t} = \dfrac{3.5}{0.015} = 230\) N (233 N) upward. The weight is \(mg = 2.5\) N: about 1% of the contact force, so neglecting it here is reasonable. The classic error is \(8.0 - 6.0\); the velocity reversed, so the change is the sum of the speeds.

5. An egg dropped onto a hard floor from 1 m breaks; the same egg dropped onto a thick pillow does not. Both eggs stop. In a short paragraph, explain why using impulse and momentum.

   Show answer

   Both eggs arrive with the same momentum and both end at rest, so the impulse \(\Delta p\) each receives is the same. Impulse is \(F_{\text{avg}}\,\Delta t\). The pillow deforms, so the egg takes much longer to stop; with the same \(\Delta p\) spread over a longer \(\Delta t\), the average force on the shell is smaller by the same factor. The hard floor stops the egg in a very short time, requiring a large force that exceeds what the shell can withstand. Airbags, crumple zones, and bending your knees on landing all work the same way: same impulse, longer time, smaller force.

6. Cart A (1.5 kg) moving at 4.0 m/s collides with cart B (2.5 kg) at rest, and they stick together. Find their common final velocity and the kinetic energy lost. Where did that energy go?

   Show answer

   No net external horizontal force acts on the two-cart system during the collision, so momentum is conserved: \(m_A v_A = (m_A + m_B)v_f \Rightarrow v_f = \dfrac{(1.5)(4.0)}{4.0} = 1.5\) m/s, forward. \(K_i = \tfrac{1}{2}(1.5)(4.0)^2 = 12\) J; \(K_f = \tfrac{1}{2}(4.0)(1.5)^2 = 4.5\) J. Lost: \(7.5\) J, about 62%: it became thermal energy, sound, and deformation of the couplers. Momentum cannot be "lost" this way; it has no other form to take.

7. An 80 kg astronaut floating at rest throws a 4.0 kg tool at 10 m/s. The astronaut's recoil speed is

   0.050 m/s 0.50 m/s 2.0 m/s 10 m/s

   0.050 m/s is off by a factor of ten: a misplaced decimal, or dividing the tool's 40 kg·m/s by 800 instead of 80. Check it: 80 kg at 0.050 m/s carries only 4 kg·m/s, far short of the tool's 40.

   Total momentum is zero before and after: \(0 = (4.0)(10) + (80)v \Rightarrow v = -0.50\) m/s, opposite the throw. The astronaut has 20 times the mass, so one-twentieth the speed.

   2.0 m/s comes from dividing 80 by 40 instead of 40 by 80: inverting the ratio. The heavier object must recoil slower than the tool moves, not faster.

   Equal speeds would mean equal speeds, not equal momenta; conservation of momentum requires \(m_1 v_1 = m_2 v_2\), so the 80 kg astronaut cannot move as fast as the 4.0 kg tool.

8. A 1.0 kg cart moving at 6.0 m/s collides elastically with a 2.0 kg cart at rest. Find both final velocities and verify that kinetic energy is conserved.

   Show answer

   Momentum: \((1.0)(6.0) = (1.0)v_1 + (2.0)v_2\). Energy: \(\tfrac{1}{2}(1.0)(6.0)^2 = \tfrac{1}{2}(1.0)v_1^2 + \tfrac{1}{2}(2.0)v_2^2\). Solving (substitute \(v_1 = 6.0 - 2v_2\)) gives \(v_2 = +4.0\) m/s and \(v_1 = -2.0\) m/s: the lighter cart bounces back. Check: \(p = -2.0 + 8.0 = 6.0\) kg·m/s ✓; \(K = \tfrac{1}{2}(1.0)(2.0)^2 + \tfrac{1}{2}(2.0)(4.0)^2 = 2.0 + 16 = 18\) J \(= K_i\) ✓.

9. A 1200 kg car heading east at 15 m/s collides at an intersection with an 1800 kg truck heading north at 10 m/s. The vehicles lock together. Find their velocity just after the collision and the kinetic energy lost.

   Show answer

   Conserve each component separately. East: \((1200)(15) = (3000)v_x \Rightarrow v_x = 6.0\) m/s. North: \((1800)(10) = (3000)v_y \Rightarrow v_y = 6.0\) m/s. Speed \(= \sqrt{6.0^2 + 6.0^2} = 8.5\) m/s at \(45^\circ\) north of east. \(K_i = \tfrac{1}{2}(1200)(15)^2 + \tfrac{1}{2}(1800)(10)^2 = 2.25\times10^{5}\) J; \(K_f = \tfrac{1}{2}(3000)(8.49)^2 = 1.08\times10^{5}\) J. Lost: \(1.17\times10^{5}\) J.

10. Experimental design. Two carts with spring bumpers roll on a level track; you have two motion sensors and a balance. Design a procedure to decide whether a collision between them is elastic, and state what result would show that momentum was conserved even if the collision was not elastic.

    Show answer

    Measure both masses. Place one sensor at each end of the track so each cart's velocity is recorded before and after the collision (sign matters: take one direction as positive). Compute total momentum \(m_1 v_1 + m_2 v_2\) before and after, and total kinetic energy \(\tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2\) before and after; repeat several trials. Elastic: total \(K\) after equals total \(K\) before (within uncertainty). Momentum conserved: total momentum after equals total momentum before; expected in every trial, elastic or not, because the track exerts no net horizontal force on the two-cart system. Level the track first so gravity contributes no horizontal component.

Lesson 5.1 · Unit 5 · CED topic 5.1

Rotational kinematics: angle, angular velocity, angular acceleration

A spinning wheel has no single position, velocity, or acceleration: the rim moves fast, the hub barely moves. What every point on the wheel does share is how far it has turned and how fast that angle is changing. So rotation gets its own variables, and they obey equations you already know.

Exam tip: keep \(\omega\) in rad/s inside every equation. Graders accept rpm only in a final answer, and the linking equations in the next lesson fail outright if you feed them rpm.

Lesson 5.2 · Unit 5 · CED topic 5.2

Connecting linear and rotational motion

Every point on a rigid rotating object has the same \(\omega\) and the same \(\alpha\). What differs is how far each point is from the axis, and that distance converts angular quantities into the ordinary linear ones for that point.

Common error: writing \(a_c = r\alpha\). Centripetal acceleration depends on how fast the point is moving (\(\omega\)), not on how fast it is speeding up (\(\alpha\)).

Lesson 5.3 · Unit 5 · CED topic 5.3

Torque: the rotational effect of a force

Push a door near its hinges and it barely moves; push the same way at the handle and it swings. Force alone doesn't decide rotation. What matters is the force, where it's applied, and its direction relative to the pivot: together, that's torque.

Lesson 5.4 · Unit 5 · CED topic 5.4

Rotational inertia: how mass is distributed matters

Mass measures how hard an object is to accelerate in a straight line. For rotation, the resistance depends on mass and on how far that mass sits from the axis, and distance counts twice, because it appears squared.

Quick numbers: a 2.0 kg disk of radius 0.30 m has \(I = \tfrac{1}{2}(2.0)(0.30)^2 = 0.090\) kg·m²; a hoop of the same mass and radius has 0.18 kg·m²: double, since all its mass sits at the full radius.

Lesson 5.5 · Unit 5 · CED topic 5.5

Rotational equilibrium: balancing torques

A bridge, a ladder, a shelf bracket: none of them accelerate, and none of them rotate. That's two separate conditions, and static problems need both. The trick that makes them solvable is that you get to choose the pivot.

Lesson 5.6 · Unit 5 · CED topic 5.6

Newton's second law in rotational form

Unbalanced force produces linear acceleration in proportion to mass. Unbalanced torque produces angular acceleration in proportion to rotational inertia. Once you see the analogy, the hardest rotation problems on the exam become two familiar equations linked by a string.

Unit 5 practice · 10 problems

Unit 5 practice: Torque and Rotational Dynamics

Ten problems covering the whole unit. Keep every angular speed in radians per second inside an equation, and always say which pivot you chose. Use \(g = 9.8\ \text{m/s}^2\).

1. A wheel starts from rest and reaches 1200 rpm in 5.0 s with constant angular acceleration. Find \(\alpha\) and the number of revolutions during the spin-up.

   Show answer

   Convert: \(\omega = 1200\times\dfrac{2\pi}{60} = 126\) rad/s. \(\alpha = \dfrac{\omega - \omega_0}{t} = \dfrac{125.7}{5.0} = 25\) rad/s². \(\theta = \tfrac{1}{2}\alpha t^2 = \tfrac{1}{2}(25.1)(5.0)^2 = 314\) rad \(= \dfrac{314}{2\pi} = 50\) revolutions.

2. A turntable spinning at 40 rad/s is switched off and slows at 4.0 rad/s². Through what angle does it turn before stopping, and how long does that take?

   Show answer

   Time-free equation: \(\omega^2 = \omega_0^2 + 2\alpha\,\Delta\theta \Rightarrow 0 = 40^2 + 2(-4.0)\,\Delta\theta\), so \(\Delta\theta = \dfrac{1600}{8.0} = 200\) rad ≈ 32 revolutions. Time: \(t = \dfrac{0 - 40}{-4.0} = 10\) s.

3. A wheel of radius 0.40 m turns at 15 rad/s. Point P is on the rim; point Q is 0.20 m from the axle. Find the speed and centripetal acceleration of each point, and state which of \(\omega\), \(v\), and \(a_c\) the two points share.

   Show answer

   Both points share \(\omega = 15\) rad/s (the whole wheel turns together). \(v = r\omega\): P: \((0.40)(15) = 6.0\) m/s; Q: \((0.20)(15) = 3.0\) m/s. \(a_c = r\omega^2\): P: \((0.40)(15)^2 = 90\) m/s²; Q: \((0.20)(225) = 45\) m/s². Speed and centripetal acceleration both scale with \(r\); only \(\omega\) is shared.

4. A door is 0.80 m wide. Find the torque about the hinges when you push at the outer edge with 25 N (a) perpendicular to the door, (b) at 45° to the door's surface, (c) directly toward the hinges along the door.

   Show answer

   \(\tau = rF\sin\theta\). (a) \((0.80)(25)\sin 90^\circ = 20\) N·m. (b) \((0.80)(25)\sin 45^\circ = 14\) N·m: only the perpendicular component of the push counts. (c) \(\theta = 0\), so \(\tau = 0\): a force along the line through the pivot has no lever arm and cannot rotate the door.

5. Which change, made alone, doubles the torque a person exerts on a wrench?

   doubling the force applied perpendicular to the handle at the same point applying the same force at half the distance from the bolt applying the same force at 30° to the handle instead of 90° applying the same force along the handle toward the bolt

   Torque is \(\tau = r_\perp F = rF\sin\theta\), so with \(r\) and the 90° angle fixed, doubling \(F\) doubles \(\tau\). Everything else in the expression is unchanged.

   Torque is proportional to the lever arm, so halving the distance from the bolt halves the torque. Moving the hand closer to the pivot does the opposite of what a longer wrench does.

   Only the perpendicular component of the force, \(F\sin\theta\), produces torque, and \(\sin 30^\circ = 0.5\), so this halves the torque instead of doubling it. Perpendicular (90°) is already the most effective angle.

   A force directed along the handle toward the bolt passes straight through the pivot, so its lever arm is zero and it produces no torque at all. It pushes on the bolt but does not turn it.

6. Two 2.0 kg point masses sit at the ends of a light 1.0 m rod. Find the rotational inertia about the rod's center and about one end. Explain in a sentence why they differ.

   Show answer

   \(I = \sum m_i r_i^2\). Center: each mass is 0.50 m away, \(I = 2(2.0)(0.50)^2 = 1.0\) kg·m². End: one mass at \(r = 0\), the other at 1.0 m, \(I = (2.0)(0)^2 + (2.0)(1.0)^2 = 2.0\) kg·m². Distance enters squared, so moving one mass from 0.50 m to 1.0 m adds more than moving the other from 0.50 m to 0 removes.

7. A hoop, a solid disk, and a solid sphere have the same mass \(M\) and radius \(R\). Rank their rotational inertias about their centers from largest to smallest, and explain the ranking in words without formulas.

   Show answer

   Hoop \(\gt\) disk \(\gt\) sphere: \(MR^2 \gt \tfrac{1}{2}MR^2 \gt \tfrac{2}{5}MR^2\). Rotational inertia counts each bit of mass by the square of its distance from the axis. The hoop keeps all its mass at the full radius \(R\). The disk spreads mass from the center outward, so much of it sits at small \(r\). The sphere goes further (its mass is spread in three dimensions, and slices near the poles are close to the axis), so its average \(r^2\) is smallest of the three.

8. A uniform 3.0 m plank of mass 20 kg rests on supports at both ends. A 60 kg person stands 1.0 m from the left end. Find the upward force from each support.

   Show answer

   Weights: plank \(196\) N at its center (1.5 m), person \(588\) N at 1.0 m. Pivot at the left support so \(F_L\) drops out; counterclockwise positive: \(\sum\tau = F_R(3.0) - (196)(1.5) - (588)(1.0) = 0 \Rightarrow F_R = \dfrac{294 + 588}{3.0} = 294\) N. Forces: \(F_L + F_R = 784\) N, so \(F_L = 490\) N. Check about the right end: \(F_L(3.0) = 196(1.5) + 588(2.0) = 1470 \Rightarrow F_L = 490\) N ✓. The support nearer the person carries more.

9. A 3.0 kg block hangs from a light string wrapped around a solid-disk pulley of mass 2.0 kg and radius 0.20 m on a frictionless axle. The system is released from rest. Find the block's acceleration, the tension, and the pulley's angular acceleration.

   Show answer

   Block: \(mg - T = ma\). Pulley: \(TR = I\alpha = \left(\tfrac{1}{2}MR^2\right)\dfrac{a}{R} \Rightarrow T = \tfrac{1}{2}Ma\). Combine: \(mg = \left(m + \tfrac{1}{2}M\right)a \Rightarrow a = \dfrac{(3.0)(9.8)}{3.0 + 1.0} = 7.35\) m/s² ≈ 7.4 m/s². \(T = \tfrac{1}{2}(2.0)(7.35) = 7.4\) N (less than the 29.4 N weight, as it must be). \(\alpha = \dfrac{a}{R} = \dfrac{7.35}{0.20} = 37\) rad/s².

10. Experimental design. A wheel of unknown rotational inertia spins on a low-friction axle. You have string, a set of hanging masses, a meterstick, and a motion sensor that records the falling mass's position versus time. Design a procedure to determine the wheel's rotational inertia, including the graph you would plot.

    Show answer

    Wrap the string around a hub of measured radius \(R\), hang a known mass \(m\), release from rest, and use the sensor to find the mass's acceleration \(a\) (from the slope of \(v\)–\(t\) or a fit to \(x\)–\(t\)). For the mass, \(mg - T = ma\) gives \(T = m(g - a)\). The torque on the wheel is \(\tau = TR\) and its angular acceleration is \(\alpha = a/R\). Repeat for several hanging masses and plot \(\tau\) (vertical) against \(\alpha\) (horizontal): since \(\tau = I\alpha\), the slope of the best-fit line is \(I\). A nonzero intercept would reveal a frictional torque at the axle. Keeping the string taut and vertical, and using a low-mass string, keeps the model valid.

Lesson 6.1 · Unit 6 · CED topic 6.1

Rotational kinetic energy

A spinning flywheel isn't going anywhere, yet it clearly stores energy: touch it and you'll find out. Every bit of mass in it is moving, so the object has kinetic energy even though its center of mass is at rest. Add up \(\tfrac{1}{2}mv^2\) over all the pieces and a familiar shape appears.

Lesson 6.2 · Unit 6 · CED topic 6.2

Torque and work: energy transferred by rotation

A force does work when it pushes something through a distance. A torque does work when it turns something through an angle. The work-energy theorem carries over unchanged: net work by torques equals the change in rotational kinetic energy.

Lesson 6.3 · Unit 6 · CED topic 6.3

Angular momentum and angular impulse

Linear momentum \(p = mv\) measures "how much motion" an object carries in a line, and a force acting over time changes it. Rotation has the same structure: angular momentum measures how much rotation an object carries, and torque acting over time changes it.

Lesson 6.4 · Unit 6 · CED topic 6.4

Conservation of angular momentum

If no net external torque acts on a system, nothing can change its angular momentum. The system can rearrange itself (pull mass inward, absorb a collision, spread out), and \(\omega\) will adjust so that \(L\) stays put. This is the principle behind figure skaters, collapsing stars, and cats landing on their feet.

Lesson 6.5 · Unit 6 · CED topic 6.5

Rolling without slipping

A rolling wheel is doing two things at once: its center moves forward, and it spins about that center. "Without slipping" ties the two motions together with one condition, and that single condition makes rolling problems solvable with energy alone.

Exam tip: if a problem says the surface is frictionless, the object cannot start rolling: it slides, \(\omega\) stays constant, and only translational kinetic energy trades with \(U_g\).

Lesson 6.6 · Unit 6 · CED topic 6.6

Orbits and satellites

A satellite in circular orbit is in free fall: it just moves sideways fast enough that the ground curves away beneath it. Gravity supplies exactly the centripetal force the circle requires, and that one equation fixes the speed for any given radius.

Unit 6 practice · 10 problems

Unit 6 practice: Energy and Momentum of Rotating Systems

Ten problems covering the whole unit. Write \(I = \beta MR^2\) for the given shape, keep \(\omega\) in rad/s, and remember that rolling without slipping means \(v_{\text{cm}} = R\omega\). Use \(g = 9.8\ \text{m/s}^2\), \(G = 6.67\times10^{-11}\) N·m²/kg², \(M_E = 5.97\times10^{24}\) kg, and \(R_E = 6.37\times10^{6}\) m.

1. A solid-disk flywheel of mass 20 kg and radius 0.40 m spins at 1500 rpm. How much kinetic energy does it store?

   Show answer

   \(I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(20)(0.40)^2 = 1.6\) kg·m². \(\omega = 1500\times\dfrac{2\pi}{60} = 157\) rad/s. \(K_{\text{rot}} = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(1.6)(157.1)^2 = 2.0\times10^{4}\) J (19.7 kJ).

2. A solid sphere, a solid disk, a hoop, and a frictionless sliding block are released from rest at the top of a 3.0 m high ramp; the round objects roll without slipping. Rank their speeds at the bottom from fastest to slowest, compute each, and explain the ranking in words.

   Show answer

   \(mgh = \tfrac{1}{2}mv^2(1 + \beta) \Rightarrow v = \sqrt{\dfrac{2gh}{1 + \beta}}\) with \(2gh = 58.8\) m²/s². Block (\(\beta = 0\)): 7.7 m/s. Sphere (\(\beta = \tfrac{2}{5}\)): \(\sqrt{58.8/1.4} = 6.5\) m/s. Disk (\(\beta = \tfrac{1}{2}\)): \(\sqrt{58.8/1.5} = 6.3\) m/s. Hoop (\(\beta = 1\)): \(\sqrt{58.8/2} = 5.4\) m/s. Ranking: block \(\gt\) sphere \(\gt\) disk \(\gt\) hoop. Every object converts the same \(mgh\), but a rolling object must split it between translation and rotation. The larger the share of mass far from the axis (larger \(\beta\)), the more energy goes into spinning and the less into moving forward. Mass and radius cancel: a big hoop and a small hoop tie.

3. A hoop rolls without slipping along level ground at 4.0 m/s and reaches a hill. How high up the hill does it rise? Compare with a frictionless block sliding at the same speed.

   Show answer

   All kinetic energy becomes \(U_g\): \(\tfrac{1}{2}mv^2 + \tfrac{1}{2}(mR^2)\left(\dfrac{v}{R}\right)^2 = mv^2 = mgh\), so \(h = \dfrac{v^2}{g} = \dfrac{16}{9.8} = 1.6\) m. A sliding block reaches only \(\dfrac{v^2}{2g} = 0.82\) m. The hoop's rotational kinetic energy equals its translational kinetic energy, and both convert to height because static friction does no work in pure rolling.

4. Experimental design. You are handed a metal ball and told it is either solid or hollow, but you may not cut it open. You have an inclined ramp, a meterstick, and a stopwatch. Design a procedure to decide, and state the prediction you would test.

   Show answer

   Set the ramp at a measured angle \(\theta\) (from its height and length) and release the ball from rest. Time how long it takes to roll a measured distance \(L\); repeat several times and average. Since \(L = \tfrac{1}{2}at^2\), \(a = 2L/t^2\). Energy conservation for rolling without slipping predicts \(a = \dfrac{g\sin\theta}{1 + \beta}\): solid sphere (\(\beta = \tfrac{2}{5}\)) gives \(\tfrac{5}{7}g\sin\theta\); thin hollow sphere (\(\beta = \tfrac{2}{3}\)) gives \(\tfrac{3}{5}g\sin\theta\). At \(\theta = 20^\circ\), for example, that is 2.39 m/s² versus 2.01 m/s²: a 16% difference, easily resolved with a long ramp and repeated trials. Compare the measured \(a\) with both predictions; use a gentle slope so the ball rolls rather than slips.

5. A motor applies a constant 15 N·m torque to a wheel that turns through 20 revolutions in 4.0 s. Find the work done and the average power.

   Show answer

   \(\Delta\theta = 20\times 2\pi = 126\) rad. \(W = \tau\,\Delta\theta = (15)(125.7) = 1.9\times10^{3}\) J. \(P = \dfrac{W}{\Delta t} = \dfrac{1885}{4.0} = 470\) W. Radians are dimensionless, so N·m times rad gives joules.

6. Find the angular momentum of (a) a disk with \(I = 0.30\) kg·m² turning at 12 rad/s, and (b) a 0.20 kg puck sliding in a straight line at 5.0 m/s along a path that passes 2.0 m from point O, about point O. Does the puck's angular momentum change as it moves?

   Show answer

   (a) \(L = I\omega = (0.30)(12) = 3.6\) kg·m²/s. (b) \(L = mv\,r_\perp = (0.20)(5.0)(2.0) = 2.0\) kg·m²/s. No: the perpendicular distance from O to the line of motion stays 2.0 m as the puck moves, so \(L\) about O is constant (no torque acts about O).

7. A torque on a wheel (\(I = 0.35\) kg·m², initially at rest) is graphed against time: constant at 2.0 N·m for 3.0 s, then decreasing in a straight line to zero over the next 1.0 s. Find the wheel's final angular speed.

   Show answer

   Angular impulse is the area under \(\tau\)–\(t\): rectangle \((2.0)(3.0) = 6.0\) plus triangle \(\tfrac{1}{2}(2.0)(1.0) = 1.0\), so \(\Delta L = 7.0\) kg·m²/s. \(L = I\omega \Rightarrow \omega = \dfrac{7.0}{0.35} = 20\) rad/s.

8. A merry-go-round is a solid disk of mass 150 kg and radius 2.0 m turning at 1.5 rad/s. A 40 kg child, initially at rest, drops onto the rim and holds on. Find the new angular speed and the kinetic energy lost. Why is angular momentum conserved but not kinetic energy?

   Show answer

   \(I_{\text{disk}} = \tfrac{1}{2}(150)(2.0)^2 = 300\) kg·m²; child at rim adds \(mr^2 = (40)(2.0)^2 = 160\) kg·m². The axle exerts no torque and the child–platform friction is internal, so \(L_i = L_f\): \((300)(1.5) = (460)\omega_f \Rightarrow \omega_f = 0.98\) rad/s. \(K_i = \tfrac{1}{2}(300)(1.5)^2 = 338\) J; \(K_f = \tfrac{1}{2}(460)(0.978)^2 = 220\) J; about 120 J became thermal energy as the child's shoes skidded up to speed. Angular momentum is conserved because no external torque acts; kinetic energy is not, because the friction that brings the child up to speed does negative net work: a perfectly inelastic rotational collision.

9. A spinning figure skater pulls her arms in and spins faster. Which quantity stays constant during this motion?

   her angular speed her kinetic energy her angular momentum her rotational inertia

   Her angular speed is exactly what changes: the problem says she spins faster. Because \(L = I\omega\) is fixed, lowering \(I\) forces \(\omega\) to rise.

   Her rotational kinetic energy \(K = L^2/2I\) increases as \(I\) drops; the extra energy comes from the work her muscles do pulling her arms inward. Energy is conserved overall, but not as rotational kinetic energy alone.

   The ice exerts negligible torque about her spin axis, so her angular momentum \(L = I\omega\) is conserved; pulling her arms in lowers \(I\), so \(\omega\) must rise to keep \(L\) the same.

   Her rotational inertia is what she deliberately changes: bringing mass closer to the axis reduces \(I = \sum mr^2\). Its decrease is the cause of the faster spin, not something held constant.

10. A satellite is in a circular orbit at a radius of two Earth radii, \(r = 2R_E = 1.27\times10^{7}\) m. Find its orbital speed and period. Does the answer depend on the satellite's mass?

    Show answer

    Gravity is the centripetal force: \(\dfrac{GM_E m}{r^2} = \dfrac{mv^2}{r} \Rightarrow v = \sqrt{\dfrac{GM_E}{r}} = \sqrt{\dfrac{(6.67\times10^{-11})(5.97\times10^{24})}{1.274\times10^{7}}} = \sqrt{3.13\times10^{7}} = 5.6\times10^{3}\) m/s. \(T = \dfrac{2\pi r}{v} = \dfrac{2\pi(1.274\times10^{7})}{5.59\times10^{3}} = 1.43\times10^{4}\) s ≈ 4.0 h. The satellite's mass canceled; every object at that radius orbits at that speed.

Lesson 7.1 · Unit 7 · CED topic 7.1

Defining simple harmonic motion

Pull a mass on a spring and let go, or nudge a pendulum, and it swings back and forth. The motion repeats because whenever the object leaves equilibrium, a force pulls it back: and the farther it strays, the harder the pull. That specific kind of force produces a specific kind of motion.

Lesson 7.2 · Unit 7 · CED topic 7.2

Period of a mass-spring system and a pendulum

The period of an oscillator is set by how strongly it's pulled back (stiffness) and how much it resists being moved (inertia). Two formulas cover almost every AP problem, and the exam cares as much about what is missing from them as what's in them.

On the Moon (\(g = 1.6\) m/s²), a pendulum clock runs slow by a factor of \(\sqrt{9.8/1.6} \approx 2.5\); a mass-spring clock keeps perfect time.

Lesson 7.3 · Unit 7 · CED topic 7.3

Representing simple harmonic motion

SHM traces a cosine (or sine) in time. Once you can write the position function from a graph, the velocity and acceleration graphs follow from one fact: velocity is the slope of position, and acceleration is the slope of velocity.

Lesson 7.4 · Unit 7 · CED topic 7.4

Energy of a simple harmonic oscillator

An ideal oscillator never gains or loses mechanical energy; it just shuffles it between kinetic and potential twice per cycle. That makes energy the fastest route to "how fast is it moving when it's here?", no trig, no time, just a bookkeeping equation.

Unit 7 practice · 10 problems

Unit 7 practice: Oscillations

Ten problems covering the whole unit. Watch for what the period does not depend on, and let energy do the work whenever a problem asks for speed at a position. Use \(g = 9.8\ \text{m/s}^2\).

1. A 0.80 kg block on a spring with \(k = 320\) N/m is pulled 0.050 m from equilibrium and released. Find the maximum force on the block and its maximum acceleration, and state where in the motion each occurs and where the speed is greatest.

   Show answer

   \(|F_{\max}| = kA = (320)(0.050) = 16\) N and \(|a_{\max}| = \dfrac{kA}{m} = \dfrac{16}{0.80} = 20\) m/s², both at the turning points \(x = \pm 0.050\) m, directed toward equilibrium. The speed is greatest at \(x = 0\), where the force and acceleration are zero.

2. A mass on a spring oscillates with period \(T\). If the amplitude is doubled, the new period is

   \(2T\) \(T/2\) \(T\) \(4T\)

   Doubling the period assumes the mass travels twice the distance at the same speed. But a larger amplitude also means a larger restoring force and a larger maximum speed, and for a Hooke's-law spring the two effects cancel exactly.

   A halved period would require the oscillator to speed up by more than the distance grows. In SHM the speed scales with amplitude exactly in step with the distance, so the period neither halves nor doubles.

   \(T = 2\pi\sqrt{m/k}\) contains no amplitude. A larger swing covers more distance, but the restoring force, and therefore the speed, is proportionally larger, so the round trip takes the same time; amplitude independence is a defining property of SHM.

   \(4T\) would correspond to a quantity that scales with amplitude squared, like the total energy \(\tfrac{1}{2}kA^2\). The energy does quadruple when the amplitude doubles, but the period is unaffected.

3. A 0.50 kg mass hangs from a spring with \(k = 80\) N/m. Find the period and frequency. What happens to the period if the spring is replaced by one four times as stiff?

   Show answer

   \(T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{0.50}{80}} = 2\pi\sqrt{0.00625} = 0.50\) s (0.497 s); \(f = \dfrac{1}{T} = 2.0\) Hz. Quadrupling \(k\) halves \(T\) to 0.25 s, since \(T \propto 1/\sqrt{k}\).

4. Find the period of a 0.75 m pendulum on Earth and on the Moon (\(g = 1.6\) m/s²). Would a 0.75 m pendulum with a heavier bob have a different period?

   Show answer

   \(T = 2\pi\sqrt{\dfrac{L}{g}}\). Earth: \(2\pi\sqrt{\dfrac{0.75}{9.8}} = 1.7\) s. Moon: \(2\pi\sqrt{\dfrac{0.75}{1.6}} = 4.3\) s: weaker gravity, slower swing. Bob mass does not appear: a heavier bob feels proportionally more gravitational force and has proportionally more inertia, so its period is unchanged.

5. Four mass–spring systems: W has \(m = 1\) kg, \(k = 100\) N/m; X has \(m = 2\) kg, \(k = 100\) N/m; Y has \(m = 1\) kg, \(k = 400\) N/m; Z has \(m = 4\) kg, \(k = 400\) N/m. Rank their periods from longest to shortest, and identify any ties.

   Show answer

   \(T \propto \sqrt{m/k}\): W: \(\sqrt{1/100} = 0.10\); X: \(\sqrt{2/100} = 0.141\); Y: \(\sqrt{1/400} = 0.050\); Z: \(\sqrt{4/400} = 0.10\). Ranking: \(X \gt W = Z \gt Y\). W and Z tie because quadrupling both \(m\) and \(k\) leaves the ratio unchanged (both periods are 0.63 s; X is 0.89 s and Y is 0.31 s).

6. A position–time graph for a block on a spring shows maxima of \(+0.15\) m at \(t = 0\) and at \(t = 2.0\) s, with minima of \(-0.15\) m halfway between. Write \(x(t)\), find the maximum speed, and give the block's position, velocity, and acceleration at \(t = 0.50\) s.

   Show answer

   \(A = 0.15\) m (crest height), \(T = 2.0\) s (crest to crest). Starting at a maximum: \(x(t) = (0.15\ \text{m})\cos\!\left(\dfrac{2\pi t}{2.0\ \text{s}}\right) = 0.15\cos(\pi t)\). \(v_{\max} = \dfrac{2\pi A}{T} = \dfrac{2\pi(0.15)}{2.0} = 0.47\) m/s. At \(t = 0.50\) s \(= T/4\): \(x = 0.15\cos(\pi/2) = 0\); the block passes through equilibrium heading toward \(-x\), so \(v = -0.47\) m/s and \(a = 0\).

7. A student says: "At the equilibrium position the net force on the block is zero, so the block should stop there." Explain the error in a short paragraph.

   Show answer

   Zero net force means zero acceleration, the velocity is not changing at that instant, not zero velocity. Arriving at equilibrium the block has been accelerating toward it the whole way, so it is moving at its maximum speed exactly there. With no force to slow it at that instant, it simply keeps going, overshoots, and only then does the spring force build up (in the opposite direction) to slow it and turn it around at \(x = -A\). Newton's first law: an object with zero net force keeps its velocity, whatever that velocity is.

8. A 0.40 kg block on a spring with \(k = 200\) N/m oscillates with amplitude 0.060 m. Find the total energy, the maximum speed, and the speed when \(x = 0.030\) m. Is the speed at half amplitude half the maximum speed?

   Show answer

   \(E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(200)(0.060)^2 = 0.36\) J. At \(x = 0\): \(\tfrac{1}{2}mv_{\max}^2 = E \Rightarrow v_{\max} = \sqrt{\dfrac{2(0.36)}{0.40}} = 1.3\) m/s (1.34 m/s). At \(x = 0.030\) m: \(U_s = \tfrac{1}{2}(200)(0.030)^2 = 0.090\) J, so \(K = 0.36 - 0.090 = 0.27\) J and \(v = \sqrt{\dfrac{2(0.27)}{0.40}} = 1.2\) m/s (1.16 m/s). No: it is \(\sqrt{3}/2 \approx 0.87\) of \(v_{\max}\), because energy depends on the square of position.

9. (a) A pendulum of length 1.5 m is released from rest at 40° from the vertical. Find its speed at the lowest point. (b) For the oscillator in the previous problem, at what positions are the kinetic and potential energies equal?

   Show answer

   (a) Height drop \(h = L(1 - \cos 40^\circ) = (1.5)(1 - 0.766) = 0.35\) m. Tension does no work (perpendicular to the motion), so \(mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2(9.8)(0.351)} = 2.6\) m/s. (b) \(\tfrac{1}{2}kx^2 = \tfrac{1}{2}E = \tfrac{1}{4}kA^2 \Rightarrow x = \pm\dfrac{A}{\sqrt{2}} = \pm\dfrac{0.060}{1.414} = \pm 0.042\) m: closer to the ends than to the middle.

10. Experimental design. You have a spring of unknown constant, a set of known masses, and a stopwatch. Design a procedure to measure \(k\) using oscillations, including the graph you would plot and how \(k\) comes from it. Why time many oscillations rather than one?

    Show answer

    Hang each known mass \(m\) from the spring, displace it a small distance, release, and time 10 or 20 complete oscillations; divide by the count to get \(T\). Repeat and average. Squaring \(T = 2\pi\sqrt{m/k}\) gives \(T^2 = \dfrac{4\pi^2}{k}m\), so plot \(T^2\) (vertical) against \(m\) (horizontal): a straight line through the origin with slope \(4\pi^2/k\), so \(k = \dfrac{4\pi^2}{\text{slope}}\) (for \(k = 80\) N/m the slope would be 0.49 s²/kg). Timing many cycles divides the reaction-time error by the number of cycles; amplitude does not affect \(T\), so it need not be controlled precisely.

Lesson 8.1 · Unit 8 · CED topic 8.1

Density and internal structure

A bowling ball and a beach ball can be the same size, and a gold ring and a brass ring can have the same mass. What separates them is how much matter is packed into each cubic meter. Density is the property fluids care about: it decides what floats, what sinks, and how pressure builds with depth.

Why density and not mass decides floating: an object placed in a fluid pushes fluid aside, and the fluid pushes back with a force equal to the weight of what was displaced. If the object's average density is less than the fluid's, the displaced fluid outweighs the object and it floats. That's how a steel ship floats, its hull encloses so much air that the ship's average density is well below water's, while a steel bolt, with nothing but steel inside, sinks. Ice floats because water expands about 9% when it freezes, dropping its density to 917 kg/m³.

Lesson 8.2 · Unit 8 · CED topic 8.2

Pressure, and how it grows with depth

Fluids push on every surface they touch, and they push perpendicular to the surface. The push per unit area is pressure. Dive deeper and there's more fluid stacked above you, so the pressure rises: linearly, and by an amount that depends only on depth and density, not on the shape or size of the container.

Handy benchmark: every 10.3 m of fresh water adds one atmosphere (\(1.01 \times 10^5 / (1000 \times 9.8) = 10.3\) m). At that depth absolute pressure has doubled.

Lesson 8.3 · Unit 8 · CED topic 8.3

Buoyancy and Archimedes' principle

Pressure increases with depth, so a submerged object is pushed harder on its bottom than on its top. The imbalance is an upward force, and Archimedes found its size with a single sentence: the fluid pushes up with exactly the weight of the fluid the object has shoved aside.

Lesson 8.4 · Unit 8 · CED topic 8.4

Fluid flow: continuity and Bernoulli's equation

Squeeze the end of a garden hose and the water speeds up; the same volume has to pass through a smaller opening each second. Then the moving water has more kinetic energy than before, and that energy came from somewhere: the pressure dropped. Two equations capture these ideas for an ideal fluid (incompressible, no viscosity, steady flow).

Exam tip: state your assumptions (incompressible, negligible viscosity, \(v \approx 0\) at a wide surface, equal pressures at points open to air). Graders award points for naming them.

Unit 8 practice · 10 problems

Unit 8 practice: Fluids

Ten problems covering the whole unit. State whether a pressure is gauge or absolute, and always name the assumptions behind Bernoulli's equation. Use \(g = 9.8\ \text{m/s}^2\), \(\rho_{\text{water}} = 1000\) kg/m³, and \(P_0 = 1.01\times10^{5}\) Pa.

1. A rectangular block measures 0.20 m × 0.10 m × 0.050 m and has mass 2.7 kg. Find its density and identify the likely material (aluminum 2700, iron 7870, gold 19 300 kg/m³). Would cutting the block in half change its density?

   Show answer

   \(V = (0.20)(0.10)(0.050) = 1.0\times10^{-3}\) m³. \(\rho = \dfrac{m}{V} = \dfrac{2.7\ \text{kg}}{1.0\times10^{-3}\ \text{m}^3} = 2700\) kg/m³: aluminum. Cutting it in half halves both \(m\) and \(V\); density is a property of the material, not the amount, and stays 2700 kg/m³.

2. A cargo ship floats in a freshwater river and then sails into the ocean, where the water is denser. Compared with the river, in the ocean the buoyant force on the ship is

   larger smaller the same, and the ship rides higher the same, and the ship rides lower

   It is tempting to say denser water pushes harder, but a floating object's buoyant force is fixed by its weight: \(F_b = mg\) in both cases. If the buoyant force were larger, the ship would accelerate upward until it wasn't.

   A smaller buoyant force would leave the ship's weight unbalanced, and it would sink lower until \(F_b\) matched \(mg\) again. For any floating object the buoyant force equals the weight, whatever the fluid.

   Floating means \(F_b = mg\), and the ship's weight has not changed, so the buoyant force is the same. Since \(F_b = \rho_{\text{fluid}}V_{\text{sub}}g\) and \(\rho\) is larger in seawater, the displaced volume must be smaller: the ship rides higher.

   The buoyant force is indeed the same, but riding lower would mean displacing more water, and more of a denser fluid would give a buoyant force larger than the weight. A denser fluid needs less displaced volume, so the ship rises.

3. A tank of water is 4.0 m deep and open to the air. Find the gauge pressure and the absolute pressure at the bottom, and the net force on a 0.10 m × 0.10 m window in the bottom if the air outside the window is at atmospheric pressure.

   Show answer

   Gauge: \(\rho g h = (1000)(9.8)(4.0) = 3.9\times10^{4}\) Pa. Absolute: \(P_0 + \rho g h = 1.01\times10^{5} + 3.92\times10^{4} = 1.4\times10^{5}\) Pa. Atmospheric pressure pushes on both sides of the window, so only the gauge pressure produces a net force: \(F = P_{\text{gauge}}A = (3.92\times10^{4})(0.010) = 390\) N.

4. Three open containers (a narrow cylinder, a wide cylinder, and a cone with its wide end up) are each filled with fresh water to a depth of 0.50 m. A fourth, identical to the narrow cylinder, holds oil (\(\rho = 800\) kg/m³) to the same depth. Rank the gauge pressures at the bottom of the four containers and explain.

   Show answer

   Water containers tie; oil is lowest: \(P_{\text{narrow}} = P_{\text{wide}} = P_{\text{cone}} \gt P_{\text{oil}}\). Gauge pressure is \(\rho g h\); it depends only on depth and fluid density, not on the container's shape or the total amount of fluid. Water: \((1000)(9.8)(0.50) = 4900\) Pa in all three. Oil: \((800)(9.8)(0.50) = 3900\) Pa. The wide container holds far more water but the same column height presses on each square meter of its bottom.

5. A block of wood with density 750 kg/m³ floats in fresh water. What fraction of its volume is submerged? What fraction if it floats in seawater (\(\rho = 1025\) kg/m³)?

   Show answer

   Floating: \(F_b = mg \Rightarrow \rho_{\text{fluid}}V_{\text{sub}}g = \rho_{\text{wood}}Vg\), so \(\dfrac{V_{\text{sub}}}{V} = \dfrac{\rho_{\text{wood}}}{\rho_{\text{fluid}}}\). Fresh water: \(\dfrac{750}{1000} = 0.75\) (75% under). Seawater: \(\dfrac{750}{1025} = 0.73\) (73% under): denser fluid, less of the block submerged.

6. A 3.0 kg iron block (\(\rho = 7870\) kg/m³) hangs from a string, fully submerged in water. Find the buoyant force and the tension in the string. What fraction of its weight does the block appear to lose?

   Show answer

   \(V = \dfrac{m}{\rho} = \dfrac{3.0}{7870} = 3.8\times10^{-4}\) m³. \(F_b = \rho_{\text{water}}Vg = (1000)(3.81\times10^{-4})(9.8) = 3.7\) N. Equilibrium: \(T + F_b = mg \Rightarrow T = 29.4 - 3.74 = 25.7\) N ≈ 26 N. Fraction lost \(= \dfrac{F_b}{mg} = \dfrac{\rho_{\text{water}}}{\rho_{\text{iron}}} = \dfrac{1000}{7870} = 0.13\), about 13%.

7. A student holds an inflated beach ball completely under water and lets go. It accelerates upward, then pops partly out and settles floating. In a short paragraph, explain the motion using forces.

   Show answer

   Fully submerged, the buoyant force equals the weight of a ball-sized volume of water, \(\rho_{\text{water}}Vg\), while the ball's own weight is tiny because its average density (mostly air) is far below water's. The net force is upward and large, so the ball accelerates upward: \(a = (F_b - mg)/m\). As the ball rises through the surface, less of it is submerged, \(V_{\text{sub}}\) shrinks, and \(F_b\) decreases. It overshoots (it has upward momentum), then oscillates and settles where \(F_b = mg\), with only a small fraction submerged, equal to the ratio of the ball's average density to the water's.

8. Water flows at 1.5 m/s through a pipe of diameter 6.0 cm that narrows to a diameter of 2.0 cm. Find the speed in the narrow section and the volume flow rate.

   Show answer

   Continuity \(A_1v_1 = A_2v_2\), and area scales with diameter squared: \(v_2 = v_1\left(\dfrac{d_1}{d_2}\right)^2 = (1.5)\left(\dfrac{6.0}{2.0}\right)^2 = 1.5\times 9 = 14\) m/s (13.5 m/s). Flow rate: \(Q = A_1v_1 = \pi(0.030)^2(1.5) = 4.2\times10^{-3}\) m³/s ≈ 4.2 L/s, the same in both sections.

9. In a horizontal pipe, water at \(2.20\times10^{5}\) Pa moves at 3.0 m/s. Downstream the pipe's cross-sectional area is one-third as large. Find the speed and pressure there, and explain in a sentence why the pressure drops.

   Show answer

   Continuity: \(v_2 = v_1\dfrac{A_1}{A_2} = 3(3.0) = 9.0\) m/s. Bernoulli with \(y_1 = y_2\): \(P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 - v_2^2) = 2.20\times10^{5} + \tfrac{1}{2}(1000)(9.0 - 81) = 2.20\times10^{5} - 3.6\times10^{4} = 1.84\times10^{5}\) Pa. The fluid speeds up entering the narrow section, so a net force must push it forward: the pressure behind must exceed the pressure ahead. Bernoulli is energy per volume: kinetic energy per volume went up, so pressure went down.

10. (a) A large open tank has a small hole 1.8 m below the water surface. Find the speed of the water leaving the hole, stating your assumptions. (b) Experimental design: the hole is a height \(H\) above the floor. Describe how you could test \(v = \sqrt{2gh}\) by varying the water depth \(h\) and measuring where the jet lands, and what graph should be a straight line.

    Show answer

    (a) Bernoulli between the surface (1) and the hole (2): both open to air so \(P_1 = P_2\); the tank is wide so \(v_1 \approx 0\); the fluid is incompressible with negligible viscosity. Then \(\rho g h = \tfrac{1}{2}\rho v^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2(9.8)(1.8)} = 5.9\) m/s. (b) The jet leaves horizontally and falls \(H\) in time \(t = \sqrt{2H/g}\), landing at \(x = vt = \sqrt{2gh}\sqrt{2H/g} = 2\sqrt{hH}\). Fill the tank to several depths \(h\), measure the landing distance \(x\) for each (refilling to keep \(h\) steady during the measurement), and plot \(x^2\) against \(h\): the prediction is a straight line through the origin with slope \(4H\). For \(H = 0.50\) m and \(h = 1.8\) m the jet should land 1.9 m out.

Unit recap

Unit recap

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