Lesson 1.1 · Unit 1 · HS-PS2-1
A number on its own is not a physical quantity
Physics makes claims that can be checked, and a claim can only be checked if it says how much of what. "The force was 40" is not a statement anyone can test. Getting the unit right is not bookkeeping that comes after the physics; it is part of the answer, and this lesson shows how to use units as a check on work you are not sure about.
- A physical quantity is a number together with a unit. Either one alone is incomplete.
- SI is built from seven base units. For this course the ones that matter are the meter, the kilogram, the second, the ampere and the kelvin.
- Every other unit is derived from those by an equation. The newton is not a fundamental thing: it is \( \text{kg}\cdot\text{m}/\text{s}^2 \), read straight off \( F = ma \).
- Both sides of a correct equation have the same units. This is dimensional analysis, and it catches a wrong formula without your knowing the right one.
- Convert by multiplying by a fraction equal to one. Write the conversion so the unwanted unit cancels, and the arithmetic cannot go the wrong way.
- Significant figures state how precisely a quantity is known. Writing more digits than the data support is a claim you cannot back.
- A result is no more precise than the least precise input. For multiplication and division, keep the smallest number of significant figures among the inputs.
Where students lose marks: leaving a mixed unit in the middle of a calculation. Kilometers and meters in the same expression, or grams where kilograms are required, produce an answer wrong by a factor of a thousand while looking perfectly reasonable on the page. Convert everything to SI base units before the first line of arithmetic, every time.
The problem. (a) Express the newton and the joule in base units. (b) A student proposes \( v = \sqrt{2gh} \) for the speed of a falling object. Check it by units. (c) A car travels 90.0 km/h. Convert to m/s. (d) Explain how a unit mismatch destroyed a spacecraft.
Step one: derive the newton for (a). The definition comes from \( F = ma \), where \( m \) is in kilograms and \( a \) in \( \text{m}/\text{s}^2 \): \[ 1\ \text{N} = 1\ \text{kg} \cdot \text{m}/\text{s}^2 \] So a newton is not an extra fact to memorize. It is a shorthand for a combination the equation already fixed.
Step two: derive the joule. Work is force times distance, so \[ 1\ \text{J} = 1\ \text{N} \cdot \text{m} = 1\ \text{kg} \cdot \text{m}^2/\text{s}^2 \] Every derived unit unpacks this way, and being able to unpack one is what makes the next step possible.
Step three: set up the check in (b). Take the units of each symbol: \( g \) is \( \text{m}/\text{s}^2 \) and \( h \) is \( \text{m} \). The 2 is a pure number and has no units, so it cannot affect the check. Inside the root: \( \dfrac{\text{m}}{\text{s}^2} \times \text{m} = \dfrac{\text{m}^2}{\text{s}^2} \).
Step four: finish (b). Taking the square root: \( \sqrt{\dfrac{\text{m}^2}{\text{s}^2}} = \dfrac{\text{m}}{\text{s}} \), which is a speed. The formula passes. What this check can and cannot do. It would have caught \( v = \sqrt{2gh^2} \) or \( v = 2gh \) at once. It cannot catch a wrong pure number: the check passes equally for \( \sqrt{2gh} \) and \( \sqrt{7gh} \), because 2 and 7 have no units. So dimensional analysis rules out wrong formulas rather than confirming right ones, which is still worth a great deal when you are unsure.
Step five: convert in (c). Build fractions that equal one and arrange them so the unwanted units cancel: \[ 90.0\ \frac{\text{km}}{\text{h}} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}} \] The kilometers cancel against the kilometers and the hours against the hours, leaving meters over seconds, which is what was wanted.
Step six: compute (c). \( \dfrac{90.0 \times 1000}{3600} = \dfrac{90000}{3600} = 25.0\ \text{m/s} \). Worth memorizing as a check: dividing km/h by 3.6 gives m/s, so 90 becomes 25 and 36 becomes 10. If a speed in m/s comes out larger than the same speed in km/h, the conversion went the wrong way.
Step seven: answer (d). In September 1999 NASA lost the Mars Climate Orbiter as it arrived at Mars. The spacecraft had traveled for nine and a half months and was destroyed in the upper atmosphere. The mishap investigation board found that one piece of ground software reported thruster impulse in pound-force seconds, while the navigation software receiving it expected newton seconds. Neither number was wrong for its own unit.
Step eight: state what the failure shows. One pound-force is about 4.45 newtons, so every impulse figure was too small by roughly that factor. The error accumulated over the cruise, and the orbiter arrived far lower than intended. Nothing about the physics was wrong. Every equation was right and every calculation was arithmetically correct. A quantity crossed a boundary without its unit, and that alone was sufficient. The habit this should build. Write the unit beside every number on every line, including intermediate ones. It costs a few seconds and it is the only thing that catches this class of error, because the numbers themselves always look plausible.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation.
- Give the SI base unit of mass.
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The kilogram
- Express the newton in base units.
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From \( F = ma \). \( \text{kg}\cdot\text{m}/\text{s}^2 \)
- Convert 2.5 km to meters.
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2500 m
- 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
- Convert 72 km/h to m/s.
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Divide by 3.6. 20 m/s
- 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
- 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. \( \text{J} = \text{kg}\cdot\text{m}^2/\text{s}^2 \), \( \text{W} = \text{kg}\cdot\text{m}^2/\text{s}^3 \)
- 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
- 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
- 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