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Puzzled.

P3 Energy and momentum

solve motion problems with conservation of energy and momentum instead of forces.

Before you begin: P2 · Forces and Newton's laws, M3c · Vectors and components.

Videos coming soon

Three videos (part 1: work and energy; part 2: conservation of energy and power; part 3: momentum, collisions and rockets) and a worked-example video (the skateboarder) are made, and will appear here once they're uploaded to YouTube.

Chapters: part 1, work and energy (20 minutes)
Chapters: part 2, conservation of energy and power (16 minutes)
Chapters: part 3, momentum, collisions and rockets (19 minutes)
Chapters: worked example, the skateboarder (5 minutes)

144 steps about 113 min

Watch first

Notes available, video soon.

Notes

In this chapter

  • Calculate work, \(W = Fd\cos\theta\), as a dot product, and say when a force does no work or negative work.
  • Use kinetic energy, \(\tfrac12 mv^2\), the work–energy theorem, and potential energy (\(mgh\) and \(\tfrac12 kx^2\)) to find speeds without knowing the path.
  • Keep an energy ledger: friction turns kinetic energy into thermal energy, and the total never changes.
  • Work out power (\(P = W/t = Fv\)), and use watts, kilowatt-hours and efficiency.
  • Use momentum, \(p = mv\), and impulse, \(F\,\Delta t = \Delta p\), to explain airbags, recoil and rockets.
  • Solve one-dimensional collisions, elastic and inelastic, with conservation of momentum (and of energy where it applies).
Part Energy and work

Why energy?

FIRST, MAKE A PREDICTION

Two frictionless tracks start at the same height and end at the same lower height. Identical cars are released from rest. Which has the greater speed at the end?

Show the answer

Their final speeds are equal

With the same height drop and no energy losses, both lose the same potential energy per unit mass and gain the same speed.

TRY IT / FOLLOW THE REASONING

Lower the car from 10 m towards the bottom. Track potential energy, kinetic energy and speed, then change the mass while keeping the same height drop.

Follow a frictionless descent

Current height: 10 m · Mass: 2 kg

Potential energy
196 J
Kinetic energy
0 J
Speed
0 m/s
Total energy
196 J
Model assumptions

Released from rest at 10 m. No friction or air resistance; g = 9.8 m/s². Heights are above the bottom.

MAKE THE CONNECTION

Key idea

Energy needs only the start and the end of a motion, not the shape of the path between them.

A roller-coaster car is let go at the top of a hill. How fast is it going at the bottom? With forces it is a nightmare: the track bends, so the push from the track changes direction at every point. Energy needs only the start and the end, and not the shape of the track. This chapter uses two tools from earlier: \(F = ma\) and forces as vectors (P2), and components with \(\cos\theta\) (M3c).

Checkpoint: Why energy? (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A bead slides without friction down a bent wire from rest, ending 0.8 m below where it started. What do you need to know to find its speed at the bottom using energy?

Show the answer

Only the height it drops, 0.8 m.

The wire does no work, so only the drop in height matters: \(mgh = \tfrac12 mv^2\) gives \(v = \sqrt{2gh}\), whatever the shape.

Checkpoint: Why energy? (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A stone is dropped from rest from a 10 m high bridge. Ignoring air resistance, how fast is it going when it reaches the water, in metres per second?

m/s

Show the answer

14.0071410359 m/s

The energy \(mgh\) becomes \(\tfrac12 mv^2\), so \(v = \sqrt{2 \times 9.81 \times 10} \approx 14.0\) m/s, about 50 km/h.

Work

FIRST, MAKE A PREDICTION

You push hard against a wall that does not move. How much mechanical work does your force do on the wall?

Show the answer

Zero

Mechanical work requires displacement in the direction of the force; here the wall's displacement is zero.

TRY IT / FOLLOW THE REASONING

Compare pushing a box 10 m with pushing an immovable wall using the same force. Then turn a force perpendicular to a displacement and consider its contribution.

MAKE THE CONNECTION

Key idea

Work is the force along the direction of motion, times the distance moved.

Pushing a box with 50 N for 10 m does \(50 \times 10 = 500\) J of work. Pushing just as hard on a wall that doesn't move does none, however tired you get: your muscles burn energy to hold a load, but the suitcase or the wall gets nothing from it.

Work

\[ \begin{gathered} W = F\,d \\[.6em] 1\ \text{J} = 1\ \text{N m} = 1\ \text{kg m}^2/\text{s}^2 \end{gathered} \]

\(W\) is the work, \(F\) the force along the motion (N), \(d\) the distance moved (m). The unit is the joule (J). An apple (about 1 N) lifted 1 m takes about 1 J; a 70 kg person climbing a 3 m flight does about 2000 J.

Force at an angle. Only the component of the force along the displacement (the distance moved, with its direction) counts:

Work

\[ W = F\,d\cos\theta \]

where \(\theta\) is the angle between the force and the displacement.

Work

A strap pulls a suitcase with a force of 40 N at 60 degrees to the floor. It splits into 20 N along the floor, which does the work, and 34.6 N upwards, which does none. Over 20 m the work is 400 J.−46162636−46162636F = 40 N20 N along the floor60°
A 40 N strap at 60° splits into 20 N along the floor and 34.6 N upwards; only the 20 N along the floor does work.

Work

Worked number: the suitcase

A 40 N pull on a suitcase strap at 60° over 20 m: \(40 \times 20 \times 0.5 = 400\) J.

Work

The special angles:

  • \(\theta = 0^\circ\): \(\cos\theta = 1\), the full work (800 J for 40 N over 20 m).
  • \(\theta = 90^\circ\): \(\cos\theta = 0\), no work. A waiter carrying a tray level does none on it, and gravity does none on a satellite in a circular orbit.
  • \(\theta = 180^\circ\): \(\cos\theta = -1\), negative work: the force takes energy away. Friction on a sliding box does \(-150\) J over 5 m at 30 N.

The dot product. Force times distance times cosine is a way to combine two vectors into one plain number (a scalar), called the dot product:

Work

\[ \mathbf{F}\cdot\mathbf{d} = |\mathbf{F}|\,|\mathbf{d}|\cos\theta = F_x d_x + F_y d_y \]

\(|\mathbf{F}|\) and \(|\mathbf{d}|\) are the lengths of the two vectors, and the second form multiplies matching components and adds. For \(\mathbf{F} = (30, 40)\) N and \(\mathbf{d} = (6, 2)\) m: \(30\times6 + 40\times2 = 260\) J, the same as \(50 \times 6.32 \times 0.822\). If the dot product of two non-zero vectors is zero, they are at right angles.

Checkpoint: Work (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A sled is pulled 8.0 m along level snow by a rope with a 25 N pull, 37° above the horizontal. How much work does the rope do, in joules?

J

Show the answer

159.727102009 J

\(W = Fd\cos\theta = 25 \times 8.0 \times \cos37^\circ \approx 25 \times 8.0 \times 0.799 \approx 160\) J.

Checkpoint: Work (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A waiter carries a tray level across a room at a steady walking pace. How much work does the upward force of the waiter's hand do on the tray?

Show the answer

None, because the force is at right angles to the motion.

The hand pushes up and the tray moves sideways, so \(\theta = 90^\circ\) and \(W = Fd\cos90^\circ = 0\).

Checkpoint: Work (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

A force \(\mathbf{F} = (5, 12)\) N acts on an object that moves through a displacement \(\mathbf{d} = (3, 1)\) m. What is the work done, in joules, found from the components?

J

Show the answer

27 J

\(\mathbf{F}\cdot\mathbf{d} = 5\times3 + 12\times1 = 27\) J.

Kinetic energy and the work–energy theorem

FIRST, MAKE A PREDICTION

The speed of a fixed-mass object doubles. By what factor does its kinetic energy change?

Show the answer

Four

Kinetic energy depends on speed squared: ½m(2v)² = 4 × ½mv².

TRY IT / FOLLOW THE REASONING

For a 2 kg object, compare kinetic energies at 2 and 4 m/s. Calculate the additional net work needed and distinguish that increase from the final total.

MAKE THE CONNECTION

Key idea

The net work done on an object equals its change in kinetic energy, \(\tfrac12 mv^2\).

A constant net force \(F\) on a mass \(m\) over a distance \(d\) does work \(W = mad\). With \(v^2 = u^2 + 2ad\) from P1, \(ad = \tfrac12(v^2 - u^2)\), so

Kinetic energy and the work–energy theorem

\[ \begin{gathered} W_\text{net} = \tfrac12 mv^2 - \tfrac12 mu^2 \\[.6em] KE = \tfrac12 mv^2 \end{gathered} \]

\(u\) is the starting speed, \(v\) the final speed. \(KE\) is kinetic energy, the energy of motion, in joules. The work–energy theorem says the net work done on an object equals its change in kinetic energy. Positive net work speeds it up; negative net work slows it down.

Kinetic energy and the work–energy theorem

Worked number: a car and a baseball

A 1000 kg car at 20 m/s has 200 000 J (200 kJ); a 0.145 kg baseball at 40 m/s has about 116 J.

Kinetic energy and the work–energy theorem

Why speed is squared. Double the speed and the kinetic energy quadruples. Triple the speed and the energy goes up nine times.

Kinetic energy and the work–energy theorem

Worked number: braking

The same car at 40 m/s has 800 kJ. With a steady 8000 N braking force the car stops in \(200\,000/8000 = 25\) m from 20 m/s but needs \(800\,000/8000 = 100\) m from 40 m/s.

Checkpoint: Kinetic energy and the work–energy theorem (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A 0.20 kg ball is thrown at 30 m/s. What is its kinetic energy, in joules?

J

Show the answer

90 J

\(\tfrac12 \times 0.20 \times 30^2 = 90\) J.

Checkpoint: Kinetic energy and the work–energy theorem (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A cyclist triples her speed. What happens to her kinetic energy?

Show the answer

It goes up nine times.

\(KE = \tfrac12 mv^2\), so tripling \(v\) multiplies the energy by \(3^2 = 9\).

Potential energy

FIRST, MAKE A PREDICTION

You lift the same object through the same height slowly, once vertically and once using a frictionless ramp. How does the gravitational potential-energy change compare?

Show the answer

It is the same

Near Earth's surface, the change is mg times the vertical height gained, independent of route.

TRY IT / FOLLOW THE REASONING

Choose a mass and lift height. Compare a short vertical route with a longer gentle route, and explain how a smaller force over a larger distance can give the same work.

MAKE THE CONNECTION

Key idea

Work can be stored. Lifting something stores energy that comes back as motion when it falls, and only the height gained matters, not the route.

Work doesn't always speed something up. It can be stored as gravitational potential energy, and the book returns it as kinetic energy if it falls back.

Potential energy

Worked number: lifting a book

Lifting a 2 kg book slowly through 1.5 m takes a force equal to its weight, about 19.6 N, so you do about 29 J.

Potential energy

\[ \begin{gathered} PE_\text{grav} = mgh \\[.6em] PE_\text{spring} = \tfrac12 kx^2 \end{gathered} \]
  • \(m\) is the mass (kg), \(g = 9.81\ \text{m/s}^2\), \(h\) is the height above a zero level you choose (only changes in height matter).
  • \(k\) is the spring constant (N/m) and \(x\) the stretch (m).
  • The gain depends only on the height gained, not on the route: straight up, a ramp or a winding stair.

Potential energy

Worked numbers: a spring and a climber

For the spring, the force \(kx\) grows from 0, so the work is the area under a straight line: \(\tfrac12 \times 0.10 \times 20 = 1\) J for \(k = 200\) N/m and \(x = 0.10\) m, and four times that at twice the stretch.

A 60 kg climber going up 500 m gains about 294 kJ.

Potential energy

The roller coaster. Starting at rest 40 m up, \(mgh = \tfrac12 mv^2\) at the bottom. The mass cancels:

Potential energy

\[ v = \sqrt{2gh} \]

Potential energy

Energy per kilogram against the height fallen on a 40 m coaster drop. Potential energy falls from 392.4 J to 0, kinetic energy rises from 0 to 392.4 J, and the total stays at 392.4 J for every kilogram.10203040100200300400kineticpotentialtotalmetres fallen
Going down a 40 m drop, potential energy turns into kinetic energy; the total stays at 392.4 joules for every kilogram.

Potential energy

Worked number: the roller coaster

\(v = \sqrt{2 \times 9.81 \times 40} \approx 28\) m/s, about 100 km/h, whatever the shape of the track.

Half-way down (20 m up) the speed is about 19.8 m/s; over a 25 m hill (a 15 m drop) about 17.2 m/s; and no hill higher than the start can be climbed. A ball dropped from 5 m arrives at about 9.9 m/s, the same as the motion equations give, without any time or acceleration in the working. All this ignores friction and air resistance.

Checkpoint: Potential energy (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A 5.0 kg weight is lifted slowly through 2.0 m. How much gravitational potential energy does it gain, in joules?

J

Show the answer

98.1 J

\(mgh = 5.0 \times 9.81 \times 2.0 = 98.1\) J.

Checkpoint: Potential energy (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

A spring with \(k = 400\) N/m is stretched by 0.050 m. How much energy does it store, in joules?

J

Show the answer

0.5 J

\(\tfrac12 \times 400 \times 0.050^2 = 0.50\) J.

Conservation of energy

FIRST, MAKE A PREDICTION

A sliding object loses kinetic energy to friction. Has the total energy of the object and its surroundings disappeared?

Show the answer

No; energy has transferred into other forms

Mechanical energy can fall while thermal energy and other forms increase in the full system.

TRY IT / FOLLOW THE REASONING

First follow the frictionless model's constant energy total. Then imagine adding friction and draw a third account for heating rather than forcing kinetic plus potential energy to stay constant.

Follow a frictionless descent

Current height: 10 m · Mass: 2 kg

Potential energy
196 J
Kinetic energy
0 J
Speed
0 m/s
Total energy
196 J
Model assumptions

Released from rest at 10 m. No friction or air resistance; g = 9.8 m/s². Heights are above the bottom.

MAKE THE CONNECTION

Key idea

Energy can't be created or destroyed; it only changes form, or moves from one object to another.

Think of a bank account: money moves between accounts, but the total across every account is constant. The accounts are kinetic, gravitational potential, elastic, thermal, chemical and electrical energy. To do the bookkeeping, draw a boundary round a system. If nothing crosses it, the total inside is constant; if work or heat crosses it, add that to the books.

Friction makes heat. The orderly motion of all the atoms together becomes random jiggling.

Conservation of energy

Worked number: a sliding block

A 2 kg block sliding at 6 m/s has \(\tfrac12 \times 2 \times 6^2 = 36\) J. Friction of 12 N stops it in 3 m, doing \(-12 \times 3 = -36\) J of work, and 36 J of thermal energy appears in the block and floor.

A skidding 1000 kg car at 20 m/s turns 200 kJ into heat. Some goes into sound and wear, but the total is unchanged.

Conservation of energy

Worked number: using the ledger

A 2 kg block slides from rest down a rough ramp 3 m high and reaches 5 m/s: it started with \(2 \times 9.81 \times 3 \approx 59\) J, ended with \(\tfrac12 \times 2 \times 5^2 = 25\) J of kinetic energy, so about 34 J became heat.

A 0.5 kg ball dropped from 2 m and bouncing back to 1.5 m loses about 2.5 J of its 9.8 J, keeping 75%.

Conservation of energy

Myths and mistakes: energy and work
  • Work is force times distance. Only the force along the motion counts: \(Fd\cos\theta\).
  • Double the speed, double the energy. It's four times (\(v^2\)).
  • Kinetic energy can be negative. Work can; \(\tfrac12 mv^2\) never is.
  • Friction destroys energy. It turns kinetic energy into thermal energy; the total is the same.
  • We're using up the world's energy. Energy can't be used up; what runs out is useful (concentrated) energy.
  • A perpetual motion machine just needs less friction. No machine can create energy from nothing.
  • Mixing up power and energy: a kilowatt is a power, a kilowatt-hour an energy.

Checkpoint: Conservation of energy (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A 0.30 kg ball is dropped from 2.0 m and bounces back up to only 1.2 m. How much energy did the bounce turn into heat and sound, in joules?

J

Show the answer

2.3544 J

It lost \(0.30 \times 9.81 \times (2.0 - 1.2) \approx 2.35\) J of stored energy, which became thermal energy and sound.

Checkpoint: Conservation of energy (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A car skids to a stop. What has happened to its kinetic energy?

Show the answer

It became thermal energy in the tyres and the road.

Friction does negative work and the same energy appears as thermal energy (plus a little sound and wear).

Checkpoint: Conservation of energy (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

A 4.0 kg sled slides from rest down a rough hill 5.0 m high and reaches the bottom at 8.0 m/s.

Stored energy at the top, in joules.

J

Kinetic energy at the bottom, in joules.

J

Energy that became heat, in joules.

J

Show the answer
  1. 196.2 J
  2. 128 J
  3. 68.2 J

Top: \(4.0 \times 9.81 \times 5.0 = 196.2\) J. Bottom: \(\tfrac12 \times 4.0 \times 8.0^2 = 128\) J. The difference, \(68.2\) J, became thermal energy.

Power

FIRST, MAKE A PREDICTION

Two machines lift the same load through the same height. One does it in half the time. Which has the larger average useful power?

Show the answer

The faster machine, by a factor of two

The useful work is the same; doing it in half the time doubles work per second.

TRY IT / FOLLOW THE REASONING

Choose a load requiring 1000 J of useful work. Compare lift times of 10 and 5 s, then separate useful output power from the greater input power of an inefficient motor.

MAKE THE CONNECTION

Key idea

Power is how fast work is done, or energy is transferred.

Power

\[ \begin{gathered} P = \frac{W}{t} \\[.6em] 1\ \text{W} = 1\ \text{J/s} \end{gathered} \]

\(P\) is the power in watts (W), \(W\) the work or energy in joules, \(t\) the time in seconds.

Power

Worked number: climbing stairs

A 70 kg person climbing 3 m gains \(70 \times 9.81 \times 3 \approx 2060\) J: in 5 s that is about 412 W, in 15 s about 137 W. The same energy in three times the time is a third of the power.

A motor lifting 200 kg through 10 m in 20 s supplies 19 620 J / 20 s = 981 W.

Power

  • The kilowatt-hour (kWh) on your electricity bill is an energy: 1 kW for 3600 s \(= 3.6\) million J. A 60 W bulb left on for 5 h uses 0.3 kWh, or 1.08 million J.
  • Horsepower. One horsepower is about 746 W (the metric horsepower, about 735 W, is a little smaller). A 100 horsepower car can supply roughly 75 000 W at its best.

Power

More: how much power are you?

How much power are you? Food Calories on a label are kilocalories: 1 kcal = 4184 J.

  • Two thousand Calories a day is about 8.4 million J over 86 400 s, about 97 W: on average, a bright old light bulb.
  • Over a working day a typical person can sustain about 75 W of useful output.
  • A fit cyclist manages about 100 W for hours; the very best cyclists roughly 400 to 500 W for an hour.
  • A track sprinter can top 2000 W for a second or two.
  • Asleep, you give off about 80 W as heat.
  • Heating a litre of water from 20 °C to 100 °C takes about 335 000 J: a kettle does it in minutes, but a person pedalling a generator at 100 W would need nearly an hour.

Power

Force times speed. Since \(W = Fd\), \(P = Fd/t = Fv\).

Power

Worked number: a car and a motor

A car at a steady 25 m/s against a 600 N drag does \(600 \times 25 = 15\,000\) W, about 20 horsepower.

Efficiency is useful output divided by input: a motor taking in 1000 W and delivering 800 W is 80% efficient, and the other 200 W goes into heat.

Part Momentum and collisions

Power

Checkpoint: Power (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A motor lifts a 50 kg crate through 6.0 m in 12 s. What power does it supply to the crate, in watts?

W

Show the answer

245.25 W

The crate gains \(50 \times 9.81 \times 6.0 = 2943\) J in 12 s, so \(P = 2943/12 \approx 245\) W.

Checkpoint: Power (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A 100 W bulb is left on for 8.0 hours. How much energy does it use, in megajoules?

MJ

Show the answer

2.88 MJ

\(100 \times 8.0 \times 3600 = 2.88\) million J, which is 0.80 kWh.

Checkpoint: Power (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

A car moves at a steady 18 m/s against a total drag of 500 N. What power does the engine supply to overcome it, in watts?

W

Show the answer

9000 W

\(P = Fv = 500 \times 18 = 9000\) W, or 9 kW.

Momentum and impulse

FIRST, MAKE A PREDICTION

An airbag stops a passenger with the same change in momentum but over a longer time. What happens to the average stopping force?

Show the answer

Its magnitude decreases

Impulse is average force times duration. The same momentum change spread over more time requires a smaller average force.

TRY IT / FOLLOW THE REASONING

Draw two force–time rectangles with the same area, one short and tall and one long and low. Connect the areas to the same change in momentum.

MAKE THE CONNECTION

Key idea

Impulse (force times the time it acts) equals the change in momentum.

Momentum is mass times velocity, a vector measured in kg m/s:

Momentum and impulse

\[ \mathbf{p} = m\mathbf{v} \]

In a straight line, the two directions are plus and minus.

Momentum and impulse

Worked number: momentum

A 1000 kg car at 20 m/s has 20 000 kg m/s; a 10-tonne truck at the same speed has 200 000. A 0.5 kg ball moving at 12 m/s to the left has \(0.5 \times (-12) = -6\) kg m/s.

Momentum and impulse

Start with \(F = ma\) and \(a = (v - u)/\Delta t\): then

Momentum and impulse

\[ F\,\Delta t = mv - mu = \Delta p \]

The left side, force times the time it acts, is the impulse, in newton-seconds (the same as kg m/s). Impulse equals the change in momentum. If the force varies, the impulse is the area under the force–time graph, and the average force is that area divided by the time.

Momentum and impulse

Worked numbers: a kick and a hit

A 0.45 kg football kicked from rest to 25 m/s in 0.01 s: \(\Delta p = 11.25\) kg m/s, average force 1125 N.

A 0.145 kg baseball arriving at 40 m/s and hit back at 50 m/s in 0.001 s: \(\Delta p = 0.145 \times 90 = 13.05\) kg m/s, about 13 000 N.

Momentum and impulse

Newton himself wrote his second law as "the change of motion is proportional to the force", where "motion" is mass times velocity.

Airbags and crumple zones. Your change in momentum in a crash is fixed by your mass and speed, so a longer stop means a smaller force.

Momentum and impulse

Worked number: the airbag

A 70 kg person at 15 m/s has 1050 kg m/s. Stopped in 0.02 s by a hard wall that needs about 52 500 N; stopped in 0.15 s by a crumple zone and airbag, about 7000 N: 7.5 times less, needing about a metre of crumpling. Bending your knees on landing works the same way.

Checkpoint: Momentum and impulse (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

Taking right as positive, a 0.40 kg ball moves left at 5.0 m/s. What is its momentum, in kg m/s?

kg m/s

Show the answer

-2 kg m/s

\(0.40 \times (-5.0) = -2.0\) kg m/s: the minus sign says it points left.

Checkpoint: Momentum and impulse (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

A 2.0 kg cart speeds up from 3.0 m/s to 8.0 m/s in the same direction under a steady push that lasts 0.50 s. What is the force, in newtons?

N

Show the answer

20 N

\(\Delta p = 2.0 \times (8.0 - 3.0) = 10\) kg m/s, and \(F = \Delta p/\Delta t = 10/0.50 = 20\) N.

Checkpoint: Momentum and impulse (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

In a crash, an airbag makes the stop take longer. Why does that make the force on you smaller?

Show the answer

The change in momentum is the same, so spreading it over a longer time means a smaller average force.

\(F\,\Delta t = \Delta p\). With \(\Delta p\) fixed, a larger \(\Delta t\) means a smaller \(F\).

Conservation of momentum

FIRST, MAKE A PREDICTION

Two skaters initially at rest push apart on frictionless ice. What is their total momentum immediately after the push?

Show the answer

Zero

With negligible external horizontal force, the system retains its initial total momentum; the individual momenta cancel.

TRY IT / FOLLOW THE REASONING

Give the skaters different masses. Draw equal-and-opposite impulses, calculate their different speeds and add their signed momenta.

MAKE THE CONNECTION

Key idea

If no outside force acts on a system, its total momentum stays constant.

By Newton's third law, two objects that push on each other exert equal and opposite forces for the same time, so their impulses, and so their momentum changes, are equal and opposite.

Conservation of momentum

Worked number: two skaters

Two skaters, 60 kg and 40 kg, push apart from rest: total momentum stays zero, so if the 40 kg skater moves off at 3 m/s (\(-120\) kg m/s), the other has \(+120\), which is 2 m/s the other way, and 300 J of kinetic energy appear from their muscles.

Conservation of momentum

Worked number: recoil

A 4 kg rifle fires a 10 g bullet at 400 m/s: bullet momentum \(0.01 \times 400 = 4\) kg m/s, so the rifle kicks back at 1 m/s. The bullet gets 800 J of kinetic energy and the rifle 2 J: equal momenta, unequal energies.

An 80 kg astronaut throwing a 2 kg wrench at 5 m/s drifts back at \(10/80 = 0.125\) m/s.

Checkpoint: Conservation of momentum (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A 70 kg skater at rest on smooth ice throws a 0.50 kg ball forwards at 12 m/s. How fast does the skater recoil, in metres per second?

m/s

Show the answer

0.0857142857143 m/s

The ball has \(0.50 \times 12 = 6.0\) kg m/s, so the skater has \(-6.0\) kg m/s and recoils at \(6.0/70 \approx 0.086\) m/s.

Checkpoint: Conservation of momentum (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

Two skaters push apart from rest. Which quantity is equal and opposite for the two of them?

Show the answer

Their momenta.

Equal and opposite forces for the same time give equal and opposite changes in momentum.

Checkpoint: Conservation of momentum (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

Two skaters, one heavy and one light, push apart from rest. Write down which of them ends up with more kinetic energy, and why, then compare it with the answer.

Show the solution

The lighter skater. Both have momentum of the same size \(p\), and \(KE = p^2/2m\), so the smaller mass gets the larger energy. Equal momenta, unequal energies, as with the rifle and the bullet.

Collisions

FIRST, MAKE A PREDICTION

Two objects collide and stick together in an isolated system. Which quantity must remain unchanged?

Show the answer

Total momentum

With no external impulse the total momentum is conserved, although kinetic energy generally becomes other forms.

TRY IT / FOLLOW THE REASONING

Let equal masses with velocities +2 and 0 stick. Find their shared speed from momentum, then calculate kinetic energy before and after. Account for the difference.

MAKE THE CONNECTION

Key idea

Momentum is conserved in every collision with no outside force. Kinetic energy depends on how bouncy it is.

There are three kinds:

  • Elastic: no kinetic energy lost; the objects bounce apart.
  • Inelastic: some kinetic energy becomes heat, sound and deformation.
  • Perfectly inelastic: the objects stick together (the most kinetic energy lost, but momentum is still conserved).

Collisions

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

\(u\) means before and \(v\) after, with signs for direction. For elastic collisions in one dimension there is a second rule: the speed of approach equals the speed of separation, \(u_1 - u_2 = v_2 - v_1\).

Collisions

A 2 kg cart at 3 m/s clips onto a 1 kg cart at rest. Momentum is 6 kg m/s before and after. Kinetic energy falls from 9 J to 6 J.2468106momentumbefore(kg m/s)6momentumafter(kg m/s)9energybefore(J)6energyafter(J)
Two carts that clip together: momentum is 6 kg m/s before and after, but the kinetic energy falls from 9 J to 6 J.

Collisions

  • Sticking: a 2 kg cart at 3 m/s hits a 1 kg cart at rest and they clip together: \(6 = 3v\), \(v = 2\) m/s. Kinetic energy 9 J before, 6 J after: 3 J lost to heat, sound and deformation.
  • Bouncing: the same carts with springy bumpers: \(6 = 2v_1 + v_2\) and \(v_2 - v_1 = 3\) give \(v_1 = 1\) m/s, \(v_2 = 4\) m/s. Energy: \(1 + 8 = 9\) J, so it's kept.

Collisions

More collision examples
  • Newton's cradle: equal masses just swap velocities, so one ball in means one ball out.
  • A car crash: a 1500 kg car at 20 m/s hits the back of a stationary 1000 kg car and they lock: \(30\,000/2500 = 12\) m/s. Kinetic energy 300 kJ before, 180 kJ after: 120 kJ (40%) went into crushing metal.
  • Direction matters: 2 kg at 3 m/s right and 1 kg at 3 m/s left, stuck together: \(6 - 3 = 3\) kg m/s, so the 3 kg moves at 1 m/s to the right.

Collisions

Myths and mistakes: momentum and collisions
  • Momentum is only conserved in elastic collisions. It's conserved in every collision; only kinetic energy depends on bounciness.
  • Energy is lost in an inelastic collision. Kinetic energy is lost; the total energy turns into heat, sound and bent metal.
  • Forgetting direction. Momentum is a vector: use plus and minus.
  • A rocket pushes against the air or the ground. It pushes against its own exhaust.

Checkpoint: Collisions (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

A 0.50 kg ball of clay moving at 4.0 m/s hits a 1.5 kg ball of clay at rest, and they stick. How fast do they move together, in metres per second?

m/s

Show the answer

1 m/s

Before: \(0.50 \times 4.0 = 2.0\) kg m/s. After: \(2.0 \text{ kg} \times v\), so \(v = 1.0\) m/s.

Checkpoint: Collisions (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

In that collision (0.50 kg at 4.0 m/s sticks to 1.5 kg at rest, moving off at 1.0 m/s) how much kinetic energy is lost, in joules?

J

Show the answer

3 J

Before: \(\tfrac12 \times 0.50 \times 4.0^2 = 4.0\) J. After: \(\tfrac12 \times 2.0 \times 1.0^2 = 1.0\) J. So 3.0 J became heat, sound and deformation.

Checkpoint: Collisions (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

Two objects collide and stick together. Which statement is true?

Show the answer

Momentum is conserved, but some kinetic energy is lost.

Momentum survives every collision with no outside force. Sticking together loses the most kinetic energy possible, but the momentum is still all there.

Rockets

FIRST, MAKE A PREDICTION

Can a rocket accelerate in a vacuum with nothing nearby to push against?

Show the answer

Yes; it expels its own exhaust

The exhaust gains momentum one way and the rocket gains momentum the other; outside air is not required.

TRY IT / FOLLOW THE REASONING

Sketch a stationary rocket and fuel as one initial system. Give a small parcel of exhaust backward momentum and work out what balances it. Then identify the changing rocket mass.

MAKE THE CONNECTION

Key idea

A rocket pushes against its own exhaust, not against the air or the ground.

A rocket and its fuel start with total momentum zero. The engine throws hot gas backwards, so the rocket gains an equal momentum forwards: nothing outside is needed, so rockets work best in space. Since force is the rate of change of momentum, thrust is (mass of gas thrown out per second) \(\times\) (its speed). Illustrative numbers, not a real engine: 100 kg/s at 3000 m/s gives \(100 \times 3000 = 300\,000\) N. More gas per second, or faster gas, means more thrust. How fast a rocket can go for a given amount of fuel is given by the rocket equation, which needs logarithms, so it waits for the exponentials and logarithms toolkit: a faster exhaust and a bigger share of fuel both give a higher final speed.

Checkpoint: Rockets (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

A rocket throws out 250 kg of exhaust each second at 2400 m/s (illustrative numbers). What is the thrust, in newtons?

N

Show the answer

600000 N

\(250 \times 2400 = 600\,000\) N.

Checkpoint: Rockets (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Why can a rocket work in empty space, with nothing to push against?

Show the answer

It pushes against its own exhaust, which gains momentum backwards while the rocket gains it forwards.

The engine throws the gas backwards, so by conservation of momentum the rocket gains an equal momentum forwards.

Key terms

Key terms

Work
The force along the direction of motion times the distance moved, \(W = Fd\cos\theta\); measured in joules.
Joule (J)
The unit of work and energy: 1 N m = 1 kg m²/s².
Displacement
The distance moved together with its direction.
Negative work
Work done by a force pointing against the motion; it takes energy away.

Key terms

Dot (scalar) product
A way to multiply two vectors to get a plain number: \(\mathbf{F}\cdot\mathbf{d} = Fd\cos\theta = F_xd_x + F_yd_y\).
Scalar
A quantity with size but no direction.
Kinetic energy
The energy of motion, \(\tfrac12 mv^2\).
Work–energy theorem
The net work done on an object equals its change in kinetic energy.

Key terms

Potential energy
Energy stored in a position or configuration: \(mgh\) for gravity, \(\tfrac12 kx^2\) for a spring.
Conservation of energy
Energy is never created or destroyed, only changed from one form to another or moved between objects.
System
The things inside a boundary you draw to do the energy (or momentum) bookkeeping.
Thermal energy
The energy of the random jiggling of atoms; what friction turns kinetic energy into.

Key terms

Power / watt (W)
The rate of doing work or transferring energy, \(P = W/t\) / one joule per second.
Kilowatt-hour (kWh)
A unit of energy: 3.6 million joules.
Horsepower
An older unit of power, about 746 W.
Efficiency
Useful output divided by input.

Key terms

Momentum
Mass times velocity, \(p = mv\), a vector in kg m/s.
Impulse
Force times the time it acts, \(F\,\Delta t\); equal to the change in momentum.
Conservation of momentum
With no outside force, a system's total momentum stays constant.
Recoil
The backward momentum a gun or thrower gets from the thing it launches.

Key terms

Elastic / inelastic / perfectly inelastic collision
Kinetic energy kept / some lost to heat, sound and deformation / the objects stick together.
Thrust
The forward force on a rocket from throwing exhaust backwards.
Rocket equation
The formula for a rocket's final speed from its exhaust speed and fuel share (met properly after logarithms).

History and sources

History

Where a primary text wasn't read, the sources are encyclopedic or institutional pages, and the wording below is cautious.

Coriolis and "work" (1829)

Coriolis and "work" (1829). Gaspard-Gustave Coriolis, Du calcul de l'effet des machines (Paris, 1829), gave force times distance its modern name, travail. Earlier engineers used other names for the idea, and Jean-Victor Poncelet was working on it at the same time, and credited Coriolis with the word. Sources: MacTutor: Coriolis; Work (physics).

The joule (1882, 1889)

The joule (1882, 1889). William Siemens proposed the name in his British Association address in 1882, and the second International Electrical Congress in Paris adopted the joule (and the watt) in 1889, the year Joule died. Source: Joule.

"Energy", "potential energy" (1807, 1853)

"Energy", "potential energy" (1807, 1853). Thomas Young wrote in 1807 that "the product of the mass of a body into the square of its velocity may properly be termed its energy" (so without the half). The name potential energy is credited to William Rankine in 1853. Sources: History of energy; Energy Philosophy.

's Gravesande and Émilie du Châtelet (1722, 1740)

's Gravesande and Émilie du Châtelet (1722, 1740). Willem 's Gravesande dropped brass balls into soft clay and found that the dents grew with the square of the speed, not the speed (published 1722). Émilie du Châtelet used his results in her Institutions de physique (1740) to argue for Leibniz's vis viva, the "living force" going as \(mv^2\). The experiment is his, not hers. The quarrel was partly about what "force" meant, and some historians say d'Alembert (1743) settled it as a matter of definition. Sources: Stanford Encyclopedia: Émilie du Châtelet; Vis viva; Willem 's Gravesande.

The Switchback Railway (Coney Island, 16 June 1884)

The Switchback Railway (Coney Island, 16 June 1884). LaMarcus Thompson's ride was America's first intentionally constructed roller coaster: five cents, about 6 mph, gravity only. (Earlier gravity railways existed, so it is not the first ever.) Source: Hagley Museum and Library.

Joule's paddle wheel (1840s, 1850)

Joule's paddle wheel (1840s, 1850). Joule let falling weights turn a paddle wheel in a tub of water and measured how much it warmed: a given amount of work always made the same amount of heat. His careful paper (Philosophical Transactions 140, 1850) was within about one percent of the modern value. The famous honeymoon-thermometer story comes from William Thomson's recollection decades later, and no result survives. Sources: Young on Joule 1850 (PMC); James Prescott Joule; Science Museum Group: Joule's apparatus.

Mayer, Helmholtz and others (1840s)

Mayer, Helmholtz and others (1840s). Several people reached the idea independently. Julius Robert Mayer published in 1842; Hermann von Helmholtz stated it for all of physics in "Über die Erhaltung der Kraft" (Berlin, 1847), where Kraft meant what we now call energy. Sources: Conservation of energy; Julius von Mayer; Hermann von Helmholtz.

Watt and the horsepower (about 1782)

Watt and the horsepower (about 1782). James Watt, selling steam engines to mills and breweries that used horses, rounded his estimate of a strong horse's work to 33 000 foot-pounds a minute, about 746 W. The unit watt was proposed by Siemens in 1882 and adopted in 1889. Watt never used it. The story of a brewer's challenge is a legend. Source: Horsepower; Watt (unit).

Human power

Human power. About 75 W sustained over a working day; 200 to 500 W for an hour across fit people and top cyclists; over 2000 W for sprint peaks; about 80 W at rest. Sources: Human power; UCI: track sprinting, a question of watts; OpenStax College Physics 7.8. A food Calorie is a kilocalorie, and the thermochemical calorie is exactly 4.184 J (Calorie).

Newton's Principia (1687)

Newton's Principia (1687). Newton called mass times velocity the "quantity of motion" (Definition II), and his second law says the alteration of motion is proportional to the force. Source: Principia, Motte's translation (Wikisource).

Airbags and crumple zones

Airbags and crumple zones. Béla Barényi (Mercedes-Benz) patented the crumple zone in 1952, and the 1959 Fintail was the first Mercedes body developed using that patent. John Hetrick patented a compressed-air airbag in 1953 (never produced); Allen Breed's crash sensor in the late 1960s made airbags practical; Mercedes-Benz offered one with a seat-belt tensioner in the S-Class from 1981. Sources: Crumple zone; Airbag; US patent 2,649,311 (Hetrick).

The Royal Society and collisions (1668–69)

The Royal Society and collisions (1668–69). The Society took up the question of collisions; John Wallis (bodies that stick), Christopher Wren and Christiaan Huygens (bodies that bounce) each sent rules, published in 1669, and all agreed that what is conserved is a quantity of motion that counts direction. Huygens imagined a collision seen from a boat and from the shore, and showed that Descartes's rules were wrong; his full treatise appeared only in 1703, after his death. Sources: Christiaan Huygens; Royal Society picture library: Wren's paper.

Newton's cradle

Newton's cradle. Edme Mariotte demonstrated colliding suspended balls in the 1670s. The name "Newton's cradle" for the desktop toy dates from 1967, according to Wikipedia. Source: Newton's cradle.

Tsiolkovsky and the rocket equation (1903)

Tsiolkovsky and the rocket equation (1903). Konstantin Tsiolkovsky published it in 1903, but the British mathematician William Moore had worked out rocket motion in 1810 and published it in 1813. Sources: Tsiolkovsky rocket equation; William Moore.

Goddard and the New York Times (1920, 1969)

Goddard and the New York Times (1920, 1969). Robert Goddard's Smithsonian paper (title page 1919) reached the public in January 1920. On 13 January 1920 the Times ran an unsigned editorial saying he seemed to "lack the knowledge ladled out daily in high schools" and that a rocket needs something better than a vacuum to react against. Goddard had shown by experiment in 1915 that thrust works in a vacuum; on 16 March 1926 he launched the first liquid-fuel rocket from a farm in Auburn, Massachusetts (about 41 ft in 2.5 s). On 17 July 1969, the day after Apollo 11 left for the Moon, the Times printed a correction. Sources: Robert H. Goddard; NASA: Robert Goddard; Clark University: the 1926 launch. The exact Times wording is as reprinted in the 1969 correction (reprint at Crossing Wall Street; the Times archive page itself was not retrieved).

Robins and the ballistic pendulum (1742)

Robins and the ballistic pendulum (1742). Benjamin Robins described it in New Principles of Gunnery (London, 1742): a bullet embeds in a heavy hanging block and the swing gives the bullet's speed. Sources: Benjamin Robins; MacTutor: Robins.

Numbers used:

Numbers used: \(g = 9.81\ \text{m/s}^2\); 1 kcal = 4184 J; 1 horsepower about 746 W. Every number in the videos and on this page is recomputed in Python.

Worked example: the skateboarder

The problem

Problem. A skateboarder and her board have a mass of 55 kg. She starts from rest at the top of a smooth curved ramp 2.0 m high. How fast is she going at the bottom? Ignore friction and air resistance, then put friction back using a measured speed.

Solution

  1. Draw it. Point A is the top (at rest), point B the bottom. The only number from the picture is the height difference, \(h = 2.0\) m. The shape of the curve won't matter.
  2. Choose the tool. Forces would mean a changing slope and a changing acceleration at every point. Energy needs only A and B: the ramp's push is at right angles to her motion, so it does no work, and gravity's work is just the change in stored energy.

Solution

  1. Plan. Energy at the top, energy at the bottom, set them equal.
  2. Top (bottom is the zero of height): \(KE = 0\), \(PE = mgh = 55 \times 9.81 \times 2.0 \approx 1079\) J (about 1.08 kJ).
  3. Bottom: \(PE = 0\), all of it is kinetic: \(\tfrac12 mv^2\).
  4. Solve: \(mgh = \tfrac12 mv^2\). The mass cancels: \(v^2 = 2gh = 2 \times 9.81 \times 2.0 = 39.2\ \text{m}^2/\text{s}^2\), so \(v = \sqrt{39.24} \approx 6.26\) m/s, about 6.3 m/s, which is \(6.26 \times 3.6 \approx 22.5\) km/h.

Solution

  1. Check the energy: \(\tfrac12 \times 55 \times 6.26^2 \approx 1079\) J, the same as the top.
  2. Units: \(\sqrt{(\text{m/s}^2)(\text{m})} = \sqrt{\text{m}^2/\text{s}^2} = \text{m/s}\), a speed.
  3. Limits: \(h = 0\) gives \(v = 0\). The mass cancelled, so a heavier skater arrives at the same speed. It's the same formula as a ball dropped from 2.0 m (P1).

Solution

  1. Size: 22.5 km/h is a brisk cycling speed. Plausible.
  2. A taller ramp: at \(h = 4.0\) m, \(v = \sqrt{2 \times 9.81 \times 4.0} = \sqrt{78.48} \approx 8.9\) m/s: only \(\sqrt2 \approx 1.41\) times as fast, not twice, because the energy is in \(v^2\). At the half-way height (1.0 m below the top), \(v = \sqrt{2 \times 9.81 \times 1.0} \approx 4.43\) m/s.

Solution

  1. Friction back in. She is measured at 5.5 m/s at the bottom: \(KE = \tfrac12 \times 55 \times 5.5^2 \approx 832\) J. She started with 1079 J, so \(1079 - 832 \approx 247\) J, which is \(247/1079 \approx 23\%\), went into heat in the bearings and the air.

Practice

Practice

Use \(g = 9.81\) m/s² unless told otherwise.

★ Warm-ups

1.1 A 20 kg crate is pulled 6.0 m along a floor by a rope 40° above the horizontal, with a steady pull of 80 N. How much work does the rope do, in joules?

J

Solution

Only the component along the floor counts: \(80\cos40^\circ = 80 \times 0.766 \approx 61.3\) N, so \(W = Fd\cos\theta = 80 \times 6.0 \times 0.766 \approx 368\) J. Check: the same 80 N pulling along the floor would do \(80 \times 6.0 = 480\) J, and an angle between 0° and 90° must give less ✓. Units: N m = J ✓.

Problem 1.2 (★ Warm-ups)

1.2 A 1200 kg car is travelling at 15 m/s. What is its kinetic energy, in joules?

J

Solution

\(KE = \tfrac12 mv^2 = \tfrac12 \times 1200 \times 15^2 = 135\,000\) J = 135 kJ. At 30 m/s it would have \(\tfrac12 \times 1200 \times 30^2 = 540\,000\) J: double the speed, four times the energy ✓.

Problem 1.3 (★ Warm-ups)

1.3 A 1200 kg car is travelling at 30 m/s. A steady braking force of 6000 N stops it. How far does it travel while stopping, in metres?

Hint

The brakes must do negative work equal to the car's kinetic energy, so \(Fd = \tfrac12 mv^2\).

m

Solution

\(KE = \tfrac12 \times 1200 \times 30^2 = 540\,000\) J, and \(d = KE/F = 540\,000/6000 = 90\) m. Check: from 15 m/s the same car stops in 22.5 m. Double the speed gave 4 times the energy and 4 times the distance (22.5 m × 4 = 90 m) ✓.

Problem 1.4 (★ Warm-ups)

1.4 A 60 kg student runs up a flight of stairs 4.0 m high in 8.0 s. What power does she supply, in watts?

W

Solution

She gains \(mgh = 60 \times 9.81 \times 4.0 \approx 2354\) J, so \(P = W/t = 2354/8.0 \approx 294\) W. Check: walking up in 20 s would take \(2354/20 \approx 118\) W, 2.5 times less power for 2.5 times as long ✓.

Problem 1.5 (★ Warm-ups)

1.5 A 1.5 kW heater runs for 2.0 h. How much energy is that, in megajoules?

MJ

Solution

\(E = Pt = 1.5\ \text{kW} \times 2.0\ \text{h} = 3.0\) kWh, and 1 kWh is 3.6 MJ, so \(3.0 \times 3.6 = 10.8\) MJ. Check: the kWh is an energy, not a power ✓.

★★ Standard

2.1 A 3.0 kg box slides from rest down a rough ramp that is 2.5 m high, and reaches the bottom at 5.0 m/s. How much energy became heat, in joules?

Hint

Stored energy at the top, minus kinetic energy at the bottom, is what went into heat.

J

Solution

Top: \(mgh = 3.0 \times 9.81 \times 2.5 \approx 73.6\) J. Bottom: \(\tfrac12 mv^2 = \tfrac12 \times 3.0 \times 5.0^2 = 37.5\) J. Energy is conserved, so the rest is heat: \(73.6 - 37.5 \approx 36.1\) J. That is \(36.1/73.6 \approx 49\%\) of the energy. On a smooth ramp it would arrive at \(\sqrt{2 \times 9.81 \times 2.5} \approx 7.00\) m/s, faster than 5.0 m/s, as it must be ✓.

Problem 2.2 (★★ Standard)

2.2 A 1200 kg car travelling east at 15 m/s collides head-on with an 800 kg car travelling west at 10 m/s, and the two lock together. Taking east as positive, what is their velocity just afterwards, in metres per second?

m/s

Solution

Momentum before: \(1200 \times 15 + 800 \times (-10) = 18\,000 - 8000 = 10\,000\) kg m/s. After, 2000 kg move together: \(v = 10\,000/2000 = 5.0\) m/s, east. Check: the final velocity lies between the two initial velocities, towards the one with more momentum ✓.

Problem 2.3 (★★ Standard)

2.3 The same two cars (1200 kg east at 15 m/s, 800 kg west at 10 m/s) lock together. How much kinetic energy is lost, in joules?

Hint

Find the speed afterwards from momentum first, then compare the kinetic energy before and after.

J

Solution

Before: \(\tfrac12 \times 1200 \times 15^2 + \tfrac12 \times 800 \times 10^2 = 135\,000 + 40\,000 = 175\,000\) J. After (5.0 m/s): \(\tfrac12 \times 2000 \times 5.0^2 = 25\,000\) J. Lost: \(150\,000\) J, which is \(150/175 \approx 86\%\). The impulse on each car is equal and opposite: \(-12\,000\) N s on the 1200 kg car and \(+12\,000\) N s on the 800 kg car ✓.

Problem 2.4 (★★ Standard)

2.4 A 3.0 kg trolley moving at 4.0 m/s has an elastic collision with a 1.0 kg trolley at rest. How fast does the 1.0 kg trolley move afterwards, in metres per second?

Hint

Two rules for an elastic collision: momentum is conserved, and the speed of approach equals the speed of separation.

m/s

Solution

Momentum: \(3.0 \times 4.0 = 12\) kg m/s \(= 3.0v_1 + 1.0v_2\). Elastic: \(v_2 - v_1 = 4.0\). Substitute \(v_2 = v_1 + 4.0\): \(12 = 3.0v_1 + v_1 + 4.0\), so \(v_1 = 2.0\) m/s and \(v_2 = 6.0\) m/s, both forwards. Check: momentum after \(3.0 \times 2.0 + 1.0 \times 6.0 = 12\) ✓. Kinetic energy before \(\tfrac12 \times 3.0 \times 4.0^2 = 24\) J; after \(6 + 18 = 24\) J ✓.

Problem 2.5 (★★ Standard)

2.5 A 50 kg archer stands on frictionless ice and fires a 0.025 kg arrow horizontally at 60 m/s. How fast does she recoil, in metres per second?

m/s

Solution

The total momentum starts at zero. Arrow: \(0.025 \times 60 = 1.5\) kg m/s, so the archer has \(-1.5\) kg m/s: \(v = 1.5/50 = 0.030\) m/s backwards. The arrow has \(\tfrac12 \times 0.025 \times 60^2 = 45\) J and the archer \(\tfrac12 \times 50 \times 0.030^2 = 0.0225\) J: 2000 times less, exactly the mass ratio. Equal momenta, so the lighter object gets the energy ✓.

Problem 2.6 (★★ Standard)

2.6 A 0.15 kg ball hits a wall at 20 m/s and rebounds along the same line at 15 m/s. The contact lasts 5.0 ms. What is the average force on the ball, in newtons?

Hint

The direction reverses, so the change in velocity is \(20 + 15\), not \(20 - 15\).

N

Solution

Take the rebound direction as positive: the velocity goes from \(-20\) to \(+15\) m/s, a change of 35 m/s. \(\Delta p = 0.15 \times 35 = 5.25\) kg m/s, and \(F = \Delta p/\Delta t = 5.25/0.0050 = 1050\) N. The ball's weight is only \(0.15 \times 9.81 \approx 1.47\) N, so the force is about 714 times the weight. The ball loses \(30 - 16.9 \approx 13.1\) J of kinetic energy ✓.

★★★ Momentum and energy together

3.1 The ballistic pendulum. A 10 g bullet is fired into a 2.0 kg block hung on long strings. The bullet stays inside, and the block swings up by 0.10 m. How fast was the bullet going, in metres per second?

Hint

Two stages, two different laws. The collision is over in an instant, so momentum is conserved but kinetic energy is not. After it, the swing is a smooth rise with no heat, so energy is conserved.

m/s

Solution

Swing (energy). Just after the collision the block and bullet have kinetic energy \(\tfrac12(m + M)v'^2\), which becomes \((m + M)gh\) at the top. The mass cancels: \(v' = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.10} \approx 1.40\) m/s. Collision (momentum). \(mv = (m + M)v'\), with \(m + M = 2.01\) kg. \(v = 2.01 \times 1.40/0.010 \approx 282\) m/s, a typical rifle-bullet speed. The bullet's kinetic energy was \(\tfrac12 \times 0.010 \times 282^2 \approx 396\) J, and the swing keeps \(2.01 \times 9.81 \times 0.10 \approx 1.97\) J, about 0.50%. The other 394 J went into heat and the deformation of the block and bullet ✓.

Problem 3.2 (★★★ Momentum and energy together)

3.2 Slide, rough patch and spring. A 2.0 kg block slides from rest down a smooth ramp 1.5 m high onto a horizontal floor. After a rough patch 1.0 m long with \(\mu_k = 0.30\), it hits a spring with \(k = 500\) N/m. How far does it compress the spring, in metres?

Hint

Write the energy ledger once for the whole journey, from the top to the moment the block stops at maximum compression: \(mgh = \mu mg d + \tfrac12 kx^2\). Friction's work is \(\mu_k N d\), with \(N = mg\) on a level floor.

m

Solution

Start: \(mgh = 2.0 \times 9.81 \times 1.5 \approx 29.4\) J. The patch takes \(0.30 \times 2.0 \times 9.81 \times 1.0 \approx 5.89\) J as heat, leaving \(23.5\) J, so the block reaches the spring at \(\sqrt{2 \times 23.5/2.0} \approx 4.85\) m/s. At maximum compression all of it is in the spring: \(\tfrac12 kx^2 = 23.5\) J, so \(x^2 = 2 \times 23.5/500 = 0.0942\ \text{m}^2\) and \(x \approx 0.307\) m, about 31 cm. Check: with no rough patch, \(x = \sqrt{2 \times 29.4/500} \approx 0.343\) m, bigger, as it should be. The patch took 20% of the energy and the compression fell about 10%, which fits \(x \propto \sqrt{E}\) (\(\sqrt{0.80} \approx 0.89\)) ✓.

Problem 3.3 (★★★ Momentum and energy together)

3.3 Finding an unknown mass. A 1.0 kg cart at 6.0 m/s has an elastic collision with a stationary cart of unknown mass \(M\), and rebounds at 2.0 m/s. Find \(M\), in kilograms.

Hint

The first cart rebounds, so its final velocity is \(-2.0\) m/s. Use the elastic-collision rule (approach speed = separation speed) to get \(v_2\) first, then use momentum for \(M\).

kg

Solution

Elastic rule: \(v_2 - v_1 = 6.0\), so \(v_2 = 6.0 + (-2.0) = 4.0\) m/s. Momentum: \(1.0 \times 6.0 = 1.0 \times (-2.0) + M \times 4.0\), so \(8.0 = 4.0M\) and \(M = 2.0\) kg. The second cart moves at 4.0 m/s forwards. Check: momentum after \(-2.0 + 2.0 \times 4.0 = 6.0\) kg m/s ✓. Energy before \(\tfrac12 \times 1.0 \times 6.0^2 = 18\) J; after \(2.0 + 16 = 18\) J ✓. A light cart hitting a heavier one bounces back, and 2.0 kg is heavier than 1.0 kg ✓.

Check yourself

Answer these out loud, from memory. Then try again in a few days.

Check yourself

  • What is work, and how does it depend on force and displacement?

Check yourself

  • How is work calculated when a force changes along a path?

Check yourself

  • What is kinetic energy, and what does the work–energy theorem state?

Check yourself

  • What is potential energy, and how is it associated with a configuration?

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  • How are gravitational and elastic potential energy calculated near Earth's surface?

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  • What distinguishes conservative and nonconservative forces?

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  • How does conservation of mechanical energy simplify a motion problem?

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  • How can friction transform mechanical energy while total energy remains conserved?

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  • What is power, and how does it differ from energy and efficiency?

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  • How can energy diagrams identify accessible motion, turning points, and equilibrium?

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  • What is linear momentum, and how does it differ from kinetic energy?

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  • What is impulse, and how does it relate to a change in momentum?

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  • Under what conditions is the total momentum of a system conserved?

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  • How can internal forces change individual momenta without changing total momentum?

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  • What is the center of mass, and how does its motion describe a system?

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  • How do elastic, inelastic, and perfectly inelastic collisions differ?

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  • How are momentum and energy used together to analyze collisions?

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  • How do recoil, explosions, and propulsion illustrate momentum conservation?

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  • How are collisions analyzed in two dimensions and in the center-of-mass frame?

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  • Why does a rocket require a variable-mass treatment rather than a naive application of constant-mass equations?

Summary

Summary

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Next: Change and accumulation: the ideas of calculus

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Further learning

Khan Academy (found in Khan Academy's search index on 2026-10-12; Khan Academy blocks automated link checking, so tell us if one has moved):

Free textbook: OpenStax, University Physics Volume 1 (checked 2026-10-12): Chapter 7: Work and Kinetic Energy (7.1 Work, 7.2 Kinetic Energy, 7.3 Work-Energy Theorem, 7.4 Power) · Chapter 8: Potential Energy and Conservation of Energy (8.1 Potential Energy of a System, 8.2 Conservative and Non-Conservative Forces, 8.3 Conservation of Energy, 8.4 Potential Energy Diagrams and Stability, 8.5 Sources of Energy) · Chapter 9: Linear Momentum and Collisions (9.1 Linear Momentum, 9.2 Impulse and Collisions, 9.3 Conservation of Linear Momentum, 9.4 Types of Collisions, 9.5 Collisions in Multiple Dimensions, 9.6 Center of Mass, 9.7 Rocket Propulsion).

Simulations: PhET: Energy Skate Park (watch kinetic, potential and thermal energy trade places on a track, with and without friction) · PhET: Collision Lab (elastic and inelastic collisions; check momentum and kinetic energy before and after).

Next: M4 · Change and accumulation: the ideas of calculus (coming soon) · Progress