Conservation of momentum in collisions — Physics

Before this

What is momentum
Newton's third law
Forms of energy · coming soon

This leads to

Predicting a crash without knowing the forces

Two ice hockey pucks slide towards each other and collide. The collision lasts a tiny fraction of a second. During it, the forces between the pucks rise from zero to a large value and fall back again, in a way nobody could measure easily. Can you still predict how the pucks move afterwards?

Surprisingly, often you can. You do not need the forces, and you do not need the time. You need one fact: in a collision, the total momentum of the objects is the same just after as just before. This lecture shows why that is true, how to use it, and what it cannot tell you.

The last lecture promised this. It showed two skaters pushing apart, with equal and opposite momenta, and two cars whose total momentum was zero. Here is the law behind both.

Why the total cannot change

Take two objects, A and B, that collide. During the collision, A pushes on B and B pushes on A. By Newton's third law, these two forces are always equal in size and opposite in direction, at every instant. They also act for exactly the same time, because they start and stop together.

From the last lecture, force times time equals the change in momentum. So whatever change the collision makes to A's momentum, it makes an equal and opposite change to B's momentum:

Δ p_{A} = - Δ p_{B}

Add the two changes, and you get zero. The momentum moves from one object to the other, but none is created and none is lost. So the total stays the same.

Conservation of momentum

If no outside force acts on a system, its total momentum stays constant. For two objects that collide:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

Here $u_{1}$ and $u_{2}$ are the velocities before the collision, and $v_{1}$ and $v_{2}$ are the velocities after. Every velocity carries a sign, so choose a positive direction first.

A system with no overall outside force is called a closed system (some books say an isolated system). The forces between A and B are internal forces. They come in third-law pairs, so they always cancel in the total. Only an external force, from something outside the system, can change the total momentum.

What if there is friction?

Real collisions happen on roads, tables and ice, and friction is an outside force. So is momentum really conserved? Nearly, if you compare just before with just after. The forces between colliding objects are very large but last a very short time. Friction is small by comparison, and in that short time it hardly changes anything. So the law works for the collision itself. Afterwards, friction slowly takes the momentum away, as the wreck slides to a stop.

Stop and think

You drop a ball. As it falls, its momentum grows every second. Does this break the law of conservation of momentum?

Show the answer

No. The ball on its own is not a closed system. Gravity is an outside force from the Earth, and it keeps changing the ball's momentum. Now include the Earth in the system. The ball pulls the Earth up with an equal and opposite force (the third law), so the Earth gains an equal momentum upwards. The total for ball plus Earth stays at zero. The Earth's mass is so huge that its velocity change is far too small to notice. The lesson: before you use the law, decide what is in your system, and check that no outside force acts on it.

Three kinds of collision

Momentum is always conserved in a closed system. Kinetic energy, $\frac{1}{2} m v^{2}$ , which you met with the forms of energy, is a different story. In most collisions, some kinetic energy becomes heat, sound and damage. The total energy is still conserved, but less of it is left as motion. So collisions come in three kinds:

Elastic: kinetic energy is conserved as well as momentum. The objects bounce apart with no kinetic energy lost. Truly elastic collisions are rare in everyday life; hard balls such as billiard balls come close.
Inelastic: momentum is conserved, but some kinetic energy is lost. Most real collisions are of this kind.
Perfectly inelastic: the objects stick together and move as one. This loses the most kinetic energy that momentum conservation allows.

Why can't a collision lose all its kinetic energy? Because the total momentum must survive. If the total is 18 kg m/s before, something must still be moving afterwards. The stuck-together case leaves the least motion possible, but not zero, unless the total momentum was zero to begin with.

Worked examples

Worked Example 1 — Two carts that stick together

A 3 kg cart moving at 6 m/s to the right hits a 2 kg cart at rest. They stick together. Find their velocity just after. How much kinetic energy is lost?

Before: the 3 kg cart moves right at 6 m/s towards the 2 kg cart at rest. After: the carts are joined and move right together at 3.6 m/s. The total momentum is 18 kg m/s both times.

Take right as positive. Before: $3 \times 6 + 2 \times 0 = 18$ kg m/s. After, the joined carts have mass $3 + 2 = 5$ kg and one velocity $v$ . So $5 v = 18$ , and $v = 3.6$ m/s to the right.

Kinetic energy before: $\frac{1}{2} \times 3 \times 6^{2} = 54$ J. After: $\frac{1}{2} \times 5 \times 3. 6^{2} = 32.4$ J. So $54 - 32.4 = 21.6$ J is lost, which is 40% of the kinetic energy. It became heat, sound and the work done bending the coupling.

Stop and think

In Worked Example 1, how much momentum did the 3 kg cart lose, and how much did the 2 kg cart gain?

Show the answer

The 3 kg cart went from $3 \times 6 = 18$ kg m/s to $3 \times 3.6 = 10.8$ kg m/s, so it lost 7.2 kg m/s. The 2 kg cart went from 0 to $2 \times 3.6 = 7.2$ kg m/s, so it gained 7.2 kg m/s. The loss and the gain are equal, exactly as $Δ p_{A} = - Δ p_{B}$ says. Momentum was handed from one cart to the other. Kinetic energy behaves differently: the 3 kg cart lost 34.56 J, but the 2 kg cart gained only 12.96 J. The other 21.6 J did not pass to anything that moves.

Worked Example 2 — A head-on collision, with signs

A 4 kg cart moving at 5 m/s to the right meets a 6 kg cart moving at 3 m/s to the left. After the collision, the 4 kg cart moves at 2 m/s to the left. Find the velocity of the 6 kg cart.

Take right as positive. Then $u_{1} = + 5$ , $u_{2} = - 3$ and $v_{1} = - 2$ . Every leftward velocity gets a minus sign.

4 (5) + 6 (- 3) 20 - 18 10 v_{2} = 4 (- 2) + 6 v_{2} = - 8 + 6 v_{2} = 6 v_{2} = \frac{5}{3} \approx 1.67 m/s

The answer is positive, so the 6 kg cart now moves to the right. The collision reversed both carts. Is that believable? The total momentum before was $+ 2$ kg m/s, a little to the right, so after the collision the carts must share a small rightward total. They do: $4 (- 2) + 6 \times \frac{5}{3} = - 8 + 10 = + 2$ kg m/s.

One more check that momentum alone cannot give. Kinetic energy before: $\frac{1}{2} (4) (5^{2}) + \frac{1}{2} (6) (3^{2}) = 50 + 27 = 77$ J. After: $\frac{1}{2} (4) (2^{2}) + \frac{1}{2} (6) (\frac{5}{3})^{2} \approx 8 + 8.33 = 16.33$ J. The kinetic energy went down, so this is an inelastic collision, and it is possible. If the kinetic energy had gone up, the answer would be impossible: a collision cannot create energy from nowhere.

Worked Example 3 — An elastic collision between equal masses

A ball hits an identical ball at rest, head-on, in an elastic collision. Its speed is $u$ . What happens?

Call the mass $m$ . Momentum gives $m u = m v_{1} + m v_{2}$ , so $v_{1} + v_{2} = u$ . Kinetic energy is also conserved: $\frac{1}{2} m u^{2} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2}$ , so $v_{1}^{2} + v_{2}^{2} = u^{2}$ .

Square the first equation: $v_{1}^{2} + 2 v_{1} v_{2} + v_{2}^{2} = u^{2}$ . Subtract the second: $2 v_{1} v_{2} = 0$ . So one of the velocities is zero. The first ball cannot pass through the second, so $v_{1} = 0$ and $v_{2} = u$ .

The moving ball stops dead, and the other ball leaves with all its velocity. You can see this in a Newton's cradle, and in snooker when the cue ball hits a ball squarely and stops. The two balls simply swap velocities.

Now return to the two cars from the last lecture: car A, 1000 kg at 15 m/s to the right, and car B, 1500 kg at 10 m/s to the left. Their total momentum is zero.

Predict, then reveal — two cars lock together

Car A (1000 kg, 15 m/s right) and car B (1500 kg, 10 m/s left) collide head-on and lock together. Just after the collision, how does the wreck move?

The wreck stops dead. Taking right as positive, the total momentum before is 1000 × 15 + 1500 × (−10) = 15 000 − 15 000 = 0. After, the 2500 kg wreck has momentum 2500v = 0, so v = 0.

All the kinetic energy, 112 500 J from car A plus 75 000 J from car B, which is 187 500 J, becomes heat, sound and crumpled metal.

Police crash investigators use the same law in reverse. After a collision, the length of the skid marks and the road surface tell them how fast the locked wreck was moving just after the impact. Conservation of momentum then lets them work back to the speeds just before, using one more fact, such as the directions the two cars were travelling. They never need to know the forces during the crash, which is exactly the point of this lecture.

This is the most dangerous kind of crash. Each car loses all its momentum in a fraction of a second, so the forces on the people inside are very large. The lecture on impulse shows how car designers deal with that.

Now run your own collisions. Change the masses, the velocities and the kind of collision. Watch the total momentum while the carts hit each other.

Simulation — a collision on a track

Cart A starts on the left and cart B on the right. Right is positive. Choose the masses and velocities, choose sticky or bouncy, then press Play. The blue bar under the track is the total momentum. Does it ever change?

Mass of A: 3 kg

Velocity of A: 6 m/s

Mass of B: 2 kg

Velocity of B: 0 m/s

Three collisions, worked out with conservation of momentum (right is positive):

Before	Kind	After	Kinetic energy
3 kg at +6, 2 kg at 0	sticky	both at +3.6 m/s	54 J → 32.4 J
3 kg at +6, 2 kg at 0	bouncy	A at +1.2, B at +7.2 m/s	54 J → 54 J
1 kg at +6, 1 kg at 0	bouncy	A at 0, B at +6 m/s	18 J → 18 J

In every case the total momentum is the same before and after: 18 kg m/s, 18 kg m/s and 6 kg m/s.

Try the bouncy setting with equal masses and B at rest: A stops and B leaves at A's old velocity, as in Worked Example 3. Then try A light and fast against B heavy and still. The light cart bounces back, yet the total stays the same.

Three classic mistakes

Trap 1 — Dropping the signs

Velocities to the left are negative when right is positive. In Worked Example 2, writing $u_{2} = 3$ instead of $- 3$ gives $v_{2} = 7.67$ m/s, which is wrong. Write every velocity with its sign before you substitute anything.

Trap 2 — Assuming kinetic energy is conserved

Momentum is conserved in every collision in a closed system. Kinetic energy is conserved only in elastic ones. Do not use a kinetic energy equation unless the question says the collision is elastic.

Trap 3 — Forgetting the combined mass

When objects stick together, the mass after is the total mass. In Worked Example 1, $3 v = 18$ gives 6 m/s, as if the 2 kg cart were not there. The correct equation is $(3 + 2) v = 18$ .

Try it yourself

Take right as positive in every question. Check each answer by working out the total momentum before and after.

A 1200 kg car at 20 m/s runs into the back of an 800 kg car at rest. They lock together. Find their velocity just after, and the kinetic energy lost.

Hint 1
Total momentum before: $1200 \times 20 + 800 \times 0$ .

Hint 2
After, the mass is $1200 + 800 = 2000$ kg.

Full answer
Momentum before $= 24000$ kg m/s, so $2000 v = 24000$ and $v = 12$ m/s. Kinetic energy before: $\frac{1}{2} \times 1200 \times 2 0^{2} = 240000$ J. After: $\frac{1}{2} \times 2000 \times 1 2^{2} = 144000$ J. Lost: $96000$ J, which is 40%.
A 2 kg trolley moving at 4 m/s to the right meets a 3 kg trolley moving at 2 m/s to the left. They stick together. Find their velocity.

Hint 1
The 3 kg trolley's velocity is $- 2$ m/s.

Hint 2
Total momentum before: $8 - 6$ .

Full answer
Total before: $2 \times 4 + 3 \times (- 2) = 2$ kg m/s. After: $5 v = 2$ , so $v = 0.4$ m/s to the right.
A 0.05 kg ball at 10 m/s hits a 0.2 kg ball at rest, head-on. The small ball bounces straight back at 6 m/s. Find the velocity of the large ball. Was the collision elastic?

Hint 1
The small ball's velocity after is $- 6$ m/s.

Hint 2
$0.5 = 0.05 \times (- 6) + 0.2 v$ . Then compare kinetic energies.

Full answer
$0.5 = - 0.3 + 0.2 v$ , so $v = 4$ m/s to the right. Kinetic energy before: $\frac{1}{2} \times 0.05 \times 1 0^{2} = 2.5$ J. After: $\frac{1}{2} \times 0.05 \times 6^{2} + \frac{1}{2} \times 0.2 \times 4^{2} = 0.9 + 1.6 = 2.5$ J. No kinetic energy is lost, so the collision is elastic.
A 2 kg cart A at 3 m/s hits a 1 kg cart B at rest. B starts in front of A. Four students give the velocities after the collision (A, then B, in m/s): (a) 1 and 4; (b) 0 and 6; (c) 2 and 2; (d) 2.5 and 1. Which outcomes are physically possible?

Hint 1
Check three things: is the total momentum still 6 kg m/s? Is the kinetic energy no more than the 9 J before?

Hint 2
The third check: after the collision, A cannot be moving forwards faster than B, because B is in front of A.

Full answer
All four keep the momentum at 6 kg m/s. Now the energy: (a) $1 + 8 = 9$ J, (b) $0 + 18 = 18$ J, (c) $4 + 2 = 6$ J, (d) $6.25 + 0.5 = 6.75$ J. So (b) is impossible: it creates 9 J from nowhere. (d) is impossible too: A would be moving at 2.5 m/s behind B at 1 m/s, so A would have to pass through B. That leaves (a), which is the elastic outcome, and (c), where the carts stick together. Momentum alone is not enough to rule out an answer. Energy and common sense must agree too.

The bigger idea

Conservation of momentum is one of a small group of conservation laws, with conservation of energy and of electric charge. In 1918, the mathematician Emmy Noether proved a deep link between such laws and symmetry. Momentum is conserved because the laws of physics are the same everywhere: an experiment done here works the same way when it is moved somewhere else. That one fact about space is enough to guarantee that, in a closed system, the total momentum can never change.

This lecture compared momentum just before and just after a collision. The next lecture looks inside the collision itself: how big the force is, how long it lasts, and why making it last longer can save lives.

Key concepts in this lesson

In a closed system, total momentum is conserved: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
It follows from Newton's third law: the forces in a collision are equal and opposite and last the same time, so the momentum changes cancel.
Choose a positive direction first, and give every velocity its sign.
Elastic collisions conserve kinetic energy too; inelastic ones lose some; in perfectly inelastic ones the objects stick together and lose the most.
A collision can never increase the total kinetic energy, so energy is a useful check on any answer.