In a closed system, the total momentum before an interaction equals the total momentum after it. Momentum is mass × velocity, so you calculate it for each object, add the values with signs for direction and set the total before equal to the total after.
This lesson belongs to forces and momentum. It uses F = ma in the background, because a force over time changes momentum. Check the Cambridge syllabus page for whether momentum sits in the Core or Extended content for your exam year.
How do you set up a momentum problem?
A before-and-after table removes most confusion.
- Choose a positive direction and stick to it.
- Write p = mv for every object before the event, with signs. An object at rest has zero momentum.
- Write p = mv for every object after. If objects stick together, treat them as one object with the combined mass.
- Set total before = total after and solve for the unknown.
- Check units and direction, and check that the answer is reasonable.
Worked example 1: trolleys that stick together
Invented situation: trolley A has mass 2.0 kg and moves at 3.0 m/s to the right. It hits trolley B, mass 1.0 kg, which is at rest. They stick together. Find their common velocity.
Step 1, positive direction: right is positive.
Step 2, before: p(A) = 2.0 × 3.0 = 6.0 kg m/s. p(B) = 1.0 × 0 = 0. Total = 6.0 kg m/s.
Step 3, after: combined mass = 2.0 + 1.0 = 3.0 kg, velocity v. Momentum = 3.0 × v.
Step 4, equate: 3.0v = 6.0, so v = 2.0 m/s to the right.
Check: 3.0 × 2.0 = 6.0 kg m/s, the same as before.
Worked example 2: objects moving apart
Invented situation: two carts, 3.0 kg and 1.0 kg, sit together at rest with a compressed spring between them. The spring is released. The 1.0 kg cart moves left at 6.0 m/s. Find the velocity of the 3.0 kg cart.
Left is negative. Before: total momentum = 0. After: 1.0 × (−6.0) + 3.0 × v = 0, so −6.0 + 3.0v = 0 and v = 2.0. The heavier cart moves at 2.0 m/s to the right.
Check: 3.0 × 2.0 = 6.0 to the right and 1.0 × 6.0 = 6.0 to the left, so the total is zero.
The mistake to watch for
A common slip is to ignore direction when objects move towards each other.
Mistaken working: cart A, 2.0 kg at 3.0 m/s to the right, meets cart B, 1.0 kg at 4.0 m/s to the left, and they stick. Total momentum = 6.0 + 4.0 = 10 kg m/s, so v = 10 ÷ 3.0 = 3.3 m/s.
The student added the magnitudes and gave no direction. The two momenta point opposite ways.
The correction is to take right as positive: 2.0 × 3.0 + 1.0 × (−4.0) = 6.0 − 4.0 = 2.0 kg m/s. Then v = 2.0 ÷ 3.0 = 0.67 m/s to the right. Always ask whether the two objects are moving the same way or opposite ways before you add.
Check yourself
Try these, then open each answer.
1. Calculate the momentum of a 1500 kg car travelling at 12 m/s.
Show answer
p = mv = 1500 × 12 = 18 000 kg m/s in the direction of travel.
2. A 0.20 kg trolley at 5.0 m/s hits a stationary 0.30 kg trolley and they stick. Find the common speed.
Show answer
Before: 0.20 × 5.0 = 1.0 kg m/s. After: (0.20 + 0.30) × v = 0.50v. So v = 1.0 ÷ 0.50 = 2.0 m/s.
3. A 60 kg skater and a 40 kg skater stand still, then push apart. The 40 kg skater moves at 1.5 m/s. Find the speed and direction of the 60 kg skater.
Show answer
Total momentum is zero. 40 × 1.5 = 60 kg m/s one way, so the 60 kg skater has 60 kg m/s the other way. v = 60 ÷ 60 = 1.0 m/s in the opposite direction.
Where this leads next
The next lesson separates two ideas that get mixed up in force questions: balanced forces versus the absence of forces. To test everything together, use the forces and momentum practice set.
Sign errors are hard to spot in your own work. A teacher in online one-to-one Physics tuition can watch you set up a table and show exactly where a sign went wrong.