Course: Combined Science + Separate Physics (force and change in momentum is S only) | Tier: Higher | Linked equations: p = mv, F = mΔv/Δt | Lesson: Crash Test Engineer
Notes
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- Momentum = mass × velocity (p = m × v). Unit: kg m/s. Momentum is a vector, so direction matters.
- Conservation of momentum: in a closed system, the total momentum before an event is equal to the total momentum after the event.
- In an explosion, the total momentum before is zero, so afterwards the parts have equal and opposite momentum.
Solving collision questions
- Choose a positive direction (e.g. right is positive, left is negative).
- Work out the total momentum before.
- Set total momentum after equal to it.
- Solve for the unknown.
Safety features (S only)
✏️ Copy into your book (S only)
Force equals the rate of change of momentum. Air bags, seat belts, crumple zones, cycle helmets, gymnasium mats and cushioned surfaces all increase the time taken for the momentum to change. This reduces the rate of change of momentum, so the force on the person is smaller and injuries are less serious.
Questions
Q1 (H) State the principle of conservation of momentum.
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In a closed system, the total momentum before an event is equal to the total momentum after the event.
Q2 (H) Calculate the momentum of a 1200 kg car travelling at 15 m/s.
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F: p = m v
I: p = 1200 × 15
A: 18 000 kg m/s
Q3 (H) A 2.0 kg trolley moving at 3.0 m/s collides with a stationary 1.0 kg trolley. They stick together. Calculate their velocity after the collision.
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Momentum before = 2.0 × 3.0 + 1.0 × 0 = 6.0 kg m/s
Momentum after = (2.0 + 1.0) × v
6.0 = 3.0 × v
v = 2.0 m/s in the original direction
Q4 (H) A 4.0 kg gun fires a 0.010 kg bullet at 400 m/s. Calculate the recoil velocity of the gun.
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Momentum before = 0
Bullet momentum = 0.010 × 400 = 4.0 kg m/s
Gun momentum must be −4.0 kg m/s
v = −4.0 ÷ 4.0 = 1.0 m/s backwards
Q5 (H) Two ice skaters stand still, then push apart. Explain why they move off in opposite directions.
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The total momentum before is zero. Momentum is conserved, so afterwards the total must still be zero. They must have equal momentum in opposite directions.
Q6 (S only, H) Explain how a car air bag reduces injuries in a crash.
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The air bag increases the time it takes for the person’s momentum to reduce to zero. This reduces the rate of change of momentum, so the force on the person is smaller.
Q7 (S only, H) A 70 kg person moving at 10 m/s is brought to rest by a seat belt in 0.20 s. Calculate the average force on them.
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F: F = m Δv ÷ Δt
I: F = 70 × 10 ÷ 0.20
A: 3500 N
