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Momentum, Collisions and Vehicle Safety — Answers

Answer sheet with model answers and success criteria for the Year 11 forces & motion worksheet.

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Success TutoringWorksheetsAnswer sheetYear 11 · Science · Forces & Motion

Momentum, Collisions and Vehicle Safety — Answers

Part A: Momentum

Momentum is the product of mass and velocity: p = mv, measured in kg m/s. A moving object with a large mass or a high velocity has a lot of momentum and is harder to stop. Because velocity has a direction, so does momentum.

1.Calculate the momentum of a 1200 kg car travelling at 15 m/s.

Answer: 18 000 kg m/s

  1. p = mv = 1200 × 15 = 18 000 kg m/s
2.Calculate the momentum of a 60 g tennis ball travelling at 50 m/s.

Answer: 3 kg m/s

  1. 60 g = 0.06 kg
  2. p = 0.06 × 50 = 3 kg m/s
3.Which has the greatest momentum?
  1. AA 70 kg runner at 8 m/s
  2. BA 1000 kg car at rest
  3. CA 0.5 kg ball at 30 m/s
  4. DA 20 kg dog at 10 m/s

Answer: A 70 kg runner at 8 m/s

Runner: 70 × 8 = 560 kg m/s; car: 0; ball: 15 kg m/s; dog: 200 kg m/s.

4.Momentum has direction as well as size, so an object moving left can be given negative momentum.

Answer: True

Momentum is a vector quantity, like velocity.

5.A loaded truck and a small car travel side by side at the same speed. Which is harder to stop, and why?

Answer: The loaded truck, because it has a much greater mass, so at the same velocity it has much more momentum. A larger force or a longer time is needed to reduce its momentum to zero.

Part B: Conservation of momentum

Law of conservation of momentum: in a collision or explosion, when no external forces act, the total momentum before = total momentum after.

In an explosion or push-off, objects start at rest, so the total momentum is zero before and after: the momenta of the parts are equal in size and opposite in direction.

6.A 2 kg cart moving at 3 m/s collides with a stationary 1 kg cart, and they stick together. Find their velocity just after the collision.

Answer: 2 m/s

  1. Momentum before = 2 × 3 + 1 × 0 = 6 kg m/s
  2. Momentum after = (2 + 1) × v = 3v
  3. 3v = 6, so v = 2 m/s
7.Two ice skaters stand still and push off each other. The 60 kg skater moves away at 2 m/s. How fast does the 40 kg skater move, and in which direction?

Answer: 3 m/s in the opposite direction

  1. Momentum before = 0
  2. 60 × 2 = 120 kg m/s one way, so the other skater has 120 kg m/s the other way
  3. v = 120 ÷ 40 = 3 m/s, opposite direction
8.A 50 kg person jumps from a stationary 200 kg boat towards a dock at 4 m/s. What happens to the boat?

Answer: It moves away from the dock at 1 m/s

  1. Momentum before = 0
  2. Person: 50 × 4 = 200 kg m/s towards the dock
  3. Boat: 200 ÷ 200 = 1 m/s away from the dock
9.Explain, using momentum, why the lighter skater moves faster.

Answer: Both skaters must have equal momentum in opposite directions so the total stays zero. With less mass, the lighter skater needs a greater velocity to have the same momentum.

Part C: Force and stopping time

Newton's second law can be written as force = change in momentum ÷ time taken, F = Δp ÷ t. For the same change in momentum, a longer stopping time means a smaller force.

10.A 0.15 kg ball travelling at 40 m/s is caught and stopped in 0.1 s. What average force does the catcher's hand exert?

Answer: 60 N

  1. Change in momentum = 0.15 × 40 = 6 kg m/s
  2. F = 6 ÷ 0.1 = 60 N
11.The catcher moves their hands back so the ball stops in 0.3 s instead. What is the force now?

Answer: 20 N

  1. F = 6 ÷ 0.3 = 20 N
  2. Tripling the time reduces the force to one-third
12.A 1000 kg car travelling at 20 m/s hits a wall. Calculate the average force if it stops in 0.1 s, and if a crumple zone extends the stopping time to 0.5 s.

Answer: 200 000 N; 40 000 N

  1. Change in momentum = 1000 × 20 = 20 000 kg m/s
  2. 0.1 s: 20 000 ÷ 0.1 = 200 000 N
  3. 0.5 s: 20 000 ÷ 0.5 = 40 000 N
13.A 70 kg passenger moving at 20 m/s is stopped by hitting the dashboard in 0.02 s, or by an airbag and seatbelt in 0.2 s. Calculate both forces.

Answer: 70 000 N; 7000 N

  1. Change in momentum = 70 × 20 = 1400 kg m/s
  2. Dashboard: 1400 ÷ 0.02 = 70 000 N
  3. Airbag and seatbelt: 1400 ÷ 0.2 = 7000 N
14.A bicycle helmet contains a layer of foam that crushes in an impact. How does this protect the rider?
  1. AIt increases the rider's momentum
  2. BIt increases the time taken for the head to stop, reducing the force
  3. CIt makes the head stop more quickly
  4. DIt removes the need for the head to stop

Answer: It increases the time taken for the head to stop, reducing the force

The change in momentum is the same, but spreading it over a longer time means a smaller force on the skull and brain.

Part D: Explain

15.Using the idea of momentum, explain how crumple zones, seatbelts and airbags work together to reduce injuries to people in a car crash.[4 marks]

Model answer: In a crash, the car and the people inside must lose all their momentum. The change in momentum is fixed by their mass and speed, but the force depends on how quickly that change happens: F = Δp ÷ t. Crumple zones at the front and back of the car are designed to fold up in a controlled way, so the car takes longer to stop and the force on the car and passenger compartment is smaller. A seatbelt stretches slightly and holds the person in their seat, so they slow down with the car over a longer time instead of hitting the dashboard; it also spreads the force over the strong parts of the body (chest and hips). An airbag inflates and then deflates as the person moves into it, increasing the time it takes for the head and chest to stop and spreading the force over a larger area. Together, these features increase the stopping time and reduce the force, so injuries are less severe.

Success criteria:

  • The change in momentum is the same; force depends on time (F = Δp ÷ t)
  • Crumple zones increase the car's stopping time
  • Seatbelts and airbags increase the person's stopping time
  • Concludes that longer time means smaller force and less injury
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