Momentum & Collisions — Drills · Grade 12 Science

Chapter: Work, Energy & Momentum

What these drills covers

By the end of Momentum & Collisions you should be able to:

6 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: momentum, impulse, conservation of momentum, elastic collision, inelastic collision.

Problem 1

Find the momentum of a 6.0 kg object moving at 3.0 m/s.
Answer and explanation

18 kg·m/s

p = mv = 6.0 × 3.0 = 18 kg·m/s.

Problem 2

Find the momentum of a 0.20 kg ball moving at 30 m/s.
Answer and explanation

6 kg·m/s

p = mv = 0.20 × 30 = 6.0 kg·m/s.

Problem 3

A 4.0 kg cart moving at 3.0 m/s collides with and sticks to a 2.0 kg cart at rest. Find their common velocity afterward.
Answer and explanation

2 m/s

Conservation of momentum: 4.0 × 3.0 + 2.0 × 0 = (4.0 + 2.0)v′, so 12 = 6.0v′ and v′ = 2.0 m/s.

Problem 4

Kinetic energy is NOT conserved in which type of collision?
  1. An elastic collision
  2. An inelastic collision
  3. Both types
  4. Neither type
Answer and explanation

An inelastic collision

In an inelastic collision, some kinetic energy is converted to heat, sound, and deformation. In an elastic collision, kinetic energy is conserved. Momentum is conserved in both.

Problem 5

Explain, using impulse, why a car's airbag reduces the force on a passenger during a crash.
Answer and explanation

The passenger's change in momentum in the crash is fixed by their mass and speed. Since impulse FΔt equals that change in momentum, F = Δp/Δt. An airbag increases the stopping time Δt, and a larger Δt makes the average force F smaller, reducing injury.

Problem 6

A 2.0 kg cart moving right at 4.0 m/s collides head-on with a 3.0 kg cart moving left at 1.0 m/s, and the two stick together. Find the speed of the combined carts.
Answer and explanation

1 m/s

Taking right as positive: 2.0 × 4.0 + 3.0 × (−1.0) = (2.0 + 3.0)v′, so 8 − 3 = 5.0v′, giving 5 = 5.0v′ and v′ = 1.0 m/s to the right.

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