OCR A Physics · Module 3: Forces and motion
Newton’s laws & momentum: practice questions with answers
6 exam-style questions on newton’s laws & momentum. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- Using the right equation2 marks
A 1200 kg car has a driving force of 3800 N and a total resistive force of 1250 N. Calculate its acceleration.
a = m s⁻²
Show the answer and working
Answer a = 2.13 m s⁻²
Resultant force = 3800 - 1250 = 2550 N. a = F ÷ m = 2550 ÷ 1200 = 2.12 m s⁻².
- Got 3.17 m s⁻²? You used the driving force alone. The resultant force is 3800 - 1250 = 2550 N, so a = F ÷ m = 2.12 m s⁻².
- Got 4.21 m s⁻²? You added the resistive force. It opposes motion, so the resultant is 3800 - 1250 = 2550 N and a = 2.12 m s⁻².
- Using the right equation3 marks
A 0.85 kg trolley moving at 1.8 m s⁻¹ collides with a stationary 0.40 kg trolley. They stick together. Calculate their common speed after the collision.
v = m s⁻¹
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Answer v = 1.22 m s⁻¹
Momentum before = 0.85 × 1.8 = 1.53 kg m s⁻¹. After: 1.25v, so v = 1.53 ÷ 1.25 = 1.22 m s⁻¹.
- Got 0.9 m s⁻¹? You averaged the speeds. Momentum is conserved: 0.85 × 1.8 = (0.85 + 0.40)v, so v = 1.22 m s⁻¹.
- Got 3.83 m s⁻¹? You divided by the mass of the second trolley only. After the collision both move together, so divide by 1.25 kg: v = 1.22 m s⁻¹.
- Unit prefixes3 marks
A 0.16 kg ball hits a wall at 22 m s⁻¹ and rebounds at 14 m s⁻¹ along the same line. The contact lasts 8.0 ms. Calculate the average force exerted on the ball by the wall.
F = N
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Answer F = 720 N
Take the rebound direction as positive: Δv = 14 - (-22) = 36 m s⁻¹. F = mΔv ÷ Δt = 0.16 × 36 ÷ 8.0 × 10⁻³ = 720 N.
- Got 160 N? You subtracted the speeds. The ball reverses direction, so Δv = 22 + 14 = 36 m s⁻¹ and F = mΔv ÷ t = 720 N.
- Got 0.72 N? You used 8.0 ms as 8.0 s. Convert: 8.0 × 10⁻³ s, so F = 0.16 × 36 ÷ 0.0080 = 720 N.
- Rearranging equations2 marks
A 3.0 kg gun fires a 0.020 kg bullet at 400 m s⁻¹. The gun is free to move. Calculate the recoil speed of the gun.
v = m s⁻¹
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Answer v = 2.67 m s⁻¹
Total momentum is zero before and after: 3.0v = 0.020 × 400, so v = 8.0 ÷ 3.0 = 2.67 m s⁻¹.
- Got 60000 m s⁻¹? You inverted the mass ratio. Momentum conservation: 3.0 × v = 0.020 × 400, so v = 2.67 m s⁻¹.
- Got 8 m s⁻¹? This is the bullet's momentum, not the gun's speed. Divide by the gun mass: 8.0 ÷ 3.0 = 2.67 m s⁻¹.
- Using the right equation1 mark
A book rests on a table. Which force is the Newton's third law pair of the weight of the book?
- A The normal contact force from the table on the book.
- B The normal contact force from the book on the table.
- C The gravitational pull of the book on the Earth.
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Answer C: The gravitational pull of the book on the Earth.
Third law pairs act on different bodies and are the same type of force: the Earth pulls the book, and the book pulls the Earth.
- Why not A? That force acts on the same body as the weight, so they cannot be a third law pair.
- Why not B? This is the third law pair of the table's force on the book, not of the weight.
- Using the right equation1 mark
Two trolleys collide inelastically. Which quantity is conserved?
- A Momentum only
- B Kinetic energy only
- C Both momentum and kinetic energy
Show the answer and working
Answer A: Momentum only
Momentum is conserved in all collisions where no external force acts; kinetic energy is only conserved when the collision is elastic.
- Why not B? Kinetic energy is not conserved in an inelastic collision.
- Why not C? Kinetic energy is lost in an inelastic collision, so both cannot be conserved.