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AS · Practice questions · Newton's laws and momentum

Resultant force, then motion.

Six original Cambridge-style questions on applying F = ma to a resultant force, on momentum and the rate-of-change form, on third-law pairs, and on mass versus weight.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Find the resultant first.

01
Analysis
[2 marks]

A constant resultant force of 18 N acts on a 4.0 kg object initially at rest. Find its acceleration and its velocity after 5.0 s.

  • a = F / m = 18 / 4.0 = 4.5 m s⁻². ✓
  • v = u + at = 0 + 4.5 × 5.0 = 22.5 m s⁻¹. ✓
02
Analysis
[3 marks]

A 0.058 kg tennis ball is served, leaving the racket at 50 m s⁻¹ from rest. The racket is in contact with the ball for 5.0 ms. Find the change in momentum and the average force on the ball.

  • Δp = mΔv = 0.058 × 50 = 2.9 kg m s⁻¹. ✓
  • F = Δp / Δt = 2.9 / (5.0 × 10⁻³) = 580 N. ✓
03
Analysis
[3 marks]

A 1200 kg car is towed by a rope. The driving situation gives a forward pull of 3000 N while resistive forces total 1800 N. Find the acceleration of the car.

  • Resultant force = 3000 − 1800 = 1200 N. ✓
  • a = F / m = 1200 / 1200 = 1.0 m s⁻². ✓
04
Analysis
[2 marks]

A swimmer pushes backward on the wall of a pool and moves forward. State the Newton's third law pair to "the swimmer pushes backward on the wall", and explain how this makes the swimmer move.

  • The pair is "the wall pushes forward on the swimmer", equal in size and opposite in direction. ✓
  • This forward force from the wall is the resultant push on the swimmer, so by F = ma the swimmer accelerates forward. ✓
05
Analysis
[3 marks]

An astronaut has a mass of 80 kg. Taking g = 9.8 N kg⁻¹ on Earth and g = 1.6 N kg⁻¹ on the Moon, find the astronaut's weight in each place, and state the mass on the Moon.

  • On Earth: W = mg = 80 × 9.8 = 784 N. ✓
  • On the Moon: W = 80 × 1.6 = 128 N. ✓
  • Mass is unchanged at 80 kg; only the weight differs because g differs. ✓
06
Analysis
[3 marks]

A 0.50 kg ball moving at 4.0 m s⁻¹ is struck and moves off at 4.0 m s⁻¹ in the opposite direction. Find the magnitude of the change in momentum, and explain why it is not zero even though the speed is unchanged.

  • Take the new direction as positive: Δp = m(v − u) = 0.50(4.0 − (−4.0)) = 0.50 × 8.0 = 4.0 kg m s⁻¹. ✓
  • Momentum is a vector; the direction reversed, so even with equal speeds the momentum changed by 4.0 kg m s⁻¹. ✓

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