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9702 · 3.1

Momentum and Newton's laws of motion — practice questions

Practice and worked examples for 9702 Momentum and Newton's laws of motion. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A football of mass 0.45 kg is moving horizontally towards a player at 20 m/s. The player kicks the ball, causing it to move in the opposite direction with a speed of 30 m/s. If the player's boot is in contact with the ball for 0.050 s, what is the average force exerted on the ball by the player?

Show solution outline
  1. Define a positive direction and list knowns. Let the final direction of the ball be positive. Mass (m) = 0.45 kg Initial velocity (u) = -20 m/s (since it's opposite to the final direction) Final velocity (v) = +30 m/s Time interval (Δt) = 0.050 s
  2. Calculate the change in momentum (Δp). Δp = p_final - p_initial = mv - mu Δp = (0.45 kg)(+30 m/s) - (0.45 kg)(-20 m/s) Δp = 13.5 - (-9.0) Δp = 22.5 kg m/s (or 22.5 Ns)
  3. Use the impulse-momentum theorem (Newton's Second Law) to find the average force. F_avg = Δp / Δt
  4. Substitute the values and calculate the force: F_avg = 22.5 Ns / 0.050 s F_avg = 450 N
  5. State the final answer with direction: The average force exerted on the ball is 450 N in the direction of its final velocity.

Worked example 2

A 2.0 kg trolley moving at 3.0 m/s collides head-on with a stationary 1.0 kg trolley. After the collision, the 2.0 kg trolley continues in its original direction at 1.0 m/s. Calculate the velocity of the 1.0 kg trolley after the collision.

Show solution outline
  1. Identify knowns: Mass of trolley 1 (m1) = 2.0 kg Initial velocity of trolley 1 (u1) = +3.0 m/s (let's define original direction as positive) Mass of trolley 2 (m2) = 1.0 kg Initial velocity of trolley 2 (u2) = 0 m/s Final velocity of trolley 1 (v1) = +1.0 m/s Final velocity of trolley 2 (v2) = ?
  2. Apply the principle of conservation of momentum: Total momentum before = Total momentum after m1u1 + m2u2 = m1v1 + m2v2
  3. Substitute the values into the equation: (2.0 kg)(+3.0 m/s) + (1.0 kg)(0 m/s) = (2.0 kg)(+1.0 m/s) + (1.0 kg)(v2)
  4. Calculate the unknown velocity: 6.0 kg m/s + 0 = 2.0 kg m/s + (1.0 kg)v2 6.0 = 2.0 + v2 v2 = 6.0 - 2.0 v2 = +4.0 m/s
  5. State the final answer with direction: The 1.0 kg trolley moves at 4.0 m/s in the original direction of the 2.0 kg trolley.
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