Define linear momentum and give its unit.
State Newton's second law of motion in terms of momentum.
Define the impulse of a force and state what the area under a force-time graph represents.
A tennis ball of mass 0.058 kg moves at 50 m s−1. Find its momentum.
A train of mass 8.0 × 104 kg moves at 15 m s−1. Find its momentum.
A 0.45 kg football is kicked from rest and leaves the foot at 24 m s−1 after a contact time of 0.010 s. Find the average force on the ball.
A 0.060 kg ball hits a wall at 12 m s−1 and rebounds at 9.0 m s−1. The contact lasts 0.015 s. Find the average force on the ball.
A constant force of 50 N acts on a 2.0 kg trolley, initially at rest, for 0.40 s. Find (a) the impulse and (b) the final speed of the trolley.
A 60 kg passenger travelling at 12 m s−1 is brought to rest in a crash. Find (a) the change in momentum, (b) the average force if stopped in 0.080 s, and (c) the average force if an airbag extends the stopping time to 0.50 s. Comment on the difference.
Total: 26 marks. Original work by the TheLucidSTEM team. Written in the style of the papers; no past paper question is reproduced.
Answer key · full worked solutionsclick to reveal
momentum is the product of mass and velocity, p = m v; unit kg m s−1. It is a vector.
the resultant force on a body equals the rate of change of its momentum, F = Δp / Δt.
impulse is the product of force and the time for which it acts, impulse = F Δt, and equals the change in momentum. The area under a force-time graph is the impulse.
p = m v = 0.058 × 50 = 2.9 kg m s−1.
p = m v = 8.0 × 104 × 15 = 1.2 × 106 kg m s−1.
Δp = m(v − u) = 0.45 × (24 − 0) = 10.8 kg m s−1; F = Δp / Δt = 10.8 / 0.010 = 1080 N.
take the approach direction as positive: u = +12, v = −9.0 m s−1. Δp = 0.060 × (−9.0 − 12) = −1.26 kg m s−1. F = 1.26 / 0.015 = 84 N, directed away from the wall.
(a) impulse = F Δt = 50 × 0.40 = 20 N s.
(b) impulse = Δp = m v, so v = 20 / 2.0 = 10 m s−1.
(a) Δp = m v = 60 × 12 = 720 kg m s−1.
(b) F = 720 / 0.080 = 9000 N.
(c) F = 720 / 0.50 = 1440 N.
the change in momentum is the same, but the longer stopping time provided by the airbag gives a much smaller force, around six times less, which reduces the risk of injury.