Practise Changes in momentum. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy5 marks
(a) Complete the sentence. Force is equal to the rate of change of ............ Tick (✓) one box.[1]
momentum
kinetic energy
velocity
distance
(b) Which two of these safety features work by increasing the time taken for a person’s momentum to change? Tick (✓) two boxes.[2]
Air bag
Brake lights
Crash mat
Headlights
Mirror
(c) An object’s momentum changes by 12 kg m/s in 0.20 s. Calculate the force on the object. Use the equation: force = change in momentum ÷ time taken[2]
Show the answer and mark scheme
(a)Answer: momentum
(b)Answer: Air bag; Crash mat
(c)Answer: 60 N
F = 12 ÷ 0.20
60 (N)
Question 2Medium7 marks
(a) resultant force = mass × acceleration acceleration = change in velocity ÷ time taken Show how these two equations combine to give: force = change in momentum ÷ time taken[2]
(b) A footballer kicks a stationary ball of mass 0.43 kg. The ball leaves the boot at 24 m/s. Calculate the change in momentum of the ball.[2]
(c) The boot is in contact with the ball for 0.012 s. Calculate the average force of the boot on the ball. Use the Physics Equations Sheet.[2]
(d) Explain why a longer contact time, with the same change in momentum, would mean a smaller force on the ball.[1]
Show the answer and mark scheme
(a)Answer: F = m a = m (Δv ÷ t) = (m Δv) ÷ t, and m Δv is the change in momentum.
substituting a = Δv ÷ t into F = m a gives F = m Δv ÷ t
m Δv is the change in momentum (so force = change in momentum ÷ time)
(b)Answer: 10.3 kg m/s
Δp = 0.43 × 24
10.3 (kg m/s)
(c)Answer: 860 N
F = 10.32 ÷ 0.012
860 (N)
(d)Answer: Force is the rate of change of momentum, so the same change over a longer time needs a smaller force.
force = change in momentum ÷ time, so for the same change in momentum a longer time gives a smaller rate of change of momentum (a smaller force)
Question 3Hard9 marks
Modern cars have safety features including crumple zones, seat belts and air bags.
(a) Explain how these safety features reduce injuries to people in the car in a collision. Use ideas about momentum in your answer.[6]
(b) A passenger of mass 70 kg is travelling at 15 m/s when the car crashes. Without an air bag, the passenger’s head and chest would stop in 0.080 s. With the air bag, they stop in 0.40 s. Calculate the decrease in the average force on the passenger when the air bag is used. Use the Physics Equations Sheet.[3]
Show the answer and mark scheme
(a)
in a collision the car and its passengers lose a large amount of momentum in a short time
force = rate of change of momentum (F = m Δv ÷ Δt)
for the same change in momentum, increasing the time taken reduces the force
crumple zones crumple / deform, increasing the time taken for the car (and passengers) to stop
seat belts stretch slightly, increasing the time taken for the passenger to stop
air bags inflate and then compress / deflate, increasing the time taken for the head and chest to stop
seat belts and air bags also spread the force over a larger area, reducing the pressure on the body
smaller forces (and smaller decelerations) cause less severe injuries
Marked with levels of response: the full level descriptors are in the app.
(b)Answer: 10 500 N
change in momentum = 70 × 15 = 1050 (kg m/s)
forces: 1050 ÷ 0.080 = 13 125 N and 1050 ÷ 0.40 = 2625 N