Momentum and Impulse
At the 2018 World Pool Championship in Sheffield, the cue ball (0.17 kg) was struck from rest to 8 m/s. The cue tip was in contact for approximately 0.002 s, delivering an impulse of $J = \Delta p = 0.17 \times 8 = 1.36\text{ N·s}$. The average force on the cue ball during that 2 ms contact was $F = 1.36/0.002 = 680\text{ N}$ — over 400 times the weight of the ball — from a seemingly gentle tap.
Practise this lesson
Four printable worksheets — foundations to exam-style.
A 1500 kg car travelling at 20 m/s brakes gently to rest in 8 seconds. The same car hits a wall and stops in 0.08 seconds. In both cases the change in momentum is identical. Why does hitting the wall cause so much more damage?
A 2 kg object moving at 5 m/s has a momentum of:
Know
- $p = mv$ — momentum definition and unit
- $J = F\Delta t$ — impulse definition and unit
- $J = \Delta p$ — the impulse-momentum theorem
- $F = \Delta p/\Delta t$ — average force from impulse
Understand
- Why momentum is a vector — direction matters
- Why extending contact time reduces force for same $\Delta p$
- Why bouncing produces larger $\Delta p$ than sticking
- How safety devices exploit the impulse-momentum theorem
Can Do
- Calculate momentum with correct sign convention
- Calculate impulse from force and time
- Apply $J = \Delta p$ for direction reversals
- Find average force from $\Delta p$ and contact time
Core Content
Imagine stopping a golf ball travelling at 50 m/s with your hand — it stings. Now imagine stopping a bowling ball (5 kg) moving at just 1 m/s — it barely stings at all, but it is much harder to stop. Both require roughly the same effort, yet their masses and speeds are wildly different. The property they share is the product $mv$ — and stopping either one requires the same impulse.
Momentum is a vector — direction is essential. Always define a positive direction before calculating. Objects moving in the positive direction have positive momentum.
Unit: kg m/s (same as N s).
Momentum: $p = mv$ (kg m/s); momentum is a vector — always define a positive direction first; when an object bounces and reverses direction, $\Delta p = mv_f - mv_i$ uses signed velocities, giving a much larger magnitude than if the object simply stopped.
Pause — copy the highlighted definition and sign convention into your book before moving on.
True or false: A 0.5 kg ball bouncing off a wall at the same speed it arrived has a change in momentum of zero.
We just saw that momentum is the product of mass and velocity and requires a sign convention. That raises a question: what causes momentum to change, and how does the magnitude of the change relate to force and time? This card answers it → impulse $J = F\Delta t = \Delta p$; for the same $\Delta p$, shorter contact time means larger average force.
Impulse is the "dose" of force delivered over time. The same impulse can be delivered by a large force for a short time, or a small force for a long time — and both produce the same change in momentum.
Unit: N s = kg m/s.
Rearranged for average force: $F = \Delta p/\Delta t$. This means for the same momentum change, halving the contact time doubles the force — which is why hitting a wall is more dangerous than braking.
A 0.16 kg cricket ball travelling at 40 m/s is struck and returns at 35 m/s. Contact time = 2 ms. Find the average force on the ball.
Impulse-momentum theorem: $J = F\Delta t = \Delta p = mv_f - mv_i$ (N s = kg m/s); average force $F = \Delta p/\Delta t$; for direction-reversal problems always subtract signed velocities ($\Delta v = v_f - v_i$, not the difference of magnitudes); convert ms to s before calculating.
Pause — copy the highlighted theorem and the direction-reversal warning into your book before moving on.
A 0.5 kg ball travelling at 8 m/s bounces off a wall and returns at 6 m/s. Taking toward wall as positive, the change in momentum is:
For the same change in momentum, if contact time is halved, the average force is ______.
Activities
A student's impulse calculation contains four errors. Find each, explain why it is wrong, and write the correct version.
Problem: A 0.16 kg cricket ball at 40 m/s is struck and returns at 35 m/s in the opposite direction. Contact time = 2 ms. Find average force.
Student's working: "I don't need to define a positive direction — magnitudes only. $\Delta v = 40 - 35 = 5$ m/s. $\Delta p = 0.16 \times 5 = 0.8$ N s. $F = \Delta p/\Delta t = 0.8/2 = 0.4$ N."
A 75 kg driver travelling at 15 m/s is brought to rest. (a) With airbag: contact time = 0.12 s. (b) Without airbag (hits steering wheel): contact time = 0.008 s. Calculate average force in each case and explain what the difference means physiologically.
UnderstandBand 3(3 marks) 1. Explain why a car crumple zone reduces injuries in a crash even though the crumple zone doesn't reduce the total change in momentum of the occupant. Use the impulse-momentum theorem in your explanation.
ApplyBand 4(3 marks) 2. A 0.058 kg tennis ball is hit with a racquet. It arrives at 20 m/s and leaves at 28 m/s in the opposite direction. Contact time is 4 ms. Calculate the magnitude of the average force the racquet exerts on the ball.
AnalyseBand 5(4 marks) 3. A force-time graph shows a collision. The area under the graph between $t = 0$ and $t = 0.05$ s is 45 N s. The object (mass 3 kg) was initially at rest. (a) What is the impulse? (b) What is the final velocity? (c) If a second force-time graph had the same area but a flatter, wider shape — compare the peak forces and contact times of the two scenarios.
Show all answers
Short Answer — Model Answers
Q1 (3 marks): The occupant's momentum must change from $mv$ (moving at car speed) to 0 (stopped). This $\Delta p$ is fixed regardless of how the stop occurs. By the impulse-momentum theorem: $F\Delta t = \Delta p$. A crumple zone increases the collision time $\Delta t$. Since $\Delta p$ is constant, a larger $\Delta t$ means a smaller average force $F = \Delta p/\Delta t$. Reduced force means reduced injury.
Q2 (3 marks): Define positive = direction ball arrives. $v_i = +20\text{ m/s}$; $v_f = -28\text{ m/s}$. $\Delta p = 0.058 \times (-28 - 20) = 0.058 \times (-48) = -2.784\text{ kg m/s}$. $\Delta t = 4\text{ ms} = 0.004\text{ s}$. $F = \Delta p/\Delta t = -2.784/0.004 = -696\text{ N}$. Magnitude = 696 N (in direction ball arrived from).
Q3 (4 marks): (a) $J = 45\text{ N s}$ (area under F-t graph = impulse). (b) $J = \Delta p = mv_f - 0$; $v_f = J/m = 45/3 = 15\text{ m/s}$. (c) Both graphs have the same area (same impulse = 45 N s). The flatter, wider graph has a longer contact time and a lower peak force. Both produce the same final velocity of 15 m/s. This is the principle behind safety padding — same impulse, lower peak force.
The wall causes more damage because $F = \Delta p/\Delta t$ — identical momentum change divided by a much shorter time gives a much larger average force. The 2018 World Pool Championship Sheffield cue-ball example is the clearest demonstration: a gentle-looking tap over 0.002 s produces 680 N on a 0.17 kg ball — 400 times its weight — entirely because of the short contact time, not any extraordinary applied force.