Physics · Units 3 & 4

Momentum and Impulse

Understand momentum and impulse the easy way, with plain English intuition, an interactive collision simulation, conservation of momentum, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.

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Two snooker balls click together and fly apart, two cars crunch and crumple, a rocket shoves gas one way and leaps the other. In every one of these the same hidden quantity is being carefully passed around and never lost. That quantity is momentum, and once you can track it you can predict the result of almost any collision without knowing a single thing about the messy forces inside it.

What momentum really is

Momentum is just a measure of how hard something is to stop. A slow truck and a fast cricket ball can be equally tricky to halt, because momentum depends on both the mass and the velocity.

  • Momentum is mass times velocity, p=mvp = mv. Double the speed and you double the momentum; double the mass and you double it too.
  • It is a vector, so direction matters. A ball going left has the opposite sign to a ball going right, and you must keep track of that sign in every calculation.

A heavy object moving slowly can carry the same momentum as a light object moving fast. That is why a gently rolling truck and a whizzing tennis ball can each be a handful to stop.

Impulse: how momentum gets changed

To change something’s momentum you have to push on it, and the longer you push the bigger the change. That push over time is called the impulse.

Impulse is the force multiplied by how long it acts, I=F ΔtI = F\,\Delta t, and it equals the change in momentum, Δp\Delta p. This is just Newton’s second law turned inside out. A small force over a long time can deliver the same impulse as a big force over a short time, which is exactly why follow through matters in sport and why soft landings save your knees.

When objects collide

Picture two carts on a track. Before they meet, each carries its own momentum. They bump, push on each other, and move off again. Add up the momentum of both carts before the bump and it equals the total after the bump, every time, as long as nothing outside the carts pushes on them.

BEFOREAmovingBat restAFTERstuck togethercommon velocitytotal momentum before=total momentum after

The blue cart A starts with all the momentum and grey cart B is still. After they lock together they share that momentum as one combined object, shown by the red arrow. Nothing is added and nothing is lost, so the total before equals the total after.

Elastic or inelastic

Momentum survives every collision, but kinetic energy does not. Whether energy is kept tells you what kind of collision you are watching.

  • An elastic collision keeps the total kinetic energy. The objects bounce cleanly off each other, like ideal snooker balls.
  • An inelastic collision loses kinetic energy to heat, sound and bending. When the objects stick together it is called perfectly inelastic, the most lossy case of all.

In both kinds the momentum still balances perfectly. Only kinetic energy is allowed to go missing, and it goes missing whenever the objects deform or stick.

See it for yourself

Set the masses and velocities of two balls, choose how bouncy the collision is, and watch the momentum bars stay balanced before and after. Slide the elasticity down to zero to make the balls stick together, and notice the total momentum holds steady even as kinetic energy is lost.

Interactive simulation, Collision Lab Source: PhET Interactive Simulations, University of Colorado Boulder (CC BY 4.0)

How to actually solve one

Collision questions follow one reliable recipe.

  1. Choose a positive direction and write down the velocity of everything, using a minus sign for anything going the other way.
  2. Write the total momentum before, ∑pbefore=m1u1+m2u2\sum p_{\text{before}} = m_1 u_1 + m_2 u_2.
  3. Write the total momentum after. If the objects stick together they share one velocity vv, so ∑pafter=(m1+m2) v\sum p_{\text{after}} = (m_1 + m_2)\,v.
  4. Set before equal to after and solve for the unknown.

For an impulse question instead, use I=Δp=m(vf−vi)I = \Delta p = m(v_f - v_i), and remember the rebound velocity is negative. Watch the signs, because a reversed velocity is where most marks are lost.

See the recipe in action in the Worked Examples tab, then test yourself in Try It.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

Write the formula for momentum and say what kind of quantity it is.
What is impulse, and what does it equal?
When is total momentum conserved?
In which kind of collision is kinetic energy conserved?
Why does an airbag reduce the force on you in a crash?
Recall · Newton's Laws and Forces
State Newton’s second law, which impulse is built from.
Recall · Work, Energy and Power
Write the formula for kinetic energy.

Worked examples

Worked Example 1Two cars stick together

A 10001000 kg car moving at 2020 m/s collides with and sticks to a stationary 15001500 kg car. Find their common velocity just after the collision.

  1. 1

    No external force acts during the bump, so total momentum is the same before and after. Before, only the first car is moving.

    pbefore=m1u1=1000×20=20000 kg m/sp_{\text{before}} = m_1 u_1 = 1000 \times 20 = 20000 \text{ kg m/s}
  2. 2

    After, the two cars are stuck together and move as one object of mass 1000+1500=25001000 + 1500 = 2500 kg at a shared velocity vv.

    pafter=(m1+m2) v=(1000+1500) v=2500 vp_{\text{after}} = (m_1 + m_2)\,v = (1000 + 1500)\,v = 2500\,v
  3. 3

    Set the before and after momentum equal and solve for vv.

    20000=2500 v  ⟹  v=200002500=8.0 m/s20000 = 2500\,v \implies v = \frac{20000}{2500} = 8.0 \text{ m/s}
Answer
v=8.0 m/s in the original directionv = 8.0 \text{ m/s in the original direction}
Worked Example 2A ball rebounds off a wall

A 0.160.16 kg ball hits a wall at 8.08.0 m/s and rebounds at 6.06.0 m/s in the opposite direction. Find the magnitude of the impulse on the ball.

  1. 1

    Pick a direction as positive. Call the way in ++, so the rebound velocity is negative.

    vi=+8.0 m/s,vf=−6.0 m/sv_i = +8.0 \text{ m/s}, \qquad v_f = -6.0 \text{ m/s}
  2. 2

    Impulse equals the change in momentum, which is mass times the change in velocity.

    I=m(vf−vi)=0.16(−6.0−8.0)I = m(v_f - v_i) = 0.16(-6.0 - 8.0)
  3. 3

    Work out the bracket first, then multiply by the mass.

    I=0.16×(−14)=−2.24 N sI = 0.16 \times (-14) = -2.24 \text{ N s}
  4. 4

    The minus sign just means the impulse points back the way the ball came. The question asks for the magnitude.

    ∣I∣=2.24 N s|I| = 2.24 \text{ N s}
Answer
∣I∣=2.24 N s|I| = 2.24 \text{ N s}

Practice questions

Practice test

Try it yourself

6 questions, 10 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.A 2.02.0 kg trolley moves at 3.03.0 m/s. Its momentum is:

1mark
Need a hint?
Momentum is mass times velocity, p=mvp = mv.

Q2.A net force of 5.05.0 N acts on a cart for 4.04.0 s. The impulse delivered to the cart is:

1mark
Need a hint?
Impulse is force times the time it acts, I=F ΔtI = F\,\Delta t.

Q3.A 4.04.0 kg object moving at 5.05.0 m/s collides with and sticks to a stationary 6.06.0 kg object. Their common velocity just after the collision is:

1mark
Need a hint?
Total momentum before equals total momentum after. Before, only one object moves.

Q4.In which type of collision is kinetic energy conserved?

1mark
Need a hint?
Think about which kind of collision loses energy to heat, sound and deformation.

Q5.A 0.500.50 kg ball travelling at 4.04.0 m/s is struck and its velocity reverses to 4.04.0 m/s in the opposite direction. Calculate the magnitude of the impulse on the ball. Show your working.

3marks

Work this on paper. The worked solution appears once you submit.

Q6.In a crash test, the 5.05.0 kg head of a dummy hits an inflated airbag at 1212 m/s and is brought to rest. (a) Calculate the magnitude of the impulse the airbag delivers to the head. (b) Justify the use of airbags by comparing the average force on the head with that in a collision at the same speed against just the steering wheel. No calculation is needed.

3marks

Work this on paper. The worked solution appears once you submit.

VCAA 2025 Physics Exam, Section B Q2

Frequently asked questions

What is the difference between momentum and impulse?
Momentum is how much motion an object has right now, mass times velocity. Impulse is the change in that momentum, delivered by a force acting over a time. Apply an impulse and you change the momentum by exactly that amount.
Is momentum always conserved?
Total momentum is conserved whenever there is no external force on the system, which is the case during a quick collision or explosion. The forces the objects push on each other are internal and cancel, so the total before equals the total after.
Why does kinetic energy disappear when cars stick together?
In an inelastic collision some kinetic energy turns into heat, sound and the work of bending metal, so it is no longer motion energy. Momentum still balances because it does not care about those losses, but kinetic energy does not add up the same before and after.