Is momentum a vector or a scalar?

A vector. It points the same way as the velocity. In Year 11 problems are along one line, so the direction is shown by a + or − sign, and your final answer should state the direction in words.

What is the difference between elastic and inelastic collisions?

Momentum is conserved in both. In an elastic collision the total kinetic energy is also the same before and after. In an inelastic collision some kinetic energy is transformed into heat, sound and deformation, so the total KE after is less. If the objects stick together, the collision is perfectly inelastic.

Is momentum conserved in an inelastic collision?

Yes, as long as no net external force acts on the colliding objects. Inelastic means kinetic energy is not conserved; momentum still is.

What is the difference between impulse and momentum?

Momentum (p = mv) is what a moving object has. Impulse is the change in momentum, Δp = FnetΔt, caused by a net force acting for a time. They have the same unit: kg m s^−1 is the same as N s.

Do Year 11 momentum questions include collisions in two dimensions?

No. In the 2025 Year 11 course, momentum is one-dimensional: everything moves along a single line, and direction is handled with + and − signs. Collisions at angles are not part of Year 11.

Why does a car's crumple zone reduce the force on the people inside in a crash?

It makes the collision last longer, so the same change in momentum needs a smaller average force.

It reduces the change in momentum of the car.

It absorbs momentum, so momentum is no longer conserved.

It increases the impulse on the car.

A 2.0 kg trolley moving at 3.0 m s^−1 hits a stationary 1.0 kg trolley and they stick together. Which statement is correct?

They move off at 2.0 m s^−1 and 3.0 J of kinetic energy is transformed into other forms.

They move off at 1.5 m s^−1 and kinetic energy is conserved.

They move off at 2.0 m s^−1 and kinetic energy is conserved.

They move off at 3.0 m s^−1 and no energy is transformed.

For the total momentum of two colliding objects to be conserved, which condition must hold?

There must be no net external force on the two objects.

The objects must stick together.

The objects must have the same mass.

Year 11 Physics · Fundamentals of mechanics · Momentum and energy · 13 min read

Momentum and collisions

Momentum is what makes a moving object hard to stop. A truck and a tennis ball at the same speed have very different momentum, and in a collision the total momentum of the objects involved is shared out but never lost. That one rule, conservation of momentum, solves most Year 11 collision questions.

This guide covers p = mv, impulse and force–time graphs (and why crumple zones work), conservation of momentum, how to tell an elastic collision from an inelastic one, the practical, a quick quiz and a full worked example. As in the Year 11 course, everything moves along one line.

What you need to know

State what conservation of momentum requires: an isolated system, with no net external force acting.
Explain the reason collisions conserve momentum.
Calculate and compare total kinetic energy before and after a collision to classify it as elastic or inelastic.
Use p = mv and Δp = F_netΔt for momentum problems.
Use Σmv_before = Σmv_after for conservation of momentum problems, adding kinetic energy conservation where it applies.
Use Σ½mv²_before = Σ½mv²_after in elastic collisions.
Analyse momentum conservation in collisions by conducting a practical investigation.

The equations you will use:

p = mvMomentum in kg m s⁻¹. A vector: along a line, its sign gives its direction.

Δp = F_netΔt = mv − muImpulse in N s (the same unit as kg m s⁻¹). It equals the area under a force–time graph.

Σmv_before = Σmv_afterConservation of momentum, for an isolated system (no net external force).

Σ½mv²_before = Σ½mv²_afterElastic collisions only. Kinetic energy is a scalar, so there are no signs in it.

Dot points: PY11-FM-43, PY11-FM-44, PY11-FM-45, PY11-FM-46, PY11-FM-47, PY11-FM-48, PY11-FM-49 (2025 Physics syllabus, Year 11), in our own words.

Mini lesson

Momentum: mass in motion

Momentum is mass times velocity: p = mv, measured in kg m s⁻¹. Because velocity is a vector, so is momentum. It points the same way the object is moving.

In Year 11 every momentum problem is along one line, so a direction is just a sign. Choose which way is positive, write it down, and give anything moving the other way a negative velocity.

With right as positive: p = 2.0 × 3.0 = +6.0 kg m s⁻¹ and p = 1.5 × (−4.0) = −6.0 kg m s⁻¹. The total momentum of these two is zero, even though both are moving.

Impulse and force–time graphs

A net force acting for a time changes an object's momentum. The change is called the impulse: Δp = F_netΔt = mv − mu. Its unit is the newton second (N s), which is the same as kg m s⁻¹.

Real collision forces aren't constant: they rise and fall over a few milliseconds. On a force–time graph, the impulse is the area under the curve, and F_net in Δp = F_netΔt is the average force over the contact time.

Watch the signs when something bounces. Taking 'towards the wall' as positive, a 0.50 kg ball that hits a wall at 8.0 m s⁻¹ and rebounds at 6.0 m s⁻¹ has Δp = 0.50 × (−6.0 − 8.0) = −7.0 N s, not −1.0 N s, because its velocity reverses.

Why crumple zones and airbags work

In a crash a car and the people in it must lose all their momentum, so Δp is fixed by the mass and the speed. Rearranging gives F_net = Δp ÷ Δt: if the stop takes longer, the average force is smaller.

A crumple zone is a front or rear section built to fold up in a controlled way, which stretches the stopping time from a few milliseconds to tens of milliseconds. Airbags, the foam in helmets and soft landing mats use the same idea. None of them changes the impulse; they spread it over more time.

For example, stopping a 70 kg person from 15 m s⁻¹ takes an impulse of 1050 N s. Over 0.010 s that is an average force of 105 000 N; over 0.15 s it is 7000 N.

Both collisions bring the car to rest, so the impulse (the area under each curve) is the same. Spreading it over a longer time lowers the peak and the average force.

Conservation of momentum

In an isolated system (no net external force), the total momentum before an interaction equals the total momentum after: Σmv_before = Σmv_after. It holds for objects that bounce apart, objects that stick together, and explosions and recoil, where objects start at rest and are pushed apart.

Why collisions conserve momentum: by Newton's third law, during the contact A pushes on B exactly as hard as B pushes on A, in the opposite direction. Both forces act for the same contact time, so the impulses are equal and opposite. Whatever momentum A loses, B gains, and the total can't change.

Friction and other outside forces break the rule, but a collision is so short that they barely change the momentum during it. That is why the model works well for real collisions.

Newton's third law: the force on A and the force on B are equal and opposite and last for the same time, so their impulses cancel and the total momentum is unchanged.

Elastic or inelastic? Follow the kinetic energy

Momentum is conserved in every collision of an isolated system. Kinetic energy is not. To classify a collision, calculate the total kinetic energy (½mv² for each object) before and after, and compare.

Elastic: total KE after = total KE before. Use Σ½mv²_before = Σ½mv²_after alongside momentum, but only when the question says the collision is elastic.
Inelastic: total KE after is less than before. The missing energy has been transformed into heat, sound and the work done deforming the objects.
Perfectly inelastic: the objects stick together and move off with one common velocity. This loses the most KE that momentum conservation allows.
Total KE can only go up if stored energy is released, as in an explosion or a compressed spring pushing two trolleys apart.

The practical: testing conservation of momentum

A typical investigation uses two trolleys or gliders on a level track. Measure each mass on a balance and each velocity before and after with light gates, a motion sensor or slow-motion video against a ruler. Then compare the total momentum before and after.

Level the track (or tilt it slightly to cancel friction) so outside forces are small.
Repeat each collision and average the results to reduce random error.
Expect the totals to agree within a few per cent, not exactly. Friction makes the momentum after a little smaller.
Change one thing at a time (masses, speeds, bumpers: Velcro to stick, springs to bounce) and calculate KE too, so the same data classify each collision.

Common mistakes

Adding the momenta in a head-on collision as if both objects move the same way.Pick a positive direction first. An object moving the opposite way has a negative velocity, so its momentum is negative.
Working out the change in momentum of a 0.50 kg ball that hits a wall at 8.0 m s⁻¹ and bounces back at 6.0 m s⁻¹ as 0.50 × (6.0 − 8.0) = −1.0 N s.The velocity reverses, so v = −6.0 m s⁻¹: Δp = 0.50 × (−6.0 − 8.0) = −7.0 N s.
Saying momentum is lost in an inelastic collision.Momentum is conserved in every collision of an isolated system. In an inelastic collision it is the kinetic energy that drops.
Saying a crumple zone reduces the impulse or the change in momentum.The change in momentum is the same either way (the car still stops). The crumple zone makes the stopping time longer, so the average force is smaller.
Using kinetic energy conservation to solve every collision.Use Σ½mv² only for elastic collisions. Otherwise momentum alone gives the velocities, and you then calculate KE to classify the collision.
Putting a negative velocity into ½mv² and getting a negative kinetic energy.Kinetic energy is a scalar and never negative: squaring the velocity removes the sign.

Worked example

On a level air track, glider P (mass 0.40 kg) moves to the right at 0.60 m s⁻¹. It collides head-on with glider Q (mass 0.20 kg), which is moving to the left at 0.90 m s⁻¹. After the collision Q moves to the right at 0.70 m s⁻¹.

(a) Calculate the velocity of P after the collision. (3 marks)

(b) Show whether the collision is elastic or inelastic. (2 marks)

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. P's momentum before (about 0.24 kg m s⁻¹ right) is only a little bigger than Q's (about 0.18 kg m s⁻¹ left), so the total is small and to the right. Afterwards Q alone carries 0.14 kg m s⁻¹ to the right, more than the total. So P must carry some momentum to the left: I predict P bounces back to the left, slowly, at about 0.1 to 0.2 m s⁻¹. Bouncing gliders usually lose some KE, so I expect inelastic.

K · Knowns

I sketch before and after with every value on it, then choose right as positive. Q moves left before the collision, so its velocity goes in as negative.

m_P = 0.40 kg, u_P = +0.60 m s⁻¹
m_Q = 0.20 kg, u_Q = −0.90 m s⁻¹
v_Q = +0.70 m s⁻¹ (right is +)

Knowns on the sketch, with right as positive: u_P = +0.60 m s⁻¹, u_Q = −0.90 m s⁻¹, v_Q = +0.70 m s⁻¹. Unknown: v_P.

N · Not known

Part (a) wants a velocity, so my answer needs a size and a direction. Part (b) needs the two kinetic energy totals to compare.

v_P = ? m s⁻¹ (with its direction)
total KE before and after = ? J

O · Organise

The track is level and the collision is quick, so outside forces are negligible and momentum is conserved. I write the equation in symbols first. KE is a scalar, so for part (b) the signs don't matter.

m_Pu_P + m_Qu_Q = m_Pv_P + m_Qv_Q
KE = Σ½mv², before and after

W · Work it out

I find the total momentum before first, then solve for v_P. Then I work out the four kinetic energies and compare the totals.

Before: (0.40 × 0.60) + (0.20 × (−0.90)) = 0.24 − 0.18 = +0.060 kg m s⁻¹
After: 0.40v_P + (0.20 × 0.70) = 0.40v_P + 0.14
0.40v_P + 0.14 = 0.060 → v_P = (0.060 − 0.14) ÷ 0.40 = −0.20 m s⁻¹
KE before = ½ × 0.40 × 0.60² + ½ × 0.20 × 0.90² = 0.072 + 0.081 = 0.153 J
KE after = ½ × 0.40 × 0.20² + ½ × 0.20 × 0.70² = 0.008 + 0.049 = 0.057 J
0.057 J < 0.153 J: 0.096 J of KE has been transformed, so the collision is inelastic

U · Units

Velocity in m s⁻¹ and energy in joules. The data have two significant figures, so I give v_P to two.

v_P = −0.20 m s⁻¹
KE: 0.15 J before, 0.057 J after

D · Direction

The minus sign means left, opposite to my positive direction, so P has bounced back. I say the direction in words, because that's what the marker is looking for.

v_P = 0.20 m s⁻¹ to the left

(a) P moves at 0.20 m s⁻¹ to the left. (b) The total kinetic energy falls from 0.15 J to 0.057 J, so the collision is inelastic (momentum is still conserved).

Check against the prediction. I predicted P would bounce back to the left at about 0.1 to 0.2 m s⁻¹ and that some KE would be lost. 0.20 m s⁻¹ to the left and an inelastic collision both fit. One more check: momentum after = 0.40 × (−0.20) + 0.14 = +0.060 kg m s⁻¹, the same as before.

Exam advice

What markers look for

Give every velocity, momentum and impulse a direction in words, not just a sign. NESA's feedback reminds HSC students every year to give directions for calculated vector quantities, so make it a habit now.NESA marking feedback, 2023, summarised
Write your sign convention down before any numbers ('right is positive') and keep it for the whole question, before and after the collision.CortexHSC teaching notes
Before choosing equations, ask what is conserved. HSC answers that didn't apply conservation of momentum where the situation called for it lost marks; in a Year 11 collision question it is almost always the first line.NESA marking feedback, 2025, summarised
Classify a collision with numbers: calculate both kinetic energy totals and compare them. 'It's inelastic because they bounced' earns nothing without the calculation.CortexHSC teaching notes
For 'explain why momentum is conserved', give the whole chain: third-law forces are equal and opposite, they act for the same time, so the impulses are equal and opposite and the total momentum doesn't change.CortexHSC teaching notes
Show the formula, the substitution and the answer with a unit (kg m s⁻¹ or N s for momentum). Correct working earns marks even when the final number slips.General advice from NESA marking feedback, 2021 to 2025, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2023 · 2025.

Where this leads in Year 12

Momentum comes back in Year 12. Conservation of momentum and energy together show how Chadwick's experiments proved the neutron exists, collision experiments in particle accelerators are used to find quarks, leptons and bosons, and special relativity changes the momentum of objects moving close to the speed of light.

FAQ

Is momentum a vector or a scalar?: A vector. It points the same way as the velocity. In Year 11 problems are along one line, so the direction is shown by a + or − sign, and your final answer should state the direction in words.
What is the difference between elastic and inelastic collisions?: Momentum is conserved in both. In an elastic collision the total kinetic energy is also the same before and after. In an inelastic collision some kinetic energy is transformed into heat, sound and deformation, so the total KE after is less. If the objects stick together, the collision is perfectly inelastic.
Is momentum conserved in an inelastic collision?: Yes, as long as no net external force acts on the colliding objects. Inelastic means kinetic energy is not conserved; momentum still is.
What is the difference between impulse and momentum?: Momentum (p = mv) is what a moving object has. Impulse is the change in momentum, Δp = F_netΔt, caused by a net force acting for a time. They have the same unit: kg m s⁻¹ is the same as N s.
Do Year 11 momentum questions include collisions in two dimensions?: No. In the 2025 Year 11 course, momentum is one-dimensional: everything moves along a single line, and direction is handled with + and − signs. Collisions at angles are not part of Year 11.

Related guides

Explore more of CortexHSC

CortexHSC is a free Year 11 Physics study app built by a NSW Physics teacher. Pick up where this guide leaves off:

Free to start, no card needed. Some features need a free account; a few are part of a paid pass.