Newton's laws, momentum and impulse: VCE Physics Unit 3

Syllabus dot point

“Investigate and apply theoretically and practically Newton's three laws of motion in situations where two or more coplanar forces act along a straight line and in two dimensions; apply the concepts of momentum and impulse, including the conservation of momentum in one and two dimensions, and distinguish between elastic and inelastic collisions”

11 min readUpdated 21 May 2026

Quick answer

Newton's laws give $F_{net} = ma$ on each axis, impulse is the area under a force-time graph equal to $\Delta p$ , and momentum is conserved (vectorially) in every isolated collision while kinetic energy is conserved only in elastic ones.

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What this dot point is asking
The answer
Examples in context
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Exam-style questions

What this dot point is asking

VCAA wants you to apply Newton's three laws when more than one force acts, in one or two dimensions, and to use momentum ( $p = mv$ ) and impulse ( $J = F \Delta t = \Delta p$ ) including the conservation of momentum in collisions. You also need to distinguish elastic collisions (kinetic energy conserved) from inelastic collisions (kinetic energy not conserved).

The answer

Newton's three laws

First law (inertia): An object continues at rest or in uniform straight-line motion unless acted on by a net external force.
Second law: The net force on an object equals its mass times its acceleration: $F_{net} = ma$ . In two dimensions, resolve forces along perpendicular axes and apply $F_{net} = ma$ on each axis independently.
Third law: When object A exerts a force on object B, object B exerts an equal and opposite force on A. The two forces act on different objects, so they never cancel on a single free-body diagram.

Momentum and impulse

Momentum is a vector: $p = mv$ , units kg m/s.

Impulse is the change in momentum: $J = \Delta p = F_{net} \Delta t$ , units N s (same as kg m/s).

On a force-time graph, impulse is the area under the curve. This is how VCAA tests variable forces (for example a ball striking a bat).

To reduce the average force in a collision, increase the contact time $\Delta t$ for a given $\Delta p$ . This is why airbags, crumple zones and bent knees on landing reduce injury.

Conservation of momentum

In an isolated system (no net external force), total momentum is conserved:

\sum p_{before} = \sum p_{after}

In two dimensions, momentum is conserved independently along each axis. Resolve the velocity vectors into x and y components and apply conservation on each axis.

Elastic vs inelastic collisions

	Momentum	Kinetic energy
Elastic	conserved	conserved
Inelastic	conserved	not conserved
Perfectly inelastic (stick together)	conserved	not conserved (maximum KE lost)

In VCE, almost all real collisions are inelastic. Steel ball bearings come close to elastic; cars colliding, clay landing on a board, and bullets embedding in blocks are all inelastic.

Worked example

A $0.50$ kg ball moving east at $4.0$ m/s collides with a stationary $0.30$ kg ball. After the collision the $0.50$ kg ball moves east at $1.0$ m/s. Find the velocity of the $0.30$ kg ball and classify the collision.

Step 1: Apply conservation of momentum to find the unknown velocity

Because there is no net external horizontal force, total momentum before equals total momentum after. Taking east as positive, the stationary ball contributes zero initial momentum.

(0.50)(4.0) + 0 = (0.50)(1.0) + (0.30) v

2.0 = 0.5 + 0.30 v, \quad \text{so } v = 5.0 \text{ m/s east.}

Step 2: Check whether kinetic energy is conserved to classify the collision

For an elastic collision, total kinetic energy must be the same before and after. Calculating both totals and comparing them gives a definitive classification.

Kinetic energy before: $\dfrac{1}{2}(0.50)(4.0)^2 = 4.0$ J.

Kinetic energy after: $\dfrac{1}{2}(0.50)(1.0)^2 + \dfrac{1}{2}(0.30)(5.0)^2 = 0.25 + 3.75 = 4.0$ J.

KE is conserved, so the collision is elastic.

Common traps

Treating Newton's third law as cancelling forces: The action-reaction pair acts on two different objects. In a free-body diagram of one object, only one of the pair appears.
Forgetting that momentum is a vector: When a ball rebounds, $\Delta p$ uses signed velocities. A ball striking a wall at $8$ m/s and rebounding at $6$ m/s changes momentum by $m(8 - (-6)) = 14m$ , not $2m$ .
Assuming kinetic energy is conserved: Only momentum is automatically conserved in an isolated collision. KE is conserved only if the collision is explicitly elastic.
Confusing impulse with force: Impulse has units N s, force has units N. Always include $\Delta t$ .
Using speeds instead of velocity components in 2D: Conservation of momentum is applied separately to the x and y axes.

Examples in context

Example 1. Melbourne CBD Spencer Street truck-car collision. A $1500$ kg sedan travelling east at $15$ m s $^{-1}$ collides with a $5000$ kg parked delivery truck at the Spencer Street intersection. Treating as a one-dimensional inelastic collision, momentum conservation gives $1500 \times 15 + 0 = (1500 + 5000) v$ , so $v = 3.46$ m s $^{-1}$ east. Impulse on the truck is $\Delta p_{\rm truck} = 5000 \times 3.46 = 1.73 \times 10^4$ kg m s $^{-1}$ ; on the sedan, $\Delta p_{\rm sedan} = 1500 \times (3.46 - 15) = -1.73 \times 10^4$ kg m s $^{-1}$ , equal and opposite per Newton's third law. With contact time $0.1$ s, average force on each is $1.73 \times 10^5$ N.

Example 2. AFL ruck-tap impulse exchange at the MCG. Two AFL ruckmen contest a centre bounce. Player A ( $95$ kg, $4.0$ m s $^{-1}$ north) and Player B ( $90$ kg, $3.5$ m s $^{-1}$ south) meet in two dimensions. Total north momentum: $95 \times 4 - 90 \times 3.5 = 65$ kg m s $^{-1}$ . If they push apart with a brief contact, total north momentum is conserved at $65$ kg m s $^{-1}$ . If A continues at $1.0$ m s $^{-1}$ north, B must move at $(65 - 95)/90 = -0.33$ m s $^{-1}$ (still south). Impulse on each is $\Delta p_A = 95(1 - 4) = -285$ kg m s $^{-1}$ , equal and opposite to impulse on B.

Try this

Q1. State Newton's three laws of motion in one sentence each. [3 marks]

Cue. I: object at rest stays at rest unless acted on by net force. II: $\vec{F} = m\vec{a}$ . III: equal and opposite forces between interacting bodies.

Q2. A $0.4$ kg ball moving east at $25$ m s $^{-1}$ rebounds west at $20$ m s $^{-1}$ after striking a bat. Calculate (a) the change in momentum, and (b) the average force if contact lasts $0.005$ s. [4 marks]

Cue. (a) $\Delta p = 0.4(-20 - 25) = -18$ kg m s $^{-1}$ (west $18$ ). (b) $F = \Delta p/\Delta t = 3600$ N.

Q3. Refer to the Spencer Street collision. (a) Calculate the post-collision velocity. (b) Determine the impulse on the truck. (c) Apply Newton's third law to compare the impulse on the sedan. [2+2+2 marks]

Cue. (a) $3.46$ m s $^{-1}$ . (b) $1.73 \times 10^4$ kg m s $^{-1}$ east. (c) Equal magnitude, opposite direction on the sedan.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

A 1500 kg car travelling east at 12 m/s collides head-on with a 1000 kg car travelling west at 18 m/s. The two cars stick together. Calculate the velocity of the wreckage immediately after the collision and state whether the collision is elastic or inelastic.

Show worked answer →

Take east as positive. Apply conservation of momentum: $p_{before} = p_{after}$ .

$p_{before} = (1500)(12) + (1000)(-18) = 18000 - 18000 = 0$ kg m/s.

Combined mass $= 2500$ kg, so $v_{after} = 0 / 2500 = 0$ m/s. The wreckage is stationary.

The collision is inelastic because the cars stick together (a perfectly inelastic collision). Kinetic energy is not conserved: $KE_{before} = \frac{1}{2}(1500)(12)^2 + \frac{1}{2}(1000)(18)^2 = 108000 + 162000 = 270000$ J; $KE_{after} = 0$ J. The lost kinetic energy is converted to heat, sound and deformation.

Markers reward the signed setup, the conservation statement, and the explicit elastic-vs-inelastic justification using kinetic energy.

Practice3 marks

A 0.20 kg ball travelling at 8.0 m/s strikes a wall and rebounds at 6.0 m/s. The ball is in contact with the wall for 0.040 s. Calculate the average force exerted on the ball by the wall.

Show worked answer →

Impulse equals change in momentum: $J = \Delta p = F \Delta t$ .

Take the initial direction of motion as positive, so $v_i = +8.0$ m/s and $v_f = -6.0$ m/s.

$\Delta p = m(v_f - v_i) = 0.20 \times (-6.0 - 8.0) = 0.20 \times (-14.0) = -2.80$ kg m/s.

$F = \frac{\Delta p}{\Delta t} = \frac{-2.80}{0.040} = -70$ N.

The magnitude of the average force is 70 N, directed away from the wall (back toward the ball's origin).

Markers reward signs handled correctly, impulse equated to change in momentum (not just $mv$ ), and a final answer with magnitude and direction.