SACE Board SACE 2024-style practice: A 0.20\ kg ball hits a wall horizontally at 8.0\ m s^-1 and rebounds along the same line at 6.0\ m s^-1. The contact lasts 0.050\ s. (a) Calculate the magnitude of the impulse on the ball. (b) Calculate the average force exerted by the wall on the ball.

Take the initial direction as positive, so u = +8.0\ m s^-1 and v = -6.0\ m s^-1. (a) J = p = m(v - u) = (0.20)(-6.0 - 8.0) = -2.8\ N s. The magnitude is 2.8\ N s, directed away from the wall. (2 marks) (b) F_avg = p t = 2.80.050 = 56\ N. (1 mark) Markers reward using signed velocities so the rebound is captured; had the ball merely stopped, p would be only 1.6\ N s.

SACE Board SACE 2022-style practice: A photon of wavelength 508\ nm strikes a solar sail at 30.0^ from the surface and reflects at 30.0^. The magnitude of the photon momentum is 1.3110^-27\ kg m s^-1 before and after. Use a vector diagram to determine the magnitude and direction of the change in momentum of the photon.

The photon hits at 30.0^ to the surface, so its momentum makes 60.0^ with the surface normal. By symmetry the component parallel to the sail is unchanged, while the component perpendicular to the sail reverses. The perpendicular component of each photon is psin 30.0^ = (1.3110^-27)(0.500) = 6.5510^-28\ kg m s^-1. The change in momentum is twice this perpendicular component: $| p| = 2psin 30.0^ = 2(6.5510^-28) = 1.3110^-27\ kg m s^-1.$ The change is directed perpendicular to the sail surface, into the sail (along the inward normal). 1 mark for the vector diagram, 1 mark for using only the perpendicular component, 1 mark for the magnitude, 1 mark for the direction.

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SAPhysicsSyllabus dot point

How does the impulse of a force change the momentum of an object?

Relate impulse to the change in momentum of an object using $\vec{J} = \vec{F}\Delta t = \Delta \vec{p}$ , and interpret force-time graphs.

Defining momentum and impulse, the impulse-momentum theorem, reading impulse from a force-time graph, and how impulse explains safety features, with worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Momentum
Impulse and the impulse-momentum theorem
Force-time graphs
Why safety features work
How SACE assesses this

What this dot point is asking

You need to define momentum and impulse, state and apply the impulse-momentum theorem, find impulse as the area under a force-time graph, and use these ideas to explain safety features such as crumple zones and airbags.

Bouncing ball

Impulse on a ball that rebounds

A $0.20\ \text{kg}$ ball hits a wall horizontally at $8.0\ \text{m s}^{-1}$ and rebounds at $6.0\ \text{m s}^{-1}$ . Find the impulse on the ball and the average force if contact lasts $0.05\ \text{s}$ .

Take the initial direction as positive. Then $u = +8.0$ , $v = -6.0\ \text{m s}^{-1}$ .

Impulse $= \Delta p = m(v - u) = 0.20\,(-6.0 - 8.0) = -2.8\ \text{N s}$ .

The negative sign means the impulse points away from the wall. Its magnitude is $2.8\ \text{N s}$ .

Average force $= \Delta p / \Delta t = 2.8 / 0.05 = 56\ \text{N}$ .

If we had ignored the rebound and assumed the ball just stopped, $\Delta p$ would be only $1.6\ \text{N s}$ : bouncing produces a larger impulse.

Momentum

Because momentum is a vector, direction matters. A ball bouncing back reverses its momentum, so its change in momentum is larger than if it simply stopped.

Impulse and the impulse-momentum theorem

Newton's second law can be written in terms of momentum: the net force equals the rate of change of momentum,

\vec{F}_{\text{net}} = \frac{\Delta\vec{p}}{\Delta t}.

Rearranging gives the impulse-momentum theorem:

\vec{J} = \vec{F}_{\text{net}}\,\Delta t = \Delta\vec{p} = m\vec{v} - m\vec{u}.

Impulse and momentum change have the same units, $\text{N s} = \text{kg m s}^{-1}$ . The key insight is the trade-off between force and time: for a fixed change in momentum, increasing the contact time decreases the average force.

Force-time graphs

When a force varies during a collision, the impulse is the area under the force-time graph:

J = \text{area under the } F\text{-}t \text{ graph}.

For a triangular force pulse, the area is

\tfrac{1}{2}(\text{base})(\text{height})

. The average force is the constant force that would give the same area (same impulse) over the same time, so a tall narrow spike and a low broad pulse can deliver the same impulse.

Why safety features work

Crumple zones, airbags, helmet padding and crash mats all work the same way. In a collision the change in momentum (the impulse) is essentially fixed: the occupant must be brought to rest. By extending the contact time $\Delta t$ , these devices reduce the average force $F = \Delta p / \Delta t$ on the body. A longer, gentler deceleration replaces a short, violent one, reducing injury.

The same principle explains why a cricketer "gives" with the ball when catching, and why falling onto a soft surface hurts less than onto concrete.

How SACE assesses this

SACE Stage 2 impulse questions come in two main styles: a numerical calculation of impulse or average force (often involving a rebound, so signs matter), and a vector-diagram question where a photon or ball changes direction and you must resolve the momentum into components. For the calculation style, always assign a positive direction first and substitute signed velocities, then state the magnitude and direction of the answer. For the vector style, draw $\vec{p}_i$ and $\vec{p}_f$ tip-to-tail, identify $\Delta\vec{p} = \vec{p}_f - \vec{p}_i$ , and resolve into the parallel and perpendicular components before combining. Markers consistently reward a clear sign convention and a final answer stated with both magnitude and direction.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20243 marksA

0.20\ \text{kg}

ball hits a wall horizontally at

8.0\ \text{m s}^{-1}

and rebounds along the same line at

6.0\ \text{m s}^{-1}

. The contact lasts

0.050\ \text{s}

. (a) Calculate the magnitude of the impulse on the ball. (b) Calculate the average force exerted by the wall on the ball.

Show worked answer →

Take the initial direction as positive, so $u = +8.0\ \text{m s}^{-1}$ and $v = -6.0\ \text{m s}^{-1}$ .

(a) $J = \Delta p = m(v - u) = (0.20)(-6.0 - 8.0) = -2.8\ \text{N s}$ . The magnitude is $2.8\ \text{N s}$ , directed away from the wall. (2 marks)

(b) $F_\text{avg} = \dfrac{\Delta p}{\Delta t} = \dfrac{2.8}{0.050} = 56\ \text{N}$ . (1 mark) Markers reward using signed velocities so the rebound is captured; had the ball merely stopped, $\Delta p$ would be only $1.6\ \text{N s}$ .

SACE 20224 marksA photon of wavelength

508\ \text{nm}

strikes a solar sail at

30.0^\circ

from the surface and reflects at

30.0^\circ

. The magnitude of the photon momentum is

1.31\times10^{-27}\ \text{kg m s}^{-1}

before and after. Use a vector diagram to determine the magnitude and direction of the change in momentum of the photon.

Show worked answer →

The photon hits at $30.0^\circ$ to the surface, so its momentum makes $60.0^\circ$ with the surface normal. By symmetry the component parallel to the sail is unchanged, while the component perpendicular to the sail reverses.

The perpendicular component of each photon is $p\sin 30.0^\circ = (1.31\times10^{-27})(0.500) = 6.55\times10^{-28}\ \text{kg m s}^{-1}$ .

The change in momentum is twice this perpendicular component:

|\Delta \vec{p}| = 2p\sin 30.0^\circ = 2(6.55\times10^{-28}) = 1.31\times10^{-27}\ \text{kg m s}^{-1}.

The change is directed perpendicular to the sail surface, into the sail (along the inward normal). 1 mark for the vector diagram, 1 mark for using only the perpendicular component, 1 mark for the magnitude, 1 mark for the direction.