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Momentum and impulse (SACE Stage 2 Physics)

Syllabus dot point

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

9 min read

Quick answer

Momentum is p⃗=mv⃗\vec{p} = m\vec{v} (a vector, kg m/s). Impulse is the product of force and the time it acts, J⃗=F⃗Δt\vec{J} = \vec{F}\Delta t, and equals the change in momentum, J⃗=Δp⃗\vec{J} = \Delta\vec{p}. So the same change in momentum can be produced by a large force for a short time or a small force for a long time.

Jump to a section
  1. What this dot point is asking
  2. Momentum
  3. Impulse and the impulse-momentum theorem
  4. Force-time graphs
  5. Why safety features work
  6. How SACE assesses this
  7. Exam-style questions

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 kg0.20\ \text{kg} ball hits a wall horizontally at 8.0 m s−18.0\ \text{m s}^{-1} and rebounds at 6.0 m s−16.0\ \text{m s}^{-1}. Find the impulse on the ball and the average force if contact lasts 0.05 s0.05\ \text{s}.

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

Impulse =Δp=m(v−u)=0.20 (−6.0−8.0)=−2.8 N s= \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 N s2.8\ \text{N s}.

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

If we had ignored the rebound and assumed the ball just stopped, Δp\Delta p would be only 1.6 N s1.6\ \text{N s}: bouncing produces a larger impulse.

Momentum

Momentum

The momentum of an object is the product of its mass and velocity:

p⃗=mv⃗\vec{p} = m\vec{v}

It is a vector pointing in the direction of motion, measured in kg m s−1\text{kg m s}^{-1} (equivalently N s).

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,

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

Rearranging gives the impulse-momentum theorem:
J⃗=F⃗net Δt=Δp⃗=mv⃗−mu⃗.\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, N s=kg m s−1\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=area under the F-t graph.J = \text{area under the } F\text{-}t \text{ graph}.

For a triangular force pulse, the area is 12(base)(height)\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 Δt\Delta t, these devices reduce the average force F=Δp/ΔtF = \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 p⃗i\vec{p}_i and p⃗f\vec{p}_f tip-to-tail, identify Δp⃗=p⃗f−p⃗i\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.

Common errors
Treating momentum as a scalar
Momentum is a vector. For a rebound you must subtract the final and initial momenta as vectors, including signs.
Forgetting the rebound doubles the effect
A ball that bounces back has a larger momentum change than one that merely stops, so the impulse and force are larger.
Confusing impulse with force
Impulse is force multiplied by time (N s), not a force (N). The area under a force-time graph is an impulse.

Mixing up "reduce the force" with "reduce the momentum change." Safety devices do not reduce the momentum change; they extend the time so the force is smaller.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2024-style3 marks
A 0.20 kg0.20\ \text{kg} ball hits a wall horizontally at 8.0 m s−18.0\ \text{m s}^{-1} and rebounds along the same line at 6.0 m s−16.0\ \text{m s}^{-1}. The contact lasts 0.050 s0.050\ \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 m s−1u = +8.0\ \text{m s}^{-1} and v=−6.0 m s−1v = -6.0\ \text{m s}^{-1}.

(a) J=Δp=m(v−u)=(0.20)(−6.0−8.0)=−2.8 N sJ = \Delta p = m(v - u) = (0.20)(-6.0 - 8.0) = -2.8\ \text{N s}. The magnitude is 2.8 N s2.8\ \text{N s}, directed away from the wall. (2 marks)

(b) Favg=ΔpΔt=2.80.050=56 NF_\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, Δp\Delta p would be only 1.6 N s1.6\ \text{N s}.

SACE 2022-style4 marks
A photon of wavelength 508 nm508\ \text{nm} strikes a solar sail at 30.0∘30.0^\circ from the surface and reflects at 30.0∘30.0^\circ. The magnitude of the photon momentum is 1.31×10−27 kg m s−11.31\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∘30.0^\circ to the surface, so its momentum makes 60.0∘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 psin⁡30.0∘=(1.31×10−27)(0.500)=6.55×10−28 kg m s−1p\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:

∣Δp⃗∣=2psin⁡30.0∘=2(6.55×10−28)=1.31×10−27 kg m s−1.|\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.

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Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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