SACE Board SACE 2023-style practice: A 1200\ kg car moving east at 18.0\ m s^-1 collides head-on with a 1800\ kg van moving west at 12.0\ m s^-1. The two vehicles lock together. (a) Calculate the common velocity immediately after the collision. (b) Show that the collision is inelastic.

Take east as positive and apply conservation of momentum, m_1 u_1 + m_2 u_2 = (m_1 + m_2)v. (a) sum p_before = (1200)(+18.0) + (1800)(-12.0) = 21600 - 21600 = 0\ kg m s^-1. So (1200 + 1800)v = 0, giving v = 0\ m s^-1: the wreck is momentarily stationary. (2 marks) (b) Compare kinetic energy. Before: E_k = 12(1200)(18.0)^2 + 12(1800)(12.0)^2 = 1.94410^5 + 1.29610^5 = 3.2410^5\ J. After: E_k = 12(3000)(0)^2 = 0\ J. All 3.2410^5\ J is converted to heat, sound and deformation, so the collision is inelastic. (2 marks) Markers reward the signed momentum sum and the explicit before/after E_k comparison.

SACE Board SACE 2022-style practice: A 0.16\ kg ball moving right at 4.0\ m s^-1 collides elastically with a stationary 0.40\ kg ball. After the collision the 0.16\ kg ball rebounds at 1.7\ m s^-1. Calculate the velocity of the 0.40\ kg ball and confirm the collision is consistent with being elastic.

Take right as positive. Conservation of momentum: m_1 u_1 = m_1 v_1 + m_2 v_2. $(0.16)(4.0) = (0.16)(-1.7) + (0.40)v_2.$ $0.64 = -0.272 + 0.40 v_2 v_2 = 0.9120.40 = 2.28\ m s^-1\ right.$ Elastic check: before E_k = 12(0.16)(4.0)^2 = 1.28\ J. After E_k = 12(0.16)(1.7)^2 + 12(0.40)(2.28)^2 = 0.231 + 1.040 = 1.27\ J. The kinetic energy is conserved to rounding, consistent with an elastic collision. Markers award the signed momentum equation, the value of v_2, and the kinetic-energy comparison.

§-Syllabus dot point

SAPhysicsSyllabus dot point

Why is the total momentum of an isolated system the same before and after a collision, even when kinetic energy is lost?

Apply conservation of momentum to elastic and inelastic collisions in one dimension, and distinguish them using kinetic energy.

How conservation of momentum solves one-dimensional collisions, the difference between elastic and inelastic collisions, and why momentum is conserved while kinetic energy may not be.

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
Why momentum is conserved
Elastic vs inelastic collisions
Solving a collision
Recoil and explosions

What this dot point is asking

You need to apply the law of conservation of momentum to one-dimensional collisions, write a correct before-and-after momentum equation (with signs for direction), and use kinetic energy to decide whether a collision is elastic or inelastic.

Perfectly inelastic collision

A 1200 kg car runs into a stationary 1800 kg car and they lock together

The first car moves east at $15\ \text{m s}^{-1}$ ; the second is at rest. Find their common velocity and the kinetic energy lost.

Momentum (east positive): $m_1 u_1 = (m_1 + m_2)v$

$1200 \times 15 = (1200 + 1800)v$

$18000 = 3000\,v \implies v = 6.0\ \text{m s}^{-1}\ \text{east}.$
Kinetic energy before: $\tfrac{1}{2}(1200)(15)^2 = 1.35 \times 10^5\ \text{J}$ .
Kinetic energy after: $\tfrac{1}{2}(3000)(6.0)^2 = 5.4 \times 10^4\ \text{J}$ .
Lost: $1.35\times10^5 - 5.4\times10^4 = 8.1 \times 10^4\ \text{J}$ , converted to heat, sound and deformation. Momentum is conserved; kinetic energy is not, so this is inelastic.

Why momentum is conserved

Newton's third law says the forces two colliding bodies exert on each other are equal and opposite. During the collision these are the only significant forces (the collision is brief), so the impulse on one body is the exact negative of the impulse on the other. Equal and opposite impulses mean equal and opposite momentum changes, so the total momentum change of the system is zero.

The "isolated" condition matters. If an external force such as friction or an applied push acts during the interaction, momentum is not conserved for the bodies alone. In most SACE collision problems we treat the interaction as brief enough that external forces are negligible over the contact time.

Elastic vs inelastic collisions

The dividing line is kinetic energy, not momentum (momentum is conserved in both).

To classify a collision, compute the total $E_k = \tfrac{1}{2}mv^2$ before and after. If they match, it is elastic; if $E_k$ after is less, it is inelastic. You can never decide elasticity from momentum alone, because momentum is conserved either way.

Solving a collision

Draw before and after, mark velocities with signs.
Write the momentum equation and substitute known values.
If the objects stick together (perfectly inelastic), use one final velocity $v$ for both.
To classify, compare the total kinetic energy before and after.

Most SACE collision questions are one-dimensional, so a single signed equation is enough. The hardest step is sign bookkeeping: a velocity to the left is negative if you chose right as positive, and forgetting this flips the answer.

Recoil and explosions

An "explosion" is the reverse of a perfectly inelastic collision: a system starts at rest (total momentum zero) and separates into pieces. Because total momentum stays zero, the pieces fly apart with equal and opposite momenta. This is how recoil works: a fired projectile and the gun gain momenta of equal magnitude in opposite directions, so a light bullet leaves fast while the heavier gun recoils slowly.

The same algebra handles a person stepping off a stationary boat, or a trolley releasing a spring-loaded block: the total momentum is zero before and after, so $m_1 v_1 = -m_2 v_2$ .

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 20234 marksA

1200\ \text{kg}

car moving east at

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

collides head-on with a

1800\ \text{kg}

van moving west at

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

. The two vehicles lock together. (a) Calculate the common velocity immediately after the collision. (b) Show that the collision is inelastic.

Show worked answer →

Take east as positive and apply conservation of momentum, $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ .

(a) $\sum p_\text{before} = (1200)(+18.0) + (1800)(-12.0) = 21600 - 21600 = 0\ \text{kg m s}^{-1}$ .
So $(1200 + 1800)v = 0$ , giving $v = 0\ \text{m s}^{-1}$ : the wreck is momentarily stationary. (2 marks)

(b) Compare kinetic energy. Before: $E_k = \tfrac{1}{2}(1200)(18.0)^2 + \tfrac{1}{2}(1800)(12.0)^2 = 1.944\times10^5 + 1.296\times10^5 = 3.24\times10^5\ \text{J}$ .
After: $E_k = \tfrac{1}{2}(3000)(0)^2 = 0\ \text{J}$ .
All $3.24\times10^5\ \text{J}$ is converted to heat, sound and deformation, so the collision is inelastic. (2 marks) Markers reward the signed momentum sum and the explicit before/after $E_k$ comparison.

SACE 20223 marksA

0.16\ \text{kg}

ball moving right at

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

collides elastically with a stationary

0.40\ \text{kg}

ball. After the collision the

0.16\ \text{kg}

ball rebounds at

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

. Calculate the velocity of the

0.40\ \text{kg}

ball and confirm the collision is consistent with being elastic.

Show worked answer →

Take right as positive. Conservation of momentum: $m_1 u_1 = m_1 v_1 + m_2 v_2$ .

(0.16)(4.0) = (0.16)(-1.7) + (0.40)v_2.

0.64 = -0.272 + 0.40 v_2 \Rightarrow v_2 = \dfrac{0.912}{0.40} = 2.28\ \text{m s}^{-1}\ \text{right}.

Elastic check: before $E_k = \tfrac{1}{2}(0.16)(4.0)^2 = 1.28\ \text{J}$ .
After $E_k = \tfrac{1}{2}(0.16)(1.7)^2 + \tfrac{1}{2}(0.40)(2.28)^2 = 0.231 + 1.040 = 1.27\ \text{J}$ .
The kinetic energy is conserved to rounding, consistent with an elastic collision. Markers award the signed momentum equation, the value of $v_2$ , and the kinetic-energy comparison.