Conservation of momentum in collisions (SACE Stage 2 Physics)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

In any isolated system (no net external force), total momentum is conserved: $\sum p_{\text{before}} = \sum p_{\text{after}}$ . This holds for every collision. Kinetic energy is only conserved in an elastic collision; in an inelastic collision some kinetic energy is transformed into heat, sound and deformation, but momentum is still conserved.

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What this dot point is asking
Why momentum is conserved
Elastic vs inelastic collisions
Solving a collision
Recoil and explosions
Exam-style questions

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.

Conservation of momentum

For an isolated system (no net external force):

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where

u

are initial velocities and

v

are final velocities. Momentum is a vector, so assign

+

and

-

to directions and keep them consistent.

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).

Elastic and inelastic

Elastic collision: total kinetic energy is conserved as well as momentum. The objects bounce apart with no loss of $E_k$ .
Inelastic collision: momentum is conserved but total kinetic energy decreases - some is converted to heat, sound and permanent deformation.
Perfectly inelastic collision: the objects stick together and move off with a common velocity. This loses the maximum possible kinetic energy.

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 $m_1 v_1 = -m_2 v_2$ .

Common errors

Confusing the two conservation laws: Momentum is conserved in every isolated collision; kinetic energy is only conserved in elastic ones. Never assume kinetic energy is conserved unless told the collision is elastic.
Dropping the signs: Momentum is a vector. A ball bouncing back has a velocity opposite in sign, and forgetting this halves or doubles your answer.
Treating "sticking together" as elastic: A perfectly inelastic collision (objects join) loses the most kinetic energy possible, yet momentum is still conserved.
Using speed instead of velocity: Always include direction; otherwise the before-and-after equation can give the wrong total.

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 2023-style4 marks

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 2022-style3 marks

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.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Stage 2 Physics subject outline (2025 onwards) — SACE Board
Stage 2 Physics: external assessment — SACE Board