A 0.15 kg cricket ball travelling at 30 m s^-1 is struck and returns at 40 m s^-1 in the opposite direction. Contact time is 2.0 ms. Calculate the impulse and the average force.

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Topic 1: Linear motion and force

Define linear momentum and impulse, and apply the principle of conservation of momentum to one-dimensional collisions and explosions, distinguishing between elastic and inelastic collisions

A focused answer to the QCE Physics Unit 2 dot point on momentum and impulse. Defines $p = mv$ and $J = F \Delta t = \Delta p$ , walks through conservation of momentum in one-dimensional collisions and explosions, and distinguishes elastic from inelastic by whether kinetic energy is conserved. Works the QCAA two-cart collision standard problem.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-29

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

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What this dot point is asking
Momentum
Impulse
Conservation of momentum
Elastic and inelastic collisions
Explosions
Examples in context
Try this

What this dot point is asking

QCAA wants you to define momentum and impulse, link them through Newton's second law, and apply conservation of momentum to one-dimensional interactions. The QCAA EA tests this both as a calculation (two-body collision) and as a qualitative discussion (elastic versus inelastic, identifying isolated systems).

Momentum

Linear momentum is the product of mass and velocity:

\vec{p} = m \vec{v}

SI unit: kg m s $^{-1}$ (equivalently N s). Momentum is a vector. Sign matters.

Impulse

Impulse is the product of net force and the time it acts, and it equals the change in momentum:

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

This follows directly from $F = ma = m \Delta v / \Delta t$ . For a variable force, impulse equals the area under the force-time graph.

Impulse is why crumple zones, airbags, padded gloves and bent knees on landing reduce injury: stretching out $\Delta t$ reduces the peak force needed to deliver the same $\Delta p$ .

Conservation of momentum

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

\sum m_i \vec{v}_{i,\text{before}} = \sum m_i \vec{v}_{i,\text{after}}

For a one-dimensional collision between two bodies, this becomes:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Conservation of momentum holds in every type of collision, including the most dramatic inelastic ones.

Elastic and inelastic collisions

Type	Momentum	Kinetic energy	Example
Elastic	Conserved	Conserved	Ideal billiard balls, hard atomic collisions
Inelastic	Conserved	Not conserved (some becomes heat, sound, deformation)	Most real collisions
Perfectly inelastic	Conserved	Maximum $KE$ loss	Bodies stick together

For perfectly inelastic collisions where the bodies stick:

m_1 u_1 + m_2 u_2 = (m_1 + m_2) v

Explosions

An explosion is the time-reverse of a perfectly inelastic collision. A single body at rest separates into two bodies with equal and opposite momenta:

0 = m_1 v_1 + m_2 v_2

The lighter fragment moves faster, in the opposite direction to the heavier fragment.

Perfectly inelastic car collision

A $1500$ kg car travelling east at $20$ m s $^{-1}$ collides head-on with a stationary $1000$ kg car. They lock together after impact. Find the velocity immediately after, and the kinetic energy lost.

Step 1: Set a sign convention and identify the collision type

The two cars lock together, so this is a perfectly inelastic collision. Take east as positive. Before the collision the moving car has momentum $p = mv = (1500)(20) = 30\,000$ kg m s $^{-1}$ ; the stationary car contributes zero. The combined mass after is $1500 + 1000 = 2500$ kg.

Step 2: Apply conservation of momentum to find the joint velocity

Because the collision time is short, external forces deliver negligible impulse and total momentum is conserved. Setting total momentum before equal to total momentum after:

(1500)(20) + (1000)(0) = (2500)\,v

v = \frac{30\,000}{2500} = 12 \text{ m s}^{-1} \text{ east.}

Step 3: Calculate kinetic energy before and after

Kinetic energy is not automatically conserved. Compute each side separately:

KE_{\text{before}} = \tfrac{1}{2}(1500)(20)^2 = 300\,000 \text{ J}

KE_{\text{after}} = \tfrac{1}{2}(2500)(12)^2 = 180\,000 \text{ J}

Step 4: Find the energy lost and classify the collision

The energy difference is the amount converted to heat, sound and deformation:

KE_{\text{lost}} = 300\,000 - 180\,000 = 120\,000 \text{ J}

Momentum is conserved (both sides equal $30\,000$ kg m s $^{-1}$ ), but kinetic energy is not, so the collision is inelastic.

Final answer: joint velocity $= 12$ m s $^{-1}$ east; kinetic energy lost $= 120\,000$ J.

Examples in context

Example 1. A Cairns light-rail tram (mass $30000 \text{ kg}$ ) at $8.0 \text{ m s}^{-1}$ couples with a stationary trailer ( $15000 \text{ kg}$ ). Conservation of momentum gives $30000 \times 8.0 = 45000 \times v$ , so $v = 5.33 \text{ m s}^{-1}$ . Kinetic energy drops from $9.6 \times 10^5 \text{ J}$ to $6.4 \times 10^5 \text{ J}$ - the collision is inelastic, with $3.2 \times 10^5 \text{ J}$ dissipated as sound, heat and coupler-spring storage. QCAA Unit 2 EA Paper 2 uses precisely this two-cart structure.

Example 2. A safety barrier on the Bremer River bridge deforms over $0.40 \text{ m}$ when struck by a $1500 \text{ kg}$ truck at $16 \text{ m s}^{-1}$ . Impulse changes momentum from $24000 \text{ kg m s}^{-1}$ to zero. Stopping time is $t = 2 \times 0.40 / 16 = 0.050 \text{ s}$ (assuming constant deceleration), giving average force $F = \Delta p / t = 480 \text{ kN}$ . Doubling crumple length to $0.80 \text{ m}$ halves the average force, an impulse-time tradeoff QCAA Unit 2 IA1 data-test scenarios examine explicitly.

Try this

Q1. Define linear momentum and impulse, and state their SI units. [2 marks]

Cue. $p = mv$ , units $\text{kg m s}^{-1}$ ; $J = F\Delta t = \Delta p$ , units $\text{N s}$ .

Q2. A $0.15 \text{ kg}$ cricket ball travelling at $30 \text{ m s}^{-1}$ is struck and returns at $40 \text{ m s}^{-1}$ in the opposite direction. Contact time is $2.0 \text{ ms}$ . Calculate the impulse and the average force. [3 marks]

Cue. $\Delta p = 0.15 \times (40+30) = 10.5 \text{ N s}$ ; $F = J/t = 5250 \text{ N}$ .

Q3. A Cairns tram ( $30000 \text{ kg}$ ) at $8.0 \text{ m s}^{-1}$ couples with a stationary trailer ( $15000 \text{ kg}$ ). (a) Calculate the joint velocity post-coupling. (b) Determine the kinetic energy before and after, and the energy dissipated. (c) Classify the collision and justify. [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

Cue. (a) $5.33 \text{ m s}^{-1}$ ; (b) $9.6 \times 10^5 \text{ J}$ , $6.4 \times 10^5 \text{ J}$ , loss $3.2 \times 10^5 \text{ J}$ ; (c) inelastic, KE not conserved.

Exam-style practice questions

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

Year 11 SAC5 marksA

0.20

kg ball travelling at

8.0

m s

^{-1}

to the right collides head-on with a stationary

0.30

kg ball. After the collision the

0.20

kg ball moves at

1.6

m s

^{-1}

to the left. (a) Find the velocity of the

0.30

kg ball after the collision. (b) Is the collision elastic? Justify.

Show worked answer →

Take right as positive.

(a) Conservation of momentum. $p_{\text{before}} = p_{\text{after}}$ .

$(0.20)(8.0) + (0.30)(0) = (0.20)(-1.6) + (0.30) v$

$1.6 = -0.32 + 0.30 v$

$v = 1.92 / 0.30 = 6.4$ m s $^{-1}$ to the right.

(b) Kinetic energy check.

$KE_{\text{before}} = \frac{1}{2}(0.20)(8.0)^2 = 6.4$ J.

$KE_{\text{after}} = \frac{1}{2}(0.20)(1.6)^2 + \frac{1}{2}(0.30)(6.4)^2 = 0.256 + 6.144 = 6.4$ J.

$KE$ is conserved, so the collision is elastic.

Markers reward the explicit sign convention, both kinetic-energy substitutions, and the conclusion that follows from the comparison.