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Momentum, impulse and conservation in collisions (QCE Physics Unit 2)

Syllabus dot point

“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”

9 min read

Quick answer

Momentum p⃗=mv⃗\vec{p} = m \vec{v} is conserved in any isolated one-dimensional collision or explosion, kinetic energy is only conserved in elastic collisions, and impulse J⃗=F⃗Δt=Δp⃗\vec{J} = \vec{F} \Delta t = \Delta \vec{p} explains why stretching the collision time (airbags, crumple zones) reduces the peak force on a body.

Jump to a section
  1. What this dot point is asking
  2. Momentum
  3. Impulse
  4. Conservation of momentum
  5. Elastic and inelastic collisions
  6. Explosions
  7. Examples in context
  8. Try this
  9. Exam-style questions

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:

p⃗=mv⃗\vec{p} = m \vec{v}

SI unit: kg m s−1^{-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:

J⃗=F⃗net Δt=Δp⃗=mv⃗f−mv⃗i\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=mΔv/ΔtF = 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 Δt\Delta t reduces the peak force needed to deliver the same Δp\Delta p.

Conservation of momentum

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

∑miv⃗i,before=∑miv⃗i,after\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:

m1u1+m2u2=m1v1+m2v2m_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 KEKE loss Bodies stick together

For perfectly inelastic collisions where the bodies stick:

m1u1+m2u2=(m1+m2)vm_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=m1v1+m2v20 = 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 15001500 kg car travelling east at 2020 m s−1^{-1} collides head-on with a stationary 10001000 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 000p = mv = (1500)(20) = 30\,000 kg m s−1^{-1}; the stationary car contributes zero. The combined mass after is 1500+1000=25001500 + 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(1500)(20) + (1000)(0) = (2500)\,v

v=30 0002500=12 m s−1 east.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:

KEbefore=12(1500)(20)2=300 000 JKE_{\text{before}} = \tfrac{1}{2}(1500)(20)^2 = 300\,000 \text{ J}

KEafter=12(2500)(12)2=180 000 JKE_{\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:

KElost=300 000−180 000=120 000 JKE_{\text{lost}} = 300\,000 - 180\,000 = 120\,000 \text{ J}

Momentum is conserved (both sides equal 30 00030\,000 kg m s−1^{-1}), but kinetic energy is not, so the collision is inelastic.

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

Common traps
Treating kinetic energy as conserved in every collision
KEKE is only conserved in elastic collisions. Always check.
Dropping the sign on velocity
In a head-on collision the bodies have opposite signs of velocity. Mixing them up gives a momentum total that is too large by a factor of two or three.
Applying momentum conservation to a system with an external force
If a car is being pushed by a person during the collision, the system is not isolated. For typical QCAA problems, the collision time is short enough that external forces (friction, weight) deliver negligible impulse.
Confusing impulse with force
Impulse has units of N s (or kg m s−1^{-1}). Force has units of N. They are equal only when multiplied or divided by Δt\Delta t.

Examples in context

Example 1. A Cairns light-rail tram (mass 30000 kg30000 \text{ kg}) at 8.0 m s−18.0 \text{ m s}^{-1} couples with a stationary trailer (15000 kg15000 \text{ kg}). Conservation of momentum gives 30000×8.0=45000×v30000 \times 8.0 = 45000 \times v, so v=5.33 m s−1v = 5.33 \text{ m s}^{-1}. Kinetic energy drops from 9.6×105 J9.6 \times 10^5 \text{ J} to 6.4×105 J6.4 \times 10^5 \text{ J} - the collision is inelastic, with 3.2×105 J3.2 \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 m0.40 \text{ m} when struck by a 1500 kg1500 \text{ kg} truck at 16 m s−116 \text{ m s}^{-1}. Impulse changes momentum from 24000 kg m s−124000 \text{ kg m s}^{-1} to zero. Stopping time is t=2×0.40/16=0.050 st = 2 \times 0.40 / 16 = 0.050 \text{ s} (assuming constant deceleration), giving average force F=Δp/t=480 kNF = \Delta p / t = 480 \text{ kN}. Doubling crumple length to 0.80 m0.80 \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=mvp = mv, units kg m s−1\text{kg m s}^{-1}; J=FΔt=ΔpJ = F\Delta t = \Delta p, units N s\text{N s}.

Q2. A 0.15 kg0.15 \text{ kg} cricket ball travelling at 30 m s−130 \text{ m s}^{-1} is struck and returns at 40 m s−140 \text{ m s}^{-1} in the opposite direction. Contact time is 2.0 ms2.0 \text{ ms}. Calculate the impulse and the average force. [3 marks]

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

Q3. A Cairns tram (30000 kg30000 \text{ kg}) at 8.0 m s−18.0 \text{ m s}^{-1} couples with a stationary trailer (15000 kg15000 \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 m s−15.33 \text{ m s}^{-1}; (b) 9.6×105 J9.6 \times 10^5 \text{ J}, 6.4×105 J6.4 \times 10^5 \text{ J}, loss 3.2×105 J3.2 \times 10^5 \text{ J}; (c) inelastic, KE not conserved.

Exam-style questions

Questions in the style of QCAA 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.

Year 11 SAC-style5 marks
A 0.200.20 kg ball travelling at 8.08.0 m s−1^{-1} to the right collides head-on with a stationary 0.300.30 kg ball. After the collision the 0.200.20 kg ball moves at 1.61.6 m s−1^{-1} to the left. (a) Find the velocity of the 0.300.30 kg ball after the collision. (b) Is the collision elastic? Justify.
Show worked answer →

Take right as positive.

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

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

1.6=−0.32+0.30v1.6 = -0.32 + 0.30 v

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

(b) Kinetic energy check.

KEbefore=12(0.20)(8.0)2=6.4KE_{\text{before}} = \frac{1}{2}(0.20)(8.0)^2 = 6.4 J.

KEafter=12(0.20)(1.6)2+12(0.30)(6.4)2=0.256+6.144=6.4KE_{\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.

KEKE 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.

Practise this

Sources & how we know this

Based on the official QCAA syllabus and published QCAA past papers and marking guidelines. Not yet individually reviewed by a person; always check QCAA and your school's materials for current rules.

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