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How are collisions analysed using conservation of momentum?

Apply the principle of conservation of momentum to one-dimensional collisions and explosions, distinguishing elastic (kinetic energy conserved) and inelastic (kinetic energy not conserved) collisions

A focused answer to the VCE Physics Unit 2 dot point on collisions. Applies conservation of momentum in one dimension, distinguishes elastic from inelastic by whether KE is conserved, and works the VCAA SAC-style two-cart collision with energy-loss assessment.

Generated by Claude OpusReviewed by Better Tuition Academy6 min answer

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What this dot point is asking

VCAA wants you to apply conservation of momentum to one-dimensional collisions, distinguish elastic and inelastic types using kinetic energy, and analyse explosions as the time-reverse of perfectly inelastic collisions.

Momentum

p⃗=mv⃗\vec{p} = m \vec{v}

SI unit: kg m sβˆ’1^{-1} (equivalent to N s). Vector quantity.

Conservation of momentum

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

βˆ‘mivβƒ—i,before=βˆ‘mivβƒ—i,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:

m1u1+m2u2=m1v1+m2v2m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Sign matters. Pick a positive direction; bodies moving the other way get negative signs.

Elastic vs inelastic

Type Momentum Kinetic energy Example
Elastic Conserved Conserved Idealised carts on a track
Inelastic Conserved Not conserved Most real collisions
Perfectly inelastic Conserved Maximum loss Bodies stick together

For perfectly inelastic collisions where 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 initially at rest separates into two pieces:

0=m1v1+m2v20 = m_1 v_1 + m_2 v_2

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

Impulse

Impulse equals change in momentum:

J⃗=F⃗netΔt=Δp⃗\vec{J} = \vec{F}_{\text{net}} \Delta t = \Delta \vec{p}

Stretching the collision time (airbags, crumple zones, padded helmets) reduces the peak force needed to deliver the same Ξ”p\Delta p.

Worked example

A 15001500 kg car at 2020 m sβˆ’1^{-1} east collides head-on with a 10001000 kg car at rest. They stick. Find the velocity after and KE lost.

Conservation of momentum: (1500)(20)+0=(2500)v(1500)(20) + 0 = (2500) v. v=12v = 12 m sβˆ’1^{-1} east.

KE before: 12(1500)(20)2=300 000\frac{1}{2}(1500)(20)^2 = 300\,000 J.

KE after: 12(2500)(12)2=180 000\frac{1}{2}(2500)(12)^2 = 180\,000 J.

KE lost: 120 000120\,000 J (turned into heat, sound, deformation). Inelastic.

Common traps

Treating KE as always conserved. KE is conserved only in elastic collisions.

Dropping signs in head-on collisions. Bodies approaching each other have opposite signs of velocity.

Applying conservation to a non-isolated system. External forces (large friction, gravity over long times) can invalidate the assumption. For typical short-duration collisions on smooth surfaces, the system is effectively isolated.

Confusing impulse with force. Impulse has units of N s (or kg m sβˆ’1^{-1}). Force has units of N.

In one sentence

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 reduces the peak force (airbags, crumple zones).

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

Year 11 SAC5 marksCart A ($0.40$ kg) moves at $3.0$ m s$^{-1}$ east; cart B ($0.60$ kg) is stationary. After collision A moves at $0.60$ m s$^{-1}$ west and B at some velocity east. (a) Find B's velocity. (b) Determine whether the collision is elastic.
Show worked answer β†’

Take east as positive.

(a) Momentum conservation. (0.40)(3.0)+0=(0.40)(βˆ’0.60)+(0.60)vB(0.40)(3.0) + 0 = (0.40)(-0.60) + (0.60) v_B.

1.20=βˆ’0.24+0.60vB1.20 = -0.24 + 0.60 v_B.

vB=1.44/0.60=2.40v_B = 1.44/0.60 = 2.40 m sβˆ’1^{-1} east.

(b) Elasticity check.

KE before: 12(0.40)(3.0)2=1.80\frac{1}{2}(0.40)(3.0)^2 = 1.80 J.

KE after: 12(0.40)(0.60)2+12(0.60)(2.40)2=0.072+1.728=1.80\frac{1}{2}(0.40)(0.60)^2 + \frac{1}{2}(0.60)(2.40)^2 = 0.072 + 1.728 = 1.80 J.

KE is conserved. Collision is elastic.

Markers reward sign convention, momentum conservation with both directions, KE in two terms, and explicit conclusion.

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