Collisions and conservation of momentum (VCE Physics Unit 2)

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

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

Conservation of momentum

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

For a one-dimensional collision between two bodies:

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

Elastic vs inelastic

For perfectly inelastic collisions where bodies stick:

Explosions

An explosion is the time-reverse of a perfectly inelastic collision. A single body initially at rest separates into two pieces:

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

Impulse

Impulse equals change in momentum:

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

Worked example

A $1500$ kg car at $20$ m s$^{-1}$ east collides head-on with a $1000$ kg car at rest. They stick. Find the velocity after and KE lost.

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

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

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

KE lost: $120\,000$ 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 s$^{-1}$). Force has units of N.

In one sentence

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

Examples in context

Example 1. Spencer Street rail-yard shunt coupling. Two freight wagons in the Melbourne Spencer Street rail yard collide and couple. A $20\,000$ kg wagon rolling at $1.5$ m s$^{-1}$ strikes a stationary $30\,000$ kg wagon. Momentum conservation gives $20\,000 \times 1.5 + 0 = (20\,000 + 30\,000) v$, so $v = 30\,000/50\,000 = 0.60$ m s$^{-1}$. Initial KE was $\tfrac{1}{2} \times 20\,000 \times 1.5^2 = 22\,500$ J; final KE is $\tfrac{1}{2} \times 50\,000 \times 0.6^2 = 9000$ J. Lost energy of $13\,500$ J appears as heat, sound and permanent deformation in the buffers, confirming the collision is inelastic.

Example 2. AFL ruck contest at the MCG. Two ruckmen contest a centre bounce. Player A ($95$ kg) moves north at $4.0$ m s$^{-1}$; player B ($90$ kg) moves south at $3.5$ m s$^{-1}$. Treating the brief contact as a one-dimensional collision and assuming they move together immediately after, $95 \times 4.0 + 90 \times (-3.5) = (95 + 90) v$, giving $v = (380 - 315)/185 = 0.35$ m s$^{-1}$ north. Initial KE is $\tfrac{1}{2}(95)(16) + \tfrac{1}{2}(90)(12.25) = 1311$ J; final KE is $\tfrac{1}{2}(185)(0.35)^2 = 11$ J. Most energy is dissipated in muscle work and sound, confirming the contest is highly inelastic.

Try this

Q1. State the principle of conservation of momentum. [2 marks]

• Cue. In an isolated system (no external net force), total momentum before equals total momentum after any interaction.

Q2. A $1500$ kg car travelling at $20$ m s$^{-1}$ east collides head-on with a $1000$ kg car travelling at $15$ m s$^{-1}$ west. After collision the wreckage moves together. Calculate the final velocity and identify whether the collision is elastic. [4 marks]

• Cue. $1500(20) + 1000(-15) = 2500 v$, $v = 6.0$ m s$^{-1}$ east. KE not conserved (drops from $4.13 \times 10^5$ J to $4.5 \times 10^4$ J), so inelastic.

Q3. Refer to the Spencer Street shunt. (a) Calculate the joined-wagon velocity. (b) Determine the kinetic energy lost. (c) Outline two destinations for the lost energy. [2+2+2 marks]

• Cue. (a) $0.60$ m s$^{-1}$. (b) $13\,500$ J. (c) Heat and sound in buffers, plus permanent deformation.