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

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.