Physics 9702 · Paper 2

Elastic & Inelastic Collisions

Momentum is conserved in every collision. Kinetic energy is not.
Whether the kinetic energy survives is what separates the two kinds of collision.

The two kinds

Perfectly elastic	Inelastic
momentum conserved	momentum conserved
total kinetic energy conserved	total kinetic energy decreases
relative speed of approach = relative speed of separation	objects may deform or stick together
rare: gas molecules, some ball collisions	most real collisions

Same start, two endings. Total momentum is zero before and after in both cases. Kinetic energy survives only in the elastic case.

In an inelastic collision the missing kinetic energy becomes heat, sound and the energy of bending the objects. Total energy is still conserved.
Sticking together is the extreme case: it loses the largest possible amount of kinetic energy (while momentum is still conserved). It is called perfectly inelastic.

Common mistake

Writing “energy is not conserved” for an inelastic collision. Energy is always conserved. Write “kinetic energy decreases” or “kinetic energy is transferred to other forms”.

The KE test

To classify a collision, compare the total kinetic energy before and after.

Worked example

Smallest case: the motion moves to the other ball

Ball A (mass m) moves at 4.0 m s⁻¹ and hits an identical stationary ball B. A stops; B moves off at 4.0 m s⁻¹. Elastic or inelastic?

KE before: $\tfrac{1}{2}m(4.0)^2 = 8.0m$ . KE after: $\tfrac{1}{2}m(4.0)^2 = 8.0m$ .
Kinetic energy unchanged: perfectly elastic.

Answer

\text{elastic}

Worked example

Exam version: find the speed, then test

Ball X (0.240 kg) moving at 16 m s⁻¹ hits a stationary ball Y (0.480 kg). X stops. Find Y's velocity, then decide whether the collision is elastic. (9702/23/O/N/24 Q2)

Momentum: $0.240 \times 16 = 0.480\,v$ , so $v = 8.0 \text{ m s}^{-1}$ .
KE before: $\tfrac{1}{2}(0.240)(16)^2 = 30.7 \text{ J}$ . KE after: $\tfrac{1}{2}(0.480)(8.0)^2 = 15.4 \text{ J}$ .
Kinetic energy fell by about 15 J, so the collision is inelastic.

Answer

v = 8.0 \text{ m s}^{-1}, \text{ inelastic}

Your turn— tap to reveal the worked answer (9702-style)

A ball of mass $m$ moving at speed $v$ hits a stationary ball of mass $2m$ and they stick. What fraction of the kinetic energy is lost?

Momentum: $mv = 3m\,v'$ , so $v' = v/3$ .
KE before $= \tfrac{1}{2}mv^2$ ; after $= \tfrac{1}{2}(3m)(v/3)^2 = \tfrac{1}{6}mv^2$ .
Fraction lost $= \dfrac{\tfrac{1}{2} - \tfrac{1}{6}}{\tfrac{1}{2}} = \tfrac{2}{3}$ .

Answer: two thirds of the kinetic energy is lost.

The relative-speed shortcut

In a perfectly elastic collision, the speed at which the objects approach each other equals the speed at which they separate: $u_A - u_B = v_B - v_A$ .
This gives a second, linear equation. Pair it with momentum conservation and you can find both final velocities without using the squared KE equation.
The mark scheme accepts either test for “is it elastic”: compare total KE, or compare approach and separation speeds (9702/22/M/J/25 Q4(a)).

Worked example

Using the shortcut

A ball of mass m moving at 9.0 m s⁻¹ makes a perfectly elastic head-on collision with a stationary ball of mass 2m. Find both final velocities. (9702/11/O/N/25 Q10)

Momentum: $9.0m = m v_1 + 2m v_2$ , so $v_1 + 2v_2 = 9.0$ .
Relative speed: approach = 9.0, so separation $v_2 - v_1 = 9.0$ .
Add the two equations: $3v_2 = 18$ , so $v_2 = 6.0 \text{ m s}^{-1}$ and $v_1 = -3.0 \text{ m s}^{-1}$ : the light ball bounces back.

Answer

v_2 = 6.0 \text{ m s}^{-1}, \; v_1 = 3.0 \text{ m s}^{-1} \text{ backwards}

When a question says “perfectly elastic”, it is giving you a second equation: speed of approach = speed of separation. Use it instead of the KE equation.

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Elastic & Inelastic Collisions

Momentum is conserved in every collision. Kinetic energy is not.
Whether the kinetic energy survives is what separates the two kinds of collision.

The two kinds

Perfectly elastic	Inelastic
momentum conserved	momentum conserved
total kinetic energy conserved	total kinetic energy decreases
relative speed of approach = relative speed of separation	objects may deform or stick together
rare: gas molecules, some ball collisions	most real collisions

Same start, two endings. Total momentum is zero before and after in both cases. Kinetic energy survives only in the elastic case.

In an inelastic collision the missing kinetic energy becomes heat, sound and the energy of bending the objects. Total energy is still conserved.
Sticking together is the extreme case: it loses the largest possible amount of kinetic energy (while momentum is still conserved). It is called perfectly inelastic.

Common mistake

Writing “energy is not conserved” for an inelastic collision. Energy is always conserved. Write “kinetic energy decreases” or “kinetic energy is transferred to other forms”.

The KE test

To classify a collision, compare the total kinetic energy before and after.

Worked example

Smallest case: the motion moves to the other ball

Ball A (mass m) moves at 4.0 m s⁻¹ and hits an identical stationary ball B. A stops; B moves off at 4.0 m s⁻¹. Elastic or inelastic?

KE before: $\tfrac{1}{2}m(4.0)^2 = 8.0m$ . KE after: $\tfrac{1}{2}m(4.0)^2 = 8.0m$ .
Kinetic energy unchanged: perfectly elastic.

Answer

\text{elastic}

Worked example

Exam version: find the speed, then test

Ball X (0.240 kg) moving at 16 m s⁻¹ hits a stationary ball Y (0.480 kg). X stops. Find Y's velocity, then decide whether the collision is elastic. (9702/23/O/N/24 Q2)

Momentum: $0.240 \times 16 = 0.480\,v$ , so $v = 8.0 \text{ m s}^{-1}$ .
KE before: $\tfrac{1}{2}(0.240)(16)^2 = 30.7 \text{ J}$ . KE after: $\tfrac{1}{2}(0.480)(8.0)^2 = 15.4 \text{ J}$ .
Kinetic energy fell by about 15 J, so the collision is inelastic.

Answer

v = 8.0 \text{ m s}^{-1}, \text{ inelastic}

Your turn— tap to reveal the worked answer (9702-style)

A ball of mass $m$ moving at speed $v$ hits a stationary ball of mass $2m$ and they stick. What fraction of the kinetic energy is lost?

Momentum: $mv = 3m\,v'$ , so $v' = v/3$ .
KE before $= \tfrac{1}{2}mv^2$ ; after $= \tfrac{1}{2}(3m)(v/3)^2 = \tfrac{1}{6}mv^2$ .
Fraction lost $= \dfrac{\tfrac{1}{2} - \tfrac{1}{6}}{\tfrac{1}{2}} = \tfrac{2}{3}$ .

Answer: two thirds of the kinetic energy is lost.

The relative-speed shortcut

In a perfectly elastic collision, the speed at which the objects approach each other equals the speed at which they separate: $u_A - u_B = v_B - v_A$ .
This gives a second, linear equation. Pair it with momentum conservation and you can find both final velocities without using the squared KE equation.
The mark scheme accepts either test for “is it elastic”: compare total KE, or compare approach and separation speeds (9702/22/M/J/25 Q4(a)).

Worked example

Using the shortcut

A ball of mass m moving at 9.0 m s⁻¹ makes a perfectly elastic head-on collision with a stationary ball of mass 2m. Find both final velocities. (9702/11/O/N/25 Q10)

Momentum: $9.0m = m v_1 + 2m v_2$ , so $v_1 + 2v_2 = 9.0$ .
Relative speed: approach = 9.0, so separation $v_2 - v_1 = 9.0$ .
Add the two equations: $3v_2 = 18$ , so $v_2 = 6.0 \text{ m s}^{-1}$ and $v_1 = -3.0 \text{ m s}^{-1}$ : the light ball bounces back.

Answer

v_2 = 6.0 \text{ m s}^{-1}, \; v_1 = 3.0 \text{ m s}^{-1} \text{ backwards}

When a question says “perfectly elastic”, it is giving you a second equation: speed of approach = speed of separation. Use it instead of the KE equation.