2019 A/L Physics Past Paper · Question 36 (English Medium)

Answer: (4) Diagram (4)

What the question asks

We want: how each of the four balls moves after all the collisions.

That is, the speed and the direction of each one.

Relevant theory

For a system of interacting objects, the momentum remains constant, provided no external force acts on the system.

The surface is smooth. So no force acts along it.

So the momentum is the same before and after every collision.

The collisions are elastic. So the kinetic energy is the same too.

From these two rules, when $m_1$ hits a still $m_2$ with speed $u$:

$$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u$$

$$v_2 = \frac{2m_1}{m_1 + m_2}u$$

Required equations

$m_1u = m_1v_1 + m_2v_2$

$\frac{1}{2}m_1u^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

$m_1 = m_2 \Rightarrow v_1 = 0,\ v_2 = u$

Reading the options

The diagrams show how the first ball and the $2m$ ball move at the end.

We follow the collisions one by one. Then we compare.

Reasoning and calculation

At the start

Ball 1 moves to the right with $v$.

Balls 2, 3 and $2m$ are still.

The total momentum is $mv$. The total kinetic energy is $\frac{1}{2}mv^2$.

1 hits 2, then 2 hits 3

Equal masses swap their speeds.

Ball 1 stops. Ball 2 moves off with $v$.

Ball 2 hits ball 3 and stops. Ball 3 moves off with $v$.

3 hits $2m$

$$v_3 = \frac{m - 2m}{3m}v = -\frac{v}{3}$$

The minus sign means ball 3 turns back to the left.

$$v_{2m} = \frac{2m}{3m}v = \frac{2v}{3}$$

The $2m$ ball moves on to the right.

The way back

Ball 3 moves left with $\frac{v}{3}$ and hits ball 2. They swap. Ball 3 stops.

Ball 2 hits ball 1. They swap. Ball 2 stops.

Ball 1 moves off to the left with $\frac{v}{3}$.

No more collisions happen. Ball 1 goes left and $2m$ goes right.

At the end:

Ball 1: $\frac{v}{3}$ to the left.

Balls 2 and 3: still.

The $2m$ ball: $\frac{2v}{3}$ to the right.

This matches diagram (4): an arrow of $\frac{v}{3}$ to the left and an arrow of $\frac{2v}{3}$ to the right.

Checks

$mv$

$-m\frac{v}{3} + 2m\frac{2v}{3} = mv$

$\frac{1}{2}mv^2$

$\frac{1}{2}m\frac{v^2}{9} + \frac{1}{2}(2m)\frac{4v^2}{9} = \frac{1}{2}mv^2$

Conclusion

Ball 1 moves to the left with $\frac{v}{3}$.

The $2m$ ball moves to the right with $\frac{2v}{3}$.

The two in the middle are still.

So the answer is diagram (4).

Thinking it is enough for the momentum to be right.

In an elastic collision the kinetic energy must stay the same too.

(2), (3) and (5) keep the momentum. But they do not keep the kinetic energy.