2023 A/L Physics Past Paper · Question 16 (English Medium)

Answer: (2) $4$

Given information

What we know:

The system is the two trolleys, $X$ and $Y$.

Trolley $X$ has mass $m$. Trolley $Y$ has mass $M$.

Before the collision, trolley $Y$ moves with speed $u$. Trolley $X$ moves twice as fast, with speed $2u$, in the same direction.

The surface is smooth and horizontal.

After the collision the two trolleys move together. The speed of trolley $Y$ has gone up by $20\%$.

What the question asks

We want: the ratio $\dfrac{M}{m}$ of the two masses.

What each letter means:

$m$ ($\text{kg}$): the mass of trolley $X$.

$M$ ($\text{kg}$): the mass of trolley $Y$.

$u$ ($\text{m s}^{-1}$): the speed of trolley $Y$ before the collision.

$v$ ($\text{m s}^{-1}$): the common speed of the two trolleys after the collision.

Relevant theory

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

momentum of the system before collision = momentum of the system after collision

The system is the two trolleys. The quantity that stays constant is their total momentum.

Why does it stay constant here? The surface is smooth, so no outside force pushes the trolleys along the line. The only forces in the collision are the pushes between the two trolleys. Those are inside the system.

So the total momentum before the collision equals the total momentum after it.

For example, think of two sledges on smooth ice. A light sledge catches up with a heavy sledge and they lock together.

The light sledge gives away some of its push. The heavy sledge gets it. The total push stays the same, but now it moves a bigger mass.

Required equations

$p = mv$

$p_{\text{before}} = m(2u) + Mu$

$p_{\text{after}} = (m + M)\,v$

$p_{\text{before}} = p_{\text{after}}$

Applying it to this question

Both sides are in $\text{kg m s}^{-1}$, and the final answer is a ratio of masses, so no change of units is needed.

$(2m + M)u$

$p_{\text{before}} = 2mu + Mu$

$1.2u$

$v = u + 0.2u$

$1.2(m + M)u$

$p_{\text{after}} = (m + M)(1.2u)$

$2m + M = 1.2m + 1.2M$

$0.8m = 0.2M$

$4$

$\dfrac{M}{m} = \dfrac{0.8}{0.2}$

Reasoning and calculation

The speed of $Y$ goes up by just a fifth. A trolley that is heavy hardly changes its speed when a light trolley hits it.

So $Y$ must be much heavier than $X$. The ratio $\dfrac{M}{m} = 4$ says $Y$ is four times as heavy. That is option (2).

Checks

$(2 - k) / (k - 1)$

The unit: the ratio of two masses in $\text{kg}$ has no unit, so $4$ is a plain number.

Does it make sense?

Put the answer back: with $M = 4m$ the momentum before is $(2m + 4m)u = 6mu$. After, it is $5m \times 1.2u = 6mu$. The two are equal.

The momentum does not change in the collision, so this fits.

Conclusion

$\dfrac{M}{m} = 4$

The ratio is $\dfrac{M}{m} = \dfrac{0.8}{0.2} = 4$.

The answer is (2).

Write the total momentum before and after. Use the speeds as multiples of $u$, so $u$ cancels.

Then put your ratio back into both sides. They must come out equal.

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