2022 A/L Physics Past Paper · Question 24 (English Medium)

Answer: (2) $30\ \text{m}$

Given information

What we know:

The ball leaves the edge horizontally with $u = 10\ \text{m s}^{-1}$. The building is $h = 20\ \text{m}$ high. We take $g = 10\ \text{m s}^{-2}$.

The ball splits into two identical pieces, each of mass $m$. So the ball has mass $2m$.

Piece $X$ lands $10\ \text{m}$ from the building. Both pieces land at the same moment. Air resistance is ignored.

What the question asks

We want: the distance $R$ at which piece $Y$ lands.

What each letter means:

$u$ ($\text{m s}^{-1}$): the horizontal speed of the ball when it leaves the edge.

$h$ ($\text{m}$): the height of the building.

$t$ ($\text{s}$): the time the ball takes to fall to the ground.

$v_X$, $v_Y$ ($\text{m s}^{-1}$): the horizontal speeds of the pieces just after the split.

$x_X$, $x_Y$ ($\text{m}$): the horizontal distances where the pieces land. Here $x_X = 10\ \text{m}$ and $x_Y = R$.

Relevant theory

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

So the vertical motion of the object is simillar to a motion under gravity.

s = ut + \frac{1}{2}at^2

Horizontal component remains constant.

For example, think of a firework shell that bursts in the air. The pieces fly apart, but the pushes that break it are inside the shell. So the shell as a whole carries on as if nothing had burst. One piece lands nearer, the other lands farther, and the two landing places balance about where the whole shell would have landed.

First the time of fall. The ball starts with no downward speed, so $h = \tfrac{1}{2}gt^2$. The pieces land together, so the split did not change the vertical motion. The fall time is still $t$.

$$t = \sqrt{\dfrac{2h}{g}}$$

In words: the fall time depends only on the height and on $g$.

Now the split. The forces that break the ball are inside the system, so the total horizontal momentum stays the same.

$$2m\,u = m\,v_X + m\,v_Y \quad\Rightarrow\quad v_X + v_Y = 2u$$

In words: the two horizontal speeds after the split add up to twice the speed of the ball.

Let the split happen after a time $t_1$. Before it each piece moved at $u$. After it each piece moves at its own speed for the remaining time $t - t_1$.

$$x_X + x_Y = 2u\,t_1 + (v_X + v_Y)(t - t_1) = 2u\,t$$

In words: the two landing distances add up to $2ut$, whenever the split happens.

Required equations

$t = \sqrt{\dfrac{2h}{g}}$

$v_X + v_Y = 2u$

$x_X + x_Y = 2u\,t$

$R = 2u\,t - x_X$

Applying it to this question

Speed times time gives distance: $\text{m s}^{-1} \times \text{s} = \text{m}$.

$2$

$t = \sqrt{\dfrac{2 \times 20}{10}} = \sqrt{4}$

$20$

$v_X + v_Y = 2 \times 10$

$40$

$x_X + x_Y = 2u\,t = 2 \times 10 \times 2$

$30$

$R = 40 - 10$

Reasoning and calculation

The whole ball would have landed $u\,t = 20\ \text{m}$ from the building. The two identical pieces share this equally, one on each side. $X$ is $10\ \text{m}$ short of $20\ \text{m}$, so $Y$ is $10\ \text{m}$ beyond it, at $30\ \text{m}$. That is option (2).

Checks

$2 * v * t - x$

Does it make sense?

The centre of the two landing places, $\dfrac{10 + 30}{2} = 20\ \text{m}$, is exactly where the whole ball would have landed. The pieces are identical, so they balance about that point.

$X$ landed short of it, so $Y$ must land beyond it. $30\ \text{m}$ is beyond $20\ \text{m}$.

Conclusion

$R = 30\ \text{m}$

The fall time is $t = 2\ \text{s}$ and the landing distances add up to $2ut = 40\ \text{m}$. So $R = 40 - 10 = 30\ \text{m}$.

The answer is (2).

A split caused by forces inside the ball does not change the total horizontal momentum. So the two pieces land balanced about where the whole ball would have landed.

Find the fall time first, then use $x_X + x_Y = 2ut$.

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