2023 A/L Physics Past Paper · Question 32 (English Medium)

Answer: (1) Graph (1)

What the question asks

We want: the graph of the velocity $v$ against the time $t$ for the ball.

What each letter means:

$v$ ($\text{m s}^{-1}$): the velocity of the ball. We take up as positive, so a falling ball has a negative $v$.

$t$ ($\text{s}$): the time.

$g$ ($\text{m s}^{-2}$): the acceleration of gravity.

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

$K$ ($\text{J}$): the kinetic energy of the ball.

$\Delta K$ ($\text{J}$): the kinetic energy the ball loses at one bounce. It is the same at each bounce.

Relevant theory

When the velocity of a body is plotted against time, the velocity (v) - time (t) graph can be obtained.

Figure 1.7 shows that the velocity increases with time, and hence it represents acceleration by the gradient of the graph.

It should be noted that since displacement and velocity are vector quantities, it is important to represent the direction of motion graphically.

Translational kinetic energy of the body, $$W = \frac{1}{2}mv^2$$

The first quote says what a velocity-time graph is. The second quote says its gradient is the acceleration. The third quote says the direction of the motion must be shown on the graph, so a velocity can be negative.

The last quote gives the kinetic energy of a moving body, $K = \dfrac{1}{2}mv^2$.

For example, drop a rubber ball on the floor. It bounces, each bounce is lower than the one before, and in the end it lies still.

It does not bounce up for ever, because it loses some energy each time it hits the floor.

Required equations

$\dfrac{\Delta v}{\Delta t} = -g$

$K = \dfrac{1}{2}mv^2$

$\dfrac{1}{2}mv_n^2 = \dfrac{1}{2}mv_0^2 - n\,\Delta K$

Reasoning and calculation

Where does the line bend or jump?

Between two bounces the ball is in the air. The gradient is $-g$ throughout the flight, so the graph is a straight line sloping down, and the lines for the different flights are parallel.

At a bounce the ball turns round at once. The velocity jumps from a negative value to a positive one. That is a vertical jump in the graph.

$4.47$

$v_0 = \sqrt{2gh} = \sqrt{2 \times 10 \times 1}$

$0$

$v(0)$

$-4.47$

$v = -v_0$

How the bounces shrink:

The energy lost at each bounce is the same. So $v^2$ goes down in equal steps, $v_n^2 = v_0^2 - \dfrac{2n\,\Delta K}{m}$.

Take $v_0^2 = 20$ and a step of $4$, in $\text{m}^2\,\text{s}^{-2}$. Then $v_1 = 4$, $v_2 = 3.46$, $v_3 = 2.83$, $v_4 = 2$ and the next one is $0$.

The speeds fall slowly at first and then faster. After a few bounces there is no kinetic energy left, and the ball stops. The graph ends.

The graph we expect:

It starts at $0$ and goes down along a straight line. At the bounce it jumps up to a smaller positive value, and then goes down along a parallel line, crossing zero at the top of the rise.

This repeats with smaller and smaller jumps, and then it ends after a few bounces.

$sqrt(2 * g * h)$

Reading the options

Now match it with the five graphs.

Graph (1) goes down from zero, jumps up, and has parallel falling lines that get shorter and end. It matches.

Graph (2) stays at or below zero. Graph (3) has a first line steeper than the rest. Graph (4) has peaks that stay as high as the first. Graph (5) climbs and falls in a zigzag. None of these match.

Why the other options fail

Graph (2) stays at or below zero, so the ball would not rise after a bounce.

Graph (3) has a first line steeper than the rest, but the gradient is $-g$ for the whole flight.

Graph (4) has peaks that stay as high as the first, so no energy is lost.

Graph (5) climbs as well as falls, which needs an acceleration upwards.

Conclusion

$v_n^2 = v_0^2 - \dfrac{2n\,\Delta K}{m}$

The velocity goes down along a straight line, jumps up at each bounce to a smaller value, and the lines are parallel. That is graph (1).

The answer is (1).

The quotes give the velocity-time graph, the gradient as acceleration, and the kinetic energy. That the velocity jumps at a bounce and that $v^2$ falls in equal steps are our reasoning from them. We take the bounce as taking no time, and we ignore the air.

Teacher's note

Between bounces the gradient is $-g$, so the lines are parallel. At a bounce the velocity jumps up.

Equal losses of kinetic energy make $v^2$ fall in equal steps, so the ball stops after a few bounces.

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