2019 A/L Physics Past Paper · Question 31 (English Medium)

Answer: (4) $\left(\frac{3}{4}\right)^6$ m

What the question asks

We know:

The ball is let go from $h_0 = 1$ m.

At the start it is still: $u = 0$

At each bounce, its speed falls by $25\%$.

So after a bounce, the speed is $\frac{3}{4}$ of the speed before.

We want: the height the ball rises to after three bounces.

Relevant theory

The relationships between the physical properties that describe a particular motion of a body are called equations of motion.

The motion under the action of gravitational force, with negligible resistances, such as air resistance, is called motion under gravity.

The ball moves under gravity.

For motion with uniform acceleration:

$$v^2 = u^2 + 2as$$

The ball rises until its speed is zero.

So a ball that leaves the floor with speed $v$ rises to:

$$h = \frac{v^2}{2g}$$

Required equations

$v^2 = u^2 + 2as$

$h = \frac{v^2}{2g}$

Reasoning and calculation

What stays the same: $g$. It is the same in every flight.

What changes: the speed after each bounce.

Look at one bounce.

The speed becomes $\frac{3}{4}$ of what it was.

The height goes with $v^2$. That is speed times speed.

So the height becomes $\frac{3}{4} \times \frac{3}{4}$ of what it was.

$\frac{h_{n+1}}{h_n} = \left(\frac{v_{n+1}}{v_n}\right)^2 = \left(\frac{3}{4}\right)^2$

Now do it three times.

Each bounce multiplies the height by $\left(\frac{3}{4}\right)^2$.

$h_3 = h_0\left(\frac{3}{4}\right)^2\left(\frac{3}{4}\right)^2\left(\frac{3}{4}\right)^2 = 1 \times \left(\frac{3}{4}\right)^6$

Reading the options

Now match it to the options.

(4) is $\left(\frac{3}{4}\right)^6$ m. It matches.

(1) uses the speed fraction only once.

(2) is the height after one bounce only.

(3) takes one $\frac{3}{4}$ per bounce.

(5) takes three $\frac{3}{4}$s per bounce.

So only (4) fits.

Checks

$h0 * (3 / 4)^6$

Check the unit.

$\left(\frac{3}{4}\right)^6$ is only a number.

A number times $1$ m is a length in metres. Good.

Does it make sense?

$\left(\frac{3}{4}\right)^6 \approx 0.18$. So the ball rises about $0.18$ m.

Each bounce loses a lot of height.

After three bounces, only a small hop is left. That fits.

Conclusion

$h_3 = \left(\frac{3}{4}\right)^6\ \text{m}$

Each bounce: height times $\left(\frac{3}{4}\right)^2$.

Three bounces:

$$h_3 = 1 \times \left(\frac{3}{4}\right)^6\ \text{m}$$

So the answer is option (4).

Thinking the height also falls by $25\%$, like the speed.

The height goes with $v^2$.

So at each bounce the height becomes $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$ of what it was.