1. A student carries out an experiment with a tennis ball and a motion sensor connected to a laptop - Scottish Highers Physics - Question 1 - 2019

To determine the height from which the ball is released, we can use the kinematic equation:

$s = ut + \frac{1}{2} a t^2$

Where:

• $s$ is the displacement (height),
• $u$ is the initial velocity (which is $0$ because the ball is released from rest),
• $a$ is the acceleration ($-9.8 \text{ m s}^{-2}$),
• $t$ is the time ($0.50 \text{ s}$).

Substituting the values:

s = 0 \times 0.50 + \frac{1}{2} \times (-9.8) \times (0.50)^2 \ $s = 0 + \frac{1}{2} \times (-9.8) \times 0.25$ $s = -1.225 \text{ m}$

Since the height must be a positive value, the ball is released from a height of $1.225 \text{ m}$.

The change in momentum is given by:

$\Delta p = mv_f - mv_i$

Given:

• Mass ($m$) = 57.0 g = 0.057 kg,
• Initial velocity before the bounce ($v_i$) = -4.9 m s⁻¹,
• Final velocity after the bounce ($v_f$) is assumed to be equal in magnitude but opposite in direction (approximately +4.9 m s⁻¹) during the bounce.

Thus, $\Delta p = 0.057 \times (4.9) - 0.057 \times (-4.9)$ $\Delta p = 0.057 \times (4.9 + 4.9) = 0.057 \times 9.8$ $\Delta p = 0.5586 ext{ kg m s}^{-1}$

Therefore, the magnitude of the change in momentum is approximately $0.559 ext{ kg m s}^{-1}$.

To find the average force, we can use the impulse-momentum theorem, stating that the impulse (force times time) is equal to the change in momentum:

$F_{avg} \Delta t = \Delta p$

Rearranging this gives:

$F_{avg} = \frac{\Delta p}{\Delta t}$

We already calculated $\Delta p = 0.5586 ext{ kg m s}^{-1}$ and the duration of the bounce is assumed to be around $0.27 \text{ s}$ (based on the graph).

So, $F_{avg} = \frac{0.5586}{0.27} \approx 2.07 ext{ N}$

The magnitude of the average force is approximately $2.07 ext{ N}$.

To complete the acceleration graph:

• There is a constant negative acceleration between 0 s and 0.50 s due to gravity, which should be a straight line sloping downward. The value of the acceleration is around $-9.8 ext{ m s}^{-2}$.
• Between 0.50 s and 1.18 s, the acceleration should return to $0 ext{ m s}^{-2}$ as the ball reaches its maximum height before descending back down. This part of the graph will show a return to $0$ from a negative value and then stay at $0$ until the ball hits the ground again.
• The resulting sketch should reflect these observations.