Figure 10 shows a small steel ball held at a height, $h$, above the ground - Edexcel - GCSE Physics - Question 7 - 2021 - Paper 1

One possible reason for the discrepancy is that the students' reaction times when starting and stopping the stopwatch may have introduced errors. This means they may not have precisely captured the moment the ball was released and when it hit the ground.

To improve their accuracy, the students could use electronic timers that can be triggered automatically when the ball is released and when it impacts the ground, reducing human reaction time errors.

The force exerted by the floor on the box can be calculated using the change in momentum:
Force = \frac{change , in , momentum}{time}\text{.}
The change in momentum is 8.7 kg m/s and the time taken is 0.35 s:
$Force = \frac{8.7}{0.35} = 24.857 \text{ N} \approx 25 \text{ N}$

The magnitude of the force exerted by the box on the floor is 25 N. The direction of the force is opposite to the direction of the force exerted by the floor on the box, which is upwards. Thus, the direction is downwards.

First, we determine the velocity of the ball at $S$ using the formula for kinetic energy:
$v^2 = u^2 + 2as$
where, $u=0$, $a=10 \text{ m/s}^2$, and $s=3.8 \text{ m}$.
Substituting the values:
$v^2 = 0 + 2 \times 10 \times 3.8 = 76$
Now, calculating $v$:
$v = 8.7 \text{ m/s}$
Next, we can find the mass using momentum formula:
Momentum = mass \times velocity
$0.40 = m \times 8.7 \implies m = \frac{0.40}{8.7} \approx 0.046 \text{ kg}$