Figure 9 shows a small steel ball held at a height, h, above the ground - Edexcel - GCSE Physics Combined Science - Question 5 - 2021 - Paper 1

One reason the students’ value for t may be different is that their reaction time in starting and stopping the stopwatch introduces measurement error, leading to a longer recorded time.

An improvement would be to use an automatic timer or photogate system to measure the time precisely, minimizing human error associated with manual timing.

The force exerted by the floor can be calculated using the formula for force based on the change in momentum:

$\text{Force} = \frac{\text{change in momentum}}{\text{time}}$

Given:

• Momentum just before hitting the floor = 8.7 kg m/s
• Time to come to rest = 0.35 s

The change in momentum is equal to the initial momentum since final momentum is 0:

$\text{Change in momentum} = 8.7 \, \text{kg m/s}$

Thus,

$\text{Force} = \frac{8.7}{0.35} \approx 24.86 \, \text{N} \approx 25 \, \text{N}$

Magnitude: 25 N

Direction: The force acts upwards, opposite to the direction of the weight of the box, towards the floor.

The mass of the ball can be calculated using the formula for momentum:

$\text{Momentum} = \text{mass} \times \text{velocity}$

Rearranging gives:

$\text{mass} = \frac{\text{Momentum}}{\text{velocity}}$

Given that momentum at S is 0.40 kg m/s and the velocity at that point is:

• Since the height from which the ball falls does not affect the final momentum directly, we can assume the velocity at S can be derived from the free fall equations. However, without additional details on velocity, simply using momentum gives us:

Thus:

$\text{mass} = \frac{0.40}{\text{velocity}}$

To find velocity, more specific height considerations or values are needed. However, for the purpose of the exam, be sure to replace velocity with an appropriate calculation based on drop height or derived formulas from kinematics.