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 possible reason is that the students' reaction times while using the stopwatch could lead to inaccurate measurements, especially if the ball falls quickly.

An improvement could be to use a more precise timing device or system, such as sensors, to eliminate human error in timing the fall.

To calculate the force, we can use the formula:

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

Given:

• Momentum before hitting floor = 8.7 kg m/s
• Box comes to rest, so final momentum = 0 kg m/s
• Time to come to rest = 0.35 s

Change in momentum:

$\Delta p = 0 - 8.7 = -8.7 \, \text{kg m/s}$

Now substituting:

$\text{force} = \frac{-8.7}{0.35} \approx -24.8571 \, \text{N}$

Thus, rounding, we get:

The magnitude of the force is approximately 25 N.

The magnitude of the force is 25 N.

The direction of the force is upwards, opposite to the weight of the box, which is acting downwards.

To find the mass of the ball, we use the momentum formula:

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

Given:

• Momentum = 0.40 kg m/s
• Velocity at S is not provided, consider using the impulse formula to find it:
• At S, we need to relate momentum to mass:

Using specific case of gravity:

At the point just before S:

Use the formula for velocity under free fall:

$v = g t$

Plugging into momentum:

$0.40 = m \times v\text{ (where v should be velocity just before S)}$

Using the equivalent velocity based on height:

Estimate g = 10 m/s² (used earlier); thus, average height relates back to inputs:

For calculation yield, rearranging gives:

$m = \frac{0.40}{v} \rightarrow \, m = \frac{0.40}{velocity}$.

However, empirical data evaluates at that instant suggesting alterations to depth to estimate for 3.8m yields mean effects in decay. Execution brings mass around 0.04 kg that's derived.