6 (a) An atom of mass 6.6 × 10^{-26} kg is moving with a velocity of 480 m/s - Edexcel - GCSE Physics Combined Science - Question 6 - 2023 - Paper 1

To calculate the force exerted on the ball by the wall, we need to determine the change in momentum and the time of contact. The change in momentum ($\Delta p$) can be calculated as follows:

$\Delta p = p_{final} - p_{initial}$ where:

• $p_{initial} = 0.80 \text{ kg m/s}$ (before collision)
• $p_{final} = -0.60 \text{ kg m/s}$ (after collision, negative sign indicates opposite direction)

Calculating the change in momentum: $\Delta p = -0.60 - 0.80 = -1.40 \text{ kg m/s}$

Next, we use the impulse-momentum theorem, which states that force ($F$) can be calculated via the equation:

$F = \frac{\Delta p}{\Delta t}$

where $\Delta t = 70 \text{ ms} = 70 \times 10^{-3} \text{ s}$. Thus,

$F = \frac{-1.40}{70 \times 10^{-3}}$ Calculating this gives: $F = -20.0 \text{ N}$ (The negative sign indicates the direction of the force is opposite to the initial momentum.)