9 (a) An atom of mass 6.6 x 10^{-27} kg is moving with a velocity of 480 m/s - Edexcel - GCSE Physics - Question 9 - 2023 - Paper 1

Before the collision, the momentum of the ball is $0.80 \text{ kg m/s}$ and after the collision, it is $0.60 \text{ kg m/s}$ in the opposite direction.

The change in momentum ($\Delta p$) is:

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

Here, the initial momentum is $0.80 \text{ kg m/s}$ (to the right) and the final momentum is $-0.60 \text{ kg m/s}$ (to the left):

$\Delta p = -0.60 - 0.80 = -1.40 \text{ kg m/s}$

The time of contact is $70 \text{ ms} = 70 \times 10^{-3} \text{ s}$. To find the force (F), we use:

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

Substituting in the values gives:

$F = \frac{-1.40 \text{ kg m/s}}{70 \times 10^{-3} \text{ s}} = -20.0 \text{ N}$

Thus, the force exerted on the ball by the wall is $20.0 \text{ N}$ in the direction opposite to the ball's initial motion.