Dynamics: Question 7

Worked solution

Step 1: Define a positive direction and find the momentum before and after

Take the ball’s original direction of travel (toward the wall) as positive.

Momentum before the bounce: $p_i = 0.20\times8.0 = 1.6\text{ kg m s}^{-1}$

Momentum after the bounce (the ball rebounds, so its velocity is now in the negative direction): $p_f = 0.20\times(-6.0) = -1.2\text{ kg m s}^{-1}$

Step 2: Find the change in momentum

$\Delta p = p_f - p_i = -1.2 - 1.6 = -2.8\text{ kg m s}^{-1}$

The magnitude of the change in momentum is $2.8\text{ kg m s}^{-1}$. Notice this is larger than either individual momentum value, because the rebound reverses the ball’s direction rather than simply slowing it down.

Step 3: Apply force as the rate of change of momentum

$F = \frac{|\Delta p|}{\Delta t} = \frac{2.8}{0.050} = 56\text{ N}$

Why the other options are wrong

• A ($8.0\text{ N}$): comes from using the difference in speeds ($8.0-6.0=2.0\text{ m s}^{-1}$) rather than treating the rebound as a reversal, giving $\Delta p = 0.20\times2.0=0.4\text{ kg m s}^{-1}$ and $F=0.4/0.050=8.0\text{ N}$, this wrongly treats the bounce as a simple slowing down.
• B ($24\text{ N}$): comes from using only the final momentum, $1.2/0.050=24\text{ N}$, ignoring the initial momentum entirely.
• C ($32\text{ N}$): comes from using only the initial momentum, $1.6/0.050=32\text{ N}$, ignoring the change caused by the rebound.

Final answer

• The average force exerted by the wall on the ball is $\boxed{56}\text{ N}$, option D.