Answer
The magnitude of the impulse on the wall from the ball is $~~1.3~kg~m/s$
Work Step by Step
We can let the initial velocity be $v_i = 5.2~m/s$
In part (a), we found that the speed after rebounding is $3.7~m/s$
We can let the velocity after rebounding be $v_f = -3.7~m/s$
We can find the change in momentum of the ball:
$\Delta p = m~v_f-m~v_i$
$\Delta p = m~(v_f-v_i)$
$\Delta p = (0.150~kg)~(-3.7~m/s-5.2~m/s)$
$\Delta p = -1.3~kg~m/s$
The impulse on the ball from the wall is equal to the ball's change in momentum.
The magnitude of the impulse on the ball from the wall is $~~1.3~kg~m/s$
By Newton's Third Law, the impulse on the wall from the ball is equal in magnitude to the impulse on the ball from the wall.
Therefore, the magnitude of the impulse on the wall from the ball is $~~1.3~kg~m/s$
