Answer
The magnitude of the impulse on the wall from the ball is $~~2.18~kg~m/s$
Work Step by Step
We can let the initial velocity be $~~v_i = 7.8~m/s$
Then the velocity after rebounding is $~~v_f = -7.8~m/s$
We can find the change in ball's momentum:
$\Delta p = m \Delta v$
$\Delta p = m~(v_f-v_i)$
$\Delta p = (0.140~kg)~(-7.8~m/s-7.8~m/s)$
$\Delta p = -2.18~kg~m/s$
The impulse on the ball from the wall is equal to the ball's change in momentum. Therefore, the impulse on the ball from the wall is $-2.18~kg~m/s$
By Newton's Third Law, the impulse on the wall from the ball is equal in magnitude, but opposite in direction, to the impulse on the ball from the wall.
Therefore, the magnitude of the impulse on the wall from the ball is $~~2.18~kg~m/s$.
