Answer
$J = (-0.45~kg~m/s)~\hat{i}-(0.45~kg~m/s)~\hat{j}-(1.08~kg~m/s)~\hat{k}$
Work Step by Step
In part (a), we found that the change in the ball's momentum is:
$\Delta p = (-0.45~kg~m/s)~\hat{i}-(0.45~kg~m/s)~\hat{j}-(1.08~kg~m/s)~\hat{k}$
The impulse on the ball from the wall is equal to the change in the ball's momentum.
Therefore, the impulse on the ball is:
$J = (-0.45~kg~m/s)~\hat{i}-(0.45~kg~m/s)~\hat{j}-(1.08~kg~m/s)~\hat{k}$
