Answer
$\Delta p = (-0.572~kg~m/s)~\hat{j}$
Work Step by Step
We can express the ball's initial velocity in unit-vector notation:
$v_i = [(2.00 ~sin~30.0^{\circ})~\hat{i}+(2.00 ~cos~30.0^{\circ})~\hat{j}]~m/s$
$v_i = [(1.00)~\hat{i}+(1.73)~\hat{j}]~m/s$
We can express the ball's initial momentum in unit-vector notation:
$p_i = m~v_i$
$p_i = (0.165~kg)~[(1.00)~\hat{i}+(1.73)~\hat{j}]~m/s$
$p_i = [(0.165)~\hat{i}+(0.286)~\hat{j}]~kg~m/s$
We can express the ball's final momentum in unit-vector notation:
$p_f = [(0.165)~\hat{i}-(0.286)~\hat{j}]~kg~m/s$
We can express the change in the ball's momentum in unit-vector notation:
$\Delta p = p_f-p_i$
$\Delta p = (-0.572~kg~m/s)~\hat{j}$
