Answer
$\Delta P=(-3\space Kg.m/s) \hat y$
Work Step by Step
We can find the required change in momentum as follows:
$\Delta P=m(v_f^{\rightarrow}-v_i^{\rightarrow})$
$\Delta P=mv(sin 60^{\circ} \hat x- cos 60^{\circ} \hat y)-(sin 60^{\circ} \hat x+ cos 60^{\circ} \hat y)$
$\Delta P=-2m\space v\space cos 60^{\circ} \hat y$
We plug in the known values to obtain:
$\Delta P=-2(0.2)(15)(0.5)\hat y$
$\Delta P=(-3\space Kg.m/s) \hat y$
