A particle of mass m strikes a wall at an a wall at angle of incidence $60^o$ with velocity v elastically. The change in momentum is:

Magnitude of the velocity at the end of collision does not changer.

Now, change in momentum

$\begin{aligned} & \overrightarrow{\Delta P}=\overrightarrow{P_{a}}-\overrightarrow{P_{b}} \\ & \vec{P}_{b}=m \overrightarrow{m v} \& \overrightarrow{P_{a}}=\overrightarrow{m v} \\ \Delta P= & \sqrt{P_{b}^{2}+P_{a}^{2}-2 P_{b} P_{a} \cos \theta} \\ \Delta P= & \sqrt{(m v)^{2}+(m v)^{2}-2(m v) \cdot(m v) \cdot\left(\cos 60^{\circ}\right)} \\ \Delta P= & \sqrt{(m v)^{2}+(m v)^{2}-2 m v \cdot m v \cdot \frac{1}{x}}\left(\because \cos 60^{\circ}=\frac{1}{2}\right. \\ \Delta P= & \sqrt{(m v)^{2}+(m v)^{2}-(m v)^{2}} \\ \Delta P= & \sqrt{(m v)^{2}} \\ \Delta P= & m v \end{aligned}$

$\therefore$ Option (A) is correct