A mass of $100\text{g}$ strikes the wall with a speed of $5\text{m/s}$ at an angle as shown in the figure. It rebounds with the same speed. If the contact time is $2\times {10}^{-3}\text{s}$, then what is the force applied on the mass by the wall?

The correct option is C $250\sqrt{3}\text{N}$ to left
The components of initial and final momentum are shown in the figure.


Momentum of ball is only changing in the direction perpendicular to the wall.
Force exerted by the wall,
$F=\frac{\Delta P}{\Delta t}=\frac{-mv\sin {60}^{\circ }-(+mv\sin {60}^{\circ })}{2\times {10}^{-3}}$
$[$taking rightward as $+]$
$\Rightarrow F=-\frac{\sqrt{3}mv}{2\times {10}^{-3}}=-\frac{\sqrt{3}\times 100\times {10}^{-3}\times 5}{2\times {10}^{-3}}$
$\Rightarrow F=-250\sqrt{3}\text{N}$
Force is exerted on the ball towards left direction.

$\begin{array}{c}\text{Why this question?}\\ \text{Caution: To avoid any mistakes, first resolve the}\\ \text{component of initial and final momentum.}\\ \text{Tips: Momentum of body changes in the}\\ \text{direction of force applied.}\end{array}$