A 1 kg object strikes a wall with velocity 1 ms^-1 at an angle of 60° with the wall and reflects at the same angle. If it remains in contact with wall for 0.1 s, then the force exerted on the wall is :-

If we resolve the momentum vector in two components along the walls  $mv \cos {60}^{°}$, remains same before and after collision, but the component perpendicular to the wall reverses its direction.

Hence, change in momentum

$-mv\sin \theta -mv\sin \theta =-2mv\sin \theta$

force acting   $=\frac{2mv\sin \theta }{t}$

Here,   $\mathrm{m}=1\mathrm{kg}$

$\begin{array}{rl} & v=1\mathrm{m}/\mathrm{s}\\ & \theta ={60}^{\circ }\\ & t=0.1\mathrm{s}\\ & \text{So, }F=\frac{2\times 1\times 1\sqrt{3}}{2\times 0.1}=10\sqrt{3}\mathrm{N}\end{array}$