A ball of 0.5 kg collided with wall at 30^ and bounced back...

Given, $\mathrm{m}=0.5 \mathrm{~kg}, \mathrm{v}=12 \mathrm{~m} / \mathrm{s}, \Delta \mathrm{t}=1 \mathrm{~s}, \theta=30^{\circ}$ Force applied by wall on ball, $\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$ or $\quad \mathrm{F}=\frac{\left(\mathrm{p}_{\mathrm{f}}\right)_{\mathrm{H}}-\left(\mathrm{p}_{\mathrm{i}}\right)_{\mathrm{H}}}{\Delta \mathrm{t}}$ In this elastic collision, final and initial velocity will be same but direction will changed. $\therefore$ Horizontal component, \left(\mathrm{p}_{\mathrm{f}}\right)_{\mathrm{H}}=\mathrm{mv} \cos \theta$and \left(\mathrm{p}_{\mathrm{i}}\right)_{\mathrm{H}}=-\mathrm{mv} \cos \theta$$\begin{aligned} & \therefore \mathrm{F}=\frac{\mathrm{mv} \cos \theta-(-\mathrm{mV} \cos \theta)}{\Delta \mathrm{t}}=\frac{2 \mathrm{mV} \cos \theta}{\Delta \mathrm{t}} \\ & =\frac{2 \times 0.5 \times 12 \times \cos 30^{\circ}}{1}=6 \sqrt{3} \mathrm{~N} \\ & \end{aligned}$$$