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Question Number 23049 by Tinkutara last updated on 25/Oct/17

$$\mathrm{A}\mathrm{particle}\mathrm{of}\mathrm{mass}m\mathrm{strikes}\mathrm{on}\mathrm{ground}$$$$\mathrm{with}\mathrm{angle}\mathrm{of}\mathrm{incidence}45°.\mathrm{If}\mathrm{coefficient}$$$$\mathrm{of}\mathrm{restitution},e=\frac{1}{\sqrt{2}},\mathrm{find}\mathrm{the}\mathrm{velocity}$$$$\mathrm{after}\mathrm{impact}\mathrm{and}\mathrm{angle}\mathrm{of}\mathrm{reflection}.$$

Answered by ajfour last updated on 25/Oct/17

$$paralleltoground(iffrictionbe$$$$zerothen)weconservelinear$$$$momentum.$$$$letinitialspeedbe{\mathit{v}}_0,andfinal$$$$speedbe\mathit{v}.$$$$mv\sin \theta ={mv}_0\sin 45°$$$$\Rightarrow v\sin \theta =\frac{v_0}{\sqrt{2}}....\left(i\right)$$$$v_{separation}=e\left(v_{approach}\right)$$$$\Rightarrow v\cos \theta =\frac{1}{\sqrt{2}}\times v_0\cos 45°$$$$\Rightarrow v\cos \theta =\frac{v_0}{2}...\left(ii\right)$$$$\left(i\right)\mathbf{÷}\left(ii\right)gives$$$$\tan \theta =\sqrt{2}\Rightarrow \mathit{\theta }={\tan }^{-1}\sqrt{2}$$$$v^2\sin {}^2\theta +v^2\cos {}^2\theta =\frac{v_0^2}{2}+\frac{v_0^2}{4}$$$$\Rightarrow \mathit{v}=\frac{{\mathit{v}}_0\sqrt{3}}{2}.$$$$$$

Commented by Tinkutara last updated on 25/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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