Question

A particle of mass $m$ moving with $a$ speed $u$ strikes a smooth horizontal surface at an angle 45 the particle rebounds it an angle $\varphi$ with speed $v$, if coefficient of restitution is $\frac{1}{\sqrt{3}}$ , then angle of $\varphi$ is:-

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${37}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

griven particle of mass $m$, moving with speed 'u netounds wrh a bpeed

Since, the force applied at the peint of contact is in the derection of $N$, so only vertical component of velocity wull change

$\mathrm{ucos}45=\mathrm{vsin}\varphi -\left(1\right)$

$\begin{array}{c}e=\frac{V_{\text{rel aften collision}}}{V_{\text{rel befre collision}}}=\frac{v\cos \varphi }{u\sin u{5}^{\circ }}\\ \frac{1}{\sqrt{3}}=\frac{\sqrt{2}v}{u}\cdot \cos \varphi =\left(2\right)\end{array}$

$\begin{array}{cc}\text{using equation}1\text{a}2:& \frac{1}{\sqrt{6}}=\frac{1}{\sqrt{2}}\text{cot}\varphi \\ \varphi & ={60}^{\circ }\end{array}$