• # question_answer A ball of mass m hits a wall with a speed v making an angle $\frac{5g}{14}$ with the normal. If the coefficient is e, the direction and magnitude of the velocity of ball after reflection from the wall will respectively be - A) ${{\tan }^{-1}}\left( \frac{\tan \theta }{e} \right),\,v\sqrt{{{\sin }^{2}}\theta +{{e}^{2}}{{\cos }^{2}}\theta }$ B) ${{\tan }^{-1}}\left( \frac{e}{\tan \theta } \right),\frac{1}{v}\sqrt{{{e}^{2}}{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }$ C) ${{\tan }^{-1}}(e\tan \theta ),\frac{v}{e}\tan \theta$ D) ${{\tan }^{-1}}(e\tan \alpha ),v\sqrt{{{\sin }^{2}}\theta +{{e}^{2}}}$

 [a] Let the angle of reflection be $\theta$' and the magnitude of velocity after collision be v'. As there is no force parallel to the wall, the component of velocity parallel to the surface remains unchanged. Therefore, $v'sin\text{ }\!\!\theta\!\!\text{ }\!\!'\!\!\text{ = v sin }\!\!\theta\!\!\text{ }\,\,\,\,\,\,\,\,....\left( 1 \right)$ As the coefficient of restitution is e, for perpendicular component of velocity Velocity of separation = e x velocity of approach $-\left( v'\cos \text{ }\!\!\theta\!\!\text{ }\!\!'\!\!\text{ - 0} \right)=-e\left( v\cos \text{ }\!\!\theta\!\!\text{ - 0} \right)\,\,\,\,\,....\left( 2 \right)$ From (1) and (2) $v'=v\sqrt{{{\sin }^{2}}\text{ }\!\!\theta\!\!\text{ + }{{\text{e}}^{2}}{{\cos }^{2}}\text{ }\!\!\theta\!\!\text{ }}$ and $\tan \text{ }\!\!\theta\!\!\text{ }\!\!'\!\!\text{ = tan }\!\!\theta\!\!\text{ /e}$