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) \[{{\tan }^{-1}}(e\tan \alpha ),v\sqrt{{{\sin }^{2}}\theta +{{e}^{2}}}\]
Correct Answer: A
Solution :
| [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 \[\tan \text{ }\!\!\theta\!\!\text{ }\!\!'\!\!\text{ = tan }\!\!\theta\!\!\text{ /e}\] |
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