Work Energy And Power Question 320
Question: 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 -
Options:
A) $ {{\tan }^{-1}}( \frac{\tan \theta }{e} ),v\sqrt{{{\sin }^{2}}\theta +e^{2}{{\cos }^{2}}\theta } $
B) $ {{\tan }^{-1}}( \frac{e}{\tan \theta } ),\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}} $
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Answer:
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\theta\text{ }’\text{ = v sin }\theta….( 1 ) $
As the coefficient of restitution is e, for perpendicular component of velocity
Velocity of separation = e x velocity of approach
$ -( v’\cos \theta\text{ }’\text{ - 0} )=-e( v\cos \theta\text{ - 0} )….( 2 ) $
From (1) and (2)
$ v’=v\sqrt{{{\sin }^{2}}\theta\text{ + }{e^{2}}{{\cos }^{2}}\theta} $
and $ \tan \theta\text{ }’\text{ = tan }\theta\text{ /e} $
