A ball is projected from a point in a horizontal plane so as to strike a vertical wall at right angle and after rebounding from the wall and once from the horizontal plane it returns to the point of projection. Find coefficient of restitution (e) for the two collisions, assuming it to be same for both collision. Find the value of 1/e.

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Solution

$S i n c e t h e f i r s t c o l l i s i o n i s w i t h t h e v e r t i c a l w a l l, t h e t i m e o f f l i g h t i s n o t g o i n t o b e c h a n g e d, m e a n i n g t h e t i m e t a k e n t i l l t h e s e c o n d c o l l i s i o n i s \frac{2 u s i n θ}{g} . A f t e r t h e f i r s t c o l l i s i o n v_{x} = e u c o s θ A f t e r s e c o n d c o l l i s i o n v_{y} = e u s i n θ, s o t i m e o f f l i g h t (f r o m s e c o n d c o l l i s i o n t o t h e i n i t i a l p o i n t) i s \frac{2 e u s i n θ}{g}, d i s t a n c e t r a v e l l e d i s \frac{R}{2} - \frac{u s i n θ}{g} e u c o s θ . T h u s, \frac{2 u s i n θ \times u c o s θ}{2 g} - \frac{u s i n θ}{g} e u c o s θ = \frac{2 e u s i n θ}{g} e u c o s θ \Rightarrow 1 - e = 2 e^{2} \Rightarrow 2 e^{2} + e - 1 = 0 \Rightarrow e = 0.5 \Rightarrow \frac{1}{e} = 2$

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A ball is projected with a velocity $u$ at an elevation from a point distance d from a smooth vertical wall in a plane perpendicular to it. After rebounding from the wall, it returns to the point of projection, find the maximum distance $d$ for which the ball can return to the point of projection.