Question

A ball strikes a wall with a velocity $\overrightarrow{u}$ at an angle $\theta$ with the normal to the wall surface and rebounds from it at an angle $\beta$ with the surface. Then :

• A. $(\theta +\beta )<{90}^o,$ if the wall is smooth
• B. if the wall is rough, coefficient of restitution $=\frac{tan\beta }{cos\theta }$
• C. if the wall is rough, coefficient of restitution $<\frac{tan\beta }{cos\theta }$
• D. none of the above

Answer 1

Answer: (A), (C)

$(\theta +\beta )<{90}^o,$ if the wall is smooth
if the wall is rough, coefficient of restitution $<\frac{tan\beta }{cos\theta }$

coefficient of restitution : $e=\frac{\text{velocity of seperation}}{\text{velocity of approch}}$

here velocity of approch $v_a=$ normal component of velocity=$u\cos \theta$

let velocity of seperation be $v_s$

$e=\frac{v_s}{v_a}$

$e=\frac{v_s}{u\cos \theta }$

here velocity of seperation $v_s=$ $eu\cos \theta$

let velocity after rebounding be $v$

$v_s$normal component of velocity$=v\cos \alpha$

equating both values of $v_s=v\cos \alpha =eu\cos \theta$

$v=\frac{eu\cos \theta }{\cos \alpha }$

equating horizontal component of velocity

$usin\theta =v\sin \alpha$

putting $v=\frac{eu\cos \theta }{\cos \alpha }$

$usin\theta =\sin \alpha$ $\times \frac{eu\cos \theta }{\cos \alpha }$

$\tan \theta =e\tan \alpha$

$\alpha +\beta ={90}^o$

$e=\frac{\tan \theta }{\tan \alpha }=\frac{\cot \alpha }{\tan \theta }=\frac{tan\beta }{\tan \theta }$

if collision is not perfectly elastic

$e<1$

$\tan \theta <\tan \alpha$

$\theta <\alpha$

adding $\beta$ both sides

$\theta +\beta <\alpha +\beta$ and we know $\alpha +\beta ={90}^o$

$\theta +\beta <{90}^o$

if ball is rough then

$e<(\frac{\tan \theta }{\tan \alpha }=\frac{\cot \alpha }{\tan \theta }=\frac{tan\beta }{\tan \theta })$

$e<\frac{tan\beta }{\tan \theta }$