Question

A ball of mass m hits the floor with a speed v making an angle of incidence $\theta$ with the normal. The coefficient of restitution is e. The speed of reflected ball and the angle of reflection of the ball will be:

• A. $v^{\prime }=v,\theta ={\theta }^{\prime }$
• B. $v^{\prime }=\frac{v}{2},\theta =2{\theta }^{\prime }$
• C. $v^{\prime }=2v,\theta =2\theta$
• D. $v^{\prime }=\frac{3v}{2},\theta =\frac{2{\theta }^{\prime }}{3}$

Answer 1

The correct option is A $v^{\prime }=v,\theta ={\theta }^{\prime }$

Velocity of approach in the given case is the normal component of velocity.

Hence, $v_n=v\cos \theta$

By definition of coefficient of restitution, velocity of separation will be,

$v_n^{\prime }=e{v}_n=ev\cos \theta$

Tangential component of velocity will not change.

$v_t^{\prime }=v_t=v\sin \theta$

Speed of reflected ball is:

$v^{\prime }=\sqrt{v_n^2+v_t^2}$

$=\sqrt{(ev\cos \theta {)}^2+(v\sin \theta {)}^2}$

$=v\sqrt{e^2{\cos }^2\theta +{\sin }^2\theta }$

Angle with normal is given by:

${\theta }^{\prime }={\tan }^{-1}\frac{v_t^{\prime }}{v_n^{\prime }}$

${\theta }^{\prime }={\tan }^{-1}\left(\frac{v\sin \theta }{ev\cos \theta }\right)$

${\theta }^{\prime }={\tan }^{-1}\left(\frac{\tan \theta }{e}\right)$