Question

A ball of mass m hits a floor with a speed v
making an angle a with the normal N. The
coefficient of restitution is e. The angle made by
reflected ball with floor is​

Answer 1

Answer:

The angle made by the reflected ball with the horizontal floor is

$\boxed{\tan^{-1}(e\cot\alpha)}$

Explanation:

The ball makes an angle α with the normal, its speed is v

Therefore,

The speed of the ball in vertical direction = vcosα

Speed of the ball in horizontal direction = vsinα

Since the coefficient of restitution is e

Therefore after hitting the floor the vertical component of the velocity will be evcosα

The horizantal component of the velocity will remain the same i.e. vsinα

Now if the ball after hitting the floor makes an angle θ with the floor (horizontal) thenits velocity will also be in that direction

Hence

$\tan\theta=\frac{\text{vertical component}}{\text{horizontal component}}$

$\implies \tan\theta=\frac{ev\cos\alpha}{v\sin\alpha}$

$\implies \tan\theta=e\cot\alpha$

$\implies\theta=\tan^{-1}(e\cot\alpha)$

Answer 2

Given :

Mass of ball=m kg

Speed =V

Angle of incidence=α

Coefficient of resolution= e

Speed of ball in Vertical direction= v sinα

So after hitting the wall, the vertical component of speed = evcosα

The component of velocity v  along common tangent direction vsinα will remain unchanged.

Thus after collision ,

component of velocity v1 and vsinα and ev cosα

V1= √(vsinα)² + (evcosα )²

Tan β= vsinα/evcosα= tanα/e

∴Tanβ=tanα/e