Question

A body dropped freely from a height 'h' on to a horizontal plane, bounces up and down and finally comes to rest. The coefficient of restitution is 'e'. The ratio of velocities at the beginning and after two rebounds is?

Answer 1

coefficient of restitution for an elastic/inelastic collision = e =
               Relative velocity after collision / relative velocity before collision

Let us say the body has a velocity of V just before collision with the plane.

velocity of the body after rebounding is  V1 =  e * V  directed upwards.

Velocity of the body just before rebounding a second time = V1 = eV directed downwards.

Velocity of the body just after rebounding second time = e V1 = e² V

Ratio = V : 2² V  = 1 : e²

Answer 2

Answer:

$\dfrac{v''}{v}= \dfrac{e^2}{1}$

Explanation:

The ball is dropped from a height 'h'. Its initial velocity must be zero and gravity will act on it in downward direction.

According to 3rd Equation of Motion:-

$v^{2}-u^{2}=2as \\ \\v^{2} - (0)^{2}=2\times g\times h \\ \\v = \sqrt{2gh}$

After the body hits the ground, its velocity decreases to:-

$v'=e\times v$

The body will have a velocity of 'ev' after the first collision.

After rebouding for the second time, its velocity becomes:-

$v''=e\times (ev) = e^2v$

Ratio of the velocity in beginning to velocity after two rebounds would be:-

$\dfrac{v''}{v}=\dfrac{e^2\sqrt{2gh}}{\sqrt{2gh}} \\ \\ \\\dfrac{v''}{v}= \dfrac{e^2}{1}$