Question

A ball after freely falling from a height of 4.9m strikes a horizontal plane. If the coefficient of restitution is 3/4, the ball will strike second time with the plane
 $\begin{align}
  & a)1/2s \\
 & b)1s \\
 & c)3/2s \\
 & d)3/4s \\
\end{align}$

Answer 1

Hint: The coefficient of restitution is defined as the ratio of the relative velocity of separation after collision to the velocity of approach before collision. It is given in the question that the ball falls from a height of 4.9m to the horizontal plane. Hence we will determine the velocity of the ball just before hitting the plane. The velocity of the plane is zero. Hence using the definition of coefficient of restitution we will determine the final velocity of the ball after collision and using a kinematic equation after how long the ball will strike second time with the plane.

Formula used:
Formula used:
$\eta =\dfrac{S-s}{U-u}$
$V={{V}_{\circ }}+at$
${{V}^{2}}-{{V}_{\circ }}^{2}=2aH$

Complete step by step answer:
Let us say the ball hits the horizontal plane with velocity U and moves away from the plane with velocity S. let the velocity of the plane before collision be u and after collision be s. Therefore the coefficient of restitution$(\eta )$ is given by,
$\eta =\dfrac{U-u}{S-s}$
Let us say the ball falls from a height H with velocity $({{V}_{\circ }})$ and acceleration a . Therefore its final velocity V is given by,
${{V}^{2}}-{{V}_{\circ }}^{2}=2aH$
The initial velocity ${{V}_{\circ }}$ of the ball is zero falling from a height of 4.9m and is accelerating under gravity (g), hence
$\begin{align}
  & {{V}^{2}}-{{V}_{\circ }}^{2}=2gH \\
 & {{V}^{2}}-(0)=2\times 10\times 4.9 \\
 & \Rightarrow V=\sqrt{2\times 10\times 4.9} \\
 & \Rightarrow V=7\sqrt{2}m{{s}^{-1}} \\
\end{align}$
The horizontal plane is at rest i.e. u=s=0. Therefore the final velocity (S) of the ball after collision from the expression of coefficient of restitution we get,
$\begin{align}
  & \eta =\dfrac{S-s}{U-u} \\
 & \eta =\dfrac{S}{U} \\
 & \Rightarrow \dfrac{3}{4}=\dfrac{S}{7\sqrt{2}} \\
 & \Rightarrow S=\dfrac{21\sqrt{2}}{4} \\
\end{align}$
As the ball after collision has some initial velocity(S,) after collision from the plane it will rise to a particular height. As the ball gets decelerated due to gravity its final velocity on its maximum height attained will be zero. Hence using Newton’s first kinematic equation i.e. $V={{V}_{\circ }}+at$ (where
${{V}_{\circ }}$ is the initial velocity of the body and V is the final velocity of the body under acceleration ‘a’ at time t) we get,
$\begin{align}
  & V={{V}_{\circ }}+gt \\
 & \Rightarrow 0=\dfrac{21\sqrt{2}}{4}-10\times t \\
 & \Rightarrow -\dfrac{21\sqrt{2}}{4}=-10\times t \\
 & \Rightarrow \dfrac{21\sqrt{2}}{40}=t \\
 & \Rightarrow t=0.742s \\
\end{align}$
Since the time taken by the ball to reach the maximum height is equal to the time taken to reach back to the plane. Hence the total time taken (T) for the ball to reach back to the horizontal plane after hitting for the first time is equal to,
$\begin{align}
  & T=2t \\
 & \Rightarrow T=2\times 0.742 \\
 & \Rightarrow T=1.486\approx 1.5=\dfrac{3}{2}s \\
\end{align}$

So, the correct answer is “Option C”.

Note:
These above equations that we have used are applicable when the above scenario is performed in vacuum. In reality, the air resistance comes into picture which will affect the time of flight of the ball after hitting the plane. It is unless the dimensions of the ball are very small, we can neglect the force due to air resistance.