Question

A rubber ball is dropped from a height of 5m on a plane. On bouncing it rises to 1.8m. The ball loses its velocity on bouncing by a factor of –
$$\begin{array}{l}A){\displaystyle \frac{16}{25}}\\ B){\displaystyle \frac{2}{5}}\\ C){\displaystyle \frac{3}{5}}\\ D){\displaystyle \frac{9}{25}}\end{array}$$

Answer 1

Hint: Find the velocity before hitting the ground and after hitting the ground separately. Gravity acts as an accelerant when the ball is falling, and it is the reason of deceleration when the ball is going up.

Formula used:
We are going to use the simple kinematics formula,
$$s=ut+{\displaystyle \frac{1}{2}}a{t}^2$$ ………………… (1)
The other formula that we have used is,
$$v=u+at$$ …………………… (2)
We can also use the formula,
$$v^2=u^2+2as$$ …………………… (3)
Here, $$s$$ is the distance travelled by the ball,
           $$u$$ is the initial velocity,
           $$a$$ is the acceleration,
   and, $$t$$ is the time of the motion.

Complete answer:
First, let us calculate the velocity of the ball just before it hits the surface.
It is given that, the distance travelled, $$s=5m$$,
Initial velocity, $$u=0m/s$$ ,
Acceleration is $$a$$.
Putting these values in equation (1) we get,

$$\begin{array}{rl}& 5=(0)t+{\displaystyle \frac{1}{2}}a{t}^2\\ & \Rightarrow a{t}^2=10\\ & \Rightarrow t^2={\displaystyle \frac{10}{a}}\\ & \Rightarrow t=\sqrt{{\displaystyle \frac{10}{a}}}\end{array}$$
Putting the value of $$t$$ in equation (2), we will get,
$$\begin{array}{rl}& v_1=(0)+a\sqrt{{\displaystyle \frac{10}{a}}}\\ & \Rightarrow v_1=\sqrt{10a}\end{array}$$
So, the ball reached the surface at a velocity of $$v_1=\sqrt{10a}$$.
Now, we can proceed to the next part.
As the ball reaches a maximum distance of 1.8m, it attains zero velocity when it is at the top most point.
Here a is negative as it is a deceleration
Let's assume that the initial velocity while going up was,
Let $$v_2$$ be the final velocity is zero at the top most point.
So, we can write the equation (3) as,
$$\begin{array}{rl}& 0={v_2}^2-2a{s}_2\\ & \Rightarrow v_2=\sqrt{2a{s}_2}\end{array}$$
But here, $$s_2=1.8$$ .
So, the rebound velocity is given by,
$$\begin{array}{rl}& v_2=\sqrt{2a\times 1.8}\\ & \Rightarrow v_2=\sqrt{3.6a}\end{array}$$
Hence the rebound velocity is $$v_2=\sqrt{3.6a}$$.
Now, the ball loses its velocity by a factor of ,
$\text{loss = }{\displaystyle \frac{v_{1-}{v}_2}{v_1}}$
       $\begin{array}{rl}& ={\displaystyle \frac{\sqrt{10a}-\sqrt{3.6a}}{\sqrt{10a}}}\\ & =1-{\displaystyle \frac{3}{5}}={\displaystyle \frac{2}{5}}\end{array}$

So, the correct answer is “Option B”.

Note:
You should approach the problem in a step by step manner as shown in the solution. You should not go for the energy conservation laws due to the nature of collision. You can avoid extra calculation by keeping the values in abstract form like it has been shown in the solution. This factor is called the factor of restitution. When the moving object some of its kinetic energy in the collision, this is called inelastic collision. The energy conservation law does not hold in this case.