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
(A) $\dfrac{{16}}{{25}}$
(B) $\dfrac{2}{5}$
(C) $\dfrac{3}{5}$
(D) $\dfrac{9}{{25}}$

Answer 1

Hint : We can solve this question by using the equation of motion. Here our problem can be divided into two cases. By using the values given in the problem in the equation of motion and finding the velocity of each case separately and using the values of velocity we will find the loss in velocity.
Formula used:
$\Rightarrow 2ad = {v^2} - {u^2}$ ,
where,$u$ = initial velocity $5m$, $v$ = final velocity, $a$ = acceleration, $d$ = distance.

Complete step by step answer
Our problem is divided into two cases. The first case is in which the ball is dropped and the second is when it bounces back.
Starting with our first case in which the ball is dropped from a height of. Now the equation of motion can be given as
$\Rightarrow 2a{d_1} = {v_1}^2 - {u_1}^2$
$\Rightarrow {v_1}^2 = {u_1}^2 + 2a{d_1}$
Here the initial velocity $u = 0$ while the final velocity ${v_1}$ , as the ball is falling downwards then acceleration $a = g$ which acceleration due to gravity. Hence
$\Rightarrow {v_1}^2 = 0 + 2 \times 5 \times g = 10g$
$\therefore {v_1} = \sqrt {10g}$ --------------- Equation $(1)$
Similarly, for the second case, the ball bounced back at a height of $1.8m$ . Now the equation of motion can be given as
$\Rightarrow 2a{d_2} = {v_2}^2 - {u_2}^2$
$\Rightarrow {v_2}^2 = {u_2}^2 + 2a{d_2}$
Here the initial velocity $u = {v_2}$ while the final velocity ${v_2} = 0$ , as the ball is going upwards then acceleration $a = - g$ which acceleration due to gravity. Hence
$\Rightarrow 0 = v_2^2 + 2 \times 1.8 \times - g$
$\Rightarrow v_2^2 = 2 \times 1.8 \times g = 3.6g$
Therefore we get the value of ${v_2}$ as
$\Rightarrow {v_2} = \sqrt {3.6g}$ --------------- Equation $(2)$
Now by using Equation $(1)$ and Equation $(2)$ we can find the loss in velocity which can be given as
$\Rightarrow \dfrac{{{v_1} - {v_2}}}{{{v_1}}} = \dfrac{{\sqrt {10g} - \sqrt {3.8g} }}{{\sqrt {10g} }}$
On further solving the equation we get,
$\Rightarrow \dfrac{{{v_1} - {v_2}}}{{{v_1}}} = 1 - \sqrt {\dfrac{{36g}}{{100g}}} = 1 - \dfrac{3}{5}$
Which results in
$\Rightarrow \dfrac{{{v_1} - {v_2}}}{{{v_1}}} = \dfrac{2}{5}$
Hence option (B) is the correct answer.

Note
Here one has to note that the sign of acceleration due to gravity when the ball falls will be positive while when it goes upwards after bouncing then the value of acceleration due to gravity will become negative as it is going upwards against the gravity.