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A rubber ball is dropped from a height of 5 meters on a plane, where the acceleration due to gravity is not shown. On bouncing, it rises to 1.8 meters. The ball loses its velocity on bouncing by a factor ofA. $\dfrac{{16}}{{25}}$B. $\dfrac{2}{5}$C. $\dfrac{3}{5}$D. $\dfrac{9}{{25}}$

Last updated date: 20th Jun 2024
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Hint: In this question, we need to determine the factor by which there is a loss in the velocity of the ball when it bounces back. For this, we will use the relation between the velocity of the ball, the height of the ball, and the acceleration.

Complete step by step answer:Let the velocity with which the ball hits the ground be ${v_1}$ and the velocity of the ball just after hitting the ground be ${v_2}$.
Following Newton’s law, which states that the velocity of the ball with which it hits the ground is the square root of the 2 times the product of the acceleration and the height. Mathematically, $v = \sqrt {2gh}$.
Here, the ball was released from a height of 50 meters, so, substitute ${h_1} = 50$ in the equation ${v_1} = \sqrt {2g{h_1}}$ to determine the expression for the velocity of the ball just before hitting the ground in terms of acceleration due to gravity as:
${v_1} = \sqrt {2g{h_1}} \\ = \sqrt {2g\left( 5 \right)} \\ = \sqrt {10g} - - - - (i) \\$
Also, the ball bounces back by 1.8 meters, so, substitute ${h_2} = 1.8$ in the equation ${v_2} = \sqrt {2g{h_2}}$ to determine the expression for the velocity of the ball just after hitting the ground in terms of acceleration due to gravity as:
${v_2} = \sqrt {2g{h_2}} \\ = \sqrt {2g\left( {1.8} \right)} \\ = \sqrt {3.6g} - - - - (ii) \\$
Now, divide the equations (i) and (ii) as:
$\dfrac{{{v_2}}}{{{v_1}}} = \dfrac{{\sqrt {3.6g} }}{{\sqrt {10g} }} \\ = \sqrt {\dfrac{{1.8}}{5}} \\ = \sqrt {0.36} \\ = 0.6 \\$
Subtracting from 1 both sides of the above equation, we get
$1 - \dfrac{{{v_2}}}{{{v_1}}} = 1 - 0.6 \\ \dfrac{{{v_1} - {v_2}}}{{{v_1}}} = 0.4 \\ \dfrac{{\partial v}}{{{v_1}}} = \dfrac{2}{5} \\ Loss{\text{ }}in{\text{ }}velocity = \dfrac{2}{5} \\$
Hence, the loss in the velocity of the ball will be $\dfrac{2}{5}$.
Option B is correct.

Note:It is interesting to note here that the ball will always lose a factor of velocity wherever it hits a surface as its potential has been converted into the kinetic energy. Moreover, the candidates should also note here that the height of the ball will go on reducing due to loss in potential energy.