Question

A rubber ball is dropped from a height of 5 m on a plane. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of
(A) 3/5
(B) 2/5
(C) 16/5
(D) 9/25

Answer 1

If a ball is dropped from height h, the velocity with which it will strike the surface is given by $v = \sqrt {2gh}$. And if a ball is thrown up with speed u, the height it reaches is given by $H = \dfrac{{{u^2}}}{{2g}}$.

Complete step by step solution:
We know that when a ball is dropped from height h, it strikes the surface with speed
$v = \sqrt {2gh}$,
It is given that the ball is dropped from height 5 m. Substituting this value and value of acceleration due to gravity in the formula we get
$v = \sqrt {2 \times 10 \times 5}$,
$v = \sqrt {100}$,
$v = 10m/s$.
We also know that when a ball rebounds with speed u, the height it reaches is given by
$H = \dfrac{{{u^2}}}{{2g}}$,
It is given that the ball reaches height 1.8 m on rebound. Substituting this value and value of acceleration due to gravity in the formula, we get
$1.8 = \dfrac{{{u^2}}}{{2 \times 10}}$,
${u^2} = 36$,
$u = 6m/s$.
Hence the ball loses its velocity by 4m/s on rebounding.
The fractional loss can be calculated by dividing the loss in velocity with initial velocity, i.e. velocity with which it strikes the floor i.e 10m/s
Hence, fractional loss in velocity = $\dfrac{4}{{10}}$,
fractional loss in velocity $= \dfrac{2}{5}$.

Therefore, the correct answer to the question is option : B

Note: Before attempting to solve the problem, the student needs to be able to solve basic problems of motion in 1 dimension. The formulas used in this question can easily be calculated by the use of three equations of motion, one should just know that when a ball is released from some height its initial velocity can be taken as zero and when the ball reaches a maximum height of its journey, the velocity at the highest point can be taken as zero. Taking initial velocity zero and applying second equation of motion given the result, $v = \sqrt {2gh}$