Question

Two balls of same mass are dropped from the same height h. on to the floor. The first ball bounces to a height $h/4$ after the collision $\&$ the second ball to a height $h/16.$ The impulse applied by the first $\&$ second ball on the floor are $l_1$ and $l_2$ respectively. Then

• A. $5l_1=6l_2$
• B. $6l_1={51}_2$
• C. $3l_1=2l_2$
• D. $2l_1=l_2$

Answer 1

The correct option is A $5l_1=6l_2$

Given,

First ball height is $h/4$

Second ball height is $h/16$

We know that

$v_0=\sqrt{2gh}$

Here,

$v_0$ is velocity

$g$ is acceleration due to gravity

$h$ is height of particle at velocity $v_0$

Velocity of particle first

$v_1=\sqrt{2g\times \frac{h}{4}}{v}_1=\frac{1}{2}{v}_0$

Velocity of particle second

$v_2=\sqrt{2g\times \frac{h}{16}}{v}_1=\frac{1}{4}{v}_0$

Now impulse is change in momentum therefore,

Impulse for first particle is

$I_1=m{v}_1-(-m{v}_0)=m(\frac{v_0}{2}+v_0)=\frac{3}{2}m{v}_0$

Impulse for secound particle is

$I_2=m{v}_2-(-m{v}_0)=m(\frac{v_0}{4}+v_0)=\frac{5}{4}m{v}_0$

Ratio of particle first and second is,

$\frac{I_1}{I_2}=\frac{3/2}{5/4}\frac{I_1}{I_2}=\frac{3\times 4}{2\times 5}=\frac{6}{5}5I_1=6I_2$

Hence, correct option is $\left(A\right)$