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 collection & the second ball to a height h/16. the impulse applied by the first & second ball on the floor are $I_1$ and $I_2$ respectively . then

• A. $5I_1=6I_2$
• B. $6I_1=5I_2$
• C. $3I_1=2I_2$
• D. $2I_1=I_2$

Answer 1

The correct option is C $3I_1=2I_2$

When the ball is dropped from height h its velocity $v_1=\sqrt{2gh}$

Now velocity of the ball after the bounce, where height $h=\frac{h}{4}$,velocity$v_{1a}=\sqrt{2g\frac{h}{4}}$

Now impulse $I_1=F_{1}t=mat=m(v_1-v_{1a})=m(\sqrt{2gh}-\sqrt{2g\frac{h}{4}})=m(v_1-\frac{1}{2}{v}_1)=\frac{1}{2}m{v}_1$

Similarly for the ball when reaches a height of $\frac{h}{4}$ after bounce,velocity$v_{2a}=\sqrt{2g\frac{h}{16}}$ and velocity of the ball before

bounce$v_2=\sqrt{2gh}$

Similarly the impulse of the second ball$I_2=\frac{3}{4}{v}_{2}m$

Given mass of both the ball are same.

We can compare both the impulses as

$\frac{4I_2}{3v_2}=\frac{2I_2}{v_1}$

Now as both $v_2=v_1=\sqrt{2gh}$

$3I_1=2I_2$