Question

Two balls of same mass are dropped from the same height $h$ on to the floor. The first ball bounces to height $h/4$ after the collision and the second ball to height $h/16$. The impulse applied by the first & second ball on the floor is $I_1$ and $I_2$ respectively. Then:

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

Answer 1

Answer: (D)

$3I_1=2I_2$

We know $v=\sqrt{2gh}$

when $h=\frac{h}{4}{v}_1^{{}^{\prime }}=\frac{1}{2}\sqrt{2gh}=\frac{1}{2}{v}_1\text{where}{v}_1=\sqrt{2gh}$

When $h=\frac{h}{16}{v}_2=\sqrt{2g\frac{h}{16}}=\frac{1}{4}{v}_2;where{v}_2=\sqrt{2gh}$

Now impulse $I_1=m(v_1-v_1^{{}^{\prime }})=\frac{1}{2}m{v}_{1}and{I}_2=m(v_2-v_2^{{}^{\prime }})=\frac{3}{4}m{v}_2\frac{I_1}{I_2}=\frac{2}{3}=>3I_1=2I_2$

(Considering velocity to be same for both the balls)