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 o the floor are $I_1$ and $I_2$ respectively. Then

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

Answer 1

The correct option is A $5I_1=6I_2$

Velocity before strike is $u=\sqrt{29h}$

I

bounces to h/4

so velocity after bouncing was ins

given by $\frac{h}{4}=\frac{v^2}{29}$

$\Rightarrow V=\sqrt{\frac{9h}{2}}=\frac{u}{2}$

chance in velocity $=-\frac{u}{2}-u$

= final - initial.

final taken as -ve because

direction changed.

so chores in momentum $△P=M△V=-\frac{3mu}{2}$

impulse $=F△t=△P=\frac{-3mu}{2}=I_1$

II ball $u=\sqrt{29u}$

bonces to $h/16\Rightarrow \frac{h}{10}=\frac{v^2}{29}$

$\Rightarrow v=\frac{\sqrt{29h}}{4}=\frac{u}{4}$

$\Rightarrow △v=\frac{-u}{4}-u=\frac{-5u}{4}$

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

$\Rightarrow △p=\frac{-5mll}{4}=I_2$