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 and the second ball into a height $h/16$. The impulse applied by the first and second ball on 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 $5I_1=6I_2$

For first ball;

$V_i=\sqrt{2gh}$

$V_f=\sqrt{\frac{2gh}{4}}=\frac{V_i}{4}\times 2$

$P_i=m_i{v}_i$

$P_f=\frac{m_i{v}_i}{4}=-\frac{P_i}{4}\times 2$

$\frac{P_f-P_i}{\Delta t}\times \Delta t=Impulse$

$I_1=\frac{3}{2}{m}_i{v}_i$

For second ball;

$P_i=m_i{v}_i$

$P_f=-\frac{m_i{v}_i}{4}$

$P_f-P_i=-\frac{5}{4}{m}_i{v}_i=I_2$

So, $\frac{I_1}{I_2}=\frac{\frac{3}{2}}{\frac{5}{4}}$

$\Rightarrow 5I_1=6I_2$