Question

Two identical balls $A$ and $B$ are released from the position shown in Fig. 1.205. They collide elastically with each other on the horizontal portion. The ratio of heights attained by $A$ and $B$ after collision is (neglect friction)

• A. $1:4$
• B. $2:1$
• C. $4:13$
• D. $2:5$

Answer 1

The correct option is C $4:13$

In elastic collision velocity will interchanged after collision if masses are identical.

Now, $v_1=u_2,v_2=u_1$

$\Rightarrow v_1=u_2=\sqrt{2gh}$ and $v_2=u_1=\sqrt{8gh}$

Now, A goes $h$ height lenf but B goes above height $h$ after height $h$ B follow projective.

When Particle B is at heigh $h$ then assume velocity $v.$

$\Rightarrow \frac{1}{2}m{v_2}^2=mgh+\frac{1}{2}m{v}^2$

$\Rightarrow v=\sqrt{6gh}$

It will follow projectile of velocity $v$ and angke $=60°$

maximum height $H=\frac{u^2{\sin }^2\theta }{2g}=\frac{6gh{\sin }^{2}60°}{2\times g}$

$\Rightarrow H=3\times \frac{3}{4}h$

$=\frac{9}{4}h$

$\Rightarrow$ Total height of B $=h+H=1+\frac{9}{4}h=\frac{13}{4}h$

Ratio $=\frac{h}{\frac{13h}{4}}=\frac{4}{13}.$

Hence, the answer is $\frac{4}{13}.$