Question

A particle $P$ is sliding down a frictionless hemispherical bowl. It passes the point $A$ at $t=0$. At this instant of time, the horizontal component of its velocity is $v$. A bead $Q$ of the same mass as $P$ is ejected from $A$ at $t=0$ along the horizontal string $AB$, with the speed $v$. Friction between the bead and the string may be neglected. Let $t_P$ and $t_Q$ be the respective times taken by $P$ and $Q$ to reach the point $B$. Then:

• A. $t_P<t_Q$
• B. $t_P=t_Q$
• C. $t_P>t_Q$
• D. $\frac{t_P}{t_Q}=\frac{\text{length of arc ACB}}{\text{length of AB}}$

Answer 1

The correct option is A

$t_P<t_Q$

At $A$ the horizontal speeds of both the masses is the same. The velocity of $Q$ remains the same in horizontal as no force is acting in the horizontal direction. But in case of $P$ as shown at any intermediate position, the horizontal velocity first increases (due to $N\sin \theta$), reaches a max value at $O$ and then decreases. Thus it always remains greater than $v$. Therefore $t_P<t_Q$.