Question
A bead $P$ sliding on a frictionless semi-circular string $(ACB)$ and it is at point $S$ at $t = 0$ and at this instant the horizontal component of its velocity is $v$. Another bead $Q$ of the same mass as $P$ is ejected from point $A$ at $t = 0$ along the horizontal string $AB$, with the speed $v$, friction between the beads and the respective strings may be neglected in both cases. Let $t_{p}$ and $t_{Q}$ be the respective times taken by beads $P$ and $Q$ to reach the point $B$, then the relation between $t_{p}$ and $t_{Q}$ is
Options
Solution
(4) $t_{p} = t_{Q}$
Sol. Horizontal displacement of Q is more then P .
$$X_{Q} > X_{P}$$
Horizontal component of velocity is same
So $t_{p} = \frac{x_{p}}{v}$
$$\begin{matrix} & t_{Q} = \frac{x_{Q}}{v} \\ & t_{Q} > t_{p} \end{matrix}$$
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