A bead $P$ sliding on a frictionless semi-circular string ( $A C B$ ) and it is at point $S$ at $t =

Question

A bead $P$ sliding on a frictionless semi-circular string ( $A C B$ ) 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 $A B$, 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

Professor PC Thomas Classes · Answer

Bead Q travels along the straight horizontal path with a constant speed $v$, so its travel time is $t_Q = \frac{AB}{v}$. Bead P travels along the semi-circular path. As it slides down into the dip, gravity accelerates it, increasing its speed above $v$ such that its horizontal velocity component $v_x$ is consistently greater than or equal to $v$ throughout the motion. Since P covers the same horizontal distance $AB$ but with an average horizontal speed greater than $v$, it will reach point B faster than Q. Therefore, $t_P < t_Q$.