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,

Let  $l$  be the length of the chord AB. The time taken by the particle $Q$  and the particle $P$  to reach the point  $B$ are$t_Q=l/v$ , and $t_P=l/{\overline{v}}_h$ ,where ${\overline{v}}_h$  is the average horizontal velocity of the particle  $P.$

The forces acting on the particle  $P$ are its weight mg and normal reaction  $N$. The weight is always in downward vertical direction and hence cannot affect horizontal velocity  $v_h$ . However,  $N$  has a horizontal component that accelerates particle $P$  from $A$  to  $C$ and retards it from  $C$ to $B$ . By symmetry,   $P$ will have same velocity on points symmetrically located about  $C$. At  $B,$ velocity is again $v_h$  and at every point between $A$  and  B  it is greater than  $v_h$ . Thus, the average horizontal velocity of P  is more than its starting value i.e.,  ${\overline{v}}_h>v.$ Hence, $t_P<t_Q$  .