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 :

Step 1: Analyze the motion of bead $Q$.

The bead $Q$ moves along the horizontal string $AB$ with constant speed $v$ because there is no friction.

$$(v_Q)_x = v$$

Step 2: Analyze the initial state of particle $P$.

At point $A$, the horizontal component of the velocity of particle $P$ is $v$.

$$(v_P)_x(0) = v$$

Step 3: Apply conservation of energy to particle $P$.

As $P$ slides down the bowl, gravity does positive work, increasing its speed $s$ above its initial speed $u$.

$$\frac{1}{2} m s^2 = \frac{1}{2} m u^2 + mgh$$ $$s > u$$

Step 4: Compare the horizontal velocity components.

Except at the endpoints $A$ and $B$, the horizontal velocity component of $P$ remains strictly greater than $v$.

$$(v_P)_x > v$$

Since the horizontal velocity of $Q$ is constant, we have the following inequality throughout the motion:

$$(v_P)_x > (v_Q)_x$$

Step 5: Compare the travel times.

Both particles undergo the same horizontal displacement $AB$.

$$t_P < t_Q$$

Answer: $t_P < t_Q$