A particle P is sliding down a frictionless h | vedclass

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Question

(A) For particle $Q$, the velocity is constant at $v$ along the horizontal direction, so the time taken to cover the horizontal distance $AB$ is ${t_Q

Vedclass · Accepted Answer

${t_P} < {t_Q}$

Vedclass · Answer

(A) For particle $Q$, the velocity is constant at $v$ along the horizontal direction, so the time taken to cover the horizontal distance $AB$ is ${t_Q} = \frac{AB}{v}$.For particle $P$, the horizontal component of velocity $v_x$ is given by $v_x = v \cos 	heta$, where $	heta$ is the angle the velocity vector makes with the horizontal. As the particle slides down the bowl, it gains speed due to gravity. The horizontal component of its velocity $v_x$ at any point is greater than or equal to the initial horizontal component $v$ because the particle accelerates as it moves towards the lowest point $C$ and then decelerates, but its speed remains higher than the initial speed $v$ throughout the path until it reaches $B$. Since the horizontal component of velocity for $P$ is always greater than or equal to $v$ throughout the motion, the average horizontal velocity of $P$ is greater than $v$. Therefore, the time taken by $P$ to cover the same horizontal distance $AB$ is less than the time taken by $Q$. Thus, ${t_P} < {t_Q}$.

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