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, ${t_P} < {t_Q}$.

Column $I$	Column $II$
$(A) U_1(x) = \frac{U_0}{2} \left[1 - \left(\frac{x}{a}\right)^2\right]^2$	$(P)$ The force acting on the particle is zero at $x = a$.
$(B) U_2(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2$	$(Q)$ The force acting on the particle is zero at $x = 0$.
$(C) U_3(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2 \exp \left[-\left(\frac{x}{a}\right)^2\right]$	$(R)$ The force acting on the particle is zero at $x = -a$.
$(D) U_4(x) = \frac{U_0}{2} \left[\frac{x}{a} - \frac{1}{3}\left(\frac{x}{a}\right)^3\right]$	$(S)$ The particle experiences an attractive force towards $x = 0$ in the region $\|x\| < a$.
	$(T)$ The particle with total energy $\frac{U_0}{4}$ can oscillate about the point $x = -a$.

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