Work Energy Power

JEE Main level, single correct, medium

39 of 183

Question 39 of 183

A bead PP sliding on a frictionless semi-circular string (ACB)(A C B) and it is at point SS at t=0t =0 and at this instant the horizontal component of its velocity is vv. Another bead QQ of the same mass as PP is ejected from point AA at t=0t=0 along the horizontal string ABA B, with the speed vv, friction between the beads and the respective strings may be neglected in both cases. Let tPt_{P} and tQt_{Q} be the respective times taken by beads PP and QQ to reach the point BB, then the relation between tPt_{P} and tQt_{Q} is

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Answer: (A) tP<tQ{ }^{t_{P} \lt t_{Q}}

Horizontal displacement of Q is more than P.
XQ>XPX_Q \gt X_P
Horizontal component of velocity is same

So
tp=xpvt_p = \dfrac{x_p}{v}
tQ=xQvt_Q = \dfrac{x_Q}{v}
tQ>tpt_Q \gt t_p

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