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Question

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 speed v. Friction between the bead and the string may be neglected. Let tP and tQ be the respective times taken by P and Q to reach the point B. Then:

A
tp<tQ
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B
tp=tQ
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C
tp>tQ
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D
tptQ=Length of arc ACBLength of chord AB
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Solution

The correct option is A tp<tQ
As the particle P moves from A to C, its horizontal velocity increases from v. Again the horizontal velocity decreases to v as P moves form C to B. So the horizontal velocity of P remains greater than or equal to v. But the horizontal velocity of bead Q remains constant equal to v. For same horizontal displacement AB, P takes smaller time than Q i.e.,
tp<tQ.

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