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 $t_{P}$ and $t_{Q}$ be respective times taken by P and Q to reach point B, then:

t_{P} < t_{Q}

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t_{P} = t_{Q}

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t_{P} > t_{Q}

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\frac{t_{P}}{t_{Q}} = \frac{l e n g t h o f a r c A B C}{l e n g t h o f c h o r d A B}

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Solution

The correct option is A $t_{P} < t_{Q}$
Horizontal displacement of both particles is equal.
For particle Q, the horizontal velocity is always equal to v.
For particle P, at the start the motion will be accelerated one, causing increase in velocity.
Thus $t_{P} < t_{Q}$

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