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

Options

A. (a)`t_(P) lt t_(o)`
B. (b)`t_(P) = t_(o)`
C. (c)`t_(p) gt t_(o)`
D. (d)`(t_(p))/(t_(o)) = ("length of area ACB")/("length of chord AB")`

Answer

A

Explanation

If we consider the motion of the particle P along curved path ABC, motion between AC will be an accelerated one while between CB a retarded one. But in both case horizontal component of its velocity will be greater than or equal to v . Now considering the motion of particle Q. In this case the velocity of particle Q is always equal to v. Horizontal displacements for both the particles are equal. Therefore, `t_(P) lt t_(Q)`.