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 direction with the 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 prove that tp < tq

Answer 1

according to question, a particle P is sliding down a frictionless hemispherical bowl. so, motion of the particle P between A to C will be an accelerated one while C to B will be a retarded one. but in any case of horizontal component of its velocity will be greater than or equal to v ( as ,v_p = v/cosa or v' = vseca > v) . on the other hand in case particle Q, it is always equal to v.

so, it is clear that $v_p> v=v_Q$

here horizontal displacement for both particle is same. e.g., S

then, time taken by particle p, $t_p=S/v_p$

and time taken by particle Q, $t_Q=S/v_Q$

as $v_p> v_Q$

so, $t_p< t_Q$ or, we can write it $t_p< t_Q$ hence proved

Answer 2

Answer:

Explanation: in this question the dispalcemennt is equal.

And will assume that the slide is frictionless .