 # A particle P is sliding down a frictionless hemisphere 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 horizontal spring AB with the speed v. Friction between bead and the string may be neglected. Let t(p) and t(q) be the respective times taken by P and Q tp reach the point B then A. t(p)=t(q) B. t(p)>t(q) C. t(p) 11 years ago

the mass of the particle P and the bead Q is the same and it is also given that they have the same velocity.

the object is a hemispherical bowl .

Let D1 be distance travelled by the particle P with mass m and velocity v.

D1 = vt

and D2 be the distance travelled by the bead Q with mass m and velocity v

since D2 is actually the displacement of Particle P and distance of bead B

we know displacement <= distance

displacement of P < distance travelled by P

==> distance travelled by bead Q < distance travelled by P

since velocity is the same.

time taken to travel a smaller distance is less than time taken to travel a larger distance.

therefore time taken by Particle P > time taken by bead Q

option B t(P) > t(Q) is the final answer

11 years ago

SINCE THE VELOCITY COMPONENT IN THE HORIZONTAL DIRECTION IS SAME FOR BOTH THE BEADS AND THERE IS NO FRICTION IN BOTH THE PATHS . THEREFORE T(P)=T(Q).

3 years ago
At A the horizontal speeds of both the masses are the same. The velocity of Q remains the same in horizontal as no force is acting in the horizontal direction. But in case of P as shown at any intermediate position, the horizontal velocity first increases (due to force Nsintheta), reaches a max value at O and then decreases. Thus it always remains greater than v. Therefore tp