Question

A particle of mass m is projected with speed root Rg/4
from top of a smooth hemisphere as shown in.
figure. If the particle starts slipping from the highest
point, then the horizontal distance between the
point where it leaves contact with sphere and the
point at which the particle was placed is​
ans: Rroot 7 by 4

Answer 1

Answer:

R√7/4.

Explanation:

For the particle to stay till the point where it goes out of the hemisphere, it must have some centripetal force which is fc = mv^2/r while the point from where the particle will get out of the hemisphere will be  mgcosθ. So, mgcosθ = mv^2/r.

The conservation of the linear momentum of the body is given as mgR+1/2mu^2 =mgRcosθ+1/2mv^2.

On substituting the values we will get that gR+gR/2*4=gRcosθ+1/2gRcosθ which on solving we will get cosθ = 3/4, so the value of sinθ will be √7/4. Hence, the value of horizontal distance of the projection as Rsinθ = R√7/4.