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 the sphere and the point at which the particle was placed is

Answer 1

Answer:

R√7/4.

Explanation:

The centripetal force towards the center of the sphere will be fc = mv^2/r and from making the component of the point of detachment of the particle from the sphere is 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.