Question

Mass 2m is kept on a smooth circular track of mass m which is kept on a smooth horizontal surface. The circular track is given a horizontal velocity v= root 2gr towards left and released The maximum height reached by 2m will be

Answer 1

Answer:

The answer is R/3.

Explanation:

K.Ei - K.Ef = P.Ef

Block will attain a height on a circular track until both move with same velocity.

Vi = √2gR

Vf = √2gR/3 (According to momentum of conservation)

1/2 m(u^2)  = 1/2 (m + 2m) v^2 + 2mgh

1/2 m(u^2) =  3m/2 x v^2 x 2mgh

After cancellation:

2gR/2 = 3/2 V^2 + 2gh

2h = Rg - 3/2 (2Rg/g)

2h = Rg - Rg/3

h = R/3

Thus the height is h = R/3.