A ball of mass m suspended from a rigid support by an inextensible massless string is released from a height h above its lowest point. At its lowest point it collides elastically with a block of mass 2m at rest on a frictionless surface. Neglect the dimensions of the ball and the block. After the collision the ball rises to a maximum height of–

Options

A. `h/3`
B. `h/2`
C. `h/8`
D. `h/9`

Answer

D

Explanation

As `mgh = 1/2 mv^2`
`implies v =sqrt(2gh)` , velocity of ball before collision
Using law of conservation of momentum
`mv =mv_1 +2mv_2`
`implies v_1 +2v_2 = v " "...(i)`
`v_1 and v_2` are velocity of ball and block after collision as coefficient of restitution (e)
`e=(v_2-v_1)/(v-0)`
`implies v_2 -v_1 = v " "...(iii) ` [ as e = 1 ]
From (i) + (ii)
`v_2 =v/3 and v_1 =v/3`
Hence, after collision maximum height
`H_(max) =(v_1^2)/(2g) =(((sqrt(2gh)/3))/(2g))^2`
`H_(max) =h/9`