A ball of mass 'm' moving with a horizontal velocity 'v' strikes the bob of mass 'm' of a pendulum at rest. During this collision, the ball sticks with the bob of the pendulum. The height to which the combined mass raises is (g = acceleration due to gravity).

Options

A. `v^(2)/(4g)`
B. `v^(2)/(8g)`
C. `v^(2)/g`
D. `v^(2)/(2g)`

Answer

B

Explanation

From the conservation of momentum
`mv + m xx 0 = (m+m)v^(')`
`rArr v^(') = (mv)/(m +m) rArr v^(') = v/2`
and we know, `h=u^(2)/(2g)` `[therefore v^(2) - u^(2) = 2gh]`
Here, `u=v^(')`
So, `h=(v^(')^(2)/(2g)` rArr v^(2)/(8g)` `{therefore v^(') = v/2}`