A small bob mass `'m'` is suspended by a massless string from a cart of the same mass `'m'` as shown in the figure. The friction between the cart and horizontal ground is negligible. The bob is given a velocity `V_(0)` in horizontal direction as shown. The maximum height attained by the bob is , `(` initially whole system `(` bob `+` string `+` cart `)` was at rest `)`.

Options

A. `(2V_(0)^(2))/(g)`
B. `(V_(0)^(2))/(g)`
C. `(V_(0)^(2))/(4g)`
D. `(V_(0)^(2))/(2g)`

Answer

C

Explanation

By linear momentum conservation in horizontal direction `=` for `(` bob `+` string `+` cart `)`
`mV_(0)=(m+m)v`
`v=(V_(0))/(2)`
By mechanical energy conservation for (bob `+` string `+` cart `+` earth)
`(1)/(2)mV_(0)^(2)+0+0=(1)/(2)(2m)v^(2)+mgh +0`
`(1)/(2)mV_(0)^(2)-(1)/(2)(2m)(V_(0)^(2))/(4)=mgh`
Solving it,
`h=(V_(0)^(2))/(4g)`