The blocks A and B shown in figure have masses `M_A=5kg` and `M_B=4kg`. The system is released from rest. The speed of B after A has travelled a distance `1m` along the incline is

Options

A. `(sqrt(3))/(2) sqrt(g)`
B. `(sqrt(3))/(4)sqrt(g)`
C. `(sqrt(g))/(2 sqrt(3))`
D. `(sqrt(g))/(2)`

Answer

C

Explanation

If `A` moves down the incline by `1` metre, `B`shall move up by `(1)/(2)` metre. If the speed of `B` is
`v` then the speed of `A` be `2v`.
From conservation of energy :
Grain in `K.E` = loss in `P.E`.
`(1)/(2) m_(A)(2v)^(2)+(1)/(2) m_(B)v^(2) = m_(A)g xx (3)/(5) -m_(B) g xx (1)/(2)`
Solving we get `v = (1)/(2) sqrt((g)/(3))`.