The masses of the block A and B are `m and M` Between A and B there is a constant force `F` but B can slide frictionlessly on the horizontal surface A is set in motion with velocity `v_(0)` while B is at rest . What is the distance moved by A relative to B before they move with the same velocity

Options

A. `(mMv_(0)^(2))/(F(m - M))`
B. `(mMv_(0)^(2))/(2F(m - M))`
C. `(mMv_(0)^(2))/(F(m + M))`
D. `(mMv_(0)^(2))/(2F(m + M))`

Answer

D

Explanation

Free body diagram

Equation of motion
`a_(B) = (F)/(M)` (in + x direction)
`a_(A) = (F)/(m) (m - x direction)`
Relative acceleration of A w.r.t, B :
`bar a_(A,B)= bar a_(A) - bar a_(B) = (F)/(m) = - F((m +M)/(mM))`
(along - x direction )
initial relative velocity of A w.r.t.B
`a_(A,B) = v_(0)`
Final relative velocity of A w.r.t.B = 0
Using `v^(2) = a^(2) + 2ax`
`0 = v_(0)^(2) - 2(F(m + M))/(nM) g rArr S = (Mmv_(0)^(2))/(2F(m +M))`