Two blocks `A` and `B` of masses in and `2m` respectively placed on a smooth floor are connected by a spring. A third body `C` of mass `m` moves with velocity `v_(0)` along the line joining `A` and `B` and collides elastically with `A`. At a certain instant of time after collision it is found that the instantaneous velocities of `A` and `B` are same then:

Options

A. `m(v_(0)^(2))/(x_(0)^(2))`
B. `m(v_(0))/(2x_(0))`
C. `2m(v_(0))/(x_(0))`
D. `(2)/(3)m((v_(0))/(x_(0)))^(2)`

Answer

D

Explanation

Initial momentum of the system block (c )=`mv_(0)` .After striking with A,the block C comes to rest and now both block A and B moves with velocity v when compression in spring is `x_(o)`.
By the law of conservation of linear momentum
`mv_(0)=(m+2m)vimpliesv=(v_(0))/(3)`
By the law of conservation of linear momentum
`mv_(0)=(m+2m)vimpliesv=(v_(0))/(3)`
By the law of conservaion of energy K.E. of block C=K.E. of system
`(1)/(2)mv_(0)^(2)=(1)/(2)(3m)((v_(0))/(3))^(2)+(1)/(2)kx_(0)^(2)`
`implies(1)/(2)mv_(0)^(2)-(1)/(6)mv_(0)^(2)+(1)/(2)kx_(0)^(2)`
`implies(1)/(2)kx_(0)^(2)-(1)/(2)mv_(0)^(2)-(1)/(6)mv_(0)^(2)=(mv_(0)^(2))/(3)`
`therefore k=(2)/(3)m((v_(0))/(x_(0)))^(2)`