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. the common velocity of `A` and `B` at time `t_(0)` is `v//3`
B. the spring constant is `k=(mv_(0)^(2))/(2x_(0)^(2))`
C. the spring constant is `k=(2mv_(0)^(2))/(3x_(0)^(2))`
D. none of these

Answer

A, C, D

Explanation

From conservation of linear momentum
`mv_(0)=(m+2m)impliesv=(v_(0))/3`
From energy conservation
`1/2mv_(0)^(2)=1/2(m+2m)v^(2)+1/2Kx_(0)^(2)`
`K=(2mv_(0)^(2))/(3x_(0)^(2))`
so, choices a annd c are correct and choices b and d are wrong.