Two masses `M_(1) and M_(2)` are connected with each other with the help of a massless spring of 'spring-constant' k. Mass `M_(1)` is fixed to a wall and the system is at rest on the frictionless floor. `M_(2)` is displaced by distance x and released. Velocity of CM of the system when `M_(1)` is detached from wall is given by

Options

A. `((kM_(1))^(1/2)x)/(M_(1)+M_(2))`
B. `(x(kM_(2))^(-1//2))/(M_(1)+M_(2))`
C. `(M_(2)""^(1//2)kx)/(M_(1)+M_(2))`
D. `((M_(1)k)^(-1//2)x)/(M_(1)-M_(2))`

Answer

B

Explanation

From conservation of momentum
`(M_(1)+M_(2))v_(CM)=M_(1) times 0+M_(2)v_(2)," ie, "v_(CM)=(M_(2)v_(2))/(M_(1)+M_(2))`
From conversation of energy
`1/2kx^(2)=1/2M_(2)v_(2)^(2) or v_(2)=sqrt(K/M_(2))x" "therefore v_(CM)=x/(M_(1)+M_(2))(kM_(2))^(1//2)`