A block C of mass m is moving with velocity `v_(0)` and collides elastically with block A of mass m and connected to another block B of mass `2m` through spring constant k. What is k if `x_(0)` is compression of spring when velocity of A and B is same?

Options

A. `(a)(mv_(0)^(2))/(x_(0)^(2))`
B. `(b)(mv_(0)^(2))/(2x_(0)^(2))`
C. `(c)(3mv_(0)^(2))/(2x_(0)^(2))`
D. `(d)(2mv_(0)^(2))/(3x_(0)^(2))`

Answer

D

Explanation

Their common velocity would be when C strikes A, A accquires the velocity of C and C beocmes stationary. A forces B to move and ultimately both acquires common velocity with spring compressed by `x_(0)`.
`v=(mv_(0)/(m+2m)` = `v_(0)/3)`
Now applying conservation of mechanical energy
`1/2mv_(0)^(2)=1/2kx_(0)^(2) + 1/2(3m)(v_(0)/3)^(2) rArr k = 2/3(mv_(0)^(2)/(x_(0)^(2)`