yes its 3x√k/6m
with velocity of m as x√2k/3m and of 2m as x√k/6m
two blocks A and B of mass m and 2m are conncected by a massless spring of force constant k.They are placed on smooth horz. plane.Spring is stretched by an amount x and then released.Calculate relative velocities of blocks when spring comes to original length..
yes its 3x√k/6m
with velocity of m as x√2k/3m and of 2m as x√k/6m
Force acting on the two masses is always equal.. hence.
am = 2a2m
Hence vm = 2v2m
initial potnential energy = 1/2kx2
1/2mvm2+1/2.2m.v2m2
1/2kx2 = 1/2mvm2+1/2.2m.v2m2
1/2kx2 = 1/2 m 4v2m2+1/2.2m.v2m2
1/2kx2 = 3 m.v2m2
1/2kx2/3m = v2m2
v2m= √k/6m x
Relative velocity is v2m +2vm
3x √k/6m