Adarsh Pal
Last Activity: 4 Years ago
Since the velocity of all the blocks is the same at maximum compression.
Apply conservation of momentum, it can be written as,
mv0+2mv0=4mv1+mv0
v1=2v0
Thus, the velocity of block B just after the collision is 2v0.
Apply conservation of momentum, just before and at the time of compression it is given as,
mv0+2mv0=5m×v
v=53v0
Apply the conservation of energy, just after collision and at the time of maximum compression, it is given as,
21mv02+21×4mv12=21×5mv2+21kx2
21mv02+21×4m(2v0)2=21×5m(53v0)2+21kx2
21kx2=2mv02(1+1−2545)
x2=5kmv02
x=5kmv02
Thus, the maximum compression of the spring after the collision is 5kmv02.
