Question

Two blocks A & B of mass m and 2m respectively are connected by a spring of spring constant k. the masses are moving to the right with a uniform velocity v' each, the heavier mass leading the lighter one. The spring has its natural length during this motion . Block B collides heea on with a third block C of mass 2m at rest , the collision being completely elastic.
1) then tell the velocity of block B just after collision.

2) the velocity of the COM of system of blocks A , B and C. is:-

3) maximum compression of the spring after collision is:-
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Answer 1

At maximum compression velocity of the all the blocks is the same.
Applying conservation of momentum,

mv + 2mv = 4mv1 + mv [velocity of the rear mass does not change immediately as it is connected with the spring and v1 is the velocity of block B after collision]

3mv-mv - 4mv1
v1 = v/2
i) hence velocity of the block B after collision = v/2

appling momentum conservation just before collision and at the time of maximum compression
mv + 2mv = 5m x v2 [ where v2 is the velocity of the COM of system of blocks
A , B and C]
ii) v2 = 3v/5
the velocity of the COM of system of blocks A , B and C = 3v/5

iii) Applying conservation of energy at the intial moment that is just after collision and at the time of maximum compression,

1/2 mv² + 1/2*4mv1² = 1/2*5mv2² +1/2kx²
1/2 mv² + 1/2*4m*v²/4 = 1/2*5m*9/25v² + 1/2kx²
mv² (1+1 - 45/25) = 1/2 kx²
mv² * 1/5 = 1/2 kx²
mv² = 5kx²
x² = mv²/5k
x = √(mv²/5k)