Answer:
The momentum of block B = -50 Kg m/s
Explanation:
Given,
The initial momentum of block A, MU = -100 Kg m/s
The final momentum of block A, MV = -200 Kg m/s
Consider the block B is initially at rest.
Therefore, the initial momentum of block B, mu = -150 Kg m/s
According to the laws of conservation of linear momentum, the momentum of the body before impact is equal to the momentum of the body after impact.
MU + mu = MV + mv
-100 + (-150) = (-200) + mv
mv = -250 - 200
= -50 Kg m/s
Hence, the momentum of the block B after impact is, mv = -50 Kg m/s
