Answer:
The momentum of block B = 27 Kg m/s
Explanation:
Given,
The initial momentum of block A, MU = 15 Kg m/s
The final momentum of block A, MV = -12 Kg m/s
Consider the block B is initially at rest.
Therefore, the initial momentum of block B, mu = 0
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
15 + (0) = (-12) + mv
mv = 15 + 12
= 27 Kg m/s
Hence, the momentum of the block B after impact is, mv = 27 Kg m/s
