Answer:
5.94m/s
Step-by-step explanation:
Using the law of conservation of momentum;
mAuA+ mBuB = mAvA+mBvB
Substitute the given values
1.4(3.7) + 0.55(0) = 1.4(2.3) + 0.55vB
5.18 + 0 = 3.22 + 0.55vB
5.18 - 3.22 = 0.55vB
1.96 = 0.33vB
vB = 1.96/0.33
vB = 5.94m/s
Hence the velocity of Ball B be after the collision is 5.94m/s