question: a bullet of mass 10g moving with velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900g.What is the velocity acquired door the block?

answer:4.4m/s
reason: let the final velocity of the bullet embedded in it be v
according to the law of conservation of momentum,
total momentum before collision = total momentum after collision
= 0.001* 400+0.9*0/0.001 + 0.9 = 4/0.91 = 400/91 = 4.4 m/s

mass of bullet= 10 g = 0.01 kg
velocity of bullet=400 m/s
mass of wooden box= 900 g =0.9 kg
velocity of wooden block= ?

formula= Mb 1*Vb 1=Mb 2*Vb 2
0.01*400=0.9*Vb2
=0.01*400/0.9=4.4444444 m/s
ans=4.4 m/s