a bullet of mass 40g strikes a stationary wooden block of mass 6kg horizontally at 500m per second.the bullet goes through the block,and emerges from the other side at 200m per second.a)Calculate the velosity of the block once the bullet emerges from it?

Answer 1

2m/second in the original direction of the bullet

Answer 2

Take the direction of bullet to a block as postive
M1vi1+m2vi2=m1vf1+m2vf2
(0,04)(500)+(6)(0)=(0.04)(200)+(6)(vf2)
20=8+6(vf2)
20-8=6(vf2)
6(vf2)=12
.:vf2=2m.s