**A 7.6 g bullet is stopped in a block of wood (mw = 3.4 kg). The speed of the bullet-plus-wood combination immediately after the collision is 0.51 m/s.**

**Answer:**

The sum, of the momenta of bodies mV before collision is equal to the sum after the collision

m1V1 + m2V2 = m1 V1’ + m2 V2’

the conditions of the problem are:

Mass of bullet m1 = 7,6 g = 0,0076 kg

Mass of the block m2 = 3,4 kg

Speed of the block before the collision V2 = 0

The bullet speed equal to the speed of the block after the shock V1 ‘= V2’ = V = 0.51 m / s.

then we have:

m1V1 + m2V2 = (m1 + m2) V

Solving for the velocity of the bullet V1:

V1=[(m1 + m2)V- (m2V2)]/ m1, where m2V2 = 0

So that:

V1=(m1 + m2)V/ m1 = (0,0076 + 3,4).(0,51)/0,0076 = 228,6 m/s