Answer:
u= 200 m/s
Explanation:
Given that
Mass of bullet ,m= 50 gm
Assume that mass of block ,M= 1.2 kg
Lets take speed of the bullet before collision = u m/s
The speed of the system after collision ,v= 8 m/s
There is no any external force ,that is why linear momentum of the system will be conserve.
Linear momentum ,P = mass x velocity
m u = (M+m)v
0.05 x u = (1.2 + 0.05 ) x 8
u= 200 m/s
Therefore the speed of the bullet just before the collision is 200 m/s.