Question

bullet of mass 10 gram moving with hundred metres per second is embedded in a block of 1 kg with initially at rest the final velocity of the system will be

Answer 1

Here we would have to apply the law of conservation of momenta .
Acc to it momenta before collision is equal to momenta after collision
So in the following equation :
Mass of bullet = M1
Mass of block=M2
Velocity of bullet=U1 (initial)
Velocity of block = U2(initial)
Velocity of the entire system = V( as now the bullet is embedded into the block and will have the same velocity as the block )

Acc to law of conservation of momenta :
M1.U1 + M2.U2 = (M1 + M2)V
As the block is stationary momentum of block is zero
So 0.01 kg.100m/s = (0.01 + 1 ) V
1 kg m/s = 1.01.V
1/1.01 = V
0.99 m/s = V

Answer 2

Answer:

1 m/s is the answer

Explanation:

thanks