Answer:
The momentum of the system, bullet and rifle just after the bullet leaves the barrel is zero, 3.6 kg m/s and -3.6 kgm/s respectively
Step-by-step explanation:
Given that,
Mass of the rifle = 6 kg
Mass of the bullet = 12 g
Velocity of bullet = 300 m/s
We need to calculate the momentum of bullet
Using formula of momentum
![P = mv](https://img.qammunity.org/2020/formulas/physics/high-school/sf82jq0v1hccf3tbythbvi54tvskpd6rhs.png)
Where, m = mass of bullet
v = velocity of bullet
Put the value into the formula
![P=12*10^(-3)*300](https://img.qammunity.org/2020/formulas/physics/college/css101ys84847iquuyx6i3s8yno4e9cafb.png)
![P=3.6\ kg m/s](https://img.qammunity.org/2020/formulas/physics/college/itbje22wvf9h56ikvu5x5fxo6eu5l4kbg0.png)
The momentum of the bullet is 3.6 kgm/s.
The momentum of rifle is opposite direction of momentum of bullet and equal in magnitude.
The momentum of rifle = - 3.6 kgm/s
Therefore, The momentum of system must be zero.
Hence, The momentum of the system, bullet and rifle just after the bullet leaves the barrel is zero, 3.6 kg m/s and -3.6 kgm/s respectively.