According to law of conservation of momentum "if there is no force acting on a system, the momentum of the system remains unchanged."
Generalizing the situation " if a group of bodies are exerting force on each other, their total momentum remains conserved before and after the interaction provided there is no external force acting on them."
v2 Example : 7
A rifle of mass 5 kg fires a bullet of mass 40 gm. The bullet leaves the barrel of the rifle with a velocity 200 m/s. If the bullet takes 0.004 s to move through the barrel, calculate the following:
(i) recoil velocity of the rifle and
(ii) the force experienced by the rifle due to its recoil. Solution.
(i) Given mass of the rifle, m1
= 5 kg
Mass of the bullet, m2
= 40 gm = 0.04 kg
Initial velocities, u1
= 0, u2
After firing velocity of the bullet, v2
= 200 m/s
Velocity of the rifle, v1
Applying the law of conservation of momentum, we get
or 0 + 0 = 5 × v1
+ 0.04 × 200
= – [0.04 × 200]/5 = –1.6 m/s
(ii) Initial momentum of the rifle = 0
Final momentum of the rifle = 5 kg × (–1.6) = –B kg-m/s
Time interval = 0.004 s
∴ Force =
= –2000 N