Ex

  Physics (NTSE/Olympiad)  

2. Force & Laws Of Motion

Law Of Conservation Of Momentum

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."
i.e. m1u1 + m2u2 = m1v1 + m2v2

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 = 0
After firing velocity of the bullet, v2 = 200 m/s
Velocity of the rifle, v1 = ?
Applying the law of conservation of momentum, we get
m1u1 + m2u2 = m1v1 + m2v2
or 0 + 0 = 5 × v1 + 0.04 × 200
or v1 = – [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

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