Q.(a) Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it. OR
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Start your 14-day free trial to unlock the full solution →Newton's third law makes the internal forces between two interacting bodies equal and opposite, proving that their total momentum stays constant; applied to a gun-bullet system starting at rest, this gives the gun's recoil velocity as V = mv/M in the direction opposite to the bullet.
Proof of the law of conservation of linear momentum:
Consider two bodies A and B forming an isolated system (no external force acting on the system), interacting with each other (e.g. during a collision or explosion) over a time interval Δt.
Let F(AB) be the force exerted by A on B, and F(BA) be the force exerted by B on A. By Newton's third law:
F(AB) = -F(BA)
By Newton's second law, force equals rate of change of momentum:
F(BA) = dp(A)/dt and F(AB) = dp(B)/dt
So:
dp(B)/dt = -dp(A)/dt
dp(A)/dt + dp(B)/dt = 0
d(p(A) + p(B))/dt = 0
This means the total momentum of the system, p(A) + p(B), does not change with time — it remains constant, as long as no external force acts on the system. This is the law of conservation of linear momentum.
Application — recoil velocity of a gun:
Before firing, the gun (mass M) and bullet (mass m) are both at rest, so the total initial momentum of the system is zero:
p_initial = 0
After firing, the bullet moves forward with velocity v, and the gun recoils backward with velocity V. Since the gun+bullet system is isolated during firing (internal explosive force only), momentum is conserved: …
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