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Q.Using Newton's third law of motion prove the law of conservation of linear momentum for a system of two interacting particles. A 30 g bullet leaves a rifle with a velocity of 300m/s and rifle recoils with a velocity of 0.60m/s. Find the mass of the rifle. (4+1=5) OR Derive the expression for the centripetal force acting on a particle of mass (m) moving with a velocity

(v) in a circular path of radius (r). A stone 0.50 kg tied to the end of a string is whirled round in a circle of radius 2m with a speed of 50 rev/min in a horizontal plane, what is the tension in the string? (4+1=5)
Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Subjective· 5mImportance★★★★★
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Newton's third law implies that the total momentum of two interacting particles is conserved if no external force acts; applying this to the rifle-and-bullet system gives a rifle mass of 15 kg.

(This is the primary part of the question; the alternative — deriving centripetal force and finding the tension in a whirled string — is not required since the primary is fully answerable.)

Part 1 — proof of conservation of linear momentum (4 marks):

Consider two particles, 1 and 2, that interact with each other (e.g., by collision or mutual attraction) and are isolated from any external force. Let F₁₂ be the force exerted on particle 1 by particle 2, and F₂₁ be the force exerted on particle 2 by particle 1.

By Newton's third law:

F₁₂ = −F₂₁

By Newton's second law, for each particle:

F₁₂ = dp₁/dt (rate of change of momentum of particle 1)

F₂₁ = dp₂/dt (rate of change of momentum of particle 2)

Substituting into the third-law relation:

dp₁/dt = −dp₂/dt

dp₁/dt + dp₂/dt = 0

d(p₁ + p₂)/dt = 0

This means the total momentum of the system, p₁ + p₂, does not change with time — it is conserved, as long as no external force acts on the two-particle system. This is the law of conservation of linear momentum.

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