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If a body of mass m moves with velocity u on a rough surface and stops after travelling a distance S due to friction, then frictional force F = ma = μR = μmg = μmg = μg. Using v² = u² - 2aS, v = 0, u² = 2μgS. Find the ratio of their linear momenta.

1. 1/2 (1)v1² = 1/2 (9)v2²
2. v1²/v2² = 9 or v1/v2 = 3
3. Ratio of their linear momenta = m1v1/m2v2 = 1/9 × (3) = 1/3
4. E = 1/2 mv². Hence, mv = (2mE)^(1/2)

Correct answer: Ratio of their linear momenta = m1v1/m2v2 = 1/9 × (3) = 1/3

Solution

The ratio of linear momenta is derived from the relationship between velocity and momentum. Given the velocity ratio v1/v2 = 3, the momentum ratio m1v1/m2v2 simplifies to 1/3.

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