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JEE MainPhysicsCenter of Mass, Momentum and Collision

Two particles of masses m and 2m are moving in the same straight line with velocities 2v and v respectively. The ratio of the kinetic energy of the center of mass to the kinetic energy of the system relative to the center of mass is

Options

  1. A1:8
  2. B9:1
  3. C8:9
  4. D8:1

Correct answer

D. 8:1

Step-by-step solution

The velocity of the center of mass is given by: v_ cm = m(2v) + 2m(v) m + 2m = 4v 3 The kinetic energy of the center of mass is: K_ cm = 1 2 (3m)v_ cm ^2 = 1 2 (3m) ( 4v 3 )^2 = 8 3 mv^2 The total kinetic energy of the system is: K_ total = 1 2 m(2v)^2 + 1 2 (2m)(v)^2 = 2mv^2 + mv^2 = 3mv^2 By Koenig's theorem, the kinetic energy of the system relative to the center of mass is: K_ rel = K_ total - K_ cm = 3mv^2 - 8 3 mv^2 = 1 3 mv^2 The required ratio is: K_ cm K_ rel = 8 3 mv^2 1 3 mv^2 = 8 1 Thus, the ratio is 8:

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