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
- A1:8
- B9:1
- C8:9
- 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: