A moving particle of mass m makes a head on elastic collision with a particle of mass 2m which is initially at rest. The fraction of the kinetic energy is lost by the colliding particle is

Engineering

Physics

Newtons Second Law of Motion

Collision

Momentum and Energy Conservation for System of Particles

Question

A moving particle of mass m makes a head on elastic collision with a particle of mass 2m which is initially at rest. The fraction of the kinetic energy is lost by the colliding particle is

1/9

2/3

1/3

8/9

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In an elastic collision, both momentum and kinetic energy are conserved. Let the initial velocity of mass m be u, and mass 2m be at rest (0).

Let v₁ and v₂ be the final velocities. Using conservation of momentum:

m u + 0 = m v_{1} + 2 m v_{2}

And conservation of kinetic energy:

\frac{1}{2} m u^{2} + 0 = \frac{1}{2} m {v_{1}}^{2} + \frac{1}{2} (2 m) {v_{2}}^{2}

Solving these, the final velocity of the colliding particle (m) is:

v_{1} = \frac{m - 2 m}{m + 2 m} u = - \frac{1}{3} u

Its final kinetic energy is $\frac{1}{2} m {(\frac{u}{3})}^{2} = \frac{1}{9} (\frac{1}{2} m u^{2})$ . The fraction lost is $1 - \frac{1}{9} = \frac{8}{9}$ .

Final Answer: 8/9