A particle of mass m and momentum \overset{\rightarrow}{P} moves an a smooth horizontal table and collides directly and perfectly elastically with a similar particle (of mass m) having momentum - \overset{\rightarrow}{2 P}. The loss (–) or gain (+) in the kinetic energy of the first particle in the collision is

Engineering

Physics

Newtons Second Law of Motion

Collision

Momentum and Energy Conservation for System of Particles

Question

A particle of mass m and momentum $\vec{P}$ moves an a smooth horizontal table and collides directly and perfectly elastically with a similar particle (of mass m) having momentum $- \vec{2 P}$ . The loss (–) or gain (+) in the kinetic energy of the first particle in the collision is

Zero

$- \frac{p^{2}}{4 m}$

$+ \frac{p^{2}}{2 m}$

$+ \frac{3 p^{2}}{2 m}$

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In a perfectly elastic collision between two identical masses (m), kinetic energy and momentum are conserved. Let initial momenta be $\vec{P}$ and $- 2 \vec{P}$ . For identical masses in 1D elastic collision, particles exchange velocities. So final momentum of first particle becomes $- 2 P$ .

Initial KE of first particle: $\frac{P^{2}}{2 m}$ . Final KE: $\frac{{(2 P)}^{2}}{2 m} = \frac{4 P^{2}}{2 m}$ . Change: $\frac{4 P^{2}}{2 m} - \frac{P^{2}}{2 m} = \frac{3 P^{2}}{2 m}$ , which is a gain.

Final answer: $+ \frac{3 p^{2}}{2 m}$