Question

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

• A. $+\frac{p^2}{2m}$
• B. $-\frac{p^2}{4m}$
• C. $+\frac{3p^2}{2m}$
• D. Zero.

Answer 1

The correct option is C $+\frac{3p^2}{2m}$

Let $\mid \overrightarrow{p}\mid =mv=p$

In an elastic head-on collision with equal masses, velocities are interchanged.
Gain in kinetic energy of first

$=\frac{1}{2}m\left[\right(2v{)}^2-v^2]$

$=\frac{3}{2}m{v}^2=\left(\frac{3}{2}m\right)(\frac{p}{m}{)}^2=\frac{3p^2}{2m}$