Question

A particle moving with kinetic energy equal to $3J$ makes an elastic head on collision with a stationary particle which has twice its mass, the maximum elastic potential energy of the system (in Joule) will be equal to

Answer 1

In a head on collision between two particles the K.E. becomes minimum and the P.E. becomes maximum at the instant when the two move with common velocity.
The momentum and total energy are conserved at any instant.
Let $m$ and $u$ be the mass and initial velocity of the first particle, 2m the mass of the stationary particle.
Let $v$ be the common velocity
$\frac{1}{2}m{u}^2=3J$
$mu=(m+2m)v$
$v=\frac{u}{3}$
$\therefore$ minimum K.E of the system$=\frac{1}{2}3m{\left(\frac{u}{3}\right)}^2$
$=\left(\frac{1}{2}m{u}^2\right)\frac{1}{3}=1J$
$\therefore$ maximum P.E.$=2J$