Question

A body of mass $m_1$ moving at a constant speed undergoes an elastic collision with a body of mass $m_2$ initially at rest. The ratio of the kinetic energy of mass $m_1$ after the collision to that before the collision is

• A. ${\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2$
• B. ${\left(\frac{m_1+m_2}{m_1-m_2}\right)}^2$
• C. ${\left(\frac{2m_1}{m_1+m_2}\right)}^2$
• D. ${\left(\frac{2m_2}{m_1+m_2}\right)}^2$

Answer 1

The correct option is A ${\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2$

Let $u_1,u_2$ be the speeds of masses $m_1$ and $m_2$ before the collision and $v_1,v_2$ be the speeds of those masses after the collision.
Here $u_2=0$.
From the conservation of linear momentum,
$m_1{u}_1=m_1{v}_1+m_2{v}_2$
From the equation of restitution
$e=\frac{v_2-v_1}{u_1-u_2}$
$1=\frac{v_2-v_1}{u_1}$
After the solving the above equations
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1$
and $v_2=\left(\frac{2m_1}{m_1+m_2}\right)u_1$
$\therefore KE$ of $m_1$ after collision $=\frac{1}{2}{m}_1{v}_1^2$
$=\frac{1}{2}{m}_1{\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2{u}_1^2$.
KE of $m_1$ before collision
$=\frac{1}{2}{m}_1{u}_1^2$. The ratio of the two is ${\left(\frac{m_1-m_2}{m_1+m_2}\right)}^2$.
Hence the correct choice is (a).