Question

A body of mass m­₁ 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.
• B.
• C.
• D.

Answer 1

The correct option is A

Let $u_1$ be the speed of mass $m_1$ before the collision.
Here $u_2=0$. Therefore, the speeds of masses $m_1$ and $m_2$ after the collision respectively are
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1$
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)