Question

A body of mass m moving with velocity $v$ collides head on with another body of mass $2m$ which is initially at rest. The ratio of $K.E.$ of colliding body before and after collision will be :$[e=1]$

• A. $1:1$
• B. $2:1$
• C. $4:1$
• D. $9:1$

Answer 1

The correct option is D $9:1$

Conserving momentum
$mv=m{v}_1+2m{v}_2$
$v=v_1+2v_2$

$e=\frac{v_2-v_1}{v}=1$ as collision is head on.

$v_2-v_1=v$
$v_1+2v_2=v$

$v_2=\frac{2}{3}v$

$v_1=\frac{-1}{3}v$

so K.E ratio$=\frac{\frac{1}{2}\times m\times v^2}{\frac{1}{2}\times m\times (\frac{v}{3}{)}^2}$

$=9:1$