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$KE$ of colliding bodies before collision $=\large\frac{1}{2} $$mv^2$

After collision the mass $=m+2m=3m$

Velocity becomes $v'=\bigg(\large \frac{m_!-m_2}{m_1+m_2}\bigg)v$

$\qquad= \large\frac{m}{3m} $$v=\large\frac{v}{3}$

KE after collision $=\large\frac{1}{2} $$m \bigg(\large\frac{v}{3}\bigg)^2$

$\qquad=\large\frac{1}{9} \frac{1}{2}$$ mv^2$

$\large\frac{KE _{before}}{KE_{after}}=\large\frac{\Large\frac{1}{2} \large mv^2}{\Large\frac{1}{9} \times \frac{1}{2} \large mv^2}$

$\qquad=9:1$

Hence d is the correct answer.

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