A body of mass m moving with velocity v elastic collides head on … - Vidyadip

MCQ

A body of mass $m$ moving with velocity $v$ elastic collides head on with another body of mass $2\,\,m$ which is initially at rest. The ratio of $K.E.$ of the colliding body before and after collision will be

A
$1 : 1$
B
$2 : 1$
C
$4 : 1$
✓
$9 : 1$

✓

Answer

Correct option: D.

$9 : 1$

d
Velocity of first body after collision is $:-$

$\mathrm{v}_{1}=\left(\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right) \mathrm{u}_{1}+\frac{2 \mathrm{m}_{2} \mathrm{u}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$

$=\left(\frac{m-2 m}{m+2 m}\right) v+\frac{2 \times(2 m) \times 0}{m+2 m}=\frac{-v}{3}$

$\mathrm{KE}_{i}=\frac{1}{2} \mathrm{mv}^{2}, \quad \mathrm{KE}_{\mathrm{f}}=\frac{1}{2} \mathrm{m}\left(\frac{-\mathrm{v}}{3}\right)^{2}=\frac{1}{2} \frac{\mathrm{mv}^{2}}{9}$

$\frac{\mathrm{KE}_{\mathrm{i}}}{\mathrm{KE}_{\mathrm{f}}}=\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{2} \frac{\mathrm{mv}^{2}}{9}}=9$

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