Question 94 of 130

A body of mass $M$ moving at speed ${V}_{0}$ collides elastically with a mass $m$ at rest. After the collision, the two masses move at angles ${\theta }_{1}$ and ${\theta }_{2}$ with respect to the initial direction of motion of the body of mass $M.$. The largest possible value of the ratio $\dfrac{M}{m},$ for which the angles ${\theta }_{1}$ and ${\theta }_{2}$ will be equal, is :

A$3$B$4$C$2$D$1$

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Answer: (A) $3$

Linear momentum conservation
(i) $y$-direction
$m{v}_{2}\sin \theta =m{v}_{1}\sin \theta$
$m{v}_{2}=m{v}_{1}$ ...(1)
(ii) $x$-direction
$m{v}_{0}=m{v}_{2}\cos \theta +m{v}_{1}\cos \theta$ ...(2)
from (1) and (2)
$m{v}_{0}=m{v}_{2}\cos \theta +m{v}_{2}\cos \theta$
$m{v}_{0}=2m{v}_{2}\cos \theta$ ...(3)
(iii) $e=1$given
$\dfrac{{v}_{2}-{v}_{1}\cos 2\theta }{{v}_{0}\cos \theta }=1$
${v}_{2}-{v}_{1}\cos 2\theta ={v}_{0}\cos \theta$ ...(4)
${v}_{2}-\dfrac{{m}_{2}2}{M}\cos 2\theta ={v}_{0}\cos \theta$
${v}_{2}\left[1-\dfrac{m}{M}\cos 2\theta \right]={v}_{0}\cos \theta$
$\dfrac{m{v}_{0}}{2m\cos \theta }\left[1-\dfrac{m}{M}\cos 2\theta \right]={v}_{0}\cos \theta$
$M-m\cos 2\theta =2m\cos 2\theta$
$M=2m{\cos }^{2}\theta +m\left(2{\cos }^{2}\theta -1\right)$
$\dfrac{M}{m}=4{\cos }^{2}\theta -1$
${\left(\dfrac{M}{m}\right)}_{\max }=4-1=3$