Physics - Center of Mass Question with Solution | TestHub

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
$\frac{v_2-v_1\cos 2\theta }{v_0\cos \theta }=1$
$v_2-v_1\cos 2\theta =v_0\cos \theta$       ...(4)
$v_2-\frac{m_{2}2}{M}\cos 2\theta =v_0\cos \theta$
$v_2\left[1-\frac{m}{M}\cos 2\theta \right]=v_0\cos \theta$
$\frac{m{v}_0}{2m\cos \theta }\left[1-\frac{m}{M}\cos 2\theta \right]=v_0\cos \theta$
$M-m\cos 2\theta =2m\cos 2\theta$
$M=2m{\cos }^2\theta +m\left(2{\cos }^2\theta -1\right)$
$\frac{M}{m}=4{\cos }^2\theta -1$
${\left(\frac{M}{m}\right)}_{\max }=4-1=3$