Ball A of mass m strikes a stationary ball B of mass M and undergoes an elastic collision. After collision ball A has a speed one third of its initial speed.  The ratio of $\frac{M}{m}$ is             

From conservation of linear momentum and kinetic energy, we can show that

$v_1^{-}=\left(\frac{m-M}{m+M}\right)v=\left(\frac{1-x}{1+x}\right)v........\left(1\right)$

Here,    $x=\frac{M}{m}$

Now speed is given one-third i.e., $v_1^{\text{'}}=\pm \frac{v}{3}$

 Substituting this in equation (1), we get $x=\frac{M}{2}=\frac{1}{2}$ and 2