VITEEE 2009 Paper

Mass of the first body $m_{1}=5$ kg and for elastic collision coefficient of restitution, e = 1. Let initially body $m_{1}$ moves with velocity v after collision velocity becomes $\left(\frac{u}{10}\right).$ Let after collision velocity of M block becomes $\left(v_{2}\right).$ By conservation of momentum $m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$ or $\;\;5u + M\times 0 = 5\times \frac{u}{10} + Mv_{2}$ or $\;\;5u = \frac{u}{2} + Mv_{2}$...(i) Since, $v_{1}-v_{2}=-e(u_{1}-u_{2})$ or $\;\;\frac{u}{10} - v_{2} = -1(u)$ or $\;\;\frac{u}{10} + u = v_{2}$ or $\;\;\frac{11u}{10} = v_{2}$...(ii) Substituting value of $v_{2}$ in Eq.(i) from Eq.(ii) $5u = \frac{u}{2} + M\left(\frac{11u}{10}\right)$ or $\;\;5 - \frac{1}{2} = M\left(\frac{11}{10}\right)$ ⇒ $M = \frac{9 \times 10}{2 \times 11} = \frac{45}{11} = 4.09\,\mathrm{kg}$