TS EAMCET 18-July-2022 Shift 2 Solved Paper

Let the mass of the moving particle $=m$ $\therefore$ The mass of the stationary particle $=\;\frac{1}{n} \times m=\;\frac{m}{n}$Also, let the speed of moving particle $=v$ Now, by the law of conservation of linear momentumSituation after collision $\therefore \;\; m v+ \left(\;\frac{m}{n} \times 0\right) =m v'+\;\frac{m}{n} v'$ . $\Rightarrow \;\; m v=m \left(v'+\;\frac{v'}{n}\right)$ $\Rightarrow$ $v=v'+\;\frac{v'}{n} \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (i)$Also, coefficient of restitution, $e=\;\frac{- (v''-v')}{ (0-v') }$Since, it is elastic collision, $\therefore \;\; e=1$ $\Rightarrow \;\; 1=\;\frac{- (v''-v')}{-v} \Rightarrow v''=v+v'$ $\Rightarrow \;\; v'=v''-v\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (ii)$From Eqs. (i) and (ii), we get $v=v''+\;\frac{v''}{n}$ $v'= \left[\;\frac{2 v}{1+\;\frac{1}{n}}\right] \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (iii)$Now, kinetic energy transferred to the stationary particle $=(\;\text{ kinetic energy }\;)_{\;\text{ final }\;}-(\;\text{ kinetic energy }\;)_{\;\text{ initial }\;}$ $\Rightarrow \;\; K_{f}'-K_{i}'=\;\frac{1}{2} \times \;\frac{m}{n} \times v''^{2}\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (iv)$Also, initial kinetic energy of mass $m$ $=\;\frac{1}{2} m v^{2} \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (v)$$\therefore$ The fraction of ${KE}$ that gets transferred to the stationary particle from the moving particle $\;=\;\frac{\;\frac{1}{2} \times \frac{m}{n} \times v'^2}{\frac{1}{2} m v^2}=\;\frac{v'^2}{n v^2}=\;\frac{1}{n} \left(\;\frac{v'}{v}\right)^2$ $\;=\;\frac{1}{n} \left(\;\frac{2 v}{ \left(1+\;\frac{1}{n}\right) v}\right)^2$ $\;=\;\frac{1}{n} \;\frac{4 v^{2}}{ \left(1+\;\frac{1}{n}\right)^{2} v^{2}}=\;\frac{4}{n \left(\;\frac{n+1}{n}\right)^{2}}=\;\frac{4 n}{(1+n)^{2}}$