QJEEM19Ph8a78: JEE Main 2019 Physics - Class 11 JEEM19Ph8a78 Solution: An alpha-particle of mass m suffers 1-dimensional elastic…

We have following collision, where mass of $\alpha$ particle $=m$ and mass of nucleus $=M$


Let $\alpha$ particle rebounds with velocity $v_1$ , then Given :

final energy of $\alpha=36 \%$ of initial energy

$\begin{aligned}\Rightarrow \frac{1}{2} m v_1^2 =0.36 \times \frac{1}{2} m v^2 \\\\\Rightarrow v_1 = 0.6 v .......(i)\end{aligned}$

As unknown nucleus gained $64 \%$ of energy of $\alpha$ , we have

$\begin{aligned}& \frac{1}{2} M v_2^2=0.64 \times \frac{1}{2} m v^2 \\& \Rightarrow v_2=\sqrt{\frac{m}{M}} \times 0.8 v .........(ii)\end{aligned}$

From momentum conservation, we have

$m v=M v_2-m v_1$

Substituting values of $v_1$ and $v_2$ from Eqs. (i) and (ii), we have

$\begin{array}{rlrl}m v =M \sqrt{\frac{m}{M}} \times 0.8 v-m \times 0.6 v \\\\\Rightarrow 16 m v =\sqrt{m M} \times 0.8 v \\\\\Rightarrow 2 m =\sqrt{m M} \\\\\Rightarrow 4 m^2 =m M \Rightarrow M=4 m\end{array}$