JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 8)
An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is
2m
4m
1.5m
3.5m
Explanation
We have following collision, where mass of $\alpha$ particle $=m$ and mass of nucleus $=M$
_12th_January_Evening_Slot_en_8_1.png)
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
$$ \begin{array}{rlrl} m v =M \sqrt{\frac{m}{M}} \times 0.8 v-m \times 0.6 v \\\\ \Rightarrow 16 m v =\sqrt{m M} \times 0.8 v \\\\ \Rightarrow 2 m =\sqrt{m M} \\\\ \Rightarrow 4 m^2 =m M \Rightarrow M=4 m \end{array} $$
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
$$ \begin{array}{rlrl} m v =M \sqrt{\frac{m}{M}} \times 0.8 v-m \times 0.6 v \\\\ \Rightarrow 16 m v =\sqrt{m M} \times 0.8 v \\\\ \Rightarrow 2 m =\sqrt{m M} \\\\ \Rightarrow 4 m^2 =m M \Rightarrow M=4 m \end{array} $$
Comments (0)
