Question

A neutron moving with a velocity $v$, collides elastically with an atom of mass number $A$. The collision is head on. If the initial kinetic energy of a neutron is $e$, then the kinetic energy of neutron after the collision is :

• A. ${\left(\frac{1+A}{1-A}\right)}^{2}E$
• B. ${\left(\frac{1-A}{1+A}\right)}^{2}E$
• C. $\left(\frac{1+A}{1-A}\right)E$
• D. $\left(\frac{1-A}{1+A}\right)E$

Answer 1

Answer: (B)

In elastic collision, kinetic energy and momentum both are conserved.
Let $v_1$ be velocity after collision
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2$
Given, $u_1=v,u_2=0,m_1=1,m_2=A$
$\therefore v_1=\left(\frac{1-A}{1+A}\right)v$ Initial kinetic energy of neutron $E=\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}\times 1\times {\left(\frac{1-A}{1+A}\right)}^2$
$\therefore \frac{E}{E}={\left(\frac{1-A}{1+A}\right)}^2$
$\Rightarrow E={\left(\frac{1-A}{1+A}\right)}^{2}E$
Final KE of neutron
$E=\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}\times 1\times {\left(\frac{1-A}{1+A}\right)}^2$
$\therefore \frac{E}{E}={\left(\frac{1-A}{1+A}\right)}^2$
$\Rightarrow E={\left(\frac{1-A}{1+A}\right)}^{2}E$