Question

A neutron makes a head-on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is

• A. 16/81
• B. 8/9
• C. 8/2
• D. 2/3

Answer 1

Answer: (B)

Let the two balls of mass $m_1$ and $m_2$ collide each other elastically with velocities $u_1$ and $u_2$. Their velocities become $v_1$ and $v_2$ after the collision.
Applying conservation of linear momentum, we get
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$ ......(i)
Also from conservation of kinetic energy
$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$ .....(ii)
Solving Eqs. (i) and (ii), we get
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2$.....(iii)
and $v_2=\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2+\left(\frac{2m_1}{m_1+m_2}\right)u_1$......(iv)
On taking approximate value the mass of deuteron is twice the mass of neutron
Given $u_1=u,u_2=0,w,=m,m_2=2m$
Velocity of neutron $v_1=\left(\frac{m-2m}{m+2m}\right)u=-\frac{u}{3}$
Velocity of deuteron $v_2=\frac{2mu}{m+2m}=\frac{2}{3}u$
Fractional energy loss$=\frac{\frac{1}{2}m{u}^2-\frac{1}{2}m{\left(-\frac{u}{3}\right)}^2}{\frac{1}{2}m{u}^2}$
$=1-\frac{1}{9}=\frac{8}{9}$