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. 44782
• C. 44775
• D. 44595

Answer 1

Answer: (B)

44782

Answer 2

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 $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ......(i) Also from conservation of kinetic energy $\frac{1}{2}m_1u^2_1+\frac{1}{2}m_2u^2_2=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$ .....(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$=\frac{\frac{1}{2}mu^2-\frac{1}{2}m\left(-\frac{u}{3}\right)^2}{\frac{1}{2}mu^2}$ $=1-\frac{1}{9}=\frac{8}{9}$