Question

A neutron makes a head on collision with a stationary deutron . Find the fractional energy loss during the collision.

Answer 1

Fractional energy loss=(change in kinetic energy)÷(initial kinetic energy)
My solution is in attachment. For further simplification put
u1 = (v2-v1) and solve.

Hope it helps!!
Pls mark it as the brainliest.

Answer 2

The fractional loss is $\bold{\frac{8}{9}}$

Solution:

$m_{1} \text { and } m_{2}$ are the masses of neutron and deuteron.

$u_{1} \text { and } u_{2}$ are the initial velocities of neutron and deuteron.

$v_{1} \text { and } v_{2}$ are the final velocities of neutron and deuteron.

Since neutron and deuteron collide with each other and the collision takes place elastically.

Linear momentum conservation is given by:

$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2} \rightarrow(1)$

Kinetic energy conservation is given by:

$\frac{1}{2} m_{1} u_{1}+\frac{1}{2} m_{2} u_{2}=\frac{1}{2} m_{1} v_{1}+\frac{1}{2} m_{2} v_{2} \rightarrow(2)$

On solving, we get,

$v_{1}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) u_{1}+\left(\frac{2 m_{2}}{m_{1}+m_{2}}\right) u_{2} \rightarrow(3)$

$v_{2}=\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right) u_{1}+\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right) u_{2} \rightarrow(4)$

Let $u_{1}=u \text { and } u_{2}=0$

$\therefore v_{1}=\left[\frac{m-2 m}{m+2 m}\right] u=-\left(\frac{m u}{3 m}\right)=-\left(\frac{u}{3}\right)$

$\therefore v_{2}=\left[\frac{2 m}{m+2 m}\right] u=\frac{2 u}{3}$

The gain of kinetic energy by deuteron is equal to the loss of kinetic energy by the neutron.

$\text{Kinetic Energy}=K E=\frac{1}{2} m u^{2}$

$K E_{d}=\frac{1}{2} \times 2 m \times\left(\frac{2 u}{3}\right)^{2}=\frac{1}{2} \times 2 m \times \frac{4 u^{2}}{9}$

$\Rightarrow K E_{d}=\frac{4 m u^{2}}{9}$

$\therefore \text { Fractional loss }=\frac{K E_{d}}{K E}=\frac{\frac{4}{9} m u^{2}}{\frac{1}{2} m u^{2}}$

$\Rightarrow \text { Fractional loss }=\frac{8}{9}$