It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is P_d; while for its similar collision with carbon nucleus at rest, fractional loss of energy is P_c.  Then ratio of P_d to P_c will be

Let initial speed of neutron is $v_0$ and kinetic energy is $K$

For 1st collision

Using momentum conservation principle,

$$\begin{gathered} m{v}_0=m{v}_1+2m{v}_2 \\ {v}_1+2v_2=v_0 ...\left(1\right) \end{gathered}$$

As $e=1$ so 

$v_2-v_1=v_0 ...\left(2\right)$

From equation $\left(1\right) and \left(2\right)$ we get

$v_2=\frac{2v_0}{3}, v_1=-\frac{v_0}{3}$

Fraction loss of energy is $p_d=\frac{{\displaystyle \frac{1}{2}}m{v}_0^2-{\displaystyle \frac{1}{2}}m{\left(-{\displaystyle \frac{v_0}{3}}\right)}^2}{{\displaystyle \frac{1}{2}}m{v}_0^2}$

$p_d=\frac{8}{9}=0.89$

For 2nd collision

Using momentum conservation principle,

$$\begin{gathered} m{v}_0=m{v}_1+12m{v}_2 \\ {v}_1+12v_2=v_0 ...\left(1\right) \end{gathered}$$

As $e=1$ so 

$v_2-v_1=v_0 ...\left(2\right)$

From equation $\left(1\right) and \left(2\right)$ we get

$v_2=\frac{2v_0}{13}, v_1=\frac{-11v_0}{13}$

Fraction loss of energy is $p_c=\frac{{\displaystyle \frac{1}{2}}m{v}_0^2-{\displaystyle \frac{1}{2}}m{\left(-{\displaystyle \frac{11v_0}{13}}\right)}^2}{{\displaystyle \frac{1}{2}}m{v}_0^2}=\frac{48}{169}=0.28$

The value of $p_{d }and p_c$ are respectively $0.89 and 0.28$. 

Ratio $\frac{p_{d } }{p_c}=3.17$