Respuesta :
0.274348
Explanation
Step 1
in a head-on elastic collision between a small projectile, in this case the neutron, and a much more massive target (atom), the projectile will bounce back with essentially the same speed and the massive target will be given a very small velocity,In elastic head-on collision, the energy of the system and total momentum is conserved
it can be represented by the expression
[tex]\begin{gathered} mv_i=mv_{f+MV_f} \\ where \\ m\text{ is the mass of the neuton } \\ v_i=\text{ initial velocity of the neutron} \\ v_f=\text{ final }velocity \\ M\text{ is the mass of the atom} \\ V_{F\text{ }}is\text{ the final velocity of the atom} \end{gathered}[/tex]then, let's set the equations
[tex]\begin{gathered} mv_i=mv_{f+MV_f} \\ \text{subtract mv}_f\text{ in both sides} \\ mv_{i-}\text{mv}_f=mv_{f+MV_F}-\text{mv}_f \\ mv_{i-}\text{mv}_f=MV_F \\ \text{factorize m } \\ m(v_i-v_f)=MV_F \\ \text{divide both sides by m} \\ \frac{m(v_i-v_f)}{m}=\frac{MV_F}{m} \\ (v_i-v_f)=\frac{M}{m}V_F \\ as\text{ the ratio ot the mas iss 12.5} \\ (v_i-v_f)=12.5V_{}\Rightarrow equation\text{ (1)} \end{gathered}[/tex]Step 2
the kinetic energy
[tex]\begin{gathered} \frac{1}{2}mv^2_i=\frac{1}{2}mv^2_f+\frac{1}{2}MV^2 \\ \frac{1}{2}mv^2_i-\frac{1}{2}mv^2_f=\frac{1}{2}MV^2 \\ \frac{1}{2}m(v^2_i-v^2_{_f})=\frac{1}{2}MV^2_{} \\ v^2_i-v^2_{_f}=\frac{MV^2}{m} \\ v^2_i-v^2_{_f}=12.5V^2_{}\Rightarrow equation(2) \end{gathered}[/tex]combinde
a)
[tex]\begin{gathered} (v_i-v_f)=12.5V_{}\Rightarrow equation\text{ (1)} \\ v^2_i-v^2_{_f}=12.5V^2_{}\Rightarrow equation(2) \\ v^2_i-v^2_{_f}=(v_i-v_f)(v_i+v_f) \\ \text{hence} \\ (v_i-v_f)(v_i+v_f) \\ v_1+v_f=V\Rightarrow equation(3) \end{gathered}[/tex]Step 3
solve the equaions
[tex]\begin{gathered} from\text{ (1) and (3)} \\ (v_i-v_f)=12.5V \\ v_f=-12.5V+v_i \\ v_{_f}=V-v_1 \\ -12.5V+v_1=V-v_1 \\ -13.5V=-2v_1 \\ \text{divide both sides by -2} \\ \frac{-13.5V}{-2}=\frac{-2v_1}{-2} \\ 6.75V=v_i \\ v_i=6.75V \end{gathered}[/tex]Calculate the fraction of the neutron's kinetic energy transferred to the atomic nucleus.
Let
[tex]\begin{gathered} \frac{M}{m}=12.5 \\ \text{fraction transfered to the atom} \\ \frac{\frac{1}{2}MV^2}{\frac{1}{2}mv^2_i} \\ \frac{\frac{1}{2}MV^2}{\frac{1}{2}mv^2_i}=12.5\frac{V^2}{(6.75V)^2}=\frac{12.5}{45.5625}=0.274348 \\ \end{gathered}[/tex]therefore, the answer is 0.274348
I hope this helps you
