Question

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 ${p_c}$. The values of ${p_d}$ and ${p_c}$ are respectively.
A) $\left( {0,0} \right)$
B) $\left( {0,1} \right)$
C) $\left( {.89,.28} \right)$
D) $\left( {.28,.89} \right)$

Answer 1

Hint: Use the law of momentum conservation to get the answer. For an impact happening between object 1 and object 2 out of a disconnected framework, the complete momentum of the two bodies before the crash is equivalent to the total momentum of the two bodies after the crash. That is, the momentum lost by object 1 is equivalent to the momentum picked up by object 2.

Complete step by step answer:
The law of momentum protection can be expressed as follows.
For an impact happening between object 1 and article 2 out of a disconnected framework, the complete momentum of the two bodies before the crash is equivalent to the all out energy of the two bodies after the crash. That is, the momentum lost by object 1 is equivalent to the momentum picked up by object 2.
Also, Newton's third law of movement is normally applied to crashes between two bodies. In an impact between two articles, the two bodies experience forces that are equivalent in greatness and inverse in bearing.
When a neutron collides with deuterium and with carbon then energy is equal to one.
According to conservation of momentum if $m$ is mass of neutron and $u$ is initial velocity of neutron initial velocity of deuterium is zero.
Mass of deuterium is $2m$, $v$ is final velocity of neutron and ${v^,}$ is final velocity of deuterium.
$mu + 0 = mv + 2m{v^,}$
$u = v + 2{v^,}$
According to newton's law of collision
$e = - 1 = \dfrac{{{v^,} - v}}{{0 - u}}$
$ \Rightarrow {v^,} = u + v$
Solving these equations we get, $v = \dfrac{{ - u}}{3}$
Now the fractional loss in kinetic energy of neutron
${P_d} = \dfrac{{\dfrac{1}{2}m{u^2} - \dfrac{1}{2}m{v^2}}}{{\dfrac{1}{2}m{u^2}}}$
Inserting value of v
${P_d} = \dfrac{{\dfrac{1}{2}m{u^2} - \dfrac{1}{2}m{{\left( {\dfrac{{ - u}}{3}} \right)}^2}}}{{\dfrac{1}{2}m{u^2}}}$
After solving
${P_d} = 0.89$
For second case when collision of neutron and carbon and mass of carbon is $12m$
$\Rightarrow$ $mu + 0 = mv + 12m{v^,}$
$\Rightarrow$ $u = v + 12{v^,}$
According to newton's law of collision
$\Rightarrow$ $e = - 1 = \dfrac{{{v^,} - v}}{{0 - u}}$
$ \Rightarrow {v^,} = u + v$
Solving these equations, $v = \dfrac{{ - 11u}}{{13}}$
Now the fractional loss in kinetic energy of neutron
$\Rightarrow$ ${P_d} = \dfrac{{\dfrac{1}{2}m{u^2} - \dfrac{1}{2}m{v^2}}}{{\dfrac{1}{2}m{u^2}}}$
Inserting value of v
$\Rightarrow$ ${P_c} = \dfrac{{\dfrac{1}{2}m{u^2} - \dfrac{1}{2}m{{\left( {\dfrac{{ - 11u}}{{13}}} \right)}^2}}}{{\dfrac{1}{2}m{u^2}}}$
After solving
${P_c} = 0.28$
Option C is correct.

Note: During a collision all the forces acting on the bodies are a part of internal forces and hence their resultant is zero, i.e no external force acts and hence the momentum is always conserved for the system, however if energy is conserved or not depends on the type of collision.