Chapter 40, Problem 70P

(a)

To determine

To show: The speed of the nucleus in the centre of mass frame is $\frac{m{v}_{\text{L}}}{\left(m+M\right)}$ .

Explanation of Solution

Introduction:

The elastic head-on collision is a type of collision in which the two bodies collide in such a way that target is at rest and its velocity is increased by twice the initial velocity after the collision.

All the elastic collisions follow the conservation laws like Conservation law of momentum and Conservation law of energy.

Write the expression for the conservation of momentum to the elastic head-on collision.

  $\left(m+M\right)V=m{v}_{\text{L}}$

Here, $m$ is the mass of the neutron, $v_{\text{L}}$ is the velocity of the neutron moving in the laboratory frame of reference, $M$ is the mass of the stationary nucleus and $V$ is the velocity of the nucleus after collision.

Simplify the above expression for the velocity of the nucleus after collision.

  $V=\left(\frac{m}{m+M}\right)v_{\text{L}}$

Conclusion:

Thus, the speed of the nucleus in the centre of mass frame is $\frac{m{v}_{\text{L}}}{\left(m+M\right)}$ .

(b)

To determine

The speed of the stationary nucleus in the centre of mass framebefore and after the collision.

Explanation of Solution

Introduction:

The elastic head-on collision is a type of collision in which the two bodies collide in such a way that target is at rest and its velocity increased by twice the initial velocity after the collision.

All the elastic collisions follow the conservation laws like Conservation law of momentum and Conservation law of energy and even all the elastic collisions are symmetric for the centre of mass frame of the system.

Since, the neutron with a speed makes an elastic head-on collision with a stationary nucleus, in laboratory frame of reference.

Therefore,

Before collision: The initial velocity of the centre of mass is equal to the velocity of the neutron.

Write the expression for the initial velocity of the nucleus with respect to centre of mass before the collision.

  $V_{\text{Mi}}=V$

Here, $V_{\text{Mi}}$ is initial velocity of the nucleus with respect to centre of mass before the collision and $V$ is the velocity of the nucleus, before collision.

After collision: The final velocity of the nucleus with respect to centre of mass after the collision

is equal to the velocity of the nucleus.

Write the expression for the final velocity of the nucleus with respect to centre of mass after the collision.

  $V_{\text{Mf}}=-V$

Here, $V_{\text{Mf}}$ is final velocity of the nucleus with respect to centre of mass after the collision and $-V$ is the velocity of the neutron after collision.

Conclusion:

Thus, the speed of the stationary nucleus with respect to the centre of mass framebefore and after the collision is $V$ and $-V$ .

(c)

To determine

The speed of the stationary nucleus in the laboratory frame after the collision.

Explanation of Solution

Introduction:

In laboratory reference frame, the elastic collision of two bodies takes place in such a way thatthe bodies continue to move with their same respective velocities after collision, but in opposite direction.

The law conservation of momentum of two particles in elastic collision states the momentum of the particles before collision must be equal to the momentum of the particles after collision.

Write the expression for the conservation of momentum in laboratory frame of reference.

  $m{v}_{\text{L}}=m{v}_{\text{f}}+M{V}_{\text{Mf}}$ …… (1)

Here, $m$ is the mass of the neutron, $v_{\text{L}}$ is the velocity of the neutron moving in the laboratory frame of reference, $M$ is the mass of the stationary nucleus, $v_{\text{f}}$ is the final velocity of the neutron and $V_{\text{Mf}}$ is the final velocity of the nucleus with respect to centre of mass after the collision.

The law conservation of energy of the two particles in elastic collision states the kinetic energy of the particles before collision must be equal to the kinetic energy of the particles after collision.

Write the expression for the initial and the final velocities from the law conservation of kinetic energy in laboratory frame of reference.

  $V_{\text{Mf}}-v_{\text{f}}=-\left(V_{\text{Mi}}-v_{\text{L}}\right)$

Here, $V_{\text{Mi}}$ is the initial velocity of the nucleus with respect to centre of mass before the collision.

Since, initially the nucleus is stationary so its initial velocity of the nucleus with respect to centre of mass before the collision is $0$ .

Substitute $0$ for $V_{\text{Mi}}$ in the above expression and simplify for the final velocity of the neutron.

  $v_{\text{f}}=V_{\text{Mf}}-v_{\text{L}}$

Substitute $V_{\text{Mf}}-v_{\text{L}}$ for $v_{\text{f}}$ in equation (1) and simplify for the final velocity of the nucleus with respect to centre of mass after the collision.

  $\begin{array}{c}m{v}_{\text{L}}=m\left(V_{\text{Mf}}-v_{\text{L}}\right)+M{V}_{\text{Mf}}\\ m{v}_{\text{L}}=m{V}_{\text{Mf}}-m{v}_{\text{L}}+M{V}_{\text{Mf}}\\ 2m{v}_{\text{L}}=\left(m+M\right)V_{\text{Mf}}\\ V_{\text{Mf}}=\left(\frac{2m}{m+M}\right)v_{\text{L}}\end{array}$

Conclusion:

Thus, the speed of the stationary nucleus in the laboratory frame after the collision is $\left(\frac{2m}{m+M}\right)v_{\text{L}}$ .

(d)

To determine

The energy of the nucleus after the collision in the laboratory frame is $\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{L}}\right)}^2\right)$ .

Explanation of Solution

Introduction:

In laboratory reference frame, the elastic collision of two bodies takes place in such a way that the bodies continue to move with their same respective velocities after collision, but in opposite direction.

The law conservation of momentum of two particles in elastic collision states the momentum of the particles before collision must be equal to the momentum of the particles after collision.

Write the expression for the conservation of momentum in laboratory frame of reference.

  $m{v}_{\text{L}}=m{v}_{\text{f}}+M{V}_{\text{Mf}}$ …… (1)

Here, $m$ is the mass of the neutron, $v_{\text{L}}$ is the velocity of the neutron moving in the laboratory frame of reference, $M$ is the mass of the stationary nucleus, $v_{\text{f}}$ is the final velocity of the neutron and $V_{\text{Mf}}$ is the final velocity of the nucleus with respect to centre of mass after the collision.

The law conservation of energy of the two particles in elastic collision states the kinetic energy of the particles before collision must be equal to the kinetic energy of the particles after collision.

Write the expression for the initial and the final velocities from the law conservation of kinetic energy in laboratory frame of reference.

  $V_{\text{Mf}}-v_{\text{f}}=-\left(V_{\text{Mi}}-v_{\text{L}}\right)$

Here, $V_{\text{Mi}}$ is the initial velocity of the nucleus with respect to centre of mass before the collision.

Since, initially the nucleus is stationary so its initial velocity of the nucleus with respect to centre of mass before the collision is $0$ .

Substitute $0$ for $V_{\text{Mi}}$ in the above expression and simplify for the final velocity of the neutron.

  $v_{\text{f}}=V_{\text{Mf}}-v_{\text{L}}$

Substitute $V_{\text{Mf}}-v_{\text{L}}$ for $v_{\text{f}}$ in equation (1) and simplify for the final velocity of the nucleus with respect to centre of mass after the collision.

  $\begin{array}{c}m{v}_{\text{L}}=m\left(V_{\text{Mf}}-v_{\text{L}}\right)+M{V}_{\text{Mf}}\\ m{v}_{\text{L}}=m{V}_{\text{Mf}}-m{v}_{\text{L}}+M{V}_{\text{Mf}}\\ 2m{v}_{\text{L}}=\left(m+M\right)V_{\text{Mf}}\\ V_{\text{Mf}}=\left(\frac{2m}{m+M}\right)v_{\text{L}}\end{array}$

Write the expression for the kinetic energy of the nucleus after the collision in the laboratory frame.

  $K_{\text{M}}=\frac{1}{2}M{\left(V_{\text{Mf}}\right)}^2$

Here, $K_{\text{M}}$ is the kinetic energy of the nucleus after the collision in the laboratory frame.

Substitute $\left(\frac{2m}{m+M}\right)v_{\text{L}}$ for $V_{\text{Mf}}$ in the above expression.

  $\begin{array}{c}{K}_{\text{M}}=\frac{1}{2}M{\left(\frac{2m}{m+M}{v}_{\text{L}}\right)}^2\\ =\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{L}}\right)}^2\right)\end{array}$

Conclusion:

Thus, the energy of the nucleus after the collision in the laboratory frame is $\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{L}}\right)}^2\right)$ .

(e)

To determine

To show: The fraction of energy lost by the neutron in the head-on elastic collision of neutron and the stationary nucleus is $\frac{4\left(m/M\right)}{{\left(1+m/M\right)}^2}$ .

Explanation of Solution

The law conservation of momentum of two particles in elastic collision states the momentum of the particles before collision must be equal to the momentum of the particles after collision.

Write the expression for the conservation of momentum in laboratory frame of reference.

  $m{v}_{\text{L}}=m{v}_{\text{f}}+M{V}_{\text{Mf}}$ …… (1)

Here, $m$ is the mass of the neutron, $v_{\text{L}}$ is the velocity of the neutron moving in the laboratory frame of reference, $M$ is the mass of the stationary nucleus, $v_{\text{f}}$ is the final velocity of the neutron and $V_{\text{Mf}}$ is the final velocity of the nucleus with respect to centre of mass after the collision.

The law conservation of energy of the two particles in elastic collision states the kinetic energy of the particles before collision must be equal to the kinetic energy of the particles after collision.

Write the expression for the initial and the final velocities from the law conservation of kinetic energy in laboratory frame of reference.

  $V_{\text{Mf}}-v_{\text{f}}=-\left(V_{\text{Mi}}-v_{\text{L}}\right)$

Here, $V_{\text{Mi}}$ is the initial velocity of the nucleus with respect to centre of mass before the collision.

Since, initially the nucleus is stationary so it’s initial velocity of the nucleus with respect to centre of mass before the collision is $0$ .

Substitute $0$ for $V_{\text{Mi}}$ in the above expression and simplify for the final velocity of the neutron.

  $v_{\text{f}}=V_{\text{Mf}}-v_{\text{L}}$

Substitute $V_{\text{Mf}}-v_{\text{L}}$ for $v_{\text{f}}$ in equation (1) and simplify for the final velocity of the nucleus with respect to centre of mass after the collision.

  $\begin{array}{c}m{v}_{\text{L}}=m\left(V_{\text{Mf}}-v_{\text{L}}\right)+M{V}_{\text{Mf}}\\ m{v}_{\text{L}}=m{V}_{\text{Mf}}-m{v}_{\text{L}}+M{V}_{\text{Mf}}\\ 2m{v}_{\text{L}}=\left(m+M\right)V_{\text{Mf}}\\ V_{\text{Mf}}=\left(\frac{2m}{m+M}\right)v_{\text{L}}\end{array}$

Write the expression for the kinetic energy of the nucleus after the collision in the laboratory frame.

  $K_{\text{M}}=\frac{1}{2}M{\left(V_{\text{Mf}}\right)}^2$

Here, $K_{\text{M}}$ is the kinetic energy of the nucleus after the collision in the laboratory frame.

Substitute $\left(\frac{2m}{m+M}\right)v_{\text{L}}$ for $V_{\text{Mf}}$ in the above expression.

  $\begin{array}{c}{K}_{\text{M}}=\frac{1}{2}M{\left(\frac{2m}{m+M}{v}_{\text{L}}\right)}^2\\ =\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{L}}\right)}^2\right)\end{array}$

Write the expression for the kinetic energy of the system of the particles before collision.

  $K_{\text{mi}}=\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{L}}\right)}^2\right)$

Here, $K_{\text{mi}}$ is the kinetic energy of the system of the particles before collision.

Write the expression for the kinetic energy of the system of the particles after collision.

  $K_{\text{mf}}=\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{f}}\right)}^2\right)$

Here, $K_{\text{mf}}$ is the kinetic energy of the system of the particles after collision

Write the expression for the energy lost by the neutron in the elastic collision.

  $\frac{-\Delta E}{E}=\frac{-\left(K_{\text{mf}}-K_{\text{mi}}\right)}{\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{L}}\right)}^2\right)}$

Substitute $\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{L}}\right)}^2\right)$ for $K_{\text{mi}}$ and $\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{f}}\right)}^2\right)$ for $K_{\text{mf}}$ in the above expression.

  $\begin{array}{c}\frac{-\Delta E}{E}=\frac{-\left(\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{f}}\right)}^2\right)-\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{L}}\right)}^2\right)\right)}{\frac{4mM}{\left(m+M\right)}\left(\frac{1}{2}m{\left(v_{\text{L}}\right)}^2\right)}\\ =\frac{\frac{1}{2}m\left({\left(v_{\text{f}}\right)}^2-{\left(v_{\text{L}}\right)}^2\right)}{\frac{1}{2}m{\left(v_{\text{L}}\right)}^2}\\ =\frac{\left({\left(v_{\text{f}}\right)}^2-{\left(v_{\text{L}}\right)}^2\right)}{{\left(v_{\text{L}}\right)}^2}\end{array}$

Substitute $V_{\text{Mf}}-v_{\text{L}}$ for $v_{\text{f}}$ in the above expression.

  $\frac{-\Delta E}{E}=\frac{\left({\left(V_{\text{Mf}}-v_{\text{L}}\right)}^2-{\left(v_{\text{L}}\right)}^2\right)}{{\left(v_{\text{L}}\right)}^2}$

Substitute $\left(\frac{2m}{m+M}\right)v_{\text{L}}$ for $V_{\text{Mf}}$ in the above expression.

  $\frac{-\Delta E}{E}=\frac{\left({\left(\left(\frac{2m}{m+M}\right)v_{\text{L}}-v_{\text{L}}\right)}^2-{\left(v_{\text{L}}\right)}^2\right)}{{\left(v_{\text{L}}\right)}^2}$

Simplify the above expression as:

  $\begin{array}{c}\frac{-\Delta E}{E}=\frac{-{\left(v_{\text{L}}\right)}^2\left({\left(\left(\frac{2m}{m+M}\right)-1\right)}^2-1\right)}{{\left(v_{\text{L}}\right)}^2}\\ =-{\left(\left(\frac{2m}{m+M}\right)-1\right)}^2-1\\ =-{\left(\frac{2m-m-M}{m+M}\right)}^2-1\end{array}$

Simplify further the above expression.

  $\begin{array}{c}\frac{-\Delta E}{E}=\frac{4mM}{{\left(m+M\right)}^2}\\ =\frac{4\left(\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$M$}\right.\right)}{{\left(1+\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$M$}\right.\right)}^2}\end{array}$

Conclusion:

Thus, the fraction of energy lost by the neutron in the head-on elastic collision of neutron and the stationary nucleus is $\frac{4\left(m/M\right)}{{\left(1+m/M\right)}^2}$ .