Chapter 14, Problem 14.29P

(a)

To determine

Show that in center of mass frame, the individual kinetic energies of the two particles are separately conserved in elastic collision.

Answer to Problem 14.29P

The individual kinetic energy of two particles is separately conserved in CM frame.

Explanation of Solution

Consider the elastic scattering of a particle of mass $m_1$ with an initial velocity $m_2$ with mass $m_2$ at rest in laboratory frame. After collision the velocity of particles would be $v_1$ and $v_2$ respectively.

The center of mass velocity of the system is

    $v=\frac{m_1{u}_1}{m_1+m_2}$

The velocity of particle of mass $m_1$ in the CM frame is,

    $\begin{array}{c}u{\text{'}}_1=u_1-v\\ =u_1-\frac{m_1{u}_1}{m_1+m_2}\\ =\frac{m_2{u}_1}{m_1+m_2}\end{array}$

The velocity of particle $m_2$ in the CM frame is,

    $\begin{array}{c}u{\text{'}}_2=u_2-v\\ =0-\frac{m_1{u}_1}{m_1+m_2}\\ =-\frac{m_1{u}_1}{m_1+m_2}\end{array}$

The particles $m_1$ and $m_2$ move with velocities $v_1^{\text{'}}$ and $v_2^{\text{'}}$ respectively after the collision in center of mass frame.

Velocity of mass $m_1$ before and after collision in the center of mass frame will be same as the collision is elastic.

  $v_1^{\text{'}}=u_1^{\text{'}}$

Therefore,

    $v_1^{\text{'}}=\frac{m_2{u}_1}{m_1+m_2}$

Velocity of mass $m_2$ after collision in CM frame is

    $v_2^{\text{'}}=u_2^{\text{'}}$

Therefore,

    $v_2^{\text{'}}=-\frac{m_1{u}_1}{m_1+m_2}$

The kinetic energies of $m_1$ and $m_2$ before and after collision in CM frame is given by

Before collision,

    $\begin{array}{c}{T}_1=\frac{1}{2}{m}_1{\left(u_1^{\text{'}}\right)}^2\\ =\frac{1}{2}{m}_1{\left(\frac{m_2{u}_1}{m_1+m_2}\right)}^2\\ =\frac{1}{2}\frac{m_1{m}_2{}^2}{{\left(m_1+m_2\right)}^2}{u}_1{}^2\end{array}$

    $\begin{array}{c}{T}_2=\frac{1}{2}{m}_2{\left(u_2^1\right)}^2\\ =\frac{1}{2}\frac{m_2{m}_1{}^2}{\left(m_1+m_2\right)}{u}_1{}^2\end{array}$

After collision,

    $\begin{array}{c}{T}_1=\frac{1}{2}{m}_1{\left(v_1^{\text{'}}\right)}^2\\ =\frac{1}{2}{m}_1{\left(\frac{m_2{u}_1}{m_1+m_2}\right)}^2\\ =\frac{1}{2}\frac{m_1{m}_2{}^2}{{\left(m_1+m_2\right)}^2}{u}_1{}^2\end{array}$

    $\begin{array}{c}T{\text{'}}_2=\frac{1}{2}{m}_2{\left(v_2^1\right)}^2\\ =\frac{1}{2}{m}_2{\left(-\frac{m_1{u}_1}{m_1+m_2}\right)}^2\\ =\frac{1}{2}\frac{m_2{m}_1{}^2}{\left(m_1+m_2\right)}{u}_1{}^2\end{array}$

Conclusion:

Thus, the individual kinetic energy of two particles is separately conserved in CM frame.

(b)

To determine

Reason why the result in sub-part (a) is not true in laboratory frame.

Answer to Problem 14.29P

In lab frame, particles acquire kinetic energy only after collision.

Explanation of Solution

In laboratory frame, the kinetic energies of the particles are not conserved separately because the particle initially at rest has no rest kinetic energy before collision, but acquires kinetic energy after collision.

Conclusion:

In lab frame, particles acquire kinetic energy only after collision.

(c)

To determine

Show that the fractional energy lost by the projectile in lab frame is $\frac{\Delta E}{E}=\frac{4\lambda }{{\left(1+\lambda \right)}^2}{\sin }^2\left({\theta }_{\text{cm}}/2\right)$

Answer to Problem 14.29P

The energy lost by the projectile is $\frac{\Delta E}{E}=\frac{4\lambda }{{\left(1+\lambda \right)}^2}{\sin }^2\left({\theta }_{\text{cm}}/2\right)$.

Explanation of Solution

The relation between velocities in the center of mass and the laboratory frames is shown in figure 1.

From the cosine law of triangle, the velocities of the particle are related as

    $\begin{array}{c}{v}_1^2=u_1^{\text{'}2}+v^2+2u_1^{\text{'}}v\cos {\theta }_{cm}\\ v_1=\sqrt{u_1^{\text{'}}{}^2+{\left(\frac{m_1{u}_1}{m_1+m_2}\right)}^2+2\left(u_1^{\text{'}}\right)\left(\frac{m_1{u}_1}{m_1+m_2}\right)\cos {\theta }_{cm}}\\ =\sqrt{{\left(\frac{m_2{u}_1}{m_1+m_2}\right)}^2+{\left(\frac{m_1{u}_1}{m_1+m_2}\right)}^2+2\left(\frac{m_2{u}_1}{m_1+m_2}\right)\left(\frac{m_1{u}_1}{m_1+m_2}\right)\cos {\theta }_{cm}}\\ =\frac{u_1}{\left(m_1+m_2\right)}\sqrt{\left(m_2{}^2+m_1{}^2+2m_1{m}_2\cos {\theta }_{cm}\right)}\end{array}$

The fractional loss in kinetic energy of the particle in the lab frame is

    $\begin{array}{c}\frac{\Delta }{E}=\frac{T_{\text{initial}}-T_{\text{final}}}{T_{\text{initial}}}\\ =\frac{\frac{1}{2}{m}_1{u}_1{}^2-\frac{1}{2}{m}_1{v}_1{}^2}{\frac{1}{2}{m}_1{u}_1{}^2}\\ =\frac{\frac{1}{2}{m}_1{u}_1{}^2-\frac{1}{2}{m}_1{\left(\frac{u_1}{\left(m_1+m_2\right)}\sqrt{\left(m_2{}^2+m_1{}^2+2m_1{m}_2\cos {\theta }_{cm}\right)}\right)}^2}{\frac{1}{2}{m}_1{u}_1{}^2}\\ =\frac{2m_1{m}_2-2m_1{m}_2\cos {\theta }_{cm}}{{\left(m_1+m_2\right)}^2}\end{array}$

Further solve,

    $\frac{\Delta E}{E}=\frac{2\frac{m_1}{m_2}\left(1-\cos {\theta }_{cm}\right)}{{\left(\frac{m_1}{m_2}+1\right)}^2}$

Substitute $\frac{m_1}{m_2}=\lambda$ and $1-\cos {\theta }_{cm}=2{\sin }^2\left(\frac{{\theta }_{cm}}{2}\right)$

    $\begin{array}{c}\frac{\Delta E}{E}=\frac{2\lambda \left(2{\sin }^2\left(\frac{{\theta }_{cm}}{2}\right)\right)}{{\left(\lambda +1\right)}^2}\\ =\frac{4\lambda }{{\left(\lambda +1\right)}^2}{\sin }^2\left(\frac{{\theta }_{cm}}{2}\right)\end{array}$

Hence proved.

Conclusion:

The energy lost by the projectile is $\frac{\Delta E}{E}=\frac{4\lambda }{{\left(1+\lambda \right)}^2}{\sin }^2\left({\theta }_{\text{cm}}/2\right)$.

(d)

To determine

What sort of collision gives the largest energy loss?

Answer to Problem 14.29P

Masses of both the projectile particle and target particle should be approximately equal for the appreciable energy lost in nuclear reactions.

Explanation of Solution

Differentiate $\frac{\Delta E}{E}$ with respect to $\lambda$ and equate to zero

    $\begin{array}{c}\frac{d\left[\frac{\Delta E}{E}=\right]}{d\lambda }=0\\ \frac{d}{d\lambda }\left(\frac{4\lambda }{{\left(1+\lambda \right)}^2}{\sin }^2\left({\theta }_{\text{cm}}/2\right)\right)=0\end{array}$

The value of ${\sin }^2\frac{{\theta }_{cm}}{2}$ is a constant and not equal to zero.

Thus

    $\begin{array}{c}\frac{d}{d\lambda }\left(\frac{4\lambda }{{\left(1+\lambda \right)}^2}\right)=0\\ \frac{\left[{\left(\lambda +1\right)}^2\frac{d}{d\lambda }\left(4\lambda \right)-4\lambda \frac{d}{d\lambda }\left({\left(\lambda +1\right)}^2\right)\right]}{{\left(1+\lambda \right)}^4}=0\end{array}$

Conclusion:

Masses of both the projectile particle and target particle should be approximately equal for the appreciable energy lost in nuclear reactions.