Question

Redo the analysis of the Rutherford experiment, this time using the concept of the centre-of-momentum reference frame. Let m = the mass of the alpha particle and M = the mass of the gold nucleus. Consider the specific case of the alpha particle rebounding straight back. The incoming alpha particle has a momentum p1 , the outgoing alpha particle has a momentum p3 , and the gold nucleus picks up a momentum p4 . (a) Determine the velocity of the centre of momentum of the system. (b) Transform the initial momenta to that frame (by subtracting the centre-of-momentum velocity from the original velocities). (c) Show that if the momenta in the centre-of-momentum frame simply turn around (180◦), with no change in their magnitudes, both momentum and energy conservation are satisfied, whereas no other possibilitysatisfies both conservation principles. (Try drawing some other
momentum diagrams.) (d) After the collision, transform back to
the original reference frame (by adding the center-of-momentum
velocity to the velocities of the particles in the center-of-mass
frame). Although using the center-of-momentum frame may be
conceptually more difficult, the algebra for solving for the final
speeds is much simpler

Short answer

1. The velocity in center of momentum frame is $\frac{{\stackrel{\rightharpoonup }{v}}_1}{1+{\displaystyle \frac{M}{m}}}$.
2. The momentum of both alpha and gold particle is ${\stackrel{\rightharpoonup }{p}}_1=m{\stackrel{\rightharpoonup }{v}}_1\frac{m{\stackrel{\rightharpoonup }{v}}_1}{{\displaystyle \frac{m}{M}}+1}$ and ${\stackrel{\rightharpoonup }{p}}_4=M{\stackrel{\rightharpoonup }{v}}_4=-\frac{M{\stackrel{\rightharpoonup }{v}}_1}{{\displaystyle \frac{M}{M}+1}}$respectively.
3. No, the system will not be conserved because only momentum will be conserved and kinetic energy of the system will not be conserved.

1. The velocities in original frame of reference are role="math" localid="1657865951935" $\frac{-2{\stackrel{\rightharpoonup }{v}}_1}{1+{\displaystyle \frac{M}{m}}}and$$\frac{-2{\stackrel{\rightharpoonup }{v}}_1}{{\displaystyle 1+\frac{m}{M}}}$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The mass of the alpha particle is m.
• The mass of the gold nucleus, M.
• The momentum of incoming alpha particle, outgoing alpha particle and the momentum picked up by the gold nucleus is $p_{1,}{p}_{3,}{p}_4$.

Concept/Significance of Centre of Momentum frame.

The centre of Momentum frame is physics is an imaginary reference frame in which the total momentum adds up to be zero, i.e., it vanishes.

(a) Determination of velocity of centre of momentum of the system.

The expression for velocity in centre of momentum frame for two particles is given by,

$\stackrel{\rightharpoonup }{v}=\frac{m_1{\stackrel{\rightharpoonup }{v}}_1+m_2{\stackrel{\rightharpoonup }{v}}_2}{m_1+m_2}$ ...(i)

Here, m₁and m₂ denotes the masses of the particles andlocalid="1657866356536" ${\stackrel{\rightharpoonup }{v}}_1,{\stackrel{\rightharpoonup }{v}}_2=0$, are the velocities of the articles.

Rewrite equation (i) for the system constituted by alpha and gold particle by substituting the velocities as ${\stackrel{\rightharpoonup }{v}}_1,{\stackrel{\rightharpoonup }{v}}_2=0$, and masses as $m_1=mand{m}_2=M$

$$\begin{gathered} \stackrel{\rightharpoonup }{v}=\frac{m{\stackrel{\rightharpoonup }{v}}_1+M\left(0\right)}{m+M} \\ =\frac{m{\stackrel{\rightharpoonup }{v}}_1}{m+M} \\ =\frac{{\stackrel{\rightharpoonup }{v}}_1}{1+{\displaystyle \frac{M}{m}}} \end{gathered}$$

Thus, the velocity in centre of momentum frame is $=\frac{{\stackrel{\rightharpoonup }{v}}_1}{1+{\displaystyle \frac{M}{m}}}$.

(b) Transformation of initial momenta in the given frame.

For alpha particle, initial velocity is ${\stackrel{\rightharpoonup }{v}}_1$. By subtracting velocity corresponding to centre of momentum frame, the new velocity of alpha particle is given by,

${\stackrel{\rightharpoonup }{v}}_1={\stackrel{\rightharpoonup }{v}}_1-\stackrel{\rightharpoonup }{v}$

localid="1657877957162" $$\begin{gathered} ={\stackrel{\rightharpoonup }{v}}_1-\frac{{\stackrel{\rightharpoonup }{v}}_1}{1+{\displaystyle \frac{M}{m}}} \\ ={\stackrel{\rightharpoonup }{v}}_1\times \frac{{\displaystyle \frac{M}{m}}}{1+{\displaystyle \frac{m}{M}}}\left(\frac{{\displaystyle \frac{{\displaystyle \frac{m}{M}}}{m}}}{M}\right) \\ =\frac{{\stackrel{\rightharpoonup }{v}}_1}{{\displaystyle \frac{m}{M}}+1} \end{gathered}$$

So, the new velocity of alpha particle is $=\frac{{\stackrel{\rightharpoonup }{v}}_1}{{\displaystyle \frac{m}{M}+1}}$

Similarly, for gold particle, initial velocity ${\stackrel{\rightharpoonup }{v}}_4=0$

role="math" localid="1657879518138" $$\begin{gathered} {\stackrel{\rightharpoonup }{v}}_4=0-\frac{{\stackrel{\rightharpoonup }{v}}_1}{1+{\displaystyle \frac{M}{m}}} \\ =-\frac{{\stackrel{\rightharpoonup }{v}}_1}{1+{\displaystyle \frac{M}{m}}} \end{gathered}$$

Thus, momentum of both alpha and gold particle is ${\stackrel{\rightharpoonup }{p}}_1=m\stackrel{\rightharpoonup }{v_1}=\frac{m{\stackrel{\rightharpoonup }{v}}_1}{{\displaystyle \frac{m}{M}}+1}$and${\stackrel{\rightharpoonup }{p}}_1=M\stackrel{\rightharpoonup }{v_1}=\frac{M{\stackrel{\rightharpoonup }{v}}_1}{{\displaystyle \frac{M}{m}+1}}$ respectively.

(c) Evaluation if the momenta in the centre-of-momentum frame simply turn around (180◦) will both momentum and energy conservation are satisfied

The momentum conservation law states that the magnitude as well as the direction of the centre of mass momentum should remain equal. For Energy conservation, the kinetic energy must remain same.

For momentum conservation of the two-particle system, the equation to be satisfied is given by,

${\stackrel{\rightharpoonup }{P}}_3+{\stackrel{\rightharpoonup }{P}}_4=\stackrel{\rightharpoonup }{0}$

If magnitude changes then the total kinetic energy also becomes variable violating the energy conservation law.

No, the system will not be conserved because only momentum will be conserved and kinetic energy of the system will not be conserved.

(d) Transformation back to the original frame of reference.

The velocities are same because the magnitudes of the momenta remain unchanged

$$\begin{gathered} {\stackrel{\rightharpoonup }{v}}_3=-{\stackrel{\rightharpoonup }{v}}_1 \\ {\stackrel{\rightharpoonup }{v}}_4=-{\stackrel{\rightharpoonup }{v}}_2 \end{gathered}$$

Add to centre of momentum velocities-

$$\begin{gathered} {\stackrel{\rightharpoonup }{v}}_3=-{\stackrel{\rightharpoonup }{v}}_3+\stackrel{\rightharpoonup }{v}=-{\stackrel{\rightharpoonup }{v}}_1+\stackrel{\rightharpoonup }{v}=-\left({\stackrel{\rightharpoonup }{v}}_1-\stackrel{\rightharpoonup }{v}\right)+\stackrel{\rightharpoonup }{v} \\ =2\stackrel{\rightharpoonup }{v}-{\stackrel{\rightharpoonup }{v}}_1 \end{gathered}$$

Simplify and put the value of $\stackrel{\rightharpoonup }{v}for{\stackrel{\rightharpoonup }{v}}_3$.

role="math" localid="1657879787684" $$\begin{gathered} {\stackrel{\rightharpoonup }{v}}_3=\frac{2{\stackrel{\rightharpoonup }{v}}_1}{1+{\displaystyle \frac{M}{m}}}-{\stackrel{\rightharpoonup }{v}}_1 \\ =-\frac{2{\stackrel{\rightharpoonup }{v}}_1}{\frac{m}{M}+1} \end{gathered}$$

Similarly solve for

$$\begin{gathered} {\stackrel{\rightharpoonup }{v}}_4+{\stackrel{\rightharpoonup }{v}}_4+\stackrel{\rightharpoonup }{v}\left({\stackrel{\rightharpoonup }{v}}_4={\stackrel{\rightharpoonup }{v}}_2\right) \\ =-\left(\stackrel{\rightharpoonup }{0}-\stackrel{\rightharpoonup }{v}\right)+\stackrel{\rightharpoonup }{v} \\ =2\stackrel{\rightharpoonup }{v} \\ =\frac{2{\stackrel{\rightharpoonup }{v}}_1}{1+\frac{m}{M}} \end{gathered}$$

Thus, the velocities in original frame of reference are and$\frac{-2{\stackrel{\rightharpoonup }{v}}_1}{{\displaystyle \frac{m}{M}}+1}and\frac{-2{\stackrel{\rightharpoonup }{v}}_1}{{\displaystyle 1+\frac{M}{m}}}$ .