91影视

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 $p_{1,}{p}_{3,}{p}_4$.

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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,

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

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.

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

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

Add to centre of momentum velocities-

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

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

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

Similarly solve for

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

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