Question

Consider two particles that experience a mutual force but no external forces. The classical equation of motion for particle 1 is ${\dot{\mathbf{v}}}_{\mathbf{1}}\mathbf{=}{\mathbf{F}}_{\mathbf{2}\mathbf{on}\mathbf{1}}\mathbf{/}{\mathbf{m}}_{\mathbf{1}}$, and for particle 2 is ${\dot{\mathbf{v}}}_{\mathbf{2}}\mathbf{=}{\mathbf{F}}_{\mathbf{1}\mathbf{on}\mathbf{2}}\mathbf{/}{\mathbf{m}}_{\mathbf{2}}$, where the dot means a time derivative. Show that these are equivalent to ${\dot{\mathbf{v}}}_{\mathbf{cm}}\mathbf{=}\mathbf{constant}$, and ${\dot{\mathbf{v}}}_{\mathbf{rel}}\mathbf{=}{\mathbf{F}}_{\mathbf{Mutual}}\mathbf{/}\mathbf{\mu }$ .Where, ${\dot{\mathbf{v}}}_{\mathbf{cm}}\mathbf{=}\left(m_1{\dot{v}}_1+m_2{\dot{v}}_2\right)\mathbf{/}\left(m_1{m}_2\right)\mathbf{,}{\mathbf{F}}_{\mathbf{Mutual}}\mathbf{=}\mathbf{-}{\mathbf{F}}_{\mathbf{ion}\mathbf{2}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\mu }\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{\left(m_1+m_2\right)}$.

In other words, the motion can be analyzed into two pieces the center of mass motion, at constant velocity and the relative motion, but in terms of a one-particle equation where that particle experiences the mutual force and has the “reduced mass” $\mathbf{\mu }$.

Short answer

The given equations are equivalent to,${\dot{v}}_{cm}=0$or ${\dot{\mathrm{v}}}_{\mathrm{cm}}=\mathrm{constant}$.

Also,

${\dot{\mathrm{v}}}_{\mathrm{rel}}=\left(\frac{1}{\mathrm{\mu }}\right){\mathrm{F}}_{\mathrm{Mutual}}$

Where, $\mathrm{\mu }=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}$

Step-by-step solution

A concept:

Centre of massis the point on a structure which characterizes the motion of the object if the whole mass of that object is concentrated on that point.

${\mathrm{m}}_1{\dot{\mathrm{v}}}_1={\mathrm{F}}_{2\mathrm{on}1}$ _…..(1)

And

${\mathrm{m}}_2{\dot{\mathrm{v}}}_2={\mathrm{F}}_{1\mathrm{on}2}$ ….. (2)

Showing v˙cm=0 or v˙cm=constant :

As you know,

$-{\mathrm{F}}_{2\mathrm{on}1}={\mathrm{F}}_{1\mathrm{on}2}$ ….. (3)

Hence, by adding equations (1) and (2) and then using (3), you get,

$m_1\stackrel{\boxed{?}}{v_1}+m_2\stackrel{\boxed{?}}{v_2}=0$ ….. (4)

If you divide eq. (4) by ${\mathrm{m}}_1+{\mathrm{m}}_2$, you get,

${\dot{\mathrm{v}}}_{\mathrm{cm}}=0$or ${\dot{\mathrm{v}}}_{\mathrm{cm}}=\mathrm{constant}$

Finding vrel:

Relative motion of an object is its motion with respect to any other moving or stationary object.

${\dot{\mathbf{v}}}_{\mathbf{rel}}\mathbf{=}\dot{{\mathbf{v}}_{\mathbf{2}}}\mathbf{-}{\dot{\mathbf{v}}}_{\mathbf{1}}$

If you rearrange and subtract equation (1) from equation (2), you get,

$$\begin{gathered} {\dot{v}}_2=-{\dot{v}}_1=\frac{F_{1on2}}{m_2}-\frac{F_{2on1}}{m_1} \\ =\left(\frac{1}{m_2}+\frac{1}{m_1}\right)F_{1on2} \\ {\dot{v}}_{rel}=\frac{\left(m_1+m_2\right)}{m_1{m}_2}{F}_{Mutual} \\ =\left(\frac{1}{\mu }\right)F_{Mutual} \end{gathered}$$

Where, $\mathrm{\mu }=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}$