Mechanics

Two Body Collisions in the Centre of Mass Frame

It is often more simple to solve problems of two body collisions by considering the frame of reference of the centre of mass (C.O.M) of the system. To do so we first establish various vector quantities to describe the position of the two bodies and the centre of mass in both the C.O.M frame and the “lab frame” – where some arbitrary point is the origin:

As shown in the diagram the position vectors of both the bodies and the centre of mass of the system with respect to the origin in the lab frame are \mathbf{r_{1}}, \mathbf{r_{2}}, \mathbf{R} respectively. Variables (thus positions, velocities etc.) in the C.O.M frame are denoted with a star. The position of the bodies in the C.O.M frame are thus \mathbf{r_{1}^{*}}, \mathbf{r_{2}^{*}} respectively. Finally \mathbf{r} represents the vector between the two bodies.

We now need to show that things we know are true in the lab frame, are also true in the C.O.M frame!

-Kinetic Energy in the C.O.M frame

From the definition of the centre of mass:

M\mathbf{R}= m_{1}\mathbf{r_{1}}+m_{2}\mathbf{r_{2}}

We can take the derivative of the equation:

M\dot{\mathbf{R}}= m_{1}\dot{\mathbf{r_{1}}}+m_{2}\dot{\mathbf{r_{2}}}= \mathbf{p_{1}}+\mathbf{p_{2}}

If we consider a situation where there are no external forces, the centre of mass will not accelerate (\ddot{\mathbf{R}}=0), meaning that the velocity of the centre of mass must be constant. Let’s now consider an elastic collision:

E_{k_{lab}}= {E_{k_{lab}}}' = \frac{1}{2}m_{1}\dot{\mathbf{r_{1}}} +\frac{1}{2}m_{2}\dot{\mathbf{r_{2}}}

We can eliminate one of the position vectors to find each body’s motion in terms of the motion of the centre of mass and the relative motion of the two bodies:

 \mathbf{r} =\mathbf{r_{1}} -\mathbf{r_{2}}  \; \therefore \mathbf{r_{1}} =\mathbf{r} + \mathbf{r_{2}}

So we can substitute into the centre of mass equation:

M\mathbf{R}= m_{1}(\mathbf{r} + \mathbf{r_{2}})+m_{2}\mathbf{r_{2}}

Collecting the terms for the position of body 2:

M\mathbf{R}= \mathbf{r_{2}}(m_{1} + m_{2})+m_{1}\mathbf{r}

Remembering that M represents the mass of the whole system, it follows that:

\mathbf{r_{2}}=\mathbf{R}- \frac{m_{1}}{M}\mathbf{r}

If we had chosen to eliminate \mathbr{r_{2}} before we would have achieved the following result similarly:

\mathbf{r_{1}}=\mathbf{R}+ \frac{m_{2}}{M}\mathbf{r}

We now have the position of both bodies in terms of the position of the centre of mass and the vector between them, thus we can substitute into the kinetic energy of the lab frame:

E_{k_{lab}}= \frac{1}{2}m_{1}\bigg (\dot{\mathbf{R}}+\frac{m_{1}}{M}\dot{\mathbf{r}} \bigg)^{2}  + \frac{1}{2}m_{2}\bigg (\dot{\mathbf{R}}-\frac{m_{2}}{M}\dot{\mathbf{r}} \bigg)^{2}

E_{k_{lab}}= \frac{1}{2}m_{1}\bigg (\mathbf{\dot{R}^{2}} + \frac{2m_{2}}{M}\dot{\mathbf{R}}\dot{\mathbf{r}} + \frac{m_{2}^{2}}{M^{2}}\mathbf{\dot{r}^{2}}\bigg)+\frac{1}{2}m_{2}\bigg (\mathbf{\dot{R}^{2}} - \frac{2m_{1}}{M}\dot{\mathbf{R}}\dot{\mathbf{r}} + \frac{m_{1}^{2}}{M^{2}}\mathbf{\dot{r}^{2}}\bigg)

Terms in \dot{\mathbf{R}}\dot{\mathbf{r}} cancel and then like terms can be collected with a bit of algebraic manipulation:

E_{k_{lab}}=  \frac{1}{2}(m_{1}+m_{2})\mathbf{\dot{R}^{2}} + \frac{1}{2}(\frac{m_{1}m_{2}}{m_{1}+m_{2}})\mathbf{\dot{r}^{2}}

We recognise that \frac{m_{1}m_{2}}{m_{1}+m_{2}} is the reduced mass of the system (\mu) and m_{1}+m_{2} is the mass of the system, thus:

E_{k_{lab}}=  \frac{1}{2}M\mathbf{\dot{R}^{2}}+\frac{1}{2}\mu\mathbf{\dot{r}^{2}}

This tells us that the kinetic energy in the lab frame is equal to the kinetic energy of the C.O.M added to the reduced mass kinetic energy of separation. In the centre of mass frame the C.O.M doesn’t move (of course), so in the C.O.M frame \dot{\mathbf{R}}= 0, so:

E_{k}^{*}= \frac{1}{2}\mu\mathbf{\dot{r}^{2}}

We now have everything we need to check whether in an elastic collision KE is conserved in the C.O.M frame. We know that because we assumed there are no external forces in the collision, \dot{\mathbf{R}} is constant, we can therefore equate the kinetic energy in the lab frame again:

{E_{k_{lab}}}' = E_{k_{lab}}=  \frac{1}{2}M\mathbf{\dot{R}^{2}}+\frac{1}{2}\mu\mathbf{\dot{{r}'}^{2}}

We proved that the kinetic energy in the centre of mass frame was equal to \frac{1}{2}\mu\mathbf{\dot{r}^{2}}, so \frac{1}{2}\mu\mathbf{\dot{{r}'}^{2}} corresponds to E_{k}^{*'}, thus:

 \frac{1}{2}M\mathbf{\dot{R}^{2}} + E_{k}^{*}= \frac{1}{2}M\mathbf{\dot{R}^{2}}+ E_{k}^{*'}

E_{k}^{*}=E_{k}^{*'}

Thus kinetic energy is conserved in the C.O.M frame in an elastic collision.

-Relation between C.O.M frame and Lab frame Variables

Lets consider an elastic collision where one of the bodies is stationary to start with in the lab frame. This, clearly, means that \mathbf{p_{2}}=0. We can therefore say that:

\mathbf{p_{2}}= m_{2}\mathbf{\dot{r}_{2}}=0

Using the relations derived before to prove the conservation of momentum in the lab frame (\mathbf{r_{2}}=\mathbf{R}- \frac{m_{1}}{M}\mathbf{\dot{r}}) we can substitute for \mathbf{\dot{r}_{2}}:

\mathbf{p_{2}}= m_{2}\mathbf{\dot{R}}- \frac{m_{1}m_{2}}{M}\mathbf{\dot{r}}=m_{2}\mathbf{\dot{R}}-\mu\mathbf{\dot{r}}=0

Recalling that \mu\mathbf{\dot{r}} is simply equal to \mathbf{p^{*}}, the following relation can be found- remember that this is only valid for \mathbf{p_{2}}=0 !:

\mathbf{\dot{R}}= \frac{\mathbf{p^{*}}}{m_{2}}

Similarly:

\mathbf{p_{1}}= m_{1}}\mathbf{\dot{R}} +\mathbf{p^{*}}=\frac{m_{1}}{m_{2}}\mathbf{p^{*}}+\mathbf{p^{*}}

And after the collision:

\mathbf{q_{1}}= m_{1}}\mathbf{\dot{R}} +\mathbf{q^{*}}=\frac{m_{1}}{m_{2}}\mathbf{p^{*}}+\mathbf{q^{*}}

Using the equations derived above, and the fact that \mathbf{p_{1}}=\mathbf{q_{1}}+\mathbf{q_2}} (for momentum to be conserved), the following “vector momentum diagram” can be drawn, showing the relations between the vector quantities being investigated. This will allow for geometry to be used to find the relations between C.O.M variables and lab variables:

As we have proved that momentum is conserved in the C.O.M frame, thus \mathbf{p^{*}}=\mathbf{q^{*}}, we thus have an isosceles triangle on the right of the diagram. The following is thus simple to infer:

\theta^{*} + 2\phi = \pi

\therefore \phi = \frac{\pi -\theta^{*}}{2}

 

We now have a relationship between the scatter angle in the C.O.M frame and the angle that body 2 scatters in the lab frame. The last thing to do is to the relate the angle at which body 1 scatters in the lab frame to the C.O.M scatter angle. It is simple to do this if we construct a right angle triangle within the diagram above as so:

Using the known ratio for the tangent of the angle \theta:

    $$ tan(\theta)= \frac{\mathbf{q^{*}}sin(\theta^{*} )}{\frac{m_{1}}{m_{2}}\mathbf{p^{*}}+\mathbf{q^{*}}cos(\theta^{*} )} $$

Remembering that \mathbf{p^{*}}=\mathbf{q^{*}}:

    $$ tan(\theta)= \frac{sin(\theta^{*} )}{\frac{m_{1}}{m_{2}}+cos(\theta^{*} )} $$