CM Frame

CM frame velocity

When applying the Conservation of momentum, there is one interesting frame where problem-solving is easier. We define the Center of Mass frame to be the one where total momentum is zero. How do we find one? First, let’s agree on the fact that if there is a frame S and S’ that moves with a velocity $\text{u}$ relative to S, then velocities in two frames can be related by:

$${\text{v}}_s={\text{v}}_{s^{\prime }}+\text{u}$$

From the definition of the CM frame, we can logically deduce:

$$\begin{array}{c}{\text{P}}^{\prime }=\sum m_i{\text{v}}_i^{\prime }=\sum m_i({\text{v}}_i-\text{u})\\ \sum m_i{\text{v}}_i-\sum m_i\cdot \text{u}=0\\ \text{P}-M\cdot \text{u}=0\\ \text{u}=\frac{\text{P}}{M}\end{array}$$

So, find the total momentum $\text{P}$ and divide it by total mass M.

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Interesting results with CM frame

As an exercise, you can try to find the final velocities in a collision of mass $m$ with velocity $v$ with a stationary object of mass $M$. You’ll quickly recall that it’s better to jump into the CM frame than solve quadratic equations, stemming from Conservation of Energy.

You can also show that Kinetic energy $E_k$ in CM frame can be related to any frame by:

E_k =E_k (CM) +\frac{1}{2}M\cdot u^2 $$(The key point in deriving this is to invoke $\sum m_i\textbf{v}_i=0$)