On Collision

20 Dec, 2023

Today when teaching mechanics, I suddenly realise that many students aren't aware of the elegancy of describing a collision in the centre of mass frame.

Centre of Mass Frame

Energy, kinetic and potential, is a frame-dependent quantity. It is overlooked by most high-school mechanics textbook that

E_{Lab} = E_{COM} + E_{rel},

where $E_{Lab}$ represents the kinetic energy in the laboratory frame, $E_{COM}$ the centre of mass frame and $E_{rel}$ the energy of centre of mass.

We can prove the result inductively by proving a two-particle case: Let two particle $m_{1}$ and $m_{2}$ moving at ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ . By differentiating the definition of centre of mass, it is not hard to find

\vec{V} = \frac{m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}}{m_{1} + m_{2}} .

Thus, ${\vec{v}}_{1}^{'} = \frac{m_{2}}{m_{1} + m_{2}} ({\vec{v}}_{2} - {\vec{v}}_{1})$ and ${\vec{v}}_{2}^{'} = \frac{m_{1}}{m_{1} + m_{2}} ({\vec{v}}_{1} - {\vec{v}}_{2})$ are the relative velocities of the two particle in the COM frame. Plug in these result:

E_{rel} = \frac{1}{2} (m + M) V^{2} = \frac{1}{2} \frac{(m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2})^{2}}{m_{1} + m_{2}}, E_{COM} = \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} ({\vec{v}}_{2} - {\vec{v}}_{1})^{2}

The second relationship requires a bit of simplification that can be easily done given the highly symmetric form. Obviously, the sum of the above yields the kinetic energy in the laboratory frame.

Collision

The coefficient of restitution is defined as the ratio of "departing speed" by "approaching speed". This definition, imo is somewhat vague, as it tells nothing concrete about the collision.

e = - \frac{v_{1} - v_{2}}{u_{1} - u_{2}} .

It is really noteworthy that the term $v_{1} - v_{2}$ is actually quite obvious in the expression $E_{COM}$ above, that should explain why people would use the relative speed of the two particles to measure the elasiticity of the collision. For it is now obvious that

E_{t} = e^{2} E_{i} .

That is, $e^{2}$ is the ratio of energy retained in the COM frame. Since we cannot create energy in collision, thus $e \leq 1$ . Similarly, it is impossible to create "negative" kinetic energy, thus $e \geq 0$ .

Energy Loss

Notice that

E_{kin} = \frac{1}{2} \frac{p^{2}}{m} .

Due to the conservation of momentum, we understand that the kinetic enery of centre of mass is preserved during the collision. All energy loss in a collision are attributed to the $(1 - e^{2}) E_{COM}$ .

#"hs-maths"