# Elastic collision COR problem.

1. Feb 25, 2010

### arteelibunao

Why is e=1 for perfectly elastic collision?

How did they arrive with that? Any derivations? Thanks.

-Artee Libunao

2. Feb 25, 2010

### Fightfish

The coefficient of restitution is defined as
$$C_{R} = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$$

For elastic collisions, it can shown from conservation of kinetic energy and conservation of momentum that relative speed of approach = relative speed of separation for the two objects colliding. Hence $$v_{2} - v_{1} = u_{1} - u_{2}$$

3. Feb 25, 2010

### Saw

If you think of the two colliding objects as a system, you realize that, in the absence of external forces (and the force arising in a collision is an internal one of the system), the system cannot accelerate: in any frame, the velocity of the center of mass must remain constant.

$$v_{CM} = \frac{{V_o M}} {{m + M}} + \frac{{v_o m}} {{m + M}} = \frac{{V_f M}} {{m + M}} + \frac{{v_f m}} {{m + M}}$$

Another expression of the same is conservation of momentum:

$$p_o = V_o M + v_o m = V_f M + v_f m = p_f$$

Or Newton's Third Law:

$$\begin{gathered} V_o M + v_o m = V_f M + v_f m_f \hfill \\ v_o m - v_f m_f = V_f M - V_o M \hfill \\ m(v_o - v_f ) = M(V_f - V_o ) \hfill \\ m(v_f - v_o ) = - M(V_f - V_o ) \hfill \\ m\Delta v = - M\Delta V \hfill \\ \end{gathered}$$

and since collision time is the same for both objects:

$$\begin{gathered} m\Delta v/\Delta t = - M\Delta V/\Delta t \hfill \\ ma = - MA \hfill \\ \end{gathered}$$

But all this only tells you how the total effect, the total acceleration, is distributed between the two objects, not its magnitude (the equality would still be true if you multiply both sides by n). To learn this, you must rely on conservation of energy and distinguish different cases depending on the characterstics of the materials involved.

In a perfectly elastic collision (an almost ideal situation), the materials may deform but recover their original size thanks to restoring forces. Another way to say it: KE turns into Potential Energy ad back again into KE. Thus the total KE of the system is conserved:

$$\begin{gathered} V_o ^2 M + v_o ^2 m = V_f ^2 M + v_f ^2 m \hfill \\ V_o ^2 M - V_f ^2 M = v_f ^2 m - v_o ^2 m \hfill \\ M(V_o ^2 - V_f ^2 ) = m(v_f ^2 - v_o ^2 ) \hfill \\ M(V_f ^2 - V_o ^2 ) = - m(v_f ^2 - v_o ^2 ) \hfill \\ M(V_f - V_o )(V_f + V_o ) = - m(v_f - v_o )(v_f + v_o ) \hfill \\ M\Delta V(V_f + V_o ) = - m\Delta v(v_f + v_o ) \hfill \\ \end{gathered}$$

If you look back to conservation of momentum, you realize that the left-hand terms of each side are the same and so you are left with:

$$\begin{gathered} V_f + V_o = v_f + v_o \hfill \\ V_f - v_f = v_o - V_o \hfill \\ V_f - v_f = - (V_o - v_o ) \hfill \\ v_{rel}^{final} = - v_{rel}^{original} \hfill \\ \end{gathered}$$

So in an elastic collision the relative velocity of separation is of the same module as the relative velocity of approach, just a different sign. In other words, the collision has "restituted" both objects to their original velocities, in terms of magnitude, i.e. the coefficient of restitution is 1:

$$v_{rel}^{final} = ev_{rel}^{original} \to CR = e = \frac{{v_{rel}^{final} }} {{v_{rel}^{original} }} = 1$$