Newton discovered that the relative velocity of separation of the two bodies after the collision is always proportional to relative velocity of approach before collision. The proportionality constant, known as the co-efficient of restitution, is expressed by $$e = - \frac{(v_1 - v_2)}{(u_1 - u_2)} \implies -\frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$$ . The negative sign indicates that the relative velocity of separation after collision and the relative velocity of approach before collision are oppositely directed.

This is the formal description of the coefficient of restitution in my book. But what troubles me is : What are these relative velocities of separation and approach, and with respect to what are they relative to?

Further, regarding $u_1, u_2$ and $v_1, v_2$; the book wrote that they are the velocities before and after collision respectively. Yes it is correct, if the motion of the bodies before and after collision are uniform; but if their motion is non-uniform, they will have several instantaneous velocities before and after collision. So, which of them should I use for a calculation? I think I will have to take the instantaneous velocities just before and after collision. Right?

The last thing I am confused about is the negative sign. I couldn't understand what the book reasoned about it . Why is not $e = \frac{v_1 - v_2}{u_1 - u_2}$ ? Are the relative velocities always oppositely directed? Really? What about the case when both the bodies travel in the same direction before and after the collision?

I am really confused. Please help.


Consider the relative velocity of one body with respect to another. That's all.

So, relative velocity of approach would mean the velocity of body 1 coming towards body 2, signified by $v_1-v_2$

And the relative velocity of separation would mean the velocity of body 1 going away from body 2, signified by $u_1-u_2$

as seen by body 2, in both the cases

Yes, take instantaneous velocities right after the collision.

And if you take the velocity of one body with respect to another, the coefficient of restitution will always be negative, or zero if your relative velocity of separation is zero; as in the case when they get stuck. But, in relative frame(of one body with respect to another), you can be sure that the other body's approaching before collision, and it'sseparating after it; unless the bodies stick to each other

  • $\begingroup$ If I take relative velocity with respect to 2nd body, $$e = \frac{v_1 - v_2}{u_1 - u_2}$$ . But the book includes the '-' sign . Why am I wrong?? $\endgroup$
    – user36790
    Nov 9 '14 at 12:05
  • $\begingroup$ Also want to know if it is mandatory for the body, which had greater velocity before impact, to have lesser velocity after impact?? $\endgroup$
    – user36790
    Nov 9 '14 at 12:08
  • $\begingroup$ I think they are using a convention which I would rather be uncomfortable with. They are keeping both $v2$ and $v1$ positive for opposite directions. But I am not sure. I would rather say you just verify that it gives you a negative number as it should. $\endgroup$
    – Cheeku
    Nov 9 '14 at 16:23

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