Is this equation for coefficient of restitution right? If so, why does it give the negative of the actual answer?

I have derived a formula for the coefficient of restitution, $$e$$, between two smooth particles, A and B, of fixed masses. I would like to confirm that it is generally true. Because it gave me the wrong answer to a question and now I am not so sure it is always true.

Deriving my Formula for the Coefficient of Restitution (First the more Common Equation)

From my understanding, the coefficient of restitution is defined as

the ratio of the magnitude of the parallel component (parallel to the line of impact) of the relative velocity of separation after collision to the magnitude of the same component of the relative velocity of approach before collision.

Let $$\vec{u_A}$$, $$\vec{u_B}$$, $$\vec{v_A}$$ and $$\vec{v_B}$$ be the initial and final velocities of A and B (as seen in our reference frame), respectively.

Notice that $$\vec{BA} = \vec{OA} - \vec{OB}$$, and differentiating both sides w.r.t. time and applying linearity, gives us $$\dot{\vec{BA}} = \dot{\vec{OA}} - \dot{\vec{OB}}$$ - i.e. the time rate of change of the position of A w.r.t. B, which is simply the velocity of A relative to B. Therefore, the velocity of approach is $$\vec{u_A} - \vec{u_B}$$ and that of separation is $$\vec{v_A} - \vec{v_B}$$.

(Note that it doesn't matter whether I choose to take the velocities of approach and separation as relative to A or B, as long as I am consistent with both. In this case, I happen to choose to take them as being relative to B.)

The line of impact is the line along which the particles exert impulse on one another. Taking $$\vec{I}$$ as the imuplse B experiences (note that it will be in the same direction as the force A exerts on B), or the impulse A experiences (it doesn't matter for our purposes, because they are parallel to one another), $$\hat{I}$$ becomes a unit vector parallel with the line of impact.

Note that, by definition, the dot product of 2 vectors gives the product of the magnitude of one with the parallel component of the other vector in the direction of the first. (If the parallel component is in the opposite direction to the first vector, the dot product will be the negative of the product of their magnitudes.) So if one is a unit vector (has a magnitude of one) then the dot product will simply give the component of the other vector in the direction of the unit vector. Therefore, the components of the velocities of approach and separation that are parallel to the line of impact are given by $$(\vec{u_A} - \vec{u_B}) \cdot \hat{I}$$ and $$(\vec{v_A} - \vec{v_B}) \cdot \hat{I}$$, respectively.

And so, the coefficient of restitution is given by: $$e = \frac{ \big\vert (\vec{v_A} - \vec{v_B}) \cdot \hat{I} \big\vert}{\big\vert (\vec{u_A} - \vec{u_B}) \cdot \hat{I} \big\vert}$$

This is a generally accepted definition of the coefficient of restitution for collisions between partciles in 2D, as Vivek has affirmed.

(This is where I go further to get to 'my formula' which I'd like to confirm is generally, always true)

I think I can also shown that $$\mathrm{sgn}\left( (\vec{v_A} - \vec{v_B}) \cdot \hat{I}\right) = - \mathrm{sgn}\left( (\vec{u_A} - \vec{u_B}) \cdot \hat{I}\right)$$.

This is because, as previously explained, $$\vec{u_A} - \vec{u_B}$$ and $$\vec{v_A} - \vec{v_B}$$ are the initial and final velocities of A, relative to B. And we know that, initially, A has to be moving towards B, and away from it finally.

We can think of the particles as smooth, solid spheres, and can ignore any rotational motion. Therefore, all points on the sphere have the same velocity at any given instant. Consider, in particular, the point, P, on the surface of A that is in contact with B at the moment of collision, and notice that we will be doing this analysis from B's frame of reference.

Just before the collision, P's velocity, $$\vec{u_A} - \vec{u_B}$$, has to have a component in the same direction as the direction from A's centre to B's, $$\vec{AB}$$. If not - if $$\vec{u_A} - \vec{u_B}$$ has a component parallel to $$\vec{AB}$$ but in the opposite direction, or none at all - the balls will not collide, because B is stationary in this frame of reference! [To imagine why, think about what would happen if a half-line was to be drawn from P (modelling its trajectory, should it be allowed keep moving with velocity $$\vec{u_A} - \vec{u_B}$$) . A collision will occur if this half-line passes through B and this happens if, and only if, there is a component of $$\vec{u_A} - \vec{u_B}$$ in the same direction as $$\vec{AB}$$.]

And just after the collision, we know that P has to move away from B, so that the A and B don't keep colliding. So in B's frame of reference, where A is moving with velocity $$\vec{v_A} - \vec{v_B}$$, the trajectory of P cannot pass through B, which implies (following a similar sort of reasoning to above) that the component of $$\vec{v_A} - \vec{v_B}$$ parallel with the $$\vec{AB}$$ must act in the opposite direction to $$\vec{AB}$$.

So if $$\vec{I}$$ is the impulse A exerts on B, (i.e. if we assume $$\hat{I} = \hat{AB}$$) then we know that $$\mathrm{sgn} \left((\vec{u_A} - \vec{u_B}) \cdot \hat{I}\right) = +1$$ and $$\mathrm{sgn} \left((\vec{u_A} - \vec{u_B}) \cdot \hat{I}\right) = -1$$, and vice versa if $$\vec{I}$$ is the impulse B exerts on A. So in, either case, $$\mathrm{sgn}\left( (\vec{v_A} - \vec{v_B}) \cdot \hat{I}\right) = - \mathrm{sgn}\left( (\vec{u_A} - \vec{u_B}) \cdot \hat{I}\right)$$

(And because $$|x| = \mathrm{sgn}(x) x$$ for any $$x \in \Bbb{R}$$) we see that we can express the coefficient of restitution as: $$e = - \frac{ (\vec{v_A} - \vec{v_B}) \cdot \hat{I} }{ (\vec{u_A} - \vec{u_B}) \cdot \hat{I} }$$ which I prefer writing as: $$-e = \frac{ (\vec{v_A} - \vec{v_B}) \cdot \hat{I} }{ (\vec{u_A} - \vec{u_B}) \cdot \hat{I} }$$

(This is 'my formula'.)

The Question where my Formula did not work

The reason I ask whether this equation is generally true is because after having used in a problem, it gave me a value of $$e<0$$, which is obviously impossible! So I fear my equation maybe wrong - specifically I fear that $$\mathrm{sgn}\left( (\vec{v_A} - \vec{v_B}) \cdot \hat{I}\right) = - \mathrm{sgn}\left( (\vec{u_A} - \vec{u_B}) \cdot \hat{I}\right)$$ may not always be true? Although I cannot see any reason why this would be the case!

The problem was:

A and B are 2 smooth particles that collide. They had initial velocities $$\vec{u_A} = \begin{pmatrix} 2 \cr 5 \end{pmatrix}$$ and $$\vec{u_B} = \begin{pmatrix} 2 \cr -2 \end{pmatrix}$$, and final velocities $$\vec{v_A} = \begin{pmatrix} 3 \cr 4 \end{pmatrix}$$ and $$\vec{v_B} = \begin{pmatrix} 0 \cr 0 \end{pmatrix}$$. A has twice the mass of B. Find the coefficient of restitution between them.

Taking $$\vec{I}$$ as the impulse A exerts on B, I found $$\hat{I} = \begin{pmatrix} -\frac{1}{\sqrt2} \cr \frac{1}{\sqrt2} \end{pmatrix}$$.

This gave me $$(\vec{u_A} - \vec{u_B}) \cdot \hat{I} = \frac{7}{\sqrt2}$$ and $$(\vec{v_A} - \vec{v_B}) \cdot \hat{I} = \frac{1}{\sqrt2}$$ which obviously means that $$\mathrm{sgn}\left( (\vec{v_A} - \vec{v_B}) \cdot \hat{I}\right) \neq - \mathrm{sgn}\left( (\vec{u_A} - \vec{u_B}) \cdot \hat{I}\right)$$ hence why I got a negative value of $$e$$. (My formula actually gave me $$- \frac 1 7$$ and the answer from the mark-scheme was $$\frac 1 7$$.)

I feel like the question-writers made a mistake and these initial and final velocities just cannot be physically possible, but I'm not 100% sure. Is it a problem with my formula for the coefficient of restitution? Does it break down in some cases? Is my idea that $$\mathrm{sgn}\left( (\vec{v_A} - \vec{v_B}) \cdot \hat{I}\right) = - \mathrm{sgn}\left( (\vec{u_A} - \vec{u_B}) \cdot \hat{I}\right)$$ just not always true?

Well, perhaps the most general definition of the coefficient of restitution, which I have read is that it is the ratio of the magnitude of the (relative) velocity of separation to the (relative) velocity of approach. Which gives, $$e=\dfrac{|v_A-v_B|}{|u_A-u_B|}$$; now with your definition, $$e=\dfrac{|(v_A-v_B)\cdot I_B|}{|(u_A-u_B)\cdot I_B|}$$, here $$I_A$$ is the line of impulse for B (or it could be replaced with that of A, since both are same), and since the line of action is along the the bisector of $$(v_A - v_B)$$ and $$(u_A- u_B)$$, the expression reduces to the general definition $$e=\dfrac{|v_A-v_B|}{|u_A-u_B|}$$.
• Hi Vivek. If by $v_A$ and $v_B$ and so on, you mean the components of the velocities parallel to the line of action of the impulse (i.e., in terms of the notation I used in my question, $\vec{v_A} \cdot \hat{I}$ and $\vec{v_B} \cdot \hat{I}$ and so on) then I do agree with you that $e = \frac{|v_A - v_B|}{|u_A - u_B|}$. Also I used $|x| = \mathrm{sgn}(x) x$ because the dot product of two vectors is a scalar, so this rule does apply - I don't think I confused the norm of a vector and the absolute value of a scalar. Mar 11 at 17:39
• Well, I am sorry for the confusion that I had created. What I wanted to mean was essentially that since you define, the coefficient of restitution as $e=\dfrac{|(v_A-v_B) \cdot I_B|}{|(u_A-u_B) \cdot I_B|}$ it doesn't matter whether, $sgn(v_A-v_B)$ and $sgn(u_A-u_B)$ are differnt. Since they have a modulus sign just outside. For instance, $\dfrac{|-4|}{|2|}$ and $\dfrac{|4|}{|2|}$ are just equal. Mar 13 at 3:00
• No worries, no confusion. I do agree with you that $e=\frac{|(\vec{v_A}-\vec{v_B})\cdot\hat{I}|}{|(\vec{u_A}-\vec{u_B})\cdot\hat{I}|}$. My question essentially boils down to whether or not this can always be simplified to $e=-\frac{(\vec{v_A}-\vec{v_B})\cdot\hat{I}}{(\vec{u_A}-\vec{u_B})\cdot\hat{I}}$ - i.e. whether we can always just remove the absolute value signs and leave a minus sign instead, for the reasons that I tried to explain in the question. (I don't think I phrased my original question very clearly, sorry - I'll rephrase it to make it clearer that this is my underlying question.) Mar 13 at 10:54