# [Physics] Question on the negativity of the coefficient of restitution

collisionconservation-lawshomework-and-exercisesmomentum

I was trying to solve a Mechanics question on Momentum.

Here is the question :

Two small smooth spheres A and B have equal radii and have masses m and km respectively. They are moving in a straight line in the same direction on a smooth horizontal table. The speed of A is $$u$$ and the speed of B is $$\frac{2u}{3}$$. Sphere A collides directly with sphere B. The coefficient of restitution between the spheres is $$\frac{4}{5}$$.

(i) Find the speed of A after the collision in terms of $$u$$ and $$k$$.

My Attempt

So I formed two equations, first the conservation of momentum:

$$mv_A+kmv_B=mu+\frac{2}{3}kmu$$

And coefficient of restitution formula :

$$v_A-v_B=\frac{4}{5}\left(u-\frac{2u}{3}\right)\tag{2}$$

Now I don't get the correct answer and the problem is that (2) is wrong according to answer book. The answer book states:

$$v_A-v_B=\frac{-4}{5}\left(u-\frac{2u}{3}\right)$$

You can see its quite similar to my equation, except that there is a – sign in front of the coefficient of restitution. Why is it negative?

The equation, $v_A-v_B = \frac{4}{5}(u-2u/3)$ is incorrect.
The proper equation for the coefficient of restitution is given by, $v_A-v_B$ = $e(u_B-u_A)$, where $u$ and $v$ are velocities along the line of impact. I believe you came across the somewhat incomplete statement that the relative speed of separation after the collision is $e$ times the initial speed of approach along the line of impact. The more accurate statement would be that the relative velocity of separation is $e$ times the negative relative velocity of approach.