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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?

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  • $\begingroup$ Please note that check my work-type questions are off-topic on this site. $\endgroup$ – Danu Sep 23 '14 at 7:39
  • $\begingroup$ No but its a question related to the negativity :/ nothing to check my work, i jus put my attempt, coz otherwise you guys will think im askin homework questions. $\endgroup$ – M.S.E Sep 23 '14 at 7:40
  • $\begingroup$ Hi Tharindu. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Sep 23 '14 at 14:00
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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.

Hope this helped.

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  • $\begingroup$ Yes thank you this helped. Yes I have been taught the relative speed of separation after the collision is e times the relative velocity of approach....I didnt know it was negative relative velocity. Thank you alot again. $\endgroup$ – M.S.E Sep 23 '14 at 8:45
  • $\begingroup$ You're welcome. I faced the same problem some time ago when collisions were taught to me for the first time. $\endgroup$ – BigBang07 Sep 23 '14 at 14:48
  • $\begingroup$ But I think the positive sign works sometimes, this is the first time I've come across the negative sign. $\endgroup$ – M.S.E Sep 24 '14 at 2:49
  • $\begingroup$ No the positive sign doesn't. The definition of the coefficient of restitution essentially states very accurately what I mentioned above. In fact there is an equivalent definition which states that the coefficient of restitution is the ratio of the impulse of reformation to the impulse of deformation. You could check this particular link. Coefficient of Restitution - Wikipedia for more information. $\endgroup$ – BigBang07 Sep 24 '14 at 17:05

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