A sphere of mass m moving with a constant velocity collides with another stationary sphere of same mass. The ratio of velocities of two spheres after the collision will be, if coefficient of restitution is e

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A sphere of mass m moving with a constant velocity collides with another stationary sphere of same mass. The ratio of velocities of two spheres after the collision will be, if coefficient of restitution is e

Physics
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is your collision elastic?
or, how is defined the "coefficient of restitution"?

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That's the only info given, for the sake of simplicity I tried solving it by considering the collision inelastic.
Oh well, the answer is \(\frac{1-e}{1+e}\)
That might help ....
if the collision is inelastic perfectly, then the final velocity of both spheres is equal to half of the initial velocity of the colliding sphere
So, you mean it can't be perfectly inelastic....Then how to find ratio of two final velocities in terms of e ?
no, sorry I'm not sure
It's ok, thnx for the effort though c:
let's suppose that our collision is perfectly inelastic, then we have this: |dw:1440942312488:dw|
using the conservation of momentum we can write this: \[\Large m{v_0} = 2mv\] and hence the final velocity is: \[\Large v = \frac{{{v_0}}}{2}\]
the two spheres are sticked together, after collision
Yeah i know that, but we need to find the ratio in terms of e. And I think I'll have to consider the collision elastic.
on the other hand, if the collision is elastic, then we have this: |dw:1440942582057:dw|
But in that case one of the final velocities will become 0, isn't it?
Yeah....
namely after collision the two spheres change their velocities
So basically, ratio will be 1:1 if the collision is inelastic...right?
And 0 if it's elastic?
the ratio is 1/2 if the collision is inelastic perfectly and 0 if that collision is elastic
So, the value of e must be in between 0-1
How, 1/2 , both the final velocities will be the same. Isn't it?
if collision is inelastic perfectly, then we can write this: \[\Large v = \frac{{{v_0}}}{2} \Rightarrow \frac{v}{{{v_0}}} = \frac{1}{2}\]
Umm..We need to find the ratio of velocities of both the ball after collision...
ok! then we have ratio=1
Thanks, I'll try to conclude to the given answer and meanwhile if something come up in your mind then please share c:
ok! :)
\[mu = mv_1 + mv_2 \implies u = v_1 + v_2\] \[e = \frac{v_2 - v_1}{u}\] solve for \(\large \frac{v_2 }{v_1} \)
By definition, from Wikipedia, we have that the coefficient of restitution is the ratio between the relative velocity after collision and the relative velocity before collision
here is the article which I read: https://en.wikipedia.org/wiki/Coefficient_of_restitution
let's suppose that part of the kinetic energy is lost due to collision, so we can write these equations: \[\Large \left\{ \begin{gathered} m{v_0} = m{u_1} + m{u_2} \hfill \\ \\ \frac{{mv_0^2}}{2} = \frac{{mu_1^2}}{2} + \frac{{mu_2^2}}{2} + \varepsilon \frac{{mv_0^2}}{2} \hfill \\ \end{gathered} \right.\] where \epsilon is such that: \[\Large 0 \leqslant \varepsilon \leqslant 1\] and \[\Large \varepsilon \frac{{mv_0^2}}{2}\] is the kinetic energy lost, due to collision, namely a fraction of the initial kinetic energy Developing those equation I get the subsequent final velocities: \[\Large {u_1} = {v_0}\frac{\varepsilon }{2},\quad {u_2} = {v_0}\left( {1 - \frac{\varepsilon }{2}} \right)\] therefore, we can write this: \[\Large \frac{{{u_2} - {u_1}}}{{{v_0}}} = 1 - \varepsilon \]

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