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I have a sphere moving forward with v (pre collision) = v1i

A second sphere colliding with the first one has its own pre collision velocity = v2i

The collision should be not totally inelastic (like, you know, the case of two human bodies colliding each other), but for my purpose I think I can consider the coefficient of restitution = 0

Using the formula

$$ I = mv_f - mv_0$$

and replacing I with F * (delta)t I woul like to know how can I compute that F, since I don't know what will be v(post collision) because it is my final unknown to compute. And I think that force should be computed with the relative velocity of the two spheres, but I have no idea at the moment on how to proceed. Any suggestion will be strongly appreciated.

UPDATE

Thank you for your suggestion ja72.

So in that case can I write something like this?

$$ F *\Delta t = \frac{ (\epsilon+1) \left(v_2-v_1\right)}{\frac{1}{m_1}+\frac{1}{m_2}} $$

But in this case, how am I supposed to evaluate final velocity of the two spheres?

Leggy7
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  • Is the collision eleastic, ineleastic or somewhere in the middle ? What is its coefficient of restitution ? – Rijul Gupta Jan 16 '14 at 12:19
  • @rijulgupta the coefficient of restitution should be something closer to 0 rather than in the middle but to avoid complex calculations, for my purpose, I can conider it as 0. I updated my question – Leggy7 Jan 16 '14 at 13:26
  • @ja72 Just read that question. I followed then the link to stackoverflow, which would be the interesting one. Unfortunately due to my lack of knowledge in physics and to the fact that is someone else code, I have difficulties in understanding – Leggy7 Jan 16 '14 at 13:54
  • Did you see the answer? It shows how to calculate the momentum transfer (impulse $J$) in general. – John Alexiou Jan 16 '14 at 13:57
  • @ja72 I updated my question – Leggy7 Jan 16 '14 at 14:24
  • How did you come to the final equation you wrote ? – Rijul Gupta Jan 16 '14 at 14:41
  • @rijulgupta thanks to ja72 suggestion which led me to this page which tells the general equation for impulse. I just wrote F * dt instead of J, asking for comfirmation – Leggy7 Jan 16 '14 at 14:46
  • Did you read the rest of the answer there ? It has been shown how to calculate the velocities ! You find impulse, change in velocities and therefore final velocities. – Rijul Gupta Jan 16 '14 at 14:52
  • woah, I misread those $$ \Delta $$ Reading fast I thought they were initial velocities and did not care too much. I'm so distract. Thank you dude, I'll wait enought points to vote all your help! – Leggy7 Jan 16 '14 at 14:56

1 Answers1

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This is how you proceed,

Since $ = 0$, bodies stick after collision

Total momentum remains conserved therefore:

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2) v_f$

Solve for $v_f$ and then,

$ I = m_1 v_f - m_1 v_1$ $ F = I/{dt}$

Addendum :

IF $e\neq 0$ use

$e = \frac{ \overrightarrow{v_{2f}} - \overrightarrow{v_{1f}}}{\overrightarrow{v_{1i}} - \overrightarrow{v_{2i}}}$

$e = \frac{\mbox{relative velocity after collision}}{\mbox{relative velocity before collision}}$

Use this along with momentum conservation, the difficulty would then be determined by how many variable's you can eliminate by using given values or otherwise.

As this question mathematically proves, Yes you can find momentum using that equation.

To find the final velocities use:

$ \Delta v = I/m$
$ v_{1f} = v_{1i} + \Delta v$

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Rijul Gupta
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  • ow god, I always ask myself why I don't answer myself correctly before asking in such cases, maybe because I'm am no enught sure of my basis in certain topics. That's a great starting point, thank you – Leggy7 Jan 16 '14 at 14:01
  • Just one more question, at this point.. if e!=0 thigs get too complicated or impossible without final velocity? – Leggy7 Jan 16 '14 at 14:02
  • Is your query resolved now ? – Rijul Gupta Jan 16 '14 at 14:31
  • I have updated my question. Can we say that your if e=0 is a particular case of that I added to my question as an update? So can I say that this is the general rule even in the case of elastic collision? (would be perfect) – Leggy7 Jan 16 '14 at 14:31
  • sorry, I am just going through this and updated my question with some information and a new doubt – Leggy7 Jan 16 '14 at 14:38