Basic collision help needed

  • Thread starter fluidistic
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  • #1
fluidistic
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Homework Statement


Some days ago and specially yesterday I realized I have a strong weakness when it comes to elastic collision problems. Unfortunately I disregarded it and I couldn't finish an exercise in a test today.


Homework Equations


Linear momentum is conserved ([tex]\vec{P_{i}}=\vec{P_f}}[/tex]) and so is the kinetic energy of the system. (Two bodies)


The Attempt at a Solution



Well in fact say you have 2 particles, [tex]m_1[/tex] and [tex]m_2[/tex]. [tex]m_2[/tex] is initially at rest while [tex]m_1[/tex] has a speed of [tex]\frac{4m}{s}[/tex].
If [tex]m_1[/tex] hits [tex]m_2[/tex], what would be the speed of [tex]m_2[/tex]?

I get that [tex]m_1v_1=m_1v_1'+m_2v_2'[/tex] [1]. One equation, two unknowns: [tex]v_1'[/tex] and [tex]v_2'[/tex].
Have I to work it out using the equation of the kinetic energy? [tex]\frac{m_1v_1^2}{2}=\frac{m_1v_1^2'}{2}+\frac{m_2v_2^2'}{2}[/tex]. I solve for [tex]v_1'[/tex] and back substitute it in [1]. Is it the way to find out [tex]v_1'[/tex] and [tex]v_2'[/tex]?
If yes then I thought about it during the test but I was too lazy to do it... (it doesn't count for my average but still, I should have done that).
Thanks for your confirmation or infirmation.
 

Answers and Replies

  • #3
fluidistic
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Thank you very much iseidthat!
That doesn't answer my question I think, but at least I've learned that [tex]v_1'=\left( \frac{m_1-m_2}{m_1+m_2}\right) v_1[/tex] and that [tex]v_2'[/tex] can be determinate by a given formula.
Ok, I'll try to answer my question by my own.
 

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