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Closest approach of particle problem  Please help 
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#1
Mar2007, 08:16 PM

P: 58

1. The problem statement, all variables and given/known data
A proton (q = 1e, m = 1u) and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, each with a speed of 0.01c. What is their distance of closest approach, as measured between their centers? [tex]e = 1.6 * 10^{19}[/tex] [tex]c = 3 * 10^8[/tex] [tex]u = 1.661 * 10^{27}[/tex] This should be a simple problem, but I wanted to know if anyone got the same answer as I did. 2. Relevant equations Conservation of energy [tex]K_i + U_i = K_f + U_f[/tex] Conservation of momentum [tex]m_1v_{1i} + m_1v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex] 3. The attempt at a solution After making my conclusion that the proton will eventually turn around and reach a 0 velocity because the bigger particle (alpha particle) will make this "collision" similar to an elastic one. I first have to find my final velocity of the alpha particle, v1. [tex]m_1v_{1i} + m_1v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex] [tex](4u)(3 * 10^6 \frac{m}{s})  (1u)(3 * 10^6 \frac{m}{s}) = (4u)v_{1f}[/tex] [tex]9.0 * 10^6u\frac{m}{s} = (4u)v_{1f}[/tex] [tex]v_{1f} = 2.25 * 10^6\frac{m}{s}[/tex] Then I plugged that velocity into the energy After plugging in and solving for my R (which is at minimum when the velocity of the proton is at 0), I get my R to be [tex]2.24 * 10^{14} m[/tex] Did anyone get this same answer? Thanks! 


#2
Mar2007, 08:35 PM

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P: 41,321

Hint: When the particles reach their distance of closest approach they will momentarily be moving together. What will their KE be at that moment?



#3
Mar2408, 05:57 PM

P: 39

DocAl, how would you determine the velocity at that moment? I only found the final velocity with respect to the Alpha particle. My final answer is approximate to the OP's but I assume you see a flaw in that line of logic. Care to elaborate?



#4
Mar2408, 06:27 PM

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P: 41,321

Closest approach of particle problem  Please help



#5
Mar2408, 06:37 PM

P: 39

Oh, so something like:
[tex](4 \times u)(3 \times 10^{6})  (1 \times u)(3 \times 10^{6}) = (4 \times u)(v_f) + (1 \times u)(v_f)[/tex] In which case [tex] K_f = \frac{1}{2}(m_{p} + m_{\alpha})v_{f}^{2}[/tex] Am I in the ballpark? 


#6
Mar2408, 06:58 PM

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P: 41,321

You got it. But be careful with signs on the RHS of your momentum equation. Write it as: [itex](m_1 + m_2)v_f[/itex].



#7
Mar2408, 07:01 PM

P: 39

On that note, I assumed on the RHS that they'd go on the opposite directions hence the different signs. Or are they actually moving in one direction together?



#8
Mar2408, 07:11 PM

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#9
Mar2408, 07:24 PM

P: 39

Very nice! But that, as you said before, happens momentarily. They do eventually separate, right? Or do they stay in this state indefinitely?



#10
Mar2408, 07:29 PM

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P: 41,321

What do you think? Consider the force that two positive charges exert on each other.



#11
Mar2408, 07:40 PM

P: 39

I'm actually thinking they'd be tailing each other for an indefinite amount of time (presuming, of course, no outside interference) by staying at the distance of closest approach. It makes sense to me this way since that is the maximum distance at which these two charges meet. But now I'm a little confused, because there does exist a repulsive force [tex]F_e[/tex] of magnitude [tex]\frac{k_{e} \times 2e \times e}{r^2}[/tex].



#12
Mar2408, 07:44 PM

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P: 41,321

The repulsive force is the key. At the distance of closest approach that force is maximumthe particles will separate. They essentially bounce off of each other.



#13
Mar2408, 07:49 PM

P: 39

Ah, I think I get the gist of it now! Thanks for the insight. Out of curiosity, does there exist a setup where two charges continue tailing each other at a distance indefinitely? It's an interesting concept to know of.



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