Closest approach of particle problem - Please help!!


by PhysicsinCalifornia
Tags: closest, particle
PhysicsinCalifornia
PhysicsinCalifornia is offline
#1
Mar20-07, 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!
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
Doc Al
Doc Al is online now
#2
Mar20-07, 08:35 PM
Mentor
Doc Al's Avatar
P: 40,886
Hint: When the particles reach their distance of closest approach they will momentarily be moving together. What will their KE be at that moment?
octahedron
octahedron is offline
#3
Mar24-08, 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?

Doc Al
Doc Al is online now
#4
Mar24-08, 06:27 PM
Mentor
Doc Al's Avatar
P: 40,886

Closest approach of particle problem - Please help!!


Quote Quote by octahedron View Post
DocAl, how would you determine the velocity at that moment?
Using conservation of momentum.
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?
Not sure what you did. At the distance of closest approach, both particles move with the same speed. (The relative speed is zero.)
octahedron
octahedron is offline
#5
Mar24-08, 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?
Doc Al
Doc Al is online now
#6
Mar24-08, 06:58 PM
Mentor
Doc Al's Avatar
P: 40,886
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].
octahedron
octahedron is offline
#7
Mar24-08, 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?
Doc Al
Doc Al is online now
#8
Mar24-08, 07:11 PM
Mentor
Doc Al's Avatar
P: 40,886
Quote Quote by octahedron View Post
Or are they actually moving in one direction together?
Think it through. If they were moving in opposite directions or had any kind of relative motion, they'd have to be approaching or separating--and thus not at the distance of closest approach.
octahedron
octahedron is offline
#9
Mar24-08, 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?
Doc Al
Doc Al is online now
#10
Mar24-08, 07:29 PM
Mentor
Doc Al's Avatar
P: 40,886
What do you think? Consider the force that two positive charges exert on each other.
octahedron
octahedron is offline
#11
Mar24-08, 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].
Doc Al
Doc Al is online now
#12
Mar24-08, 07:44 PM
Mentor
Doc Al's Avatar
P: 40,886
The repulsive force is the key. At the distance of closest approach that force is maximum--the particles will separate. They essentially bounce off of each other.
octahedron
octahedron is offline
#13
Mar24-08, 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.


Register to reply

Related Discussions
Impact Parameter, Closest Approach Advanced Physics Homework 10
Distance of closest approach Introductory Physics Homework 5
Distance of closest approach Introductory Physics Homework 0
Volume problem - Closest packing Introductory Physics Homework 4
Particle Systems- A Physically Based Modelling Approach Computing & Technology 0