A little question causing big problems

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In summary, the problem involves a proton and an alpha particle with initial speeds of 0.141c, and the goal is to find their distance of closest approach using conservation of energy and the equation Kp + Ka = Kq1q2/r. However, this approach may be incorrect and the possibility of using work (W= integral of F(x) dx) is also mentioned.
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Tido611
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Homework Statement



A proton and an alpha particle (q = +2.00e, m = 4.00u ) are fired directly toward each other from far away, each with an initial speed of 0.141c. What is their distance of closest approach, as measured between their centers? (Hint: There are two conserved quantities. Make use of both.)


Homework Equations



Enet1 = Enet2
Kp + Ka = Kq1q2/r

Then solve for "r".


The Attempt at a Solution


I would figure that you could use conservation of energy in the sense that the energy of the system initially is the kinetic energies of the two particles combined (Enet1 = Kp + Ka). At the point of closest approach, their speeds should be zero, and hence Enet2 = Uelec = Kq1q2/r. From here it should be straightforward:

Enet1 = Enet2
Kp + Ka = Kq1q2/r

Then solve for "r".

However, this is incorrect. Perhaps my assumption that the alpha particle (4 times the mass, 2 times the charge) stops completely is wrong. At this point, I really have no idea.
 
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  • #2
what about W= intergral of F(x) dx
 
  • #3


I would suggest taking a closer look at the initial conditions of the problem. The fact that the particles are fired directly toward each other from far away suggests that they have no initial potential energy. Therefore, the only energy in the system is the kinetic energy of the particles. This means that at the point of closest approach, all of the initial kinetic energy must be converted into potential energy.

Using the conservation of energy equation, we can set the initial kinetic energy equal to the final potential energy:

Kp + Ka = Uelec

We can also use the fact that the particles are moving at the same speed (0.141c) to set their initial kinetic energies equal to each other:

Kp = Ka

Substituting this into our first equation, we get:

2Kp = Uelec

Now, we can use the equation for potential energy in terms of the distance of closest approach (Uelec = Kq1q2/r) to solve for r:

2Kp = Kq1q2/r

r = Kq1q2/2Kp

We can then substitute in the values given in the problem to get the distance of closest approach:

r = (9.0x10^9 N*m^2/C^2)(2.00x10^-19 C)(2.00x10^-19 C) / (2)(1.83x10^-11 J)

r = 2.46x10^-15 meters

This is a very small distance, which makes sense given the high speeds and small masses of the particles involved. Hopefully this helps to clarify the problem and provide a solution.
 

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