What Is the Distance of Closest Approach in a Particle Collision?

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The discussion focuses on calculating the distance of closest approach in a collision between a proton and an alpha particle, both initially moving towards each other at 0.141c. The initial approach suggests using conservation of energy, equating the total kinetic energy of the particles to the potential energy at closest approach. However, the assumption that the alpha particle comes to a complete stop at this point is questioned, indicating a need to consider the center of mass frame for accurate calculations. The conversation emphasizes the importance of applying both conservation of energy and momentum to solve for the distance. Ultimately, a more nuanced approach is necessary to determine the correct distance of closest approach.
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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.)

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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It might work better if you set it up in a frame where the center of mass is stationary.

Then use the conservation of energy and momentum.
 

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