A little question causing big problems

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The problem involves calculating the distance of closest approach between a proton and an alpha particle, both initially moving towards each other at 0.141c. The conservation of energy principle is applied, equating the initial kinetic energies of the particles to the potential energy at closest approach. The assumption that the alpha particle comes to a complete stop at this point is questioned, suggesting a need to reconsider the dynamics involved. The discussion highlights the complexity of the interaction due to differing masses and charges. Clarifying these assumptions is crucial for accurately solving the problem.
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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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what about W= intergral of F(x) dx
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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