How Close Will Two High-Speed Charged Particles Get?

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AI Thread Summary
Two high-speed charged particles, each with a mass of 6.6 X 10^-27 kg and a charge of 3.2 X 10^-19 C, are approaching each other at an initial speed of 3.0 X 10^6 m/s. The discussion revolves around calculating their minimum separation while assuming no deflection from their paths. The relevant equations include energy conservation, where the total energy is the sum of electric potential energy and kinetic energy. Participants are unsure how the phrase "separated by an enormous distance" impacts the energy equation. Adjustments to the mass in the kinetic energy term are necessary for accurate calculations.
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Homework Statement


Two particles, separated by an enormous distance, approach each other. Each has an initial speed of 3.0 X 10^6 m/s. Calculate their minimum separation, assuming no deflection from their original path. The mass of a particle is 6.6 X 10^-27 kg.

Particle charge = 3.2 X 10^-19 C

Given:
V = 3.0 X 10^6m/s
mass of particle = 6.6 X 10^-27 kg
r = ?




Homework Equations


E = Eprime
Fe = kq1q2/r^2



The Attempt at a Solution


EE + EK = EEprime + EKprime
Kq1q2/r + 1/2 mv^2 = Kq1q2/r + 1/2m(vprime)^2

Dont know where to go from there
 
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How does the phrase "separated by an enormous distance" affect a term in the energy equation?

When you plug in numbers to your equation, what do you get? (I noticed you combined the kinetic energy of both particles into one term on each side, so you'll have to adjust the mass to deal with that.)
 
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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