EPE and particle collision (grade 12)

[krbs]

Homework Statement

An alpha particle moving at 3.0 x 10⁶ m/s [east] (m₂ = 6.64 x 10^-27 kg and q₂ = +3.2 x 10^-19 C) is headed directly towards a proton moving at 5.0 x 10⁶ m/s [west] (m₁ = 1.67 x 10^-27 kg, q₁ = 1.6 x 10^-19 C). Find the distance of closest approach, assuming that they start from a very far apart position.

Homework Equations

E_k + E_k = E_E

The Attempt at a Solution

$0.5(6.64 \times 10^{-27}kg)(3.0 \times 10{^6} m/s)^2\ +\ (0.5)(1.67 \times 10^{-27}kg)(-5.0 \times 10^6m/s)^2\ =\ \frac{(9.0\times10^9)(3.2 \times 10^{-19}C)(1.6\times10^{-19}C)}{r}\\ r\ =\ \frac{(9.0\times10^9)(3.2 \times 10^{-19}C)(1.6\times10^{-19}C)}{0.5(6.64 \times 10^{-27}kg)(3.0 \times 10{^6} m/s)^2\ +\ (0.5)(1.67 \times 10^{-27}kg)(-5.0 \times 10^6m/s)^2}\\ r\ =\ 9.1\times10^{-15}$

 

[mfb]

To get that close, they would have to convert their whole kinetic energy in the lab frame to potential energy - they would have to be at rest briefly. Is that possible, or do you see a conservation law that could be violated?

 

[krbs]

I'm lost, but here's my best guess. The only other conservation law I can think of is momentum. I assumed I was not violating this law because it's my understanding that it applies to the instant before and after a collision. If the particles are never at rest, then some of the energy at collision has to be kinetic. To determine the final kinetic energies I'll need their final velocities which I can determine through conservation of momentum.

 

[Ray Vickson]

Homework Statement

An alpha particle moving at 3.0 x 10⁶ m/s [east] (m₂ = 6.64 x 10^-27 kg and q₂ = +3.2 x 10^-19 C) is headed directly towards a proton moving at 5.0 x 10⁶ m/s [west] (m₁ = 1.67 x 10^-27 kg, q₁ = 1.6 x 10^-19 C). Find the distance of closest approach, assuming that they start from a very far apart position.

Homework Equations

E_k + E_k = E_E

The Attempt at a Solution

$0.5(6.64 \times 10^{-27}kg)(3.0 \times 10{^6} m/s)^2\ +\ (0.5)(1.67 \times 10^{-27}kg)(-5.0 \times 10^6m/s)^2\ =\ \frac{(9.0\times10^9)(3.2 \times 10^{-19}C)(1.6\times10^{-19}C)}{r}\\ r\ =\ \frac{(9.0\times10^9)(3.2 \times 10^{-19}C)(1.6\times10^{-19}C)}{0.5(6.64 \times 10^{-27}kg)(3.0 \times 10{^6} m/s)^2\ +\ (0.5)(1.67 \times 10^{-27}kg)(-5.0 \times 10^6m/s)^2}\\ r\ =\ 9.1\times10^{-15}$

Such problems are most easily analyzed in the so-called center-of-mass (or center-of-momentum) frame (CM), where the total momentum = 0 initially. If you use non-relatavistic mechanics, then the total momentum in a new (moving) reference frame with velocity $V$ (west) in the initial (lab) frame would be $m_1 (v_1 - V) + m_2 (v_2 - V)$ in the x-direction. Equate that to 0 to find $V$. (Note that $v_1 = 3.0 \times 10^6$ and $v_2 = - 5.0 \times 10^6$ (all in m/s), so these are velocities, not speeds. Then the initial kinetic energies in the CM frame is $\frac{1}{2} m_1 (v_1 - V)^2$ and $\frac{1}{2} m_2 (v_2 - V)^2$.

Since the collision is perfectly elastic (assuming negligible radiation due to acceleration of the charges), the total momentum = 0 and the total kinetic + potential energy remains constant at all times in the CM frame. At the closest approach the particles are both stationary in the CM frame

 

[krbs]

Thank you both for your help

 

[krbs]

Yikes. I got access to the solution to this question and now I'm confused for a different reason.

I'm only supposed to take into account magnitude, not direction of velocity? I thought one of them would be negative, since they're moving in opposite directions.

 

[TSny]

You are correct, krbs. As you say, the direction of the velocity needs to be taken into account when calculating the momentum. So, one of the velocities should be negative. The solution shown in the attachment is incorrect.

 

[krbs]

Thanks, thought I was still getting it wrong.