Problem — The two gunslingers

Problem

The two gunslingers: charged projectiles repelling each other. Two gunslingers face off at $4\ \text{m}$ distance. They simultaneously fire two charged projectiles: projectile 1 ($m_1 = 0{,}1\ \text{kg}$, $q_1 = 10^{-4}\ \text{C}$, $\vv{v}_1 = (300,\ 0)\ \text{m/s}$); projectile 2 ($m_2 = 0{,}2\ \text{kg}$, $q_2 = 2\cdot 10^{-5}\ \text{C}$, $\vv{v}_2 = (-100,\ 0)\ \text{m/s}$). Neglecting gravity, find the minimum distance of approach. Hint: switch to the centre-of-mass frame, split $E_{\text{cin,tot}}$ into $E_{\text{cin,CM}}$ and $E_{\text{cin,int}}$ (König’s theorem), then impose conservation of the internal energy with $E_{\text{pot,el}} = k\,q_1 q_2/r$.

Solution

Using $k = 8{,}99\cdot 10^9\ \text{N·m}^2/\text{C}^2$. The projectiles are charged with the same sign: they repel each other. Switch to the centre of mass (CM), whose velocity is constant (no external force).

CM velocity:

$v_{\text{CM}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{0{,}1\cdot 300 + 0{,}2\cdot(-100)}{0{,}3} = \frac{30 - 20}{0{,}3} = 33{,}3\ \text{m/s}$

Internal kinetic energy (König’s theorem). Reduced mass and relative velocity:

$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{0{,}1\cdot 0{,}2}{0{,}3} = 0{,}0667\ \text{kg},\qquad v_{\text{rel}} = 300 - (-100) = 400\ \text{m/s}$

$E_{\text{cin,int}} = \tfrac12\,\mu\, v_{\text{rel}}^2 = \tfrac12\cdot 0{,}0667\cdot 400^2 \approx 5333\ \text{J}$

Conservation of internal energy. At the minimum distance the relative velocity vanishes, so all of $E_{\text{cin,int}}$ (plus the initial potential energy) is converted into electrostatic potential energy. With $q_1 q_2 = 10^{-4}\cdot 2\cdot 10^{-5} = 2\cdot 10^{-9}\ \text{C}^2$ and $k\,q_1 q_2 = 17{,}98\ \text{J·m}$:

$E_{\text{cin,int}} + \frac{k\,q_1 q_2}{d_0} = \frac{k\,q_1 q_2}{d_{\min}},\qquad d_0 = 4\ \text{m}$

$\frac{k\,q_1 q_2}{d_0} = \frac{17{,}98}{4} = 4{,}495\ \text{J}$

Solving for $d_{\min}$:

$d_{\min} = \frac{k\,q_1 q_2}{E_{\text{cin,int}} + k q_1 q_2/d_0} = \frac{17{,}98}{5333 + 4{,}495} \approx 3{,}4\cdot 10^{-3}\ \text{m}$

$\ev{d_{\min} \approx 3{,}4\cdot 10^{-3}\ \text{m} \approx 3{,}4\ \text{mm}\quad (v_{\text{CM}} \approx 33{,}3\ \text{m/s})}$

The available internal energy ($\sim 5300\ \text{J}$) is enormous compared with the Coulomb barrier, so the projectiles approach to within a few millimetres before stopping and repelling. (The textbook key value of $1{,}83\ \text{m}$ is wrong.)

The key method is the separation of CM motion / internal motion: only the internal part of the kinetic energy ($E_{\text{cin,int}}$) is available to do work against the Coulomb repulsion, while $E_{\text{cin,CM}}$ remains unchanged and does not contribute to the approach.

Linked atoms

Topics: Elettrostatica Concepts: Energia potenziale elettrostatica · Conservazione della quantità di moto Skills: Conservazione della quantità di moto Methods: Teorema di König Objects: Proiettile