Problem — Two charges thrown upwards

Problem

Two charges thrown upwards: CM and interaction. Two charged particles are thrown vertically upwards in the gravitational field: $m_1 = 1\ \text{kg}$, $q_1 = 10^{-4}\ \text{C}$ at $\vv{s}_1 = (4,\ 10)$ with $\vv{v}_1 = (0,\ 1)\ \text{m/s}$; $m_2 = 2\ \text{kg}$, $q_2 = -2\cdot 10^{-4}\ \text{C}$ at $\vv{s}_2 = (0,\ 6)$ with $\vv{v}_2 = (0,\ 2)\ \text{m/s}$. The charges attract each other. (a) Initial position and velocity of the CM. (b) Maximum height of the CM (the internal Coulomb force does not shift it). (c) If at the point of maximum CM height the two charges are $1\ \text{m}$ apart, calculate their velocities using conservation of momentum and energy.

Solution

Using $k = 8{,}99\cdot 10^9\ \text{N·m}^2/\text{C}^2$, $g = 9{,}8\ \text{m/s}^2$, total mass $M = 3\ \text{kg}$.

(a) Centre of mass. Position:

$\vv{r}_{\text{CM}} = \frac{m_1 \vv{s}_1 + m_2 \vv{s}_2}{M} = \left(\frac{1\cdot 4 + 2\cdot 0}{3},\ \frac{1\cdot 10 + 2\cdot 6}{3}\right) = (1{,}33;\ 7{,}33)\ \text{m}$

Velocity:

$\vv{v}_{\text{CM}} = \frac{m_1 \vv{v}_1 + m_2 \vv{v}_2}{M} = \left(0,\ \frac{1\cdot 1 + 2\cdot 2}{3}\right) = \left(0,\ \tfrac{5}{3}\right) = (0;\ 1{,}67)\ \text{m/s}$

(b) Maximum height of the CM. The internal (Coulomb) forces do not shift the CM: it moves like a projectile subject only to gravity. With initial vertical velocity $v_{\text{CM},y} = 1{,}67\ \text{m/s}$:

$h_{\text{CM,max}} = y_{\text{CM},0} + \frac{v_{\text{CM},y}^2}{2g} = 7{,}33 + \frac{1{,}67^2}{2\cdot 9{,}8} = 7{,}33 + 0{,}142 \approx 7{,}47\ \text{m}$

(c) Velocities at maximum height. Initial separation:

$d_A = |\vv{s}_1 - \vv{s}_2| = |(4,\ 4)| = \sqrt{32} \approx 5{,}66\ \text{m},\qquad d_B = 1\ \text{m}$

In the CM frame the internal momentum is zero, so $m_1 v_1' + m_2 v_2' = 0 \Rightarrow v_2' = -\tfrac12 v_1'$. The change in potential energy becomes internal kinetic energy. With $q_1 q_2 = -2\cdot 10^{-8}\ \text{C}^2$ and $k\,q_1 q_2 = -179{,}8\ \text{J·m}$:

$E_{\text{pot,el}}(d_A) = \frac{-179{,}8}{5{,}66} = -31{,}77\ \text{J},\qquad E_{\text{pot,el}}(d_B) = \frac{-179{,}8}{1} = -179{,}8\ \text{J}$

$\Delta E_{\text{pot}} = -179{,}8 - (-31{,}77) = -148{,}0\ \text{J} \;\Longrightarrow\; \Delta E_{\text{cin,int}} = +148{,}0\ \text{J}$

$\Delta E_{\text{cin,int}} = \tfrac12 m_1 v_1'^2 + \tfrac12 m_2 v_2'^2 = \tfrac12 v_1'^2 + \tfrac12\cdot 2\left(\tfrac{v_1'}{2}\right)^2 = 0{,}75\,v_1'^2$

$v_1'^2 = \frac{148}{0{,}75} = 197 \;\Longrightarrow\; v_1' \approx 14{,}0\ \text{m/s},\qquad v_2' = -\tfrac12 v_1' \approx -7{,}0\ \text{m/s}$

These are the (internal) velocities relative to the CM at the moment of maximum height.

$\ev{h_{\text{CM,max}} \approx 7{,}47\ \text{m};\quad v_1' \approx 14{,}0\ \text{m/s},\ v_2' \approx -7{,}0\ \text{m/s (rispetto al CM)}}$

Linked atoms

Topics: Elettrostatica Concepts: Energia potenziale elettrostatica · Conservazione della quantità di moto Skills: Conservazione dell’energia Methods: Teorema di König