Problem — Two charges released from rest

Problem

Two charges are released from rest: $q_1 = -3\cdot 10^{-4}\ \text{C}$, $m_1 = 3\ \text{kg}$; $q_2 = 2\cdot 10^{-4}\ \text{C}$, $m_2 = 6\ \text{kg}$. Initially they are at a distance $d_A = 12\ \text{m}$ and attract each other. Find their velocities when the distance is $d_B = 3\ \text{m}$. Use two parallel tables (momentum and energy): conservation of momentum gives $v_2 = -\tfrac12 v_1$, and conservation of energy fixes $v_1$ through the change in $E_{\text{pot,el}}$.

Solution

Using $k = 8{,}99\cdot 10^9\ \text{N·m}^2/\text{C}^2$. Isolated system, starting from rest: total momentum is zero and conserved.

Momentum table:

$m_1 v_1 + m_2 v_2 = 0 \;\Longrightarrow\; 3 v_1 + 6 v_2 = 0 \;\Longrightarrow\; v_2 = -\tfrac{1}{2} v_1$

Energy table. The change in potential energy becomes kinetic energy. With $q_1 q_2 = (-3\cdot 10^{-4})(2\cdot 10^{-4}) = -6\cdot 10^{-8}\ \text{C}^2$:

$E_{\text{pot,el}}(d_A) = k\,\frac{q_1 q_2}{d_A} = 8{,}99\cdot 10^9\cdot\frac{-6\cdot 10^{-8}}{12} = -44{,}95\ \text{J}$

$E_{\text{pot,el}}(d_B) = k\,\frac{q_1 q_2}{d_B} = 8{,}99\cdot 10^9\cdot\frac{-6\cdot 10^{-8}}{3} = -179{,}8\ \text{J}$

$\Delta E_{\text{pot}} = E_{\text{pot,el}}(d_B) - E_{\text{pot,el}}(d_A) = -179{,}8 - (-44{,}95) = -134{,}85\ \text{J}$

The kinetic energy gained is $E_{\text{cin}} = -\Delta E_{\text{pot}} = +134{,}85\ \text{J}$. Substituting $v_2 = -v_1/2$:

$E_{\text{cin}} = \tfrac12 m_1 v_1^2 + \tfrac12 m_2 v_2^2 = \tfrac12\cdot 3\, v_1^2 + \tfrac12\cdot 6\left(\tfrac{v_1}{2}\right)^2 = 1{,}5\,v_1^2 + 0{,}75\,v_1^2 = 2{,}25\,v_1^2$

$v_1^2 = \frac{134{,}85}{2{,}25} = 59{,}9 \;\Longrightarrow\; v_1 = 7{,}74\ \text{m/s},\qquad v_2 = -\tfrac12 v_1 = -3{,}87\ \text{m/s}$

$\ev{v_1 \approx 7{,}75\ \text{m/s},\qquad v_2 \approx -3{,}87\ \text{m/s}}$

Links

Topics: Electrostatics Concepts: Electrostatic potential energy · Conservation of momentum Skills: Two-tables method Methods: Two parallel tables method