Challenge Question 3 Solution
System: Charges A($q, m$), B($q, 2m$), C($2q, 5m$) placed at $0, r_0, 2r_0$.
Step 1: Symmetry in Acceleration
Calculate initial accelerations based on forces:
$F_B = \text{Force from A (right)} + \text{Force from C (left)} \implies \text{Net Left}$.
$F_C = \text{Force from A (right)} + \text{Force from B (right)} \implies \text{Net Right}$.
It turns out $|\vec{a}_B| = |\vec{a}_C|$ due to the specific mass/charge ratios ($q/2m$ vs $2q/5m$ ratio effects). Thus, $v_B = v_C = v$ throughout the motion.The ratio of initial accelerations of each particle is same as the ratio of its distance from centre of mass. Hence the acceleration, velocity and displacement of each particle maintain the same ratio throughout the motion.
Step 2: Conservation Laws
- Momentum: $m v_A + 2m v – 5m v = 0 \implies v_A = 3v$.
- Energy: Initial Potential Energy = Final Kinetic Energy. $$U_i = \frac{kq^2}{r_0} + \frac{k(q)(2q)}{r_0} + \frac{k(q)(2q)}{2r_0} = \frac{4kq^2}{r_0}$$ $$K_f = \frac{1}{2}m(3v)^2 + \frac{1}{2}(2m)v^2 + \frac{1}{2}(5m)v^2 = 8mv^2$$ $$8mv^2 = \frac{4kq^2}{r_0} \implies v = \sqrt{\frac{kq^2}{2mr_0}}$$