Question

Two point charges ${10}^{-5}C$ and $-{10}^{-5}C$ are released from large separation. Their masses are 100 gm and 200 gm. If velocity of approach (in m/s) of them when they are separated by distance 3m is $\frac{x}{4}m/s$. Then x is:

Answer 1

Apply conservation of momentum

$m_1{v}_1=m_2{v}_2$

$v_1=2v_2$

Conserve energy

$0=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2-\frac{k{q}_1{q}_2}{r}$

$0=\frac{1}{2}\times 0.1v_1^2+\frac{1}{2}\times 0.2v_2^2-\frac{9\times {10}^9\times {10}^{-5}\times {10}^{-5}}{3^2}$

$\frac{1}{2}{v}_1^2+\frac{1}{2}\times 2v_2^2=1$

$2v_2^2+v_2^2=1$

$v_2=\frac{1}{\sqrt{3}}$, $v_1=\frac{2}{\sqrt{3}}$

Velocity of approach $=\sqrt{3}m/s$.