Question

One particle has a mass of 3.12 x 10-3 kg and a charge of +8.8 C. A second particle has a mass of 7.1 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.15 m, the speed of the 3.12 x 10-3 kg particle is 131 m/s. Find the initial separation between the particles.

Answer 1

Answer:$r_i=0.016119\ m\approx 16.119\ mm$

Explanation:

Given

mass of first particle is $m_1=3.12\times 10^{-3}\ kg$

mass of second particle is $m_2=7.1\times 10^{-3}\ kg$

Charge on both the particle $q=8.8\times 10^{-6}\ C$

Now final speed of first particle is $v_1=131\ m/s$

Final separation between particles is $r=0.15\ m$

As there is no external force therefore linear momentum is conserved

$0+0=m_1v_1+m_2v_2$

$0=3.12\times 10^{-3}\times 131+7.1\times 10^{-3}\times v_2$

$v_2=-\dfrac{3.12\times 10^{-3}}{7.1\times 10^{-3}}\times 131$

$v_2=-57.56\ m/s$

Conserving total energy

Initial Kinetic energy +Initial  Potential energy=Final Kinetic energy +Final Potential energy

$\Rightarrow 0+\frac{kq^2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{kq^2}{r_f}$

$\Rightarrow \frac{9\times 10^9\times 8.8^2\times 10^{-12}}{r_i}=\frac{1}{2}\times 3.12\times 10^{-3}\times 131^2+\frac{1}{2}7.1\times 10^{-3}\times (57.56)^2+\frac{9\times 10^9\times 8.8^2\times 10^{-12}}{0.15}$

$\Rightarrow \frac{0.696}{r_i}=26.771+11.76+4.646$

$\Rightarrow \frac{0.696}{r_i}=43.177$

$r_i=0.016119\ m\approx 16.119\ mm$