Question

Two 2.0 g plastic buttons each with +50 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on either side of a 5.0 g button charged to +250 nC.

All three are released simultaneously.

a. How many interactions are there that have a potential energy?

b. What is the final speed of each button?

Short answer

a. The final speed of the two plastic button's charges are +50 nC 2.4m/s

b. The final charge's speed is + 250 nC 0m/s.

Step-by-step solution

Expression of potential energy 

The expression of the potential energy in the charge is

$U=\frac{kQq}{r}$

Here, Q and q are the charges of the particles.

Kis constant andr is distance.

Part a

Let's assume the heavy button is in the middle which is button 2 and in its left the button is 1 and on the right the button is 2.

Thus, there are there types of interactions that are noticed here,

$U_{12,}{U}_{13,}{U}_{23}$are the potential energies.

Here, $U_{12,}{U}_{13,}{U}_{23}$are the potential energies in which 1 and 2, 2 and 3, and 1 and 3 are respectively.

Part b

The mass in the center is not able to move because of the symmetry but all the masses of the sides move because of the speed V.

the conservation energies and the kinetic energy masses are equals to the potential energy in the system which is,

$\begin{array}{rcl}{K}_{i+}{U}_i& =& K_f+U_f\\ \frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2& =& \frac{k{q}_1{q}_2}{r_{12}}+\frac{k{q}_1{q}_3}{r3}+\frac{k{q}_2{q}_3}{r_{23}}\\ 2\left(\frac{1}{2}\right)m{v}^2& =& k(\frac{k{q}_1{q}_2}{r_{12}}+\frac{k{q}_1{q}_3}{r3}+\frac{k{q}_2{q}_3}{r_{23}})\\ & & \end{array}$

here, $V_{i}and{V}_f$are initial and final potential, the proton mass ism and K_i and K_f are kinetic energies.

on the other hand, the distance between the charges are respectively $r_{12},r_{23},r_{13}$.

localid="1648286255495" $\begin{array}{rcl}m{v}^2& =& K\left(\frac{q_1{q}_2}{r_{12}}+\frac{q_1{q}_3}{r_{13}}+\frac{q_2{q}_3}{r_{23}}\right)\\ V& =& \frac{K\left(\frac{q_1{q}_2}{r_{12}}+\frac{q_1{q}_3}{r_{13}}+\frac{q_2{q}_3}{r_{23}}\right)}{m}\end{array}$

the substitute of 0.20m is for r₁₂ and r₂₃, 0.40 is for r₁₃ and 50nC is for q₁ and q₃, 250nC for q₃, 0.0020 kg for m and $9\times {10}^{9}J.m^2/C^2$for k in this expression,

localid="1648286267226" $\begin{array}{rcl}v& =& \sqrt{\frac{\left(9\times {10}^{9}J.m^2/C^2\right)\left({\displaystyle \frac{\left(50nC\right)\left(250nC\right)}{\left(0.20m\right)}}+{\displaystyle \frac{\left(50nC\right)\left(50nC\right)}{\left(0.40m\right)}}+{\displaystyle \frac{\left(50nC\right)\left(250nC\right)}{\left(0.20m\right)}}\right)}{\left(0.0020kg\right)}}\\ & =& 2.4m/s\end{array}$

Therefore, the final speed of two +50nC charged plastic buttons is 2.4 m/s and the final speed of the +250nC charged button is 0 m/s.