Question

ld 1.0 mm apart. One particle has a mass of 1.0 x 10-5 kg and a net charge of...
Two particles are held 1.0 mm apart. One particle has a mass of 1.0 x 10-5 kg and a net charge
of + 3.2 x 10-10 C. The other particle has a mass of 2.0 x 10-5 kg and a net charge
of + 4.8 x10-10 C. If the two particles are released, what will their speeds be when they are 2.0
mm apart?

Answer 1

Answer on Question #73860-Physics-Field Theory

Two particles are held 1.0 mm apart. One particle has a mass of $1.0 \times 10^{-5} \, \mathrm{kg}$ and a net charge of $+3.2 \times 10^{-10} \, \mathrm{C}$. The other particle has a mass of $2.0 \times 10^{-5} \, \mathrm{kg}$ and a net charge of $+4.8 \times 10^{-10} \, \mathrm{C}$. If the two particles are released, what will their speeds be when they are 2.0 mm apart?

Solution

From the conservation of energy:

$$\frac {k q _ {1} q _ {2}}{r _ {1}} = \frac {k q _ {1} q _ {2}}{r _ {2}} + \frac {M V ^ {2}}{2}$$ $$\frac {\mu V ^ {2}}{2} = k q _ {1} q _ {2} \left(\frac {1}{r _ {1}} - \frac {1}{r _ {2}}\right)$$

The speed of center of the mass is

$$V = \sqrt {(m _ {1} + m _ {2}) \frac {2 k q _ {1} q _ {2}}{m _ {1} m _ {2}} \left(\frac {1}{r _ {1}} - \frac {1}{r _ {2}}\right)}$$

The speed of first particle is

$$\begin{array}{l} v _ {1} = \frac {m _ {2}}{(m _ {1} + m _ {2})} V = \sqrt {\frac {2 k q _ {1} q _ {2} m _ {2}}{(m _ {1} + m _ {2}) m _ {1}} \left(\frac {1}{r _ {1}} - \frac {1}{r _ {2}}\right)} \\ = \sqrt {\frac {2 (9 \cdot 1 0 ^ {9}) (3 . 2 \cdot 1 0 ^ {- 1 0}) (4 . 8 \cdot 1 0 ^ {- 1 0}) (2 \cdot 1 0 ^ {- 5})}{(1 \cdot 1 0 ^ {- 5} + 2 \cdot 1 0 ^ {- 5}) 1 \cdot 1 0 ^ {- 5}} \left(\frac {1}{0 . 0 0 1} - \frac {1}{0 . 0 0 2}\right)} = 0. 3 0 \frac {m}{s} \\ \end{array}$$

The speed of second particle is

$$\begin{array}{l} v _ {2} = \frac {m _ {1}}{(m _ {1} + m _ {2})} V = \sqrt {\frac {2 k q _ {1} q _ {2} m _ {1}}{(m _ {1} + m _ {2}) m _ {2}} \left(\frac {1}{r _ {1}} - \frac {1}{r _ {2}}\right)} \\ = \sqrt {\frac {2 (9 \cdot 1 0 ^ {9}) (3 . 2 \cdot 1 0 ^ {- 1 0}) (4 . 8 \cdot 1 0 ^ {- 1 0}) (1 \cdot 1 0 ^ {- 5})}{(1 \cdot 1 0 ^ {- 5} + 2 \cdot 1 0 ^ {- 5}) 2 \cdot 1 0 ^ {- 5}} \left(\frac {1}{0 . 0 0 1} - \frac {1}{0 . 0 0 2}\right)} = 0. 1 5 \frac {m}{s} \\ \end{array}$$

Answer provided by https://www.AssignmentExpert.com