# Distance of closest approach of particles

1. Feb 3, 2009

### KillerZ

1. The problem statement, all variables and given/known data

A proton (q = 1e, m = 1u) and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, with an initial speed of 0.01c. What is their distance of closest approach, as measured between their centers? (Hint: There are two conserved quantities. Make use of both.)

2. Relevant equations

$$K_i + U_i = K_f + U_f$$

$$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

3. The attempt at a solution

How does this look?

$$e = 1.60 * 10^{-19} C$$
$$c = 3.00 * 10^8 m/s$$
$$u = 1.661 * 10^{-27} kg$$

$$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$
$$(4u)(3 * 10^8 m/s) - (u)(3 * 10^8 m/s) = (4u + u)v_{f}$$

$$v_{f} = \frac{(4u)(3 * 10^8 m/s) - (u)(3 * 10^8 m/s)}{(4u + u)}$$
$$v_{f} = 1.80 * 10^8 m/s$$

$$\frac{1}{2}(m_{1}+m_{2})v_{f}^{2} + \frac{1}{4\pi\epsilon_0}(\frac{q1q2}{\infty}) = \frac{1}{2}(m_{1}+m_{2})v_{i}^{2} + \frac{1}{4\pi\epsilon_0}(\frac{q1q2}{r})$$

$$\frac{1}{2}(4u + u)(1.80*10^8 m/s)^{2} = \frac{1}{2}(4u + u)(3.00*10^8 m/s)^{2} + \frac{1}{4\pi\epsilon_0}(\frac{2(1.60*10^{-19} C)(1.60*10^{-19} C)}{r})$$

$$4\pi\epsilon_0(\frac{1}{2}(4u + u)(1.80*10^8 m/s)^{2} - \frac{1}{2}(4u + u)(3.00*10^8 m/s)^{2}) = (\frac{2(1.60*10^{-19} C)(1.60*10^{-19} C)}{r})$$

$$r= \frac{2(1.60*10^{-19} C)(1.60*10^{-19} C)}{4\pi\epsilon_0(\frac{1}{2}(4u + u)(1.80*10^8 m/s)^{2} - \frac{1}{2}(4u + u)(3.00*10^8 m/s)^{2})}$$

$$r = -1.92 * 10^{-18} m$$

2. Feb 3, 2009

### AssyriaQ

This might sound stupid (I am tired...) but, did not the problem state that the initial speed was 0.01c? As far as I can see, you used c as initial speed.

3. Feb 4, 2009