Unequal charge and mass Attraction

Consider the mass $M_A$ to have an x coordinate of $x_1$ and the mass $M_B$ to have an x coordinate of $x_2$ .

The mass $M_A$ has its centre initially located at the origin and the mass $M_B$ has its centre located at the point (3,0).

Applying Newton's second law for each of the masses gives:

$M_A \ddot{x}_1 = \frac{GM_AM_B}{(x_2-x_1)^2} + \frac{Q_AQ_B}{4\pi(x_2-x_1)^2}$ $M_B \ddot{x}_2 = -\frac{GM_AM_B}{(x_2-x_1)^2} - \frac{Q_AQ_B}{4\pi(x_2-x_1)^2}$

Let:

$x = x_2-x_1$

Now, the equations of motion can be re-written as:

$\ddot{x}_1 =\frac{M_B}{x^2} - \frac{Q_AQ_B}{4\pi M_A x^2}$ $\ddot{x}_2 =-\frac{M_A}{x^2} + \frac{Q_AQ_B}{4\pi M_B x^2}$

Subtracting both these equations gives:

$\ddot{x}_1-\ddot{x}_2=\left[M_A+M_B - \frac{Q_AQ_B}{4\pi}\left(\frac{1}{M_A} + \frac{1}{M_B}\right)\right] \frac{1}{x^2}$

Let:

$K = M_A+M_B - \frac{Q_AQ_B}{4\pi}\left(\frac{1}{M_A} + \frac{1}{M_B}\right)$

The equation of motion simplifies to:

$\ddot{x} = \frac{K}{x^2}$ $x(0) = 3$ $\dot{x}(0) = 0$

The above equation can be re-arranged as such:

$\ddot{x} = \dot{x}\frac{d \dot{x}}{dx} = \frac{K}{x^2}$ $\implies \dot{x}\frac{d \dot{x}}{dx} = -\frac{K}{x^2}$

An extra minus sign is introduced above as x decreases $\dot{x}$ increases. Separating variables and integrating:

$\dot{x}^2 = 2K \left(\frac{1}{x}-\frac{1}{3}\right)$

$\implies \dot{x} = - \sqrt{2K \left(\frac{1}{x}-\frac{1}{3}\right)}$

Minus sign introduced here, for the same reason as stated above:

$\implies \dot{x} = - f(x)$

The particles collide at the instant when $x=2$ . Separating the variables and integrating again gives:

$\boxed{\int_{2}^{3} \frac{dx}{f(x)} = T \approx 1.219334680392212}$

Nice one. This is a great exercise for practicing numerical simulation.

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import math

# Constants

dt = 10.0**(-6.0)

MA = 7.0
MB = 11.0

QA = 17.0
QB = 23.0

R = 1.0

G = 1.0
e0 = 1.0
k = 1.0/(4.0*math.pi*e0)

###################################

# Initialize simulation

t = 0.0

xA = 0.0
xB = 3.0

xAd = 0.0
xBd = 0.0

xAdd = 0.0
xBdd = 0.0

D = xB - xA

###################################

# Run simulation

while D >= 2.0*R:

    xA = xA + xAd*dt
    xB = xB + xBd*dt

    xAd = xAd + xAdd*dt
    xBd = xBd + xBdd*dt

    D = xB - xA

    Fg = G*MA*MB/(D**2.0)
    Fe = k*QA*QB/(D**2.0)

    FA = Fg - Fe
    FB = -Fg + Fe

    xAdd = FA/MA
    xBdd = FB/MB

    t = t + dt

###################################

# Print results

print dt
print t

#>>> 
#0.0001
#1.2196
#>>> ================================ RESTART ================================
#>>> 
#1e-05
#1.21936
#>>> ================================ RESTART ================================
#>>> 
#1e-06
#1.21933699999
#>>>