$Find the antiderrivative of $\int \frac{v^2-v_o^2}{2\frac{K_e\frac{q_1q_2}{r^2}}{m} } dr$$

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$Find the antiderrivative of $\int \frac{v^2-v_o^2}{2\frac{K_e\frac{q_1q_2}{r^2}}{m} } dr$$

I'll admit I'm not the best when it comes to calculus. Anyway, I'm trying to model how close two nuclei will get to each other for a given starting velocity. Since they're charged particles, the force between them abides by Coloumb's law: $f =K_e\frac{q_1q_2}{r^2} $. Since nuclei are not quantum particles, they also abide by Newton's second law: $f=ma$. Therefore:

$ma=K_e\frac{q_1q_2}{r^2} $

Solving for a:

$a =\frac{K_e\frac{q_1q_2}{r^2} }{m} $

We know from one of the kinematic equations:

$v^2=v_o^2+2a\triangle x$

Transposing to solve for $\triangle x$:

$\triangle x=\frac{v^2-v_o^2}{2a} $

Plugging in the equation for acceleration, we get:

$\triangle x=\frac{v^2-v_o^2}{2\frac{K_e\frac{q_1q_2}{r^2}}{m} } $

Since $\triangle x$ is dependent on acceleration, which is constantly increasing as distance is decreasing, if I remember calculus correctly, I can take the definite integral with respect to r to get the total distance each nuceli travels:

$\int_{a}^{b} \frac{v^2-v_o^2}{2\frac{K_e\frac{q_1q_2}{r^2}}{m} } dr$

To get that, I'd need to take the antiderrivative:

$\int \frac{v^2-v_o^2}{2\frac{K_e\frac{q_1q_2}{r^2}}{m} } dr$

The problem is, that's the extent of my mathematical ability. We never really got to problems this complex in the sole calculus class I've taken.

So, I have two questions:
1. Since every variable except r is an independent variable, can they simply be treated as constants, and
2. What is the antiderrivative?

Calculus

Rpayne88

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Alessandro Iraci

0

Do you know the mass of both nuclei? If so, I think you can do much better by just using conservation of energy. The equations of motion get a lot more complicated than what you described.
Alessandro Iraci

0

Your approach does not work because the velocity depends on the distance, you cannot just treat it as a constant.
Alessandro Iraci

0

I don't think there is enough information to solve this problem. Does one nucleus just stay still the whole time, while the other moves towards it, or are both of them free to move? Does one start still while the other moves, or do both of them have non-zero initial velocities?
Alessandro Iraci

0

I solved the problem in a couple simple cases, let me know if that is enough or you have different specifications.

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Alessandro Iraci

0

Do you know the mass of both nuclei? If so, I think you can do much better by just using conservation of energy. The equations of motion get a lot more complicated than what you described.
Alessandro Iraci

0

Your approach does not work because the velocity depends on the distance, you cannot just treat it as a constant.
Alessandro Iraci

0

I don't think there is enough information to solve this problem. Does one nucleus just stay still the whole time, while the other moves towards it, or are both of them free to move? Does one start still while the other moves, or do both of them have non-zero initial velocities?
Alessandro Iraci

0

I solved the problem in a couple simple cases, let me know if that is enough or you have different specifications.

Answer 1

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Assuming that one particle is still and it is kept in place by some force, and that the other is moving towards it with initial distance $r_0$, initial velocity $v_0$, and mass $m$, we have that the initial energy is given by \[ E_0 = \frac{1}{2}m v_0^2 + \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_0}. \] The minimum distance is achieved when $v = 0$ (then it changes sign and the moving particle starts moving away from the other), so in that moment we have \[ E_f = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_f} \] where $r_f$ is the minimum distance we're looking for. Since the energy is preserved, $E_0 = E_f$. Solving for $r_f$, we have \[ r_f = \frac{q_1 q_2}{4 \pi \epsilon_0} \frac{1}{E_0} \] and all the variables are known, so you can just compute it.

If both particles are free to move, we can use the reference system in which the two particles are moving towards each other with velocity $\frac{v_0}{2}$, so the relative velocity is the same. I need to assume that the mass is the same, otherwise the problem becomes a lot more difficult. In this case, we have \[ E_0 = \frac{1}{4}m v_0^2 + \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_0}, \] where we get $1/4$ because we have two moving particles, each with half of the velocity as before, so $1/4$ of the kinetic energy. Because the system is symmetric, they will reach velocity $0$ simultaneously, so we get again \[ E_f = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_f} \] where $r_f$ is the minimum distance we're looking for. Since, again, the energy is preserved, $E_0 = E_f$. Solving for $r_f$, we have \[ r_f = \frac{q_1 q_2}{4 \pi \epsilon_0} \frac{1}{E_0}. \]

If the system is not symmetric, then you need to derive the equation of the motion explicitly, which is a lot more difficult. See e.g. https://math.stackexchange.com/questions/681810/one-dimensional-inverse-square-laws for the equations in the first case: with conservation of energy we can get the distance we want, but describing the motion explicitly is really difficult, you will need a ODE solver or something.

Alessandro Iraci

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$Find the antiderrivative of $\int \frac{v^2-v_o^2}{2\frac{K_e\frac{q_1q_2}{r^2}}{m} } dr$$

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