In an isolated system of two colliding protons [ say, proton1(p1) and proton2(p2)];

initially p2 is at rest, and p1 is moving with uniform horizontal velocity $\vec{u_1}=a \hat{i}$ m/s.

First way of analysing the collision:

I have considered the Columbian Force as an internal force.

For an elastic collision of two objects,final velocity p1 isgiven by;

$$v_{1}=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2} \,\,equation(A) \,\, \text { and } v_{2}=\frac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2} \,\,equation(B)$$


where, $u_1=\text{initial velocity of p1}$

$v_1=\text{final velocity of p1}$

$u_2=\text{initial velocity of p2}$

$v_2=\text{final velocity of p2}$

$m_1=m_2=m=\text{mass of proton}$

Solving for $v_1; v_1=\frac{(m-m)*a+2m*0}{m+m}=0$ and $v_2=\frac{(m-m)*0+2m*a}{m+m}=a$

In a nutshell, p1 and p2 exchange their velocities upon collision.

(Similar situation explained at https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_7B_-_General_Physics/7%3A_Momentum_Conservation/7.2%3A_Applications_of_Momentum_Conservation)

Second way of analysing the collision:

As here, the particle at rest is free to move when one particle approaches the other, due to electrostatic repulsion other will also start moving and so the velocity of first particle will decrease while of other will increase and at closest approach both will move with same velocity. This gives us $v_1=v_2=\frac{a}{2}$.

So, the question is, which one is correct?

  • $\begingroup$ according to me ideally no collision should take place because as the protons approach the each other the coulombic force would increase and if the distance between them turns out to be nearly zero force would approach to infinity $\endgroup$
    – Anonymous
    Commented Nov 16, 2020 at 7:10
  • $\begingroup$ also in your first way why didn't you considered change in their electrostatic potential energy. $\endgroup$
    – Anonymous
    Commented Nov 16, 2020 at 7:13
  • $\begingroup$ also these charges are moving you cannot ignore the magnetic effects also $\endgroup$
    – Anonymous
    Commented Nov 16, 2020 at 7:16
  • $\begingroup$ @Pranav Aggarwal , I think this was a hypothetical scenario proposed by OP to "see" what when force at a distance is present during collision. It was a simple collision question, is no need to take magnetic effects into account. $\endgroup$ Commented Nov 18, 2020 at 1:45
  • $\begingroup$ @Pranav Protons aren't point particles, so the Coulomb force is limited. And if the protons have huge relative KE you can't treat their interaction as purely electrostatic, you have to take the nuclear forces into account. $\endgroup$
    – PM 2Ring
    Commented Nov 20, 2020 at 1:38

2 Answers 2


As you have said, Coulomb's force is an internal force. Hence momentum is always conserved and both methods would give the same answer(first one).

You must realize three things:

  1. Conservation of momentum is just another way of representing Newton's Laws.
  2. This is a situation of completely elastic collision with $e=1$. Here repulsion between the protons act as the spring.

enter image description here

  1. When two bodies collide there must always come a moment, when both bodies move with the same velocity. When this happens, the potential energy within the system is maximum. In your case, after the bodies acquire equal velocities, the internal Coulombic forces would repel the protons and take them away such that they exchange their velocities.

enter image description here

But when would the collision end?

Lets say this is the initial electric potential energy of the system: $U_{electric-initial}= k\frac {q^2}{r}$

The collision would end when $U_{electric-final}=U_{electric-inital}$

This if of-course in realm of classical mechanics. In reality when two protons collide, they may create new particles.


tl;dr For infinite initial separation, the final velocities of the given particles $P_1,P_2$ would indeed be $0,\sqrt{2 m_2E}$ where $E$ is the total initial energy of the system

Certain clarifications are in order

and p1 is moving with uniform horizontal velocity

In the Coulomb scattering of two charges, it is not possible for the incoming charge to maintain a uniform velocity. It necessarily experiences acceleration in the source charge's field.

I have considered the Columbian Force as an internal force.

Regardless of your consideration, the Coulomb interaction must be considered internal for energy and momentum conservation purposes since it acts equally and oppositely on the two charges.

For an elastic collision $\ldots$

Strictly speaking the collision isn't elastic as even at non-relativistic velocities, there is some some energy loss via EM radiation. However, for our purposes, we may consider it to be negligibly small .

$\ldots$of two objects$\ldots$

For low enough energies, the initially two in number protons do maintain their count, but not necessarily at higher energies wherein new particles may be created. However at such incredibly high energies, it is also likely that the proton-proton interaction isn't purely electromagnetic anymore and so our model won't apply anyways.

$\ldots$final velocity p1 is given by$\ldots$

The formulae you have used are used in hard scattering of point objects- the object have the incoming velocities $u_i$, outgoing velocities $v_i$ and only interact at the interaction vertex. Forever before and after, they don't interact and have uniform velocities. This model isn't true for our two colliding protons.

In fact it would be more appropriate to call this proton-proton or pp scattering as opposed to the classically annotated 'collision', though that's not criminal.

Critique of "First way of analysing the collision"

The above mentioned inapplicability of the hard scattering model notwithstanding, there are further problems in your application:

  • The formulae use $u_2$ - the initial velocity of the second proton $P_2$. You have taken this to be $0$. Why? As you soon state in your second approach, the second proton is being stirred into motion due to its interaction with the incoming particle. The only time its ever at rest in the lab frame is at the start of our analysis. So, evaluating the formulae at this instant, (that's what setting $u_2=0$ implies) hardly counts as evaluating their 'final' velocities. Won't you agree that a better instant to tag as 'the moment-of-collision' would have been the moment of closest approach, if never else? Without justification for applying the formulae at the starting epoch, the treatment isn't well reasoned.
  • In soft scattering, the protons' velocities forever keep changing. The only meaningful 'initial' and 'final' velocities to talk about are those when the forces on each of them are zero. This happens when their separation is infinite. To this end you, haven't mentioned whether $P_1$ has velocity $a$ at infinity or at some finite distance from $P_2$.

Yet the final result you arrive at via your reasoning in this approach is correct. At infinite separation$^1$, the final velocities of the given particles $P_1,P_2$ would indeed be $0,\sqrt{2 m_2E}$ where $E$ is the total initial energy of the system$^2$.

Why do the hard scattering formulae work as intended? This is because when we consider the start and final states of the scattering to be those at infinite separation, wrt these epochs we can bundle the Coulomb (or any interaction(s) for that matter) into the internal machinery of the interaction vertex. The hard scattering formulae are agnostic of the internals of this collision vertex and behave as if the incoming particle collided with another particle - the only difference being the vertex is no longer a point but the entire space and interaction time isn't an instant but infinity. If this is indeed what had motivated you to this approach, it wasn't clear from your question.

Critique of "Second way of analysing the collision"

The only thing lacking here was that you stopped your analysis at the moment of closest approach. Why? As discussed in the previous section, the final velocities are those when the particles have moved infinitely far away from each other again. So after their relatively stationary moment of closest approach, their mutual repulsion drives them away. Moreover, because of the symmetry of the problem, in the COM frame, the kinematics beyond this epoch are exactly as if those up to it had been time reversed.


$^1$ Assuming the particles were initially also infinitely separated; there is a way to extend the argument to finite separations though.

$^2$ For the given config, $$E=\frac{m_1 a^2}{2}+\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{L}\tag{SI units}$$

where $L$ is the initial separation of the two charges.


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