Minimum separation between incoming proton and alpha particle

[issacnewton]

Homework Statement

A proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$ particle with velocity $v$. Initially, $\alpha$ particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be
A. $e^2/4\pi\varepsilon_0 m v^2$
B. $5e^2/4\pi\varepsilon_0 m v^2$
C. $2e^2/4\pi\varepsilon_0 m v^2$
D. $4e^2/4\pi\varepsilon_0 m v^2$

Relevant Equations

Coulomb's law of interaction between charged particles

Proton is going towards the $\alpha$ particle. So, I am thinking of using the conservation of energy as the initial kinetic energy of the proton is known and initial interaction potential energy is zero. But, we don't know the kinetic energies of proton and $\alpha$ particle when they are at their minimum separation. So, what could be possible approach here ?

Thanks

 

[PeroK]

Homework Statement: A proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$ particle with velocity $v$. Initially, $\alpha$ particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be
A. $e^2/4\pi\varepsilon_0 m v^2$
B. $5e^2/4\pi\varepsilon_0 m v^2$
C. $2e^2/4\pi\varepsilon_0 m v^2$
D. $4e^2/4\pi\varepsilon_0 m v^2$
Homework Equations: Coulomb's law of interaction between charged particles

Proton is going towards the $\alpha$ particle. So, I am thinking of using the conservation of energy as the initial kinetic energy of the proton is known and initial interaction potential energy is zero. But, we don't know the kinetic energies of proton and $\alpha$ particle when they are at their minimum separation. So, what could be possible approach here ?

Thanks

One approach is to consider a frame of reference in which the kinetic energy at the point of minimal separation is known.

 

[issacnewton]

But, we don't know the kinetic energy at the point of minimal separation.

 

[PeroK]

But, we don't know the kinetic energy at the point of minimal separation.

Can you think of a reference frame in which you would know this?

 

[issacnewton]

Would that be a center of mass frame ? And would the kinetic energy at the point of minimal separation be zero in this frame ?

 

[PeroK]

Would that be a center of mass frame ? And would the kinetic energy at the point of minimal separation be zero in this frame ?

Yes and yes It's often a useful trick in collision problems.

 

[collinsmark]

But, we don't know the kinetic energy at the point of minimal separation.

This process is a perfectly elastic collision*, so both conservation of energy and conservation of momentum apply. So once you pick a frame of reference, you know the total energy for all time, because the total energy is a constant.

The tricky bit (as @PeroK alludes to) is choosing a good frame of reference. Kinetic energy is relative (even under Galilean transformations). So you'll first need to pick a frame of reference and stick with it, and only then calculate kinetic energies.

What can you say about the center of mass of this system?

*(Although the process doesn't happen instantly, and there is always a finite separation between the particles, the process is technically a collision.)

[Edit: already beaten to the punch.]

 

[issacnewton]

Well, if the kinetic energy in the CM frame is zero, we must have both particles stationary. So, it doesn't make sense.

 

[PeroK]

Well, if the kinetic energy in the CM frame is zero, we must have both particles stationary. So, it doesn't make sense.

Why does that not make sense?

 

[issacnewton]

Since proton is approaching from an infinite distance, I would guess that in CM frame, both the proton and alpha particle have some velocities. So, total kinetic energy can not be zero

 

[PeroK]

Since proton is approaching from an infinite distance, I would guess that in CM frame, both the proton and alpha particle have some velocities. So, total kinetic energy can not be zero

Kinetic energy is not the only form energy can take!

 

[issacnewton]

I know that there will be interaction potential energy. But at infinite distances, there is no potential energy, there is only kinetic energy present. So initially, total kinetic energy can not be zero

 

[PeroK]

I know that there will be interaction potential energy. But at infinite distances, there is no potential energy, there is only kinetic energy present. So initially, total kinetic energy can not be zero

Of course not initially. The critical moment is when there is minimum separation.

 

[issacnewton]

But I am not convinced why total kinetic energy would be zero at the minimum separation ? Both could be moving. Proton will be slowing down and alpha particle will be speeding up. Can we analytically prove that total kinetic energy is zero at the minimum separation ?

 

[PeroK]

But I am not convinced why total kinetic energy would be zero at the minimum separation ? Both could be moving. Proton will be slowing down and alpha particle will be speeding up. Can we analytically prove that total kinetic energy is zero at the minimum separation ?

Think about momentum.

 

[issacnewton]

Let proton have velocity of $v_1$ and let alpha particle have velocity of $v_2$ when they are near. Now, initially proton has velocity of $v$. Now, we can think of an alpha particle as having almost 4 times mass of a proton. So conservation of linear momentum gives us
$$mv = m v_1 + 4m v_2 $$
Also, since this can be treated as the perfectly inelastic collision, the relative velocity is same except for the signs.
$$v - 0 = -(v_1 - v_2)$$
Well, solving this together, we reach the conclusion that $2 v_1 = -3v_2$. So, how does it follow from here that total kinetic energy is zero when they are near ?

 

[PeroK]

Let proton have velocity of $v_1$ and let alpha particle have velocity of $v_2$ when they are near. Now, initially proton has velocity of $v$. Now, we can think of an alpha particle as having almost 4 times mass of a proton. So conservation of linear momentum gives us
$$mv = m v_1 + 4m v_2 $$
Also, since this can be treated as the perfectly inelastic collision, the relative velocity is same except for the signs.
$$v - 0 = -(v_1 - v_2)$$
Well, solving this together, we reach the conclusion that $2 v_1 = -3v_2$. So, how does it follow from here that total kinetic energy is zero when they are near ?

What about the following argument:

In the CoM frame, total momentum is always zero. The two particles must always have equal and opposite momentum. Before the collision, the two particles are moving towards each other with equal and opposite momentum. After the collision, therefore, they must be moving away from each other with equal and opposite momentum.

At some instant, therefore, ...

 

[issacnewton]

Sorry for the late reply, was busy with work... Ok, so it becomes must easier in CoM frame. So, there is some moment, when both the momenta are zero. Hence both the kinetic energies are zero and the total initial kinetic energy has been converted into the potential energy of the interaction. Now let $v_1$ be the proton velocity in CoM frame and $v_2$ be the velocity of the alpha particle in the CoM frame. Now, $v$ is the initial proton velocity in lab frame and alpha particle is at rest initially. So, velocity of the CoM would be
$$ v_{CM} = \frac{mv + 0 }{m+4m} = \frac{v}{5} $$
Now, we can write the equations relating different velocities. $v = \frac{v}{5} + v_1$ and $0 = \frac{v}{5} + v_2$. Using this, we get $v_1 = \frac{4v}{5}$ and $v_2 = \frac{-v}{5}$. Now the initial kinetic energy would be converted into the electrical potential energy at the minimum separation. So, we can write the equation
$$ \frac{1}{2} m v_1^2 + \frac{1}{2} 4m v_2^2 = \frac{(e)(4e)}{4\pi\varepsilon_0 x} $$
where $x$ is the minimum separation of two particles when they both are at rest in CoM frame. Now, plugging the values of $v_1$ and $v_2$ and solving for $x$, we get
$$ x = \frac{5 e^2}{2\pi\varepsilon_0 m v^2} $$

So, this does not match with any of the options given. So, what am I doing wrong ?

 

[hutchphd]

What is the charge of the alpha??

 

[issacnewton]

Thanks hutch. Alpha has charge $2e$, my bad. So, that makes it
$$ x = \frac{5e^2}{\pi\varepsilon_0 m v^2} $$
But the problem assumes that the mass of an alpha particle is exactly 4 times the mass of the proton. Which is not true since proton and neutron do not have the same mass. So, person stating the problem is not careful here. We need to make some approximations here.
Thanks !

 

[hutchphd]

Its past my "good brain" hours but didn't you do that backwards?

 

[issacnewton]

I just evaluated for $x$ again after changing the charge of the alpha particle

 

[PeroK]

So, we can write the equation
$$ \frac{1}{2} m v_1^2 + \frac{1}{2} 4m v_2^2 = \frac{(e)(4e)}{4\pi\varepsilon_0 x} $$
where $x$ is the minimum separation of two particles when they both are at rest in CoM frame. Now, plugging the values of $v_1$ and $v_2$ and solving for $x$, we get
$$ x = \frac{5 e^2}{2\pi\varepsilon_0 m v^2} $$

So, this does not match with any of the options given. So, what am I doing wrong ?

Thanks hutch. Alpha has charge $2e$, my bad. So, that makes it
$$ x = \frac{5e^2}{\pi\varepsilon_0 m v^2} $$

As @hutchphd mentioned, you've bumped the charge on the alpha particle up to $8e$ now.

But the problem assumes that the mass of an alpha particle is exactly 4 times the mass of the proton. Which is not true since proton and neutron do not have the same mass. So, person stating the problem is not careful here. We need to make some approximations here.
Thanks !

Yes, that was odd. But, it was clear from the possible answers that they must be assuming that. Otherwise, you would get a non-integer factor in there.

 

[collinsmark]

So, that makes it
$$ x = \frac{5e^2}{\pi\varepsilon_0 m v^2} $$

I just evaluated for $x$ again after changing the charge of the alpha particle

I believe what @hutchphd and @PeroK are getting at is all of the given answers (in the original post) have a "4" as part of the denominator. You seem to have accidentally omitted that "4" in your answer.

 

[issacnewton]

I did not get any email for rest of the replies,. I just checked all the new messages here. Sorry for late reply. I think I made an algebraic error. I now do get
$$ x = \frac{5e^2}{4 \pi \varepsilon_0 m v^2} $$
which is option B. Thanks all.