Electrostatics (Distance of closest approach)

[Mandeep Deka]

1. A proton an an alpha-particle are projected with a velocity v=(Ke^2/mL)^0.5 each, when they are far away from each other. If their initial velocity vectors are anti parallel to each other, with a distance "L" between them, find the distance of closest approach.(mass of proton = m, charge of proton= +e, mass of alpha-particle= 4m, charge of alpha particle= +2e)
Ans: L[5+(89)^0.5]/8



2. F=kqQ/r^2, U=kqQ/r

3. Conservation of momentum and conservation of energy are the two prime theorems that are applicable in this case, as there is no external agent doing work, but they still are insufficient to eliminate the number of variables. As both the particles experience force (changing with time), towards each other they deflect from each other and at some point they are closest to each other... this is all i could sort out...

Plz help!

 

[Doc Al]

Conservation of momentum and conservation of energy are the two prime theorems that are applicable in this case, as there is no external agent doing work, but they still are insufficient to eliminate the number of variables.

Why do you think they are insufficient? Hint: What will be the speed of the particles when they reach the distance of closest approach?

Ans: L[5+(89)^0.5]/8

FYI, this answer doesn't quite make sense. Better double check it.

 

[graphene]

Could you put down your solution here? How did you obtain that value for L?

 

[Swap]

you don't even need to bother about the momentum. The required eq.(in words)
Kinetic energy of both the particles= Potential energy of the system at the closest point of approach. ( Remember the velocity at the closest point of approach is zero)

 

[Doc Al]

you don't even need to bother about the momentum. The required eq.(in words)
Kinetic energy of both the particles= Potential energy of the system at the closest point of approach. ( Remember the velocity at the closest point of approach is zero)

That is incorrect.

 

[Swap]

why?

 

[Doc Al]

why?

Consider the total momentum of the particles.

 

[Swap]

oh yes. At the time of minimum approach the velocity of proton will be 0 while the velocity of the alpha particle will be 3V right.??

 

[Doc Al]

oh yes. At the time of minimum approach the velocity of proton will be 0 while the velocity of the alpha particle will be 3V right.??

No. If that were true, they'd still be approaching each other.

 

[Swap]

no not really. once the protons velocity has changed direction the proton will move away and thus distance between them will start to increase. We can consider it as two carts of different masses coming towards each other with a spring attached to them.
If not so, it is really not possible to solve this since u r left with 3 variables.

 

[Doc Al]

no not really. once the protons velocity has changed direction the proton will move away and thus distance between them will start to increase. We can consider it as two carts of different masses coming towards each other with a spring attached to them.
If not so, it is really not possible to solve this since u r left with 3 variables.

What you are missing the relative velocity of the particles at the point of closest approach. (It is the same as the two carts with spring problem!)

 

[Swap]

can you explain more. I dun understand what do u mean by the "relative velocity of the particles at the point of closest approach".

 

[Doc Al]

can you explain more. I dun understand what do u mean by the "relative velocity of the particles at the point of closest approach".

Think about it. You made the statement that at the distance of closest approach, one particle would have speed 0 while the other had speed 3V. If that were so, the second would still be approaching the first particle. They would still be getting closer.

It's exactly the same with the two carts and spring problem. How must the speeds of the two carts relate when the spring is maximally compressed?

 

[Swap]

oh ok...I got it. That means at the point of minimum separation the velocity of both of them will be equal is it?

 

[Doc Al]

That means at the point of minimum separation the velocity of both of them will be equal is it?

Exactly!

 

[Swap]

Thanks a lot to make my concept clear. =p

 

[Mandeep Deka]

so guys is there any way to get that weird looking answer...
since the difficulty in doing the qs has been established, is there any way out of it... coz i havnt found any n nor has my teacher been so far!

so far the qs is concerned it is perfectly correct... and the ans too is!
if anyone can find the answer to that question...please do inform!

(i would specially request any student from kota institutes, like bansal, vibrant etc, to get that solution coz as far as i know, the qs has its origin in some institute of kota only!)

 

[Doc Al]

What makes you think the answer is correct? Where's your attempted solution? It seems like a fairly straightforward problem to me.

 

[Swap]

Agree with Doc al. It is a straight forward question. Moreover I got a very nice answer. It is (5/9)L. ( I don't really trust the coaching institutes u r talking abt, I have sometimes find mistake in their solutions to IIT JEE question paper.)

 

[Mandeep Deka]

even, i had thought that the answer is wrong, but i just wanted to know if anyone had some solution or if there was anyway that answer came up...
thanks for helping...

by the way there was another problem of this sort...


Two particles of equal charge and mass (q,m) are infinitely far from each other. At some instant of time one particle is projected towards other with a velocity 'v'. Find the distance of closest approach.

 

[Doc Al]

even, i had thought that the answer is wrong, but i just wanted to know if anyone had some solution or if there was anyway that answer came up...
thanks for helping...

Why don't you take a shot at it and post your solution?

by the way there was another problem of this sort...


Two particles of equal charge and mass (q,m) are infinitely far from each other. At some instant of time one particle is projected towards other with a velocity 'v'. Find the distance of closest approach.

Essentially the same basic idea. If you can solve one, you can solve the other.

 

[Mandeep Deka]

I found out the initial momentum of the system to be 3mv(direction along the v vector of the alpha particle). Let's say it to be 3mv(-i). But when they r the closest to each other, we do not know what their velocity vectors are. Let at the time when the particles are closest to each other, the velocities of the proton and alpha particle be, [v1(i)+v2(j)] and [V1(-i)+V2(-j)], respectively. Then by equating the momentum with the initial momentum, we get v2=4(V2), and 3v=v1-4(V1). Now if we try to equate the energy, we have v1,v2,V1,V2, all as variables but we can't find enough equations.

If u have solved it and found some answer, in soem different way, please put down your solution.

 

[Doc Al]

I found out the initial momentum of the system to be 3mv(direction along the v vector of the alpha particle). Let's say it to be 3mv(-i).

OK.

But when they r the closest to each other, we do not know what their velocity vectors are.

But that's easy to figure out using conservation of momentum. Have you read the discussion in this thread? The key idea is given: At the distance of closest approach, what's the relative velocity of the particles?

 

[sagardip]

Excuse me swap, what u are thinking is, the question is the same as that of the spring cart problem.But mind u in this case, the distance between the paths of the particles is L.So as they approach each other their paths will be of the shape of a rectangular hyperbola and so their velocities will not be in the same previous direction

 

[Doc Al]

But mind u in this case, the distance between the paths of the particles is L.

Ah... good point. I missed that myself in reading the problem!

Angular momentum will be conserved.

 

[Swap]

Are the two particles parallel to each other moving in opposite directions? and I don't understand what is meant by the distance between the paths. Is the perpendicular distance between the two parallel lines??

 

[Mandeep Deka]

the perpendicular distance between the particles is L and they are projected towards each other, such that their velocity vectors are anti parallel. its like in an x-y plane one particle is at (0,0) , with initial velocity v(i)(vector), and other at (infinity,L) with initial velocity, v(-i)(vector). I hope u got it!

 

[Swap]

if the path is a rectangular hyperbola and the particles are at infinity then how is it that the perpendicular distance between them is L. It should be 0. The equation of a rectangular hyperbola is xy=c^2. So if x→∞ y→0. So the perpendicular distance is 0.

 

[Mandeep Deka]

the path aint a rectangular hyperbola. Its just that the particles deflect from the trajectories and their trajectories somewhat seem to look like a rectangular hyperbola (just the shape). The distance between them is L only! i hope u got it!