Hard Momentum Conservation Impact Problem

In summary: No. That's only for a head-on collision ##l = 0##. If ##l \ne 0## then neither particle stops.How so? I might have interpreted ##l## wrongly, but, even if it's the distance separating the charges, you can't have the angle between your origin and momentum be equal to ##\pi/2## for both charges at the same time..?
  • #1
Ayesha02
49
5
Homework Statement
Two identical charged particles each carrying charge q = 0.1 mC and of mass m = 10mg are projected along two parallel lines separated by a distance l, with equal speed V0= 104 m/s in opposite directions. In the beginning electrostatic interaction between the charges can be ignored due to a large distance between them. The minimum distance between the particles is found to be 12 cm. The value of l is equal to
Relevant Equations
Linear momentum conservation
I believe momentum conservation is to be used in this sum since there's no external force, but I am not sure how to write the equation.

Can someone please help me out:)
 
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  • #2
Ayesha02 said:
Homework Statement:: Two identical charged particles each carrying charge q = 0.1 mC and of mass m = 10mg are projected along two parallel lines separated by a distance l, with equal speed V0= 104 m/s in opposite directions. In the beginning electrostatic interaction between the charges can be ignored due to a large distance between them. The minimum distance between the particles is found to be 12 cm. The value of l is equal to
Relevant Equations:: Linear momentum conservation

I believe momentum conservation is to be used in this sum since there's no external force, but I am not sure how to write the equation.

Can someone please help me out:)

Have you studied something relevant leading up to this? Do you know what the "impact parameter" is?

In any case, conservation of energy and angular momentum are your friends here.
 
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  • #3
PeroK said:
Have you studied something relevant leading up to this? Do you know what the "impact parameter" is?

In any case, conservation of energy and angular momentum are your friends here.

not exactly the term impact parameter, but yeah I've done impulse- idk if that's the same..

Moving on, coming to angular momentum I am not quite sure how to write the equation- as in about what point am i conserving angular momentum? If possible, could you write the equation?
 
  • #4
Ayesha02 said:
not exactly the term impact parameter, but yeah I've done impulse- idk if that's the same..

Moving on, coming to angular momentum I am not quite sure how to write the equation- as in about what point am i conserving angular momentum? If possible, could you write the equation?
Try angular momentum about the centre of mass.
 
  • #5
Ayesha02 said:
not exactly the term impact parameter, but yeah I've done impulse- idk if that's the same..

It's a different concept.

Like @PeroK said, write the initial angular momentum, and the final angular momentum (what does the path of the particle look like when they're really far apart again).

What sort of shape do you get? What constraints can you put on the final velocities?
 
  • #6
PeroK said:
Try angular momentum about the centre of mass.

Okay so i gather final angular momentum should be 2(m*v*6)
is that right?

Im still not able to figure out initial angular momentum
 
  • #7
Ayesha02 said:
Okay so i gather final angular momentum should be 2(m*v*6)
is that right?

Im still not able to figure out initial angular momentum

an afterthought:

why not conserve linear momentum?
 
  • #8
Ayesha02 said:
an afterthought:

why not conserve linear momentum?
Because you have a spherically symmetric potential. Working with polar coordinates ##r, \phi## and angular momentum is going to be easier.

Initial angular momentum is just ##\pm mvl/2##, isn't it?
 
  • #9
PeroK said:
Because you have a spherically symmetric potential. Working with polar coordinates ##r, \phi## and angular momentum is going to be easier.

Initial angular momentum is just ##\pm mvl/2##, isn't it?

ohh yaa!

Another thing- why don't the bodies change velocity from the initial to the final?
 
  • #10
Ayesha02 said:
ohh yaa!

Another thing- why don't the bodies change velocity from the initial to the final?

If by final you mean when they're really far apart again, then what's their potential energy? Is any external work done?
 
  • #11
etotheipi said:
If by final you mean when they're really far apart again, then what's their potential energy? Is any external work done?
Nah!
the final situation, as mentioned in the question, is when they're at min distance
 
  • #12
Ayesha02 said:
Nah!
the final situation, as mentioned in the question, is when they're at min distance

So do they have the same kinetic energy as they do initially?
 
  • #13
etotheipi said:
So do they have the same kinetic energy as they do initially?

Exactly my doubt!
why do they have the same KE finally as well?
@PeroK could you help us?
 
  • #14
Ayesha02 said:
Exactly my doubt!
why do they have the same KE finally as well?
@PeroK could you help us?

Um no it was a question but okay. Can it have the same kinetic energy?
 
  • #15
Ayesha02 said:
Nah!
the final situation, as mentioned in the question, is when they're at min distance
It's that point that you need to analyse. Apart from minimum separation, what else can you say about the motion at that point? It's implied by minimum separation.
 
  • #16
PeroK said:
It's that point that you need to analyse. Apart from minimum separation, what else can you say about the motion at that point? It's implied by minimum separation.

They can't go any further right, so shouldn't velocity be zero at that point?
 
  • #17
Ayesha02 said:
They can't go any further right, so shouldn't velocity be zero at that point?
No. That's only for a head-on collision ##l = 0##. If ##l \ne 0## then neither particle stops.
 
  • #18
archaic said:
How so? I might have interpreted ##l## wrongly, but, even if it's the distance separating the charges, you can't have the angle between your origin and momentum be equal to ##\pi/2## for both charges at the same time..?

That's just the angular momentum of one particle at the point where they're separated at infinity.
 
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  • #19
PeroK said:
No. That's only for a head-on collision ##l = 0##. If ##l \ne 0## then neither particle stops.
Ohh alright then
 
  • #20
archaic said:
How so? I might have interpreted ##l## wrongly, but, even if it's the distance separating the charges, you can't have the angle between your origin and momentum be equal to ##\pi/2## for both charges at the same time..?

Here we go again. It's a) homework helping someone else and b) sorting out your problems too!

This isn't your homework!
 
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  • #21
I'm going to step back now because I think I'm interrupting @PeroK's flow and my phone's going slightly mad with notifications.

All I will say is that you can solve the problem by applying the two conservation laws he mentioned between the points where the two are separated at infinity, traveling along two parallel lines, and the point of closest approach. Don't miss out different types of energy!
 
  • #22
Ayesha02 said:
about what point am i conserving angular momentum?
Ayesha02 said:
why not conserve linear momentum?
Same answer to both... because of the symmetry, the total linear momentum of the system is zero. As a result, it doesn't matter what axis you use for the angular momentum of the system; they will all yield the same answer.
PeroK said:
Initial angular momentum is just ##\pm mvl/2##, isn't it?
I would simply say it is mv0l.
Ayesha02 said:
why do they have the same KE finally as well?
By conservation of energy, what will the KE be when they are at separation x?
 
  • #23
haruspex said:
Same answer to both... because of the symmetry, the total linear momentum of the system is zero. As a result, it doesn't matter what axis you use for the angular momentum of the system; they will all yield the same answer.

I would simply say it is mv0l.

By conservation of energy, what will the KE be when they are at separation x?

Yes dude got it thankyou:)
 

Related to Hard Momentum Conservation Impact Problem

1. What is hard momentum conservation impact problem?

The hard momentum conservation impact problem is a concept in physics that deals with the conservation of momentum in collisions between objects. It refers to situations where the impact between two objects is so strong that traditional momentum conservation equations cannot accurately predict the outcome.

2. How is hard momentum conservation impact problem different from regular momentum conservation?

The main difference between hard momentum conservation and regular momentum conservation is the magnitude of the impact. In regular momentum conservation, the objects involved have relatively low velocities and the impact is not significant enough to cause a change in the traditional momentum conservation equations. In hard momentum conservation, the objects have high velocities and the impact is strong enough to cause a deviation from the traditional equations.

3. What factors contribute to the occurrence of hard momentum conservation impact problem?

The main factors that contribute to the occurrence of hard momentum conservation impact problem are high velocities, large masses, and strong forces. These factors result in a significant change in the momentum of the objects involved, making it difficult to accurately predict the outcome using traditional equations.

4. How do scientists approach the hard momentum conservation impact problem?

Scientists approach the hard momentum conservation impact problem by using advanced mathematical models and computer simulations. These tools allow them to accurately predict the outcome of collisions between objects with high velocities and large masses, taking into account the effects of strong forces.

5. What are some real-life examples of hard momentum conservation impact problem?

Some real-life examples of hard momentum conservation impact problem include high-speed car crashes, collisions between particles in particle accelerators, and impacts between celestial bodies such as planets or asteroids. In these scenarios, the traditional momentum conservation equations are not sufficient to accurately predict the outcome of the impact.

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