PatchworkTeapot
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A proton is traveling to the right at 2.0×107m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton.

I know that you have to use conservation of kinetic energy and conservation of momentum, but I can't find a way to relate them to solve for one unknown without needing to know the other unknown. Can anyone help me please?
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Hamoody
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What exactly are you trying to find?
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PatchworkTeapot
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Oh dear, I completely forgot to include it. What are the final velocities of the carbon atom and the proton?
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Absent Agent
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(Original post by PatchworkTeapot)
Oh dear, I completely forgot to include it. What are the final velocities of the carbon atom and the proton?
How was your attempt? Is there any specific part you don't know what to do?
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Rek'Sa
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Momentum before the collision
p1 = m1 * v1
= 12 * 2 * 10^7
= 2.4 * 10^8

Momentum after the collision equals momentum before the collision
p1 = p2
p1 = m2 * v2
2.4 * 10^8 = (12+1) * v2
v2 = 2.4 * 10^8 / 13 = 1.85 * 10^7 m/s to the right
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Rek'Sa
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So in words basically, as it is a perfectly elastic collision, no kinetic energy is loss. Therefore the momentum before the collision equals momentum after the collision (conservation of momentum). We can then calculate the momentum before the collision by multiplying the mass of the proton to its velocity (don't worry about the unit of the mass of the proton because it'll cancel out with the unit of the mass of the carbon atom), and then divide it by the sum of the mass of carbon atom and the mass of proton to get the velocity of both particles travelling in the same direction (to the right, as the symbol doesn't change). I don't know whether your question has something special (like some stuff in A-levels) or not but mine uses only GCSE knowledge.
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Absent Agent
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(Original post by Rek'Sa)
Momentum before the collision
p1 = m1 * v1
= 12 * 2 * 10^7
= 2.4 * 10^8

Momentum after the collision equals momentum before the collision
p1 = p2
p1 = m2 * v2
2.4 * 10^8 = (12+1) * v2
v2 = 2.4 * 10^8 / 13 = 1.85 * 10^7 m/s to the right
I think you have made two incorrect assumptions in your calculations. One is that the proton will travel in the same direction as the carbon atom and that it also will travel with the same speed, but correct me if I'm wrong.
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Rek'Sa
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(Original post by Mehrdad jafari)
I think you have made two incorrect assumptions in your calculations. One is that the proton will travel in the same direction as the carbon atom and that it also will travel with the same speed, but correct me if I'm wrong.
I have made the point already. In the IGCSE syllabus for CIE board, according to the textbook, that is the only way we can solve collision problems using the principle of conservation of momentum(by assuming that the 2 objects stick together during the collision) or at least that's what I only know so far as I have just started Year 10 this term. Assuming that this question is a GCSE question, and not an A Level Physics or beyond question which I do not have the knowledge of the syllabus, my solution should be correct. However, if the latter is the case, my solution may be wrong. All this not taking into account of what would happen in the real world or the chemistry of hydrogen ions and carbon atoms.
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Absent Agent
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(Original post by Rek'Sa)
I have made the point already. In the IGCSE syllabus for CIE board, according to the textbook, that is the only way we can solve collision problems using the principle of conservation of momentum(by assuming that the 2 objects stick together during the collision) or at least that's what I only know so far as I have just started Year 10 this term. Assuming that this question is a GCSE question, and not an A Level Physics or beyond question which I do not have the knowledge of the syllabus, my solution should be correct. However, if the latter is the case, my solution may be wrong. All this not taking into account of what would happen in the real world or the chemistry of hydrogen ions and carbon atoms.
Because the proton has a less mass than the carbon atom (it's also assumed that it's a carbon ion with no electrons as the electrons would disturb the collision since they would be attracted to the proton), and that the collision is perfectly elastic, during the collision both particles exert equal and opposite forces on each other, resulting in the carbon atom traveling with a particular velocity in the original direction as the proton's (because it was stationary), and with proton traveling in the opposite direction to it's original velocity because it has a less mass.

For the proton after the collision to travel in the same direction as its original velocity, it should have a greater mass than the carbon atom so that it's motion is only ******ed (and not stopped or reversed). If the proton and the carbon atom were of equal mass, then after the collision the proton would remain stationary and the carbon atom would travel with the same velocity as the initial velocity of the proton.
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testing1234568
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(Original post by Rek'Sa)
Momentum before the collision
p1 = m1 * v1
= 12 * 2 * 10^7
= 2.4 * 10^8

Momentum after the collision equals momentum before the collision
p1 = p2
p1 = m2 * v2
2.4 * 10^8 = (12+1) * v2
v2 = 2.4 * 10^8 / 13 = 1.85 * 10^7 m/s to the right
Hint: You haven't used the fact that energy is conserved. If you could answer the question without that info, why would they bother giving it to you?
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Rek'Sa
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(Original post by Mehrdad jafari)
Because the proton has a less mass than the carbon atom (it's also assumed that it's a carbon ion with no electrons as the electrons would disturb the collision since they would be attracted to the proton), and that the collision is perfectly elastic, during the collision both particles exert equal and opposite forces on each other, resulting in the carbon atom traveling with a particular velocity in the original direction as the proton's (because it was stationary), and with proton traveling in the opposite direction to it's original velocity because it has a lass man.

For the proton after the collision to travel in the same direction as its original velocity, it should have a greater mass than the carbon atom so that it's motion is only ******ed (and not stopped or reversed). If the proton and the carbon atom were of equal mass, then after the collision the proton would remain stationary and the carbon atom would travel with the same velocity as the initial velocity of the proton.
That was what I missed in my solution, thank you the lucid explanation. I suppose this explanation may be able to help the original poster too.
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Rek'Sa
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(Original post by Mehrdad jafari)
Because the proton has a less mass than the carbon atom (it's also assumed that it's a carbon ion with no electrons as the electrons would disturb the collision since they would be attracted to the proton), and that the collision is perfectly elastic, during the collision both particles exert equal and opposite forces on each other, resulting in the carbon atom traveling with a particular velocity in the original direction as the proton's (because it was stationary), and with proton traveling in the opposite direction to it's original velocity because it has a lass man.

For the proton after the collision to travel in the same direction as its original velocity, it should have a greater mass than the carbon atom so that it's motion is only ******ed (and not stopped or reversed). If the proton and the carbon atom were of equal mass, then after the collision the proton would remain stationary and the carbon atom would travel with the same velocity as the initial velocity of the proton.
By the way, I don't understand this part 'with proton traveling in the opposite direction to it's original velocity because it has a lass man'. What does lass man mean? I searched it on Google and nothing related to physics seems to appear. Is it a typo or a very technical term?
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Absent Agent
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(Original post by Rek'Sa)
By the way, I don't understand this part 'with proton traveling in the opposite direction to it's original velocity because it has a lass man'. What does lass man mean? I searched it on Google and nothing related to physics seems to appear. Is it a typo or a very technical term?
haha, sorry, I meant to say "less mass". I have been inattentive recently . Edited.
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