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# Edexcel Unit 4: Physics on the Move 6PH04 (11th June 2015) watch

• View Poll Results: What do you think the grade boundary for an A will be?
GCE-65+
3
3.57%
GCE-60-64
28
33.33%
GCE-55-59
31
36.90%
GCE-50-54
8
9.52%
GCE-45-49
1
1.19%
GCE- Lower than 44
0
0%
IAL-65+
5
5.95%
IAL-60-64
6
7.14%
IAL-55-59
2
2.38%
IAL-50-54
0
0%
IAL-45-49
0
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IAL- Lower than 44
0
0%

1. In NBA, can someone explain what area A represents - I've completely forgotten ?

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2. (Original post by BP_Tranquility)
In NBA, can someone explain what area A represents - I've completely forgotten ?

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Size of area. Ie where the flux cuts it over that space.
So BA is the product if the [component of] flux density perpendicular to area, and the size of area.

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3. (Original post by imedico10)
Flow of conventional current is left to right so electrons flow right to left making the right wing positively charged
This is wrong,
Conventional current in this case from FLHr is direction of the conductor the force on the positive charge is towards the right and that's why the right tip. Not due to the current finger in FLHR.

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4. (Original post by rabia reha)
thank u soo much...
Anytime
5. Guys I have a question about the June 2012 paper... http://qualifications.pearson.com/co...e_20120611.pdf

For question 18civ how do you calculate total energy from the graph when the values are momentum??

Also the following part, I don't understand either at all!
6. (Original post by cerlohee)
Guys I have a question about the June 2012 paper... http://qualifications.pearson.com/co...e_20120611.pdf

For question 18civ how do you calculate total energy from the graph when the values are momentum??

Also the following part, I don't understand either at all!
Hi,
As energy is shown on top of the units, as the momentum is given in terms of energy/velocity. Just add up the energies (top part) of all the products in the event, really simple. E.g. as 1cm is shown to be 10Gev/c, the energy represented by 1cm is therefore 10 GeV. The answer is 300 GeV when you add them all up.
As for part v) Two energy of the 2 top quarks would therefore be 300GeV/c^2 (as this is the total energy of all of the decay products from the two top quarks, divided by c^2. As E/c^2 = m). Therefore 1 top quark would be 150GeV/c^2.
vi) 300GeV is a very large energy, so took a long time for particle accelerators to reach this sort of energy to give rise to large mass particles such as the top quark, as E = mc^2.
7. (Original post by TheBeardedGuy)
Hi,
As energy is shown on top of the equation as the momentum is given in terms of energy/speed of light. Just add up the energies (top part) of all the products in the event, really simple. E.g. as 1cm is shown to be 10Gev/c, the energy represented by 1cm is therefore 10 GeV.
So it's energy per speed of light rather than energy divided by speed of light? I don't understand how you could equate the unit gev/c to gev :/
8. (Original post by cerlohee)
So it's energy per speed of light rather than energy divided by speed of light? I don't understand how you could equate the unit gev/c to gev :/
Energy divided by the speed of light, gives a valid unit for momentum, as you show in the previous question. But what if you don't divide by the speed of light? You don't get the momentum. You just have the energy. Don't think about equating, just think about what the units show.
9. (Original post by physicsmaths)
This is wrong,
Conventional current in this case from FLHr is direction of the conductor the force on the positive charge is towards the right and that's why the right tip. Not due to the current finger in FLHR.

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Thanks, I stand corrected
10. (Original post by imedico10)
Thanks, I stand corrected
It is a brilliant question. My favourite magnetic fields question of them all!

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11. is it only me or is june12 paper really weird/hard? i pretty much did all of the past papers available, only have 2 left and then i would have done all of the past papers in existence. but for some reason i found the june12 really hard! the thing that is worrying me is that the difficulty of the paper isn't reflected in the grade boundaries (56 for A). i did some papers where it was 51/52 for an A and they were like a walk in the park. There is something about the jun12 one tho, is it only me or did anyone else find it quite hard?
12. (Original post by TheBeardedGuy)
Energy divided by the speed of light, gives a valid unit for momentum, as you show in the previous question. But what if you don't divide by the speed of light? You don't get the momentum. You just have the energy. Don't think about equating, just think about what the units show.
Ahh that makes sense, thankyou! For the next part of the question, why do you only need to divide by two to find the mass?

(Original post by Bishoy)
is it only me or is june12 paper really weird/hard? i pretty much did all of the past papers available, only have 2 left and then i would have done all of the past papers in existence. but for some reason i found the june12 really hard! the thing that is worrying me is that the difficulty of the paper isn't reflected in the grade boundaries (56 for A). i did some papers where it was 51/52 for an A and they were like a walk in the park. There is something about the jun12 one tho, is it only me or did anyone else find it quite hard?
Yes, I stupidly did it before by chemistry exam tomorrow thinking that it'd make me more confident but the grade boundaries were so high compared to the difficult! I really struggled with everything from the thermionic emission question onwards
13. (Original post by cerlohee)
Ahh that makes sense, thankyou! For the next part of the question, why do you only need to divide by two to find the mass?

Yes, I stupidly did it before by chemistry exam tomorrow thinking that it'd make me more confident but the grade boundaries were so high compared to the difficult! I really struggled with everything from the thermionic emission question onwards
Yeah the thermionic emission question and the last question about the sum of momentum and quarks were horrible!
14. A positron with kinetic energy 2.2MeV collides with an electron at rest, and they annihilate each other.
Calculate the average energy of the two gamma photons produced as a result of the annihilation/??
HELPPP!!
15. To cerlohee

For your question about dividing by two, I think you use the fact that energy is conserved. So the total energy (after the decay) is calculated to be around 300 GeV. This means the total energy before the decay was also 300 GeV. This initial energy was comprised entirely of two (identical) top quarks, i.e. they both have the same amount of mass-energy. Thus, you divide by 2 to get that the mass of one top quark is 150 GeV/c^2. I hope that makes sense.
16. (Original post by Sal296)
A positron with kinetic energy 2.2MeV collides with an electron at rest, and they annihilate each other.
Calculate the average energy of the two gamma photons produced as a result of the annihilation/??
HELPPP!!
Calculate the total energy before (including the rest masses) and divide by two to get the average energy.
17. (Original post by ertuu)
Calculate the total energy before (including the rest masses) and divide by two to get the average energy.
the answer is 2.58x10^-13
Im not getting it :L
18. (Original post by Sal296)
A positron with kinetic energy 2.2MeV collides with an electron at rest, and they annihilate each other.
Calculate the average energy of the two gamma photons produced as a result of the annihilation/??
HELPPP!!
Is the answer 1.61MeV each per gamma photon, or 2.58x10^-13 J? If it's right I'll explain but not sure
19. (Original post by Sal296)
the answer is 2.58x10^-13
Im not getting it :L
It's easier to work in the unit you are given. So use MeV instead of what is presumably Joules.
20. (Original post by iceangel8)
Is the answer 1.61MeV each per gamma photon, or 2.58x10^-13 J? If it's right I'll explain but not sure
I think this is right. You just do: ((0.511)(2) + 2.2)/2.

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