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

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Reply 200

BP_Tranquility

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

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Reply 201

username1162972

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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Reply 202

username1162972

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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Reply 203

theoeht

Original post by rabia reha

thank u soo much...

Anytime

Reply 204

cerlohee

Guys I have a question about the June 2012 paper... http://qualifications.pearson.com/content/dam/pdf/A%20Level/Physics/2013/Exam%20materials/6PH04_01_que_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!

(edited 8 years ago)

Reply 205

TheBeardedGuy

Original post by cerlohee

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.

(edited 8 years ago)

Reply 206

cerlohee

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 :/

Reply 207

TheBeardedGuy

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.

Reply 208

imedico10

Original post by physicsmaths

Thanks, I stand corrected :smile:

Reply 209

username1162972

Original post by imedico10

Thanks, I stand corrected :smile:

It is a brilliant question. My favourite magnetic fields question of them all!

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Reply 210

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?

Reply 211

cerlohee

Original post by TheBeardedGuy

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

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

Reply 212

Bishoy

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!

Reply 213

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!!

Reply 214

ertuu

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.