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

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Original post by physicsmaths
In examiners report no one got it right so I wouldn't worry the rest of the paper was easy though.


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meh, got 51/80, yet in all the past papers ive done like from june 11(i think) to june 13 I've been scoring at least 63-65. I think the stress is getting to me
Original post by jay_em
When to use KE = p²/2m vs KE = ½mv²?

And any chance of needing protractors or compasses?


If you want to work out the momentum a particle has you can use 1/2mv^2 to work out its Ek then sub it into p^2/2m to work out its momentum. The only time I would imagine u would need a protractor is momentum diagrams.
okay so super last minute but i've confused myself

with Fleming's Left Hand Rule, are we meant to take current as conventional {+ to -} or as a flow of electrons {- to +}? because we were always taught conventional current, but in some of the exam questions the only way to get to the answer is by using the - to + direction? how am i supposed to know which direction to take?
Original post by thunderstxrms
okay so super last minute but i've confused myself

with Fleming's Left Hand Rule, are we meant to take current as conventional {+ to -} or as a flow of electrons {- to +}? because we were always taught conventional current, but in some of the exam questions the only way to get to the answer is by using the - to + direction? how am i supposed to know which direction to take?


As far as I'm aware Fleming's left hand rule always uses conventional current (+ to -)
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I. AM. FED. UP!!:bawling:
16bii pleaseeeee
June 2014 (R) QP - Unit 4 Edexcel Physics.pdf335.0KB
Quick question: Why is this not a valid conclusion on the Rutherford scattering experiment?

The nucleus is positively charged

In every single mark scheme this conclusion didn't credit marks:angry:
(edited 8 years ago)
Original post by LaurenceM
16bii pleaseeeee


Charge on sphere =100Q
Charge on ball=Q

Use F=kQ1Q2/r^2
F and r from the graph


200x10^3=8.99x10^9x100Q^2/(15x10^3)^2
Original post by bobo19966
Quick question: Why is this not a valid conclusion on the Rutherford scattering experiment?

The nucleus is positively charged

In every single mark scheme this conclusion didn't credit marks:angry:


yeah I noticed that too :/ maybe because they didn't know the charge of an alpha particle I'm guessing?
Original post by Runner104
yeah I noticed that too :/ maybe because they didn't know the charge of an alpha particle I'm guessing?


Even the textbook and revision guide said that it is a conclusion maybe you are right
Does anyone think transformers will be in the paper tomorrow, because I don't think I have seen questions on it in the past papers.
Can soeone help me with 7 e please. (the answer is 2.2 GeV)
physics q.PNG225.2KB
Original post by aero-1997
Does anyone think transformers will be in the paper tomorrow, because I don't think I have seen questions on it in the past papers.


I'm like 95% they won't ask about transformers as they haven't previously asked it but hey with that chemistry they seem like they like to **** people over
Anyone? :frown:
Original post by imedico10
Can soeone help me with 7 e please. (the answer is 2.2 GeV)


It's helpful to remember first of all that any mass of the antiproton and proton that is not converted to the mass of the products is converted to the kinetic energy of the products as mass and energy are interchangeable.

So we know initially the antiproton has an energy of 1GeV and the proton is at rest initially so has no energy, though both have a mass of 940 Mev or 0.94 Gev (I'll work in Gev to make it a bit easier calculation wise). so 0.94 x 2 = 1.88 GeV, adding the energy of the antiproton = 2.88 GeV - so this is the energy available for the mass of the products.

Now for the mass of the products, we know a neutral pion has a mass of 135 MeV/c^2, only one is produced. The mass of a charged pion is 140 Mev/c^2, we have four produced so 140 x 4 = 560 MeV in total. Total mass of products = 135 + 560 = 695 MeV/c^2 converting into GeV to give 0.695 GeV/c^2

so reactants - products, 2.88 - 0.695 = 2.185 or 2.2 GeV to 1 d.p. is available for the kinetic energy of the products. Hope this helps !
Original post by bobo19966
Quick question: Why is this not a valid conclusion on the Rutherford scattering experiment?

The nucleus is positively charged

In every single mark scheme this conclusion didn't credit marks:angry:



In the june 2014 IAL paper a mark is given for saying this.
Original post by Regina15
It's helpful to remember first of all that any mass of the antiproton and proton that is not converted to the mass of the products is converted to the kinetic energy of the products as mass and energy are interchangeable.

So we know initially the antiproton has an energy of 1GeV and the proton is at rest initially so has no energy, though both have a mass of 940 Mev or 0.94 Gev (I'll work in Gev to make it a bit easier calculation wise). so 0.94 x 2 = 1.88 GeV, adding the energy of the antiproton = 2.88 GeV - so this is the energy available for the mass of the products.

Now for the mass of the products, we know a neutral pion has a mass of 135 MeV/c^2, only one is produced. The mass of a charged pion is 140 Mev/c^2, we have four produced so 140 x 4 = 560 MeV in total. Total mass of products = 135 + 560 = 695 MeV/c^2 converting into GeV to give 0.695 GeV/c^2

so reactants - products, 2.88 - 0.695 = 2.185 or 2.2 GeV to 1 d.p. is available for the kinetic energy of the products. Hope this helps !


Thank you so much
How long are you guys planning to stay awake?
This paper will determine whether i will go to uni or not! fingers crossed, i did all what i can but you are never too prepared for physics :frown:
Original post by Hectors
meh, got 51/80, yet in all the past papers ive done like from june 11(i think) to june 13 I've been scoring at least 63-65. I think the stress is getting to me


Weird i got 72 in that paper then 57 in june 14 R. My scores fluctuate so much


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