Cheers bro was so nervous about that
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AQA Physics PHYA4  20th June 2016 [Exam Discussion Thread] watch

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 26062016 08:27

tophatmenis
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 28062016 14:06
How did people feel about the Unit 5 paper this morning?

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 06072016 16:02
Is there any demand for a Unit 4 written paper unofficial mark scheme, as I could write one up if people are interested

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 07072016 11:44
(Original post by Jay1421)
Is there any demand for a Unit 4 written paper unofficial mark scheme, as I could write one up if people are interested 
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 07072016 13:22
(Original post by Jay1421)
Is there any demand for a Unit 4 written paper unofficial mark scheme, as I could write one up if people are interested 
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 07072016 13:49
(Original post by Gifted)
Always a demand, thank you.
(Original post by C0balt)
You've got the paper on you? 
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 07072016 19:07
Unofficial Mark Scheme is here
I took screenshots of my Word document and uploaded it onto an Imgur album for ease (formatting on here was a bit of a pain) 
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 08072016 05:15
(Original post by Jay1421)
Unofficial Mark Scheme is here
I took screenshots of my Word document and uploaded it onto an Imgur album for ease (formatting on here was a bit of a pain)
If I saw that I'd start counting my marks and die in fear of missing my offer lolol
Thanks though 
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 08072016 05:30
(Original post by C0balt)
I. Must. Resist. The. Temptation. To. Click. On. The. Link.
If I saw that I'd start counting my marks and die in fear of missing my offer lolol
Thanks though
Ew exams.
Good luck! 
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 08072016 08:31
(Original post by Jay1421)
I fully understand! Whilst writing it I started to fear about meeting my offer. Jees, it's so weird that they expect 18 year olds to make life decisions, especially when one exam performance can be the difference between them being a doctor, lawyer, actor, whatever, and them falling into a pit of career misery.
Ew exams.
Good luck!
Good luck to you too! 
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 08072016 13:08
(Original post by Jay1421)
Unofficial Mark Scheme is here
I took screenshots of my Word document and uploaded it onto an Imgur album for ease (formatting on here was a bit of a pain) 
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 08072016 15:31
(Original post by Aethrell)
I thought the consensus was that there was no '' in the alpha particle momentum one?
Hope that helps! 
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 08072016 16:27
(Original post by Jay1421)
There has to be a minus there. In that case, momentum before was zero, so in order for momentum to be conserved, the 'net' momentum has to still be zero, which means the magnitudes of each momentum had to be equal and the direction had to be opposite in order to cancel them out. The question asks how you can show that momentum is conserved using V, v, N and m. In order to do that, you have to use a minus to show that the alpha particle and the recoiling nucleus are travelling in opposite directions, as that's the only way momentum can be conserved.
Hope that helps! 
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 08072016 17:12
(Original post by marioman)
Because the directions are opposite, VN  vm = 0. So VN = vm, and V = (vm)/N. No minus sign in the final equation.
When you're determining if momentum is conserved, you add them together to make zero, not minus them.
It should be VN + vm = 0, therefore VN = vm => V = (vm)/N 
crashMATHS
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 08072016 19:13
(Original post by Jay1421)
If they are in opposite directions, how can VN possibly be equal to vm? It's equal and opposite, therefore VN = vm.
When you're determining if momentum is conserved, you add them together to make zero, not minus them.
It should be VN + vm = 0, therefore VN = vm => V = (vm)/N 
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 08072016 19:37
(Original post by kingaaran)
If VN is the momentum of one particle and vm is the momentum of another particle, then VN =/= vm as it physically breaks the conservation of momentum. The momentum has to be the same: the minus sign means that they are not the same.**
At rest, the momentum is 0.
The momentum does not have to be the same, it has to have an equal magnitude. Those two points are not the same thing. Momentum, like velocity, is a vector quantity, which means it has a direction. If VN = vm, that means the velocities are both positive, which therefore suggests that they are travelling in the same direction; this isn't true, as the diagram proves.
Conservation of momentum states that the total momentum before is equal to the total momentum after.
=> The total momentum before = 0
=> The total momentum after must therefore be 0
=> therefore (VN) + (vm) = 0
=> Which in turn means that VN = vm
I think the question is quite poorly worded, it's confusing people into thinking it's looking for a scalar equation, but it can't be  momentum is vector. 
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 08072016 19:45
Bored Oxford undergrad here. Conservation of momentum states . The total momentum before is . Since no external force is applied, the total momentum afterwards is also . The question states that and are velocities, and hence vector quantities. Hence, the directions of the momenta are implicit as opposed to explicit in the notation.
So, applying conservation of linear momentum, we have:
etc.
Admittedly the question is poorly phrased, but the key point is that we're working with velocities, which are vector quantities, and hence their directions are implied.Last edited by tomg0166; 08072016 at 19:48. 
Nathanburns01
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 08072016 20:02
(Original post by tomg0166)
Bored Oxford undergrad here. Conservation of momentum states . The total momentum before is . Since no external force is applied, the total momentum afterwards is also . The question states that and are velocities, and hence vector quantities. Hence, the directions of the momenta are implicit as opposed to explicit in the notation.
So, applying conservation of linear momentum, we have:
etc.
Admittedly the question is poorly phrased, but the key point is that we're working with velocities, which are vector quantities, and hence their directions are implied.Last edited by Nathanburns01; 08072016 at 20:04. 
crashMATHS
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 08072016 20:37
(Original post by tomg0166)
Bored Oxford undergrad here. Conservation of momentum states [latex]\sum \vec{p}_{before} = \sum \vec{p}_{after}[\latex]. The total momentum before is [latex]\vec{0}[\latex]. Since no external force is applied, the total momentum afterwards is also [latex]\vec{0}[\latex]. The question states that [latex]v[\latex] and [latex]V[\latex] are velocities, and hence vector quantities. Hence, the directions of the momenta are implicit as opposed to [B]explicit[\B] in the notation.
So, applying conservation of linear momentum, we have:
[latex]\sum \vec{p}_{before} = \sum \vec{p}_{after} = \vec{0}[\latex]
[latex] \implies mv + VN = \vec{0} \\
\implies mv = VN [\latex]
etc.
Admittedly the question is poorly phrased, but the key point is that we're working with velocities, which are vector quantities, and hence their directions are implied.
I'm sorry, but you're fundamentally wrong.
*(Original post by Jay1421)
They aren't the same. They are equal in magnitude but opposite in direction. The question states that V and v are velocities, which are vector quantities. For momentum to be conserved in this case, it must total 0 when all of the individual momentums are added together.
At rest, the momentum is 0.
The momentum does not have to be the same, it has to have an equal magnitude. Those two points are not the same thing. Momentum, like velocity, is a vector quantity, which means it has a direction. If VN = vm, that means the velocities are both positive, which therefore suggests that they are travelling in the same direction; this isn't true, as the diagram proves.
Conservation of momentum states that the total momentum before is equal to the total momentum after.
=> The total momentum before = 0
=> The total momentum after must therefore be 0
=> therefore (VN) + (vm) = 0
=> Which in turn means that VN = vm
I think the question is quite poorly worded, it's confusing people into thinking it's looking for a scalar equation, but it can't be  momentum is vector.
As far as I remember, the question said that one particle recoiled at a speed V (let's call this particle A) and the other was emitted with speed v (let's call this particle B).
Now, momentum involves velocity, which is a vector quantity. Thus, when considering this 'conservation of momentum', we need to pick a direction that is consistent. I am going to pick the direction of particle B to be the positive direction. It hence follows that:
=> Momentum of particle A = (its mass)*(V) *
=> Momentum of particle B = (its mass)*v
That is consistent with the directions.
Applying the conservation of momentum: momentum before = momentum after
hence: 0 = mass*(V) + mass*v
In this context that translates at mvNV=0 => mv = NV.
Even in the later part of the question on your unofficial mark scheme you use the positive version.
If that still hasn't convinced you, let's suppose we have a stationary particle of mass 5 kg that explodes into two smaller fragment, one of mass 2 kg and the other of mass 3kg. The fragment with mass 2kg moves with speed 2 ms^(1) and the fragment with mass 3kg moves with speed v ms^(1) in the opposite direction. Find v.
Conservation of momentum tells us that: 0 = 2(2) + 3(v) => v = 4/3 ms^(1)
*
Your method: 0 = 2(2)+3v => v = 4/3 ms^(1). This comes out as negative because we have taken v to be in the wrong direction, assuming that it moves in the same direction as the other fragment.
Hence, while your result is of the correct magnitude, its sign is incorrect and, due to AQA's precise definition of the speed of the recoiling nucleus, I doubt they'd appreciate your assumption that the direction of motion of the two decay fragments is the same.Last edited by kingaaran; 08072016 at 20:38. 
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 08072016 20:39
I don't think you can jump to the conclusion that they were only concerned with the magnitudes of the velocities. They would have said that and were speeds if that were the case. Admittedly the diagram is misleading, but the use of the word 'velocities' supersedes the arrows in the diagram.
Also, the momenta will not be equal. The magnitudes of the momenta will be, but their directions will necessarily be opposite.
The problem with this question is that vectors are not treated properly in A Level physics, which seemingly creates an ambiguity. The magnitudes of the momenta will be equal, but and are not magnitudes; they're proper (if poorly notated) vector quantities. Hence and cannot be equal, since they are vectors pointing in opposite directions. I remember being confused about this when I was doing A Level physics; it's mostly down to the watered down mathematical treatments of the course, leading to a fuzzy mixing of 'vector' and 'magnitude of a vector'.
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