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    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 v and V 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 v and V are not magnitudes; they're proper (if poorly notated) vector quantities. Hence mv and NV 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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    (Original post by kingaaran)
    Interesting that a bored Oxford undergrad knows what the question says (despite it being unpublished) and has just made his account today a few minutes after I posted.

    I'm sorry, but you're fundamentally wrong.

    *

    As for your making of the second account and attempt to pull straws, here's a full explanation for you:

    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 mv-NV=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.
    Hahahahahaha.

    First of all, I'm actually a friend of Jay1421. He asked me to have a look at the question after your (and others') comments. I'm an undergraduate at Brasenose College, and I'm in the 94th percentile of my year. I'm not a user of TSR but I felt this topic is one that often causes confusion, so I thought I'd try to offer an explanation. As wonderfully in-depth as your explanation is, you misread the question. It stated that v and V are velocities. See my earlier post. Also, since you're going on to do a maths-based degree, I suggest you learn how to use LaTeX.
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    (Original post by kingaaran)
    Interesting that a bored Oxford undergrad knows what the question says (despite it being unpublished) and has just made his account today a few minutes after I posted.

    I'm sorry, but you're fundamentally wrong.

    *

    As for your making of the second account and attempt to pull straws, here's a full explanation for you:

    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 mv-NV=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.
    It's very embarrassing to make false accusations - tomg0166 is a friend of mine. I sent him the question, including the diagram, and asked him for an objective answer. The reason WHY I consulted him is for the exact thing you doubt, he attends Oxford and I figured he'd have a better understanding than myself, which has been shown.

    The fact that your reaction to being told you're wrong is to make false accusations based on arbitrary evidence says a lot about you.
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    (Original post by tomg0166)
    Hahahahahaha.

    First of all, I'm actually a friend of Jay1421. He asked me to have a look at the question after your (and others') comments. I'm an undergraduate at Brasenose College, and I'm in the 94th percentile of my year. I'm not a user of TSR but I felt this topic is one that often causes confusion, so I thought I'd try to offer an explanation. As wonderfully in-depth as your explanation is, you misread the question. It stated that v and V are velocities. See my earlier post. Also, since you're going on to do a maths-based degree, I suggest you learn how to use LaTeX.
    Indeed they are - they are velocities*in the opposite direction

    And I can use LaTex perfectly, just couldn't be asked here*
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    V and v are both velocities, so there was no need to add signs to them when calculating total momentum before and after - since velocity gives direction and magnitude already. therefore i'm fairly sure the final answer has a '-' sign.

    now, had the question said V is a speed in one direction and v is the speed in the opposite direction, then you would have to say that Vm - vN = 0 (since in the diagram, they are pointing in opposite directions to indicate the direction of the scalar values). but the question said velocity, so there was no need.
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    Its just notation - you can write the minus explicitly or you can leave it implicit. I suspect you'd get credit for both answers
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    (Original post by kingaaran)
    Indeed they are - they are velocities*in the opposite direction

    And I can use LaTex perfectly, just couldn't be asked here*
    Okay, I'm just going to try to explain this with vector notation, since that's what this question is about.

    Define  \vec{a} = \hat{i}, \vec{b} = -\hat{i}.

    This is analogous to the momenta given in the question. (Which are indeed vectors)

    Then \vec{a} = - \vec{b} \implies \vec{a} + \vec{b} = \vec{0}

    You don't need to add any minus signs in to show that  \vec{a} and \vec{b} are opposite. They are defined as so.

    AQA should have properly stated that v and V were vectors, but they are implicitly defined as so by calling them velocities (which are never scalar).
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    (Original post by kingaaran)
    Interesting that a bored Oxford undergrad knows what the question says (despite it being unpublished) and has just made his account today a few minutes after I posted.

    I'm sorry, but you're fundamentally wrong.

    *

    As for your making of the second account and attempt to pull straws, here's a full explanation for you:

    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 mv-NV=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.
    you can't make this comparison though, because in your example you take 2ms^-1 and vms^-1 to be speeds, and in the exam they were given as velocities.
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    (Original post by kingaaran)
    Interesting that a bored Oxford undergrad knows what the question says (despite it being unpublished) and has just made his account today a few minutes after I posted.

    I'm sorry, but you're fundamentally wrong.

    *

    As for your making of the second account and attempt to pull straws, here's a full explanation for you:

    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 mv-NV=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.
    The momentum of particle A does not equal (its mass)*(-V), I don't know where you're getting this from. V will already include the directional component in it, it's a VELOCITY, it HAS to have direction because it's a vector quantity. The momentum of each particle is literally (MASS) * (VELOCITY) - you can't just throw minus signs in there how you feel.

    Momentum of a = M(a)V(a)
    Momentum of b = M(b)V(b)
    Total momentum = M(a)V(a) + M(b)V(b)

    As I said, with velocity being a vector quantity, it already has it's direction in the quantity - putting a minus there is incorrect.

    Hopefully, the topic of momenta never comes up on crashMATHS.
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    (Original post by kingaaran)
    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.
    If anything, isn't your example proving that our method is correct? If the fragment with mass 2kg moves with speed 2ms^-1, and the 3kg mass moves in the opposite direction, then v SHOULD come out as negative, no? why would v come out positive? you're saying they're both speeds.
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    (Original post by kingaaran)
    Interesting that a bored Oxford undergrad knows what the question says (despite it being unpublished) and has just made his account today a few minutes after I posted.

    I'm sorry, but you're fundamentally wrong.

    *

    As for your making of the second account and attempt to pull straws, here's a full explanation for you:

    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 mv-NV=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.
    Also, just to let you know - the reason I used the positive version in the solution to the next question, the derivation, is because it makes no difference. It's an equation for Kinetic Energy, which is scalar, so the direction of the velocity is irrelevant (i.e a truck travelling at 20ms^-1 has the same kinetic energy in all directions of travel). In addition, with regards to deriving the equation, you square the V anyway, which would get rid of the minus sign, giving the same result regardless of + or - used.

    The derivation is exactly the same regardless of if you use the positive or negative 'version' of the V =(±vm)/N equation.
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    Jay1421 hey thanks for the unofficial mark scheme. Do you remember how many marks 4c was worth?
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    kingaaran hey, you posted answers to the multiple choice answers a few weeks ago, do you have a link, or remember the page number, I cant remember which it was on. Thanks
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    (Original post by boyyo)
    kingaaran hey, you posted answers to the multiple choice answers a few weeks ago, do you have a link, or remember the page number, I cant remember which it was on. Thanks
    There you are http://www.thestudentroom.co.uk/show....php?t=4177478
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    (Original post by boyyo)
    Jay1421 hey thanks for the unofficial mark scheme. Do you remember how many marks 4c was worth?
    Hi, sorry - 4c was worth 3 marks
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    (Original post by Jay1421)
    Hi, sorry - 4c was worth 3 marks
    cool thanks
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    I know this is random and people won't want to be thinking about this in the middle of their summer, but what do you think the grade boundaries for the EMPA will be? It was relatively straight-forward, right? Do people think they're going to be the highest they've ever been?
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    (Original post by tomg0166)
    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 v and V are not magnitudes; they're proper (if poorly notated) vector quantities. Hence mv and NV 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'.
    (Original post by tanyapotter)
    you can't make this comparison though, because in your example you take 2ms^-1 and vms^-1 to be speeds, and in the exam they were given as velocities.
    I don't agree with this, particularly the assumptions that v and V are not magnitudes.

    The diagram shows the direction of the final velocities of the two objects and their magnitudes, not their velocities.

    V in the case of this question denotes the magnitude of the velocity of the particle in question after the collision and not the velocity of the particle. This is very clear from the provided diagram, by the fact that they labelled the speed of the particle V and not -V.

    In any case, I strongly suspect AQA will accept either sign given the level of confusion here and the answer to this question has no bearing on the following parts.

    Edit: It's also neither necessary or indicative of intelligence to throw insults at people who have opposing opinions to you:
    Spoiler:
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    (Original post by Jay1421)
    Hopefully, the topic of momenta never comes up on crashMATHS.
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    (Original post by Euclidean)
    I don't agree with this, particularly the assumptions that v and V are not magnitudes.

    The diagram shows the direction of the final velocities of the two objects and their magnitudes, not their velocities.

    V in the case of this question denotes the magnitude of the velocity of the particle in question after the collision and not the velocity of the particle. This is very clear from the provided diagram, by the fact that they labelled the speed of the particle V and not -V.

    In any case, I strongly suspect AQA will accept either sign given the level of confusion here and the answer to this question has no bearing on the following parts.
    So you're saying V is the magnitude, the arrow is the direction, and altogether it forms the velocity, which is what the question is referring to?

    I would have thought that if that were the case, the question would very clearly have stated that V is the speed. Because speed is just a magnitude. You may be right that they'd allow either?

    By the way, have you done the EMPA? Would you happen to have any grade boundary predictions for it?
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    (Original post by tanyapotter)
    So you're saying V is the magnitude, the arrow is the direction, and altogether it forms the velocity, which is what the question is referring to?

    I would have thought that if that were the case, the question would very clearly have stated that V is the speed. Because speed is just a magnitude. You may be right that they'd allow either?

    By the way, have you done the EMPA? Would you happen to have any grade boundary predictions for it?
    I'd say that because of the way they presented the diagram, V is infact the magnitude of the velocity of whatever that particle was (memory is failing me) and that this, coupled with the direction makes up the velocity of -V.

    Hence \Delta \vec{p} = mv - MV = 0

    But I also strongly think AQA will have to accept either \pm answer simply because they didn't make their intention clear enough.

    Sorry I didn't sit the EMPA, may be worth checking past grade boundaries, looking for the grade/UMS mark you want and calculating a mean raw mark for that UMS mark.
 
 
 
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