AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread]

Interesting that a bored Oxford undergrad knows what the question says (despite it being unpublished) and has just made his account today a few minutes after I posted.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.

Indeed they are - they are velocities*in the opposite direction

And I can use LaTex perfectly, just couldn't be asked here*

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.

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.

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.

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.

@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

@Jay1421 hey thanks for the unofficial mark scheme. Do you remember how many marks 4c was worth?

Hi, sorry - 4c was worth 3 marks

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'.

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.

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.

By the way, have you done the EMPA? Would you happen to have any grade boundary predictions for it?