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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

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Original post by Chazley123
6 marker:

Talking about circular motion in the context of the mass on a string.
V is changing but speed is constant - dirrection always changing therefore accelerates
Force (centripetal) acts perpendicular to direction of motion - conditions for u.c.m
Centriptal force provided by cent acc
tension in string acts as centr - force

difficult to demonstarte horizontally:
theres mass which is acted on by gravity thus downward component makes it harder to veer perfectly horizontal,
measuring this accuracy is a hard thing to do practically

anything I may have missed?


I reckon you will get 4/6. I wrote similar but my teacher said you had to state that the tension in the string resolved vertically TSin(x) would be equal to mg if it is to remain horizontal. Also explain the 3 laws in context
Original post by QueNNch
Your argument is valid but I've also done Q 20 in past papers and it's definitely -322

Change in direction of current, making force apply in opposite direction. (Flemings LHR)


I know i've just looked, it's where the initial reading was zeroed, therefor it can have a negative value as mass already acts on the scales
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Original post by Chazley123
6 marker:

Talking about circular motion in the context of the mass on a string.
V is changing but speed is constant - dirrection always changing therefore accelerates
Force (centripetal) acts perpendicular to direction of motion - conditions for u.c.m
Centriptal force provided by cent acc
tension in string acts as centr - force

difficult to demonstarte horizontally:
theres mass which is acted on by gravity thus downward component makes it harder to veer perfectly horizontal,
measuring this accuracy is a hard thing to do practically

anything I may have missed?


You do realize it is impossible to do it perfectly horizontally... not harder.

BECAUSE: there is no vertical component of the tension to counter act the weight of the ball.
Original post by Sonnyjimisgod
I reckon you will get 4/6. I wrote similar but my teacher said you had to state that the tension in the string resolved vertically TSin(x) would be equal to mg if it is to remain horizontal. Also explain the 3 laws in context


examiners forever wanting you to go further :/
Original post by chizz1889
It also relies on momentum


The alternative answer which would oppose my thoughts is that if it has a greater mass it has more mgh and then it has a greater 1/2mv² and so there is a greater drag force. I thought the damping force was only because of drag on a pendulum, so I really don't know which solution is correct.


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Here are my answers again.
Would be very happy for any comments
ImageUploadedByStudent Room1434025155.788417.jpg


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Original post by salicional
It was mentioned a lot earlier up the thread!


Ah right, late to the party :tongue: Oh well I should pick up some marks with that however they decide to deal with it.

Original post by QueNNch
The question asked for the number of electrons on the NEGATIVE plate ONLY.

As each capacitor plate shares equal number of elections it is halved.


Ah damn, thanks! At least it's only 1 mark
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Original post by NEWT0N
I did those as well :frown: (the ones in bold)

Chin up, I'm sure you'll be fine. If that's really the total number of mistakes you made you're still looking at a high (full UMS even?) A*. I'm hoping for either just scraping full UMS but hopefully an A* in any case, although the fact that there are like 5 marks between each grade is slightly worrying


Not sure if those were the ONLY mistakes I've made, but certainly the most glaring :/

So angry with that typo kilo ohm question! I had my intercept lower and graph less steep but they ended up at the same current approaching zero??


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Original post by poee
You do realize it is impossible to do it perfectly horizontally... not harder.

BECAUSE: there is no vertical component of the tension to counter act the weight of the ball.


nice
Original post by Amanzz
Think this is incorrect. The time period of a pendulum is independent of a mass added, hence the maximum velocity is independent of the mass added. Furthermore, the only resistive force is due to the viscosity of air, an air resistive force. The ring hardly affects air resistance, and so the since the vicious force of air is dependant on velocity, which is unchanged, then the damping/viscous force is unchanged.


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Damping affects amplitude, not time period. Time period is irrelevant.
Conservation of momentum means that the ringed bob experiences less damping because it has higher mass so gives the air molecules less velocity (assuming the same surface area).
Original post by salicional
Do you think it's worth writing a complaint to AQA about the obvious typo in the capacitors question? Just in case it hasn't come to their attention that their proof-reading process is awful...


what typo?
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Original post by QueNNch
The question asked for the number of electrons on the NEGATIVE plate ONLY.

As each capacitor plate shares equal number of electrons it is halved.


THIS IS WRONG.

it is not halved.

at least I think it's not,, each plate has Q charge
question on field strength directly between two identical masses, was it zero?
Original post by Amanzz
The alternative answer which would oppose my thoughts is that if it has a greater mass it has more mgh and then it has a greater 1/2mv² and so there is a greater drag force. I thought the damping force was only because of drag on a pendulum, so I really don't know which solution is correct.


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No you were correct the first time in saying there is the same amount of drag. Force is the same as rate of change of momentum of u can remember q1 on multiple choice. This means the same amount of drag will have the same rate of change of momentum. Since velocity is independent of mass the initial velocity is the same, as is the time period. But since the mass is larger and the rate of change of momentum is equal to before means that the rate of change of velocity is reduced meaning it takes longer to lose its velocity hence less damping
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No they were asking why this scenario you're talking about is impossible.

The tension is at an angle towards the centre hence the vertical component of the tension cancels out with the weight so the resultant towards centre is the horizontal component of the tension.
Original post by NEWT0N
Doesn't tension act horizontally towards the center (and thus perpendicular to the weight?)
Original post by Chazley123
question on field strength directly between two identical masses, was it zero?


Yeah
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Original post by poee
THIS IS WRONG.

it is not halved.

at least I think it's not,, each plate has Q charge


Doesn't one plate have all the electrons?
Original post by NEWT0N
I remember getting G*m*pi*p*R/3 somewhere for the multiple choice. Anyone agree?


Yeh, i subbed in formulas/rearranged and stuff.

Close to 100% sure that's correc
Can someone make an unofficial mark scheme
Original post by NEWT0N
Doesn't tension act horizontally towards the center (and thus perpendicular to the weight?)


Yes if horizontal Tsin(x)=mg , though in practicality its unrealistic to apply this tension apparently. I didn't even mention anything to do with this though, so probably only got 4-5/6

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