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

Reply 2060

Here is PotterPhysics' answer to the first question: The answer is 800 (C)

http://www.thestudentroom.co.uk/showthread.php?p=56268219#post56268219

And the second one comes out to be -13 (A) as above :smile:

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My bad; I'll remove the error to avoid confusion. :yy:

Reply 2061

CD223

OP

10

Original post by Fvthoms

My bad; I'll remove the error to avoid confusion. :yy:

No problem! If they generally state an effect on the radius/diameter without stating the effect on the mass, then both the numerator AND denominator change when working out g or V :smile:

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Reply 2062

aprocrastinator

10

Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

The questions are 2, 6, 10, 16, 24.

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-QP-JUN12.PDF

Reply 2063

CD223

OP

10

Original post by aprocrastinator

Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

The questions are 2, 6, 10, 16, 24.

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-QP-JUN12.PDF

Question 2:
ImageUploadedByStudent Room1432810118.018798.jpg

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Reply 2064

CD223

OP

10

Original post by aprocrastinator

Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

The questions are 2, 6, 10, 16, 24.

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-QP-JUN12.PDF

Question 6:
ImageUploadedByStudent Room1432810222.548919.jpg

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Reply 2065

CD223

OP

10

Original post by aprocrastinator

Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

The questions are 2, 6, 10, 16, 24.

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-QP-JUN12.PDF

Question 10:
The answer is B.

This is because resonance curves show that a system reaches maximum amplitude (a sharp resonance peak) when the driving frequency equals the natural frequency if the damping is light.

ImageUploadedByStudent Room1432810373.619885.jpg

ImageUploadedByStudent Room1432810373.619885.jpg

When a system is damped more and more, the amplitude at resonance is reduced and moreover, the resonant frequency is slightly reduced.

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Reply 2066

CD223

OP

10

Original post by aprocrastinator

Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

The questions are 2, 6, 10, 16, 24.

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-QP-JUN12.PDF

Question 16:
The answer is C.

At the point of closest approach, the velocity is zero, not the acceleration.

If the velocity is zero, the kinetic energy is zero.

The closer the particle gets to the nucleus, the greater its potential energy becomes, ruling out A, B and D.

Question 24:
The answer is C.

When the current is at a maximum and suddenly switched off, there is the greatest change in flux linkage through the coil out of all scenarios given.

A greater change of magnetic flux linkage produces a greater emf, and as the circuit is complete, the greatest induced current reading is observed.

Hope that helps :smile:

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Reply 2067

EmiratesCaptain

2

Original post by CD223

Hahaha, I'm flattered, but no. I got rejected by Oxford after my interview in December. Being totally honest, I'm happy about it because I don't think the atmosphere would have been healthy for me - I need a place I feel I can chill out!

I'm (hopefully) going to Bath to do Computer Science in September. How about you? :smile:

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Well Oxford have missed out! Sheffield hopefully - Aerospace Engineering

Reply 2068

CD223

OP

10

Anyone care to explain why sharp points produce intense fields?

Is it something to do with:

[br]E = \dfrac{\Delta V}{\Delta d}[br]

Where there is a large potential across a small distance?

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Reply 2069

CD223

OP

10

Original post by EmiratesCaptain

Well Oxford have missed out! Sheffield hopefully - Aerospace Engineering

That's really nice of you, thank you :smile:

Good luck! What grades do you need? 😁

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Reply 2070

EmiratesCaptain

2

Original post by CD223

That's really nice of you, thank you :smile:

Good luck! What grades do you need? 😁

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No problem :smile:

AAB - The B will probably be in physics.. I find it quite hard :/

Reply 2071

CD223

OP

10

Original post by EmiratesCaptain

No problem :smile:

AAB - The B will probably be in physics.. I find it quite hard :/

Physics is my hardest subject too - which is disheartening because it's one of my more interesting subjects!

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Reply 2072

EmiratesCaptain

2

Original post by CD223

Physics is my hardest subject too - which is disheartening because it's one of my more interesting subjects!

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Haha agreed! I think unit 5 is harder than 4 though

Reply 2073

bwr19

2

Hey guys. Can anyone help me with part (c)ii of this question? I just don't understand the mark scheme. They seem to be using eV for a potential difference? I thought eV was a measurement of energy...

Thanks. Question is attached. This is from June 2012.

Screen Shot 2015-05-28 at 16.10.47.png58.0KB

Reply 2074

huniibehi

3

Yeah I was stuck on that one too, You can use work done= force x distance and use the energy they gave to find the force

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Reply 2075

huniibehi

3

Does anyone know why the particle would be positive?

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1432826346918.jpg27.3KB

Reply 2076

Jimmy20002012

4

Original post by huniibehi

Does anyone know why the particle would be positive?

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Electric field line always act from positive to negative, so if the electric field in upwards the particle must have a negative charge. I think this is why, but not 100% sure.

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Reply 2077

CD223

OP

10

Original post by bwr19

Hey guys. Can anyone help me with part (c)ii of this question? I just don't understand the mark scheme. They seem to be using eV for a potential difference? I thought eV was a measurement of energy...

Thanks. Question is attached. This is from June 2012.

Can't you do:

[br]F = \dfrac{W}{s} = \dfrac{QV}{d}[br]

Where:

[br]Q = 1.6 \times 10^{-19} C[br]V = 4500 V[br]d = 180 \times 10^{-3}m[br]

Then the minimum distance equals

[br]\dfrac{\text{Electron volts} \times 1.6 \times 10^{-19}}{F}[br]

?

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(edited 8 years ago)

Reply 2078

CD223

OP

10

Original post by huniibehi

Does anyone know why the particle would be positive?

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Is there a diagram with that question? It's just I'm unsure if you could answer it without one as all it says is "vertical" field?

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