The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

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

Scroll to see replies

What UMS marks do people think 48 is? Still really worried, hopefully I have a nice examiner marking.
does anybody know or have any idea how many marks each question on B was worth from the unofficial markscheme (first page)??
did you guys fill ur MC all in pens ? :P
Original post by Valkyrian
What UMS marks do people think 48 is? Still really worried, hopefully I have a nice examiner marking.


i cant work mine out cos i cant remember how much the questions were worth?:frown: think ill be in the same region, not happy it was really theoretical, im much better with calculations:frown:
Original post by oonic0rn
i cant work mine out cos i cant remember how much the questions were worth?:frown: think ill be in the same region, not happy it was really theoretical, im much better with calculations:frown:


Yeah same, although saying that I messed up the work done questions which is annoying, mistakes are made just normally there is 2-3 questions that I can do fairly well on to compensate with perhaps -1 or 2 marks. This time every question was horrid. Only question I did ok on was question 5.
Original post by oonic0rn
does anybody know or have any idea how many marks each question on B was worth from the unofficial markscheme (first page)??


4 marks. It's the only question with the marks on there atm! :biggrin:
Original post by RemainSilent
did you guys fill ur MC all in pens ? :P


Yeah, I think you're supposed to (can't quite remember if it said black ink or what)?
How did people attempt the work done question then?

Find V at 1 point then V at the other find the difference, then use W=QV ??
Original post by JizzaStanger
4 marks. It's the only question with the marks on there atm! :biggrin:


no sorry i meant the whole of section B for the calculations, just cant remember how many marks were awarded :tongue:
Question 21 in the MC, the answer was surely A the one with Length 1 and Width 0.2. As the F=BIL. It didn't ask for the maximum flux linkage or EMF. Therefore the one with greatest couple has to be the one with the longest side perpendicular to the field, A.
I must add my response to the motor question. I said stuff about blah blah work done against resistance so energy is lost as heat in the motor itself. But for the second reason I said because the power source was a capacitor, the current decays over time meaning the power output of the capacitor decays over time. Hence the torque of the motor decays over time to the point where it is not able to lift it anymore despite having energy left to dissipate. Also charge leakage could be something to mention as it was on the ISA.
I got 72UMS on this paper last year and I needed to improve on that by 15UMS this year to get an A overall and to get into my first choice. Looking at previous years I could drop at least 25 marks and still get the 89UMS. I dropped two or three marks on Section A and I doubt I dropped more than twenty on Section B. Hopefully I got at least 60 raw marks today.

On top of that I resat PHY6 and got at least 15UMS more by previous years' conversions. Quietly hoping for an A* but I'm happy enough knowing I've got my offer!
Reply 4772
Avatar for k9000
k9000
2
Original post by chazz1234
Question 21 in the MC, the answer was surely A the one with Length 1 and Width 0.2. As the F=BIL. It didn't ask for the maximum flux linkage or EMF. Therefore the one with greatest couple has to be the one with the longest side perpendicular to the field, A.


That's the reasoning I went with, but a couple also accounts for distance :/
Original post by CD223
*** OFFICIAL AQA PHYA4 FIELDS AND FURTHER MECHANICS JUNE 2015 EXAM DISCUSSION THREAD ***

UNOFFICIAL MARK SCHEME:

SECTION A:
1. Weight
2. Momentum of the alpha particle is equal and opposite to the nucleus Y
3. After collision T1 is at rest and T2 moves at 4.0ms^-1
4. Maximum acceleration is directly proportional to the amplitude
5. Graph with constant negative gradient passing through the origin
6. Energy of the oscillating system is 6.25*10^-3J
7. 2.33 seconds
8. 4R
9. Pi*pGMR/3
10. Field strength of Mars = 13.4NKg^-1
11. Field strength is zero at mid point
12. Incorrect that Gravitational potential is a vector quantity
13. GM/r^2, (GM/r)^1/2
14. Speed increases, period decreases
15. 4.09*10^-7 Vm^-1, upwards
16. n = 2, X = g, a = G, b = mass
17. Number of electrons 4.4*10^10
18. 1000s
19. Incorrect that capacitor stores all of energy supplied by battery (something to do with 2mJ)
20. -322g
21. Coil with 0.5L by 0.5L
22. Particle follows circular path at right angles to magnetic field
23. Fx/Fy = 1/(root 2)
24. Ip = 0.35A
25. Steel core of wire for strength.

SECTION B
1.
(a) forced vibrations are when theres vibrations that are forced that aren't at its natural frequency with a phase difference and different amplitude
resonance is when the driving frequency=natural frequency causes oscillations large amplitude pi/2 phase diff.

Forced oscillations:
When a periodic force (PF) is applied to a natural oscillator (NO) (Which has a natural frequency).
If PF has frequency much lower than NO, then NO oscillates with roughly same amplitude as periodic force. Small/no phase difference.
If PF has frequency much higher than NO, then NO oscillates with small amplitude. Variable phase difference and amplitude. Max amplitude when PF is pi/2 radians out of phase with NO and minimum amplitude when PF is in phase with NO.

(b) Which one shows greater damping?
The cone with the ring
The cone without the ring
The ring does not affect damping.
4 marks.

(1) Cone with ring shows less damping.
3 marks from:
Damping reduces the amplitude of oscillations/total energy of system (over time) (1)

Air resistance is same for both (the cone with and without ring) (1)

Reducing speed reduces kinetic energy (and therefore total energy) of cone (and therefore amplitude of oscillations) (1)

The ring increases the mass of the cone:
Application of Newton's second law to deduce that an equal force has less effect on the speed of an object with greater mass.

OR

Air resistance does work against cone. Use of E_k = (1/2)mv^2 to deduce that the same work done on a heavier object reduces its speed less.
(1 maybe 2)

2.
(a) Coulomb's law- proportional to product of magnitudes of charges, inversely proportional to square of separation..

(b) V is proportional to 1/r. V is negative for a negative point charge.
A positive test charge will have to do work against the field to reach a point in this field from infinity.

(c) Show that the magnitude of the charge is 27.8 nC (approx 30nC).

(d) Work done to move from r=0.2m to r=0.5m
= 4.5x10^-5 J

(e) Electric field strength at r= 0.4m
= 1560 V m^-1

3.
Area under the curve gives the initial charge on the capacitor before discharge (1)
[Between]

Draw curve of current for 300kOhm resistor below 150 ohm graph. Lower y intercept. Shallower gradient, longer time constant.

(a) 1.4m (height gained by 3.5N weight)

(b) Energy losses due to...
some energy converted to chemical energy or sound energy
not 100% efficient in transferring to potential energy

4.
(a) (i) When switch S is closed, the ammeter deflects and then returns to 0.
A current passes through P so a magnetic field is (suddenly) created in iron bar.
There is a change of flux through Q, inducing an current (or emf that causes a current) in Q.
After, current through P is DC/magnetic field is constant so no change of flux through Q => no current in Q (any longer).

(ii) When the variable resistance is suddenly increased,
(The ammeter) deflects in the opposite direction (to when switch S was closed) and then returns to zero.
Sudden increase of resistance causes sudden decrease of current through P. This causes a sudden decrease in the magnetic field strength so a sudden decrease in flux through Q. (When switch S was closed it was an increase so current is in opposite direction). This induces a current in Q.

(b) 0.0605 Wb turns

(c) 0.121V

5.
(a) 131 degrees

(b) 12.3N

(c) 2.01 revolutions

(d) Newton's first:
- An object will remain at rest or in uniform motion if not acted upon by an resultant force.
- Can be demonstrated by cutting the string while the ball is being spun: There is no longer a horizontal force acting on the ball, so it moves off in a straight line (in the horizontal plane) with a constant horizontal velocity (albeit under freefall in the vertical plane).

Newton's second:
-Force is rate of change of momentum
-String provides the centripetal force.
-Which acts at right angles to the ball's velocity.
-This causes a centripetal acceleration.
-Velocity is constantly changing direction, even though speed is not.
-Therefore momentum is always changing.
-Therefore there must be a resultant force.
-Resultant force is constant in magnitude since rate of change of momentum is constant in magnitude.

Newton's third:
-Every action has an equal and opposite reaction.
-The tension in the string acts on the ball.
-The ball exerts an equal and opposite force on the string.
-In real life, the Earth exerts a gravitational force downwards on the ball.
-The ball exerts a gravitational force upwards on the Earth.

Not horizontal in real life:
-The Earth exerts a gravitational force on the Earth downwards.
-This is equal to the ball's weight, W=mg.
-The ball has a non-zero mass.
-The ball does not accelerate towards the ground, so the resultant vertical force on the ball is zero.
-Therefore there is a force acting upwards on the ball equal to its weight.
-This can only be provided by the tension in the string.
-The tension only has an upwards vertical component if it is below the horizontal. (Use of T=m g sin(theta))
-A higher tension means the string can be closer to the horizontal.
-This can be achieved by spinning the ball faster, as then the string must provide a larger centripetal force and so has greater tension.

Date: 11th June 2015
Time: 09:00am
Duration: 1h 45m



can anyone please tell me what each question was worth?? even just the calculations? thanks
Original post by oonic0rn
no sorry i meant the whole of section B for the calculations, just cant remember how many marks were awarded :tongue:


Oh sorry.

1)a) 4. b) 4.

2)a) 2 b) 2 c)2 or 3 d)2 or 3 e)2 or 3.

3) Area under curve was for 2 marks I think. The sketch was 3 marks
a)2 or 3 b) 2

4)a)i) 3 ii) 2 b) 3 c) 2

5)a) 2 b)2 c)2 d) 6


Sorry I can't be more specific on some of the calculations
Original post by JizzaStanger
Oh sorry.

1)a) 4. b) 4.

2)a) 2 b) 2 c)2 or 3 d)2 or 3 e)2 or 3.

3) Area under curve was for 2 marks I think. The sketch was 3 marks
a)2 or 3 b) 2

4)a)i) 3 ii) 2 b) 3 c) 2

5)a) 2 b)2 c)2 d) 6


Sorry I can't be more specific on some of the calculations


thankyou absolute life saver!!
Original post by k9000
That's the reasoning I went with, but a couple also accounts for distance :/


Damn forgot about that just. I focused on the force not the moment. That's so annoying I thought it was a trick question, now I realise i have been done by AQA again.
Original post by chazz1234
Question 21 in the MC, the answer was surely A the one with Length 1 and Width 0.2. As the F=BIL. It didn't ask for the maximum flux linkage or EMF. Therefore the one with greatest couple has to be the one with the longest side perpendicular to the field, A.


The couple is the force on the long sides multiplied by the perpendicular distance between the forces.
F=BIL and Couple=Fd (d is perpendicular distance)
so Couple=BILd
A was L by 0.2L so
F=0.2BIL^2
B was 0.5L by 0.5L
F=0.25BIL^2
BIL^2 is constant and 0.25>0.2 so the answer is B.
That was my working anyway.
(edited 8 years ago)
Area under the curve gives the initial charge on the capacitor before discharge (1)
[Between]

Which bit was this? What was the answer; I cannot for the life of me remember this lmao.
Original post by Doomlar
Area under the curve gives the initial charge on the capacitor before discharge (1)
[Between]

Which bit was this? What was the answer; I cannot for the life of me remember this lmao.


The question right before the one that asked you to sketch the graph.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you