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

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https://gyazo.com/1c1f8ff97e2d2369eea920c54ab0a3fa

That's my solution to the momentum proof question for the people who struggled with it :smile:

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Reply 1581

particlestudent

Original post by Jay1421

I still can't tell if my momentum one was right, I got around 1 * 10^-19... which didn't seem right.

For the advantage I put that it would be faster because electricity is faster than friction (I don't even know) and then for the disadvantage, I put that to stop a car you'd require a strong electromagnet so that would result in a higher energy use, i.e less efficient? :\

Best of luck!

I got that for momentum too.... Good luck though! :smile:

Reply 1582

16Characters....

Original post by particlestudent

I got -3Q/8 for electric potential, probably wrong
And zero for that square one

I got those.

Reply 1583

Yo12345

Original post by Yo12345

Who got +5Q/8 for electric potetial, -4GM/d for gravitaional potential? Work done by 2nC charge as 0.15microC? The one with the charge moving along the electric field line that potential is constant? The one with the electric field strength at the centre of the square = 0? The pendulum one with the biggest amplitude = A? The one with Capacitance decreasing: 96V and 2.9x10^-10 charge?

Posted from TSR Mobile

EoG = no unit

Posted from TSR Mobile

Reply 1584

Mentalmirz

Original post by Yo12345

The pendulum one was B because it was the same length therefore the frequency was the same so undergoes resonance.

The charge moving along the field line wasn't constant, It was either increasing or decreasing but I can't remember which.

the capacitance one is definitely correct and I can't remember the others

Reply 1585

js.int

Original post by DesignPredator

Here are my answers to Written, the question parts might not be correct.

Written answers
1ai) 8.26 (will be a range)
ii) 2.6x10^-4

bi)28
ii)5.9x10^4
iii) bottom box ticked

2ai) Derivation
ii) 6.06x10^6 (3SF only)

b) Downward Curve staring at 9.81
Minimum at g=0 at a point slightly offset from halfway closer to Venus
Rises up to 8.87

3a) V=mv/N

b) Derivation

ci) 216
ii) 1.15x10^-19 kgms-1 OR Ns
iii) Anti electron neutrino released also so Beta minus particles have range of Kinetic Energies upon release.

4a) Small amplitude oscillation as only valid for small angular amplitudes.

b) -Set up light string with Bob on the end,
attach top of string to clamp stand boss.
-Measure length from top to centre of Bob.
-Release the Bob so it oscillates with small amplitude.
-Use stop clock to measure time taken for 20 oscillations.
-Divide this by 20 for mean T
-Plot T^2 against L
-Calc Gradient
g=(4pi^2)/gradient
-Repeat procedure

c) -Student value 4x true value
-Time period half so all values for T^2 will be 1/4 the true
-Gradient is 4x lower
-State equation or show that g is inversely proportional to gradient.

5a) Lenz's Law: Direction of induced EMF/current is always in a direction that opposes the change that caused it.
Faraday's Law: Induced EMF is proportional to rate of change of flux linkage.

b) -Change in flux linkage when current flows.
-EMF induced in wheel (or whatever it was called)
-Current induced in the wheel as good conductor.
-Current carrying conductor (wheel) opposes the field due to coil, therefore there is a force against the direction of motion.

c) Wheel not in contact with electromagnet so no wear, but in brake pads there is wear due to friction.

More energy used or less effective or heating.

For question 3(c)(iii) I put something about the beta- particle being released at relativistic speeds so E_k = (1/2)mv^2 can't be used?

Reply 1586

KD1997

Original post by CourtlyCanter

I put the one with the same length as the driver since it would resonate. B if I'm not mistaken.

yeah i didn't know what to do for the mcq, so i went for the longest one A but i'm pretty sure i got it wrong.

for the limiting facrto though i wrote what you did but I'm sad to say we're both wrong. The limitation is that the formula is only accurate for angles of displacement of 10 degrees or less since the dervivation of the formula uses the small angle approximation to say gsin theta=g

Reply 1587

rk_

@kingaaran It's me Raman compare your multiple choice answers to mine

D B B D D C D B C B C B B A A C D C C D A D C B D

Reply 1588

C0balt

Original post by particlestudent

I got -3Q/8 for electric potential, probably wrong
And zero for that square one

Got that

Reply 1589

TheLazyGuy1

Original post by particlestudent

I got -3Q/8 for electric potential, probably wrong
And zero for that square one

no thats right i got that too

Reply 1590

Yo12345

Original post by TheLazyGuy1

no thats right i got that too

Was it not 1--4 on the top. I could be mistaken but I thought you take away potential meaning it would be 5?

Posted from TSR Mobile

Reply 1591

Parallex

Original post by TheLazyGuy1

no thats right i got that too

also got those

Reply 1592

Yo12345

Original post by C0balt

Got that

Do you remember any others. Transformer one with the table was B? The one with the left hand rule was 0.0075N upwards

Posted from TSR Mobile

Reply 1593

particlestudent

Original post by C0balt

Got that

If I thought N was the original mass of nucleus and got mv/(N-m) for V and attempted to find the Ea formula, would I get 0 marks on both of them :redface:

? I hope I get some on the Ea one...

Reply 1594

particlestudent

Original post by Yo12345

Was it not 1--4 on the top. I could be mistaken but I thought you take away potential meaning it would be 5?

Posted from TSR Mobile

It's a scalar so it's the sum, 1+(-4)

Reply 1595

Mentalmirz

Original post by palacefloor

Could someone please reply to this.

For the first question the value of the resistor and time constant depended on the capacitor from the previous graph, so I worked all these out. However when I checked through I looked at the graph and worked out the capacitor to be very slightly different, will I drop any marks for not re-correcting the time constant and resistor or will it be fine?

In the second part of the question it was a different capacitor so you didn't use the graph, it was a capacitor of 470x10^-6 in the second part so the time constant will be different than the one from the graph