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

1. (Original post by chizz1889)
I think your acc graph is wrong (not sure). The distance between planets is definitely 4R as initially F is proportional to 1/4R^2 then it becomes 1/9 F so 4*9=36 root36=6 then 6-2=4 (as it says between planets so u remove each radi)
The force one i think it's wrong yeah, i didn't even want to think about it lol

I remember the graph from here.

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2. (Original post by chizz1889)
I think your acc graph is wrong (not sure). The distance between planets is definitely 4R as initially F is proportional to 1/4R^2 then it becomes 1/9 F so 4*9=36 root36=6 then 6-2=4 (as it says between planets so u remove each radi)
I think the graph is right, I did that. Max EK at equilibrium, so O acceleration, and proportional negative and positive but negative y when positive x and vice versa.
3. (Original post by Disney0702)
His graph is correct
In fact yeah that's true as acc is directly proportional to disp. fs silly mistake by me
4. (Original post by NEWT0N)
What did people put for the forced vibrations?

I remember using the cgp explanation: A system can be forced to vibrate by an external periodic force called the driving force. If the driving force does not have the same natural frequency as the system, the amplitude of oscillation will be smaller than that in free vibration.
Could also be larger?
5. (Original post by k9000)
Newton's first law at the end i put because no force in direction of velocity (always tangential) it remains at constant speed. Is that ok for the first law
I talked about when it is released for some reason but I also explained that the horizontal component of velocity is constant throughout. Did anyone else do this?
6. Yep pretty sure its 4R
7. (Original post by NEWT0N)
What did people put for the forced vibrations?

I remember using the cgp explanation: A system can be forced to vibrate by an external periodic force called the driving force. If the driving force does not have the same natural frequency as the system, the amplitude of oscillation will be smaller than that in free vibration.
Yeah I put that when the system is forced to oscillate by an external varying force at a driving frequency i.e a frequency other than it's natural frequency.
I didn't put anything about amplitude till the next bit on resonance, so talked about the phase differences above,below and at resonance and amplitude.

Will they mark it independently?
8. (Original post by kaziz)
Yep pretty sure its 4R
Yes 6R-R-R (subtract the radii of the spheres).
9. I remember getting G*m*pi*p*R/3 somewhere for the multiple choice. Anyone agree?
10. (Original post by AR_95)
yes

I said what you did but I realise its wrong

light damping has no effect on the time period because Amplitude is not related to time period, but with damping the amplitude generally increases

heavier damping should be with the ring on i think
Referring to equations v=2*pie*f*(A^2-X^2)^-2 and T= 2pie*(l/g)^2 . Frequency hence velocity are independent on the mass of the pendulum bob , therefore no damping even if ring was added
11. Glad I only need 184/300 UMS this year for an A 😛 haha
12. What did everyone put for the current through X and Y question with the iron rod? I said when resistance of variable resistor was minimum a large current reading was on the ammeter since Voltage is kept the same, and iron is a good conductor so it transfers lots of the current which is induced in the second coil? And then when the resistance was increased the current reading went down again because V was kept the same?

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13. (Original post by kaziz)
Yep pretty sure its 4R
Yeah I think it is also, I made up my own actual values to see what the right answer was and I got 4R, because I think the separation you get you have to take away the radius of the two because it asked for distance between their surfaces not centres?..wow what a way to waste my time though!
14. (Original post by wuzecheng9)
Referring to equations v=2*pie*f*(A^2-X^2)^-2 and T= 2pie*(l/g)^2 . Frequency hence velocity are independent on the mass of the pendulum bob , therefore no damping even if ring was added
That doesn't matter, damping doesn't affect time period it affects amplitude, so the amplitude may be less/more
15. (Original post by wuzecheng9)
Referring to equations v=2*pie*f*(A^2-X^2)^-2 and T= 2pie*(l/g)^2 . Frequency hence velocity are independent on the mass of the pendulum bob , therefore no damping even if ring was added
You don't get the concept of damping, damping is the effect of external forces on the system. When the system is light there is a large damping effect and then when it is heavy there is minimal effect. All due to momentum
16. (Original post by NEWT0N)
I talked about when it is released for some reason but I also explained that the horizontal component of velocity is constant throughout. Did anyone else do this?
Isn't the horizontal component constantly changing as direction is constantly changing?
17. (Original post by fruity97)
Yeah I put that when the system is forced to oscillate by an external varying force at a driving frequency i.e a frequency other than it's natural frequency.
I didn't put anything about amplitude till the next bit on resonance, so talked about the phase differences above,below and at resonance and amplitude.

Will they mark it independently?
Yes I think so. I'm not even sure if I talked about amplitude in the first one but I remember talking about phase difference between the oscillator and driver (e.g. driving frequency << natural frequency means they are in phase, etc)

They will mark it independently depending on what each part asked for. Any unrelated but correct physics will be ignored for each part. But if you talked about forced vibrations in the second one and resonance in the first one you won't get any marks, if you get my drift

EDIT: in bold
18. (Original post by Fvthoms)
Isn't the horizontal component constantly changing as direction is constantly changing?
I mentioned the velocity is constantly changing because of the centrapetal force
19. (Original post by Fvthoms)
Isn't the horizontal component constantly changing as direction is constantly changing?
Yes because a force is impressed on it. When there is no longer a force it is constant throughout its motion from release and follows a parabolic curve. The horizontal velocity now remains the same as the tangential velocity at the point of release, in accordance with Newton I.
20. (Original post by crs96)
For flux linkage i put Tm(squared) ? is that ok?

Also what did people put for the distance between the surfaces of the 2 spheres mass M radius R when F/9?

2R, 4R, 8R or 12R?
Fairly sure it was 4R.

Original force = (GM^2)/((2R)^2)
New force = (GM^2)/(9*(2R)^2) = (GM^2)/(36R^2) ---> Distance between CoM = sqrt(36R^2) = 6R

6R - 2R (ie the distance between CoM before separation) = 4R

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