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

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Goodbye firm choice, it was nice thinking it was possible to be there but now my dreams have been smashed! Never seen so many word answers in a physics test before!!


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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)


Yes I also got 4R. Was confused at first but went back to it with a few mins to go and it was all so clear :biggrin: the fact that the seperation when they were touching was 2R made it sort of confusing
I thought that paper wasn't one of the easiest but defo easier than last year. I only had to guess 2 multi choices and they where estimates:smile: I felt confident in every question they asked and especially the change in the height thing for the capacitor! Absolute babe of a question! Thanks for all the help everyone's given me and good luck on results day! Going to imagine I've completely failed this now and get straight on with unit 5, see you on the other side (unit 5 forums)


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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.
ImageUploadedByStudent Room1434022847.236156.jpg


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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.
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
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?
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Avatar for kaziz
kaziz
3
Yep pretty sure its 4R
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?
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
Glad I only need 184/300 UMS this year for an A 😛 haha
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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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!
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
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
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?
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
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
To be honest there is no point discussing the written one. It all comes down to what examiners think is right


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Original post by chizz1889
Less damping, mass is increased and so therefore is momentum. Increased momentum results in resistive forces having a lesser effect on the motion of the pendulum, thus less damping.


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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