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

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Yeah it was: damping is less with the ring.

Explanation: Cross sectional area of the cone is unaffected by the ring, so the drag force due to air resistance experienced by the cone and the cone with the ring is the same. A force results in a change in momentum, so both cones have the same change in momentum due to drag, however the cone with the ring has greater mass. The same change in momentum but a larger mass means a smaller reduction in velocity due to air resistance. This means a lower rate of reduction of velocity for the cone with the ring, and so it loses kinetic energy at a lower rate.

IMHO they didn't give enough space or marks for virtually any of the descriptive type questions.

Could you explain what was wrong with my answer then?If a system is damped the amplitude of oscillations would decrease.Using Vmax=2 x Pi x f x A we can see that A depends on only Vmax and the frequency of oscillations. Neither of these variables are affected by the mass of the pendulum bob so therefore there is no change when the mass is increased.

Reply 4441

CD223

Wasn't the electric potential 1.56x10^3 or something?

Then work done 4.15 x 10^-5?

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Reply 4442

sykik

Original post by xela238

I worked out V for each one using the V=Q/constant x r^2 then used work done = difference in V x Q to get the work done

Yh i did that now and got 4x10^ -4?? people are saying x10^-5

Reply 4443

Kennethm

the thing about the mistake is that a lot of students would have spent more time on it than other students meaning the rest of the paper could also be effected. Hate it when this happens mainly because Im so sure I would have got it correct if it was 300 ohms as opposed to 300kohms and it would have worked in my favour since i know a lot of other students wouldn't have got it correct lol.

Im sitting here trying to remember the rest of the MC Q's but i guess it's pointless.

Reply 4444

jkrolling12

Original post by xela238

I worked out V for each one using the V=Q/constant x r^2 then used work done = difference in V x Q to get the work done

Yeah same here, apparently there were two charges involved though? Can anyone remember the exact situation of the question?

Reply 4445

Sbarron

For one question I think I got a time (or frequency) as 0.5 through 90 degrees so i doubled it to 1 (as a whole cycle on a generator is 180 degrees) was this right to do???

Reply 4446

Amanzz

Original post by Tomh97

Does anyone actually know what the right answer was for the damping question?

Maybe. My teacher is Greek, those blokes practically invented physics, I asked him. He said this. So idk. Many conflicting ideas. ImageUploadedByStudent Room1434029480.999112.jpg

ImageUploadedByStudent Room1434029480.999112.jpg

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Reply 4447

SuperMushroom

Original post by CD223

Wasn't the electric potential 1.56x10^3 or something?

Then work done 4.15 x 10^-5?

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Yep thats what i got, not sure if its correct though haha

Reply 4448

DanielWall96

Original post by xela238

I worked out V for each one using the V=Q/constant x r^2 then used work done = difference in V x Q to get the work done

I used that method, however i got a negative answer for the change in V and took the modulus of the answer i got

Reply 4449

StarvingAutist

Original post by JizzaStanger

You can read about it in the attached pdf if you like. I did this just before Christmas. You'll need to have done further maths to follow the maths it but I think it's quite interesting.

I've done FM so it's sweet :smile:

thanks