AQA Physics A Unit 4- 11 June 2014

OP

Original post by RainieXD

Ah I see, I always mess up on phase difference questions :/ and yes magnetic and electric fields ...sigh

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electric fields I'm okay.

and phase difference it's always n/2 if it's velocity and displacement i think

Reply 42

Original post by RainieXD

Could someone help me with the Unit 4 2014 june paper section B, Q1 c?? I don't understand the mark scheme either :s-smilie:

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You sure you meant 2014?

Reply 43

Original post by NedStark

You sure you meant 2014?

Ah sorry, my bad - I meant the June 2013 paper

Original post by Qari

electric fields I'm okay.

and phase difference it's always n/2 if it's velocity and displacement i think

Uhmm could you explain what you mean by n/2? I saw this in the mark scheme and I have no clue what it's for..

Reply 44

OP

Original post by RainieXD

Ah sorry, my bad - I meant the June 2013 paper

Uhmm could you explain what you mean by n/2? I saw this in the mark scheme and I have no clue what it's for..

i figure it out because when displacement is 0 velocity is max and when velocity is 0 displacement is max. so the difference between when displacemnt is 0 and velocity is 0. it's normally n/2

Reply 45

Original post by RainieXD

Ah sorry, my bad - I meant the June 2013 paper

The question about the 2 pendulum?

How I did it was firstly to work the time difference out which is 0.1s. The 1.9 pendulum will complete the oscillation 0.1s faster than that 2.0 pendulum.

The 2nd oscillation will be completed 0.2s ahead and 3rd 0.3s ahead etc.

The question asks how long until the faster pendulum will come back level with the slower pendulum. If you keep going with the oscillations then by the 20th oscillation the faster pendulum will be 2.0s ahead of the slower pendulum. That's basically starting from the beginning.

So it takes the 1.90s pendulum 20 oscillations. 20x1.90= 38s

Reply 46

OP

Original post by NedStark

The question about the 2 pendulum?

How I did it was firstly to work the time difference out which is 0.1s. The 1.9 pendulum will complete the oscillation 0.1s faster than that 2.0 pendulum.

The 2nd oscillation will be completed 0.2s ahead and 3rd 0.3s ahead etc.

The question asks how long until the faster pendulum will come back level with the slower pendulum. If you keep going with the oscillations then by the 20th oscillation the faster pendulum will be 2.0s ahead of the slower pendulum. That's basically starting from the beginning.

So it takes the 1.90s pendulum 20 oscillations. 20x1.90= 38s

That's how I did it.
LCM

Reply 47

Jimmy20002012

4

Could anyone tell me, do you use Flemings left hand rule if dealing with F=BIL, and do you use Flemings Right Hand Rule if dealing with F=BQV?

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Reply 48

Original post by Jimmy20002012

Could anyone tell me, do you use Flemings left hand rule if dealing with F=BIL, and do you use Flemings Right Hand Rule if dealing with F=BQV?

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I think you always use Flemings left hand rule, my book has never covered the right hand rule (CGP) so I guess it's not necessary

Reply 49

Original post by NedStark

The question about the 2 pendulum?

How I did it was firstly to work the time difference out which is 0.1s. The 1.9 pendulum will complete the oscillation 0.1s faster than that 2.0 pendulum.

The 2nd oscillation will be completed 0.2s ahead and 3rd 0.3s ahead etc.

The question asks how long until the faster pendulum will come back level with the slower pendulum. If you keep going with the oscillations then by the 20th oscillation the faster pendulum will be 2.0s ahead of the slower pendulum. That's basically starting from the beginning.

So it takes the 1.90s pendulum 20 oscillations. 20x1.90= 38s

Original post by Qari

i figure it out because when displacement is 0 velocity is max and when velocity is 0 displacement is max. so the difference between when displacemnt is 0 and velocity is 0. it's normally n/2

Omg that took me a whilleeeee to understand :s-smilie:

but thanks guys! i finally get it :smile:

I was getting confused as to why 20t was used and not 10, but I guess at 10 it would mean they would be half way out of phase?.. Anyways thanks so much! :smile:

Reply 50

OP

Original post by RainieXD

Omg that took me a whilleeeee to understand :s-smilie:

but thanks guys! i finally get it :smile:

I was getting confused as to why 20t was used and not 10, but I guess at 10 it would mean they would be half way out of phase?.. Anyways thanks so much! :smile:

I used LCM. As in the lowest common multiple. When is it that both pendulum will have the same time

Reply 51

OP

Original post by NedStark

I think you always use Flemings left hand rule, my book has never covered the right hand rule (CGP) so I guess it's not necessary

Revision Guide or Text Book?

Reply 52

Original post by Qari

I used LCM. As in the lowest common multiple. When is it that both pendulum will have the same time

Ohhh I see, I'll do that then. The mark scheme just confused me over lol

Reply 53

Original post by Qari

Revision Guide or Text Book?

The text book version but even then it's still dummed down a bit. I've also done all the past papers and never required the right hand rule but don't take my word on this.

Reply 54

Anyone know how to do MC of june 2013 Q 13? The one with masses M and 4M, i try to add g=Gm/r^2 for both M and 4M and then equate to 0, but then I just end up with a long formula like this :frown:

5y^2 + d^2 -2dy=0
I'm sure it shouldn't be this hard for a MC question :s-smilie:

..?

Reply 55

spiruel

For those wondering about the pendulum question - if you memorise this formula and possibly learn how to derive it, you will not need to worry.

PhysicsCALL76LY.pdf

Reply 56

OP