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

1. (Original post by CD223)
For objects to appear weightless, R (the reaction force) always equals zero, right? Meaning only weight in this case provides the centripetal force?

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Yes, that's it, just at the weightlessness moment

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2. (Original post by Mehrdad jafari)
Yes, that's it, just at the weightlessness moment

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Thank you! For that question JJBinn posted I simply used Kepler's third law to find the time period for a point on the earth's equator. Why does this work? Your method seems to make more sense but my method still somehow worked.

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3. (Original post by CD223)
Thank you! For that question JJBinn posted I simply used Kepler's third law to find the time period for a point on the earth's equator. Why does this work? Your method seems to make more sense but my method still somehow worked.

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I see what you mean. I simply considered W=mg because at the equator ( essentially on the surface of the earth) the gravitational field strength is 9.8 N/kg which is basically derived from g=GM/r2. If you calculate the g at the surface of the earth it will give you 9.8 N/kg

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4. (Original post by Mehrdad jafari)
I see what you mean. I simply considered W=mg because at the equator ( essentially on the surface of the earth) the gravitational field strength is 9.8 N/kg which is basically derived from g=GM/r2. If you calculate the g at the surface of the earth it will give you 9.8 N/kg

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I think I used Kepler's third law because it would mean that the gravitational force would be the only force providing the centripetal force (as R=0) at the surface of the earth if the object is weightless.

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5. (Original post by CD223)
I think I used Kepler's third law because it would mean that the gravitational force would be the only force providing the centripetal force (as R=0) at the surface of the earth if the object is weightless.

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Yeah that's correct because the gravitational force is the weight of the object. The weight of an object at any distance is W=GMm/r2 but at the surface the gravitational can be considered as simply W=mg where g could be GM/r2

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6. Thanks for your responses guys. CD, I tried using T^2=4pi^2r^3/GM and was going to rearrange for T with M as 0, but obviously M is on the bottom of the fraction so it wouldn't work? (This is Keplers third law right?)

I think I will just remember as a rule at the equator, for an object to appear weightless you set mg=mv^2/r.
7. (Original post by Mehrdad jafari)
Yeah that's correct because the gravitational force is the weight of the object. The weight of an object at any distance is W=GMm/r2 but at the surface the gravitational can be considered as simply W=mg where g could be GM/r2

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Makes sense god I prefer this to unit 5 haha. May I ask - are you definitely able to get hold of the EMPA MS soon? Just wondered how you get it so quickly/if you've broken up from college

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8. (Original post by JJBinn)
Thanks for your responses guys. CD, I tried using T^2=4pi^2r^3/GM and was going to rearrange for T with M as 0, but obviously M is on the bottom of the fraction so it wouldn't work? (This is Keplers third law right?)

I think I will just remember as a rule at the equator, for an object to appear weightless you set mg=mv^2/r.
Kepler's third law states

So, for an object at the equator, . This equation gives you T=1.4 hours.

Never set m equal to zero as the mass never becomes zero.

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9. (Original post by CD223)
Kepler's third law states

So, for an object at the equator, . This equation gives you T=1.4 hours.

Never set m equal to zero as the mass never becomes zero.

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Oh man I was setting M to zero in that equation because I was so fixated on the fact it was weightless.

On a side note, for this equation, is the mass representative of both objects? Ie gravitational attraction occurs between 2 masses, do you add the masses together for example if it was a satellite orbiting the earth, in using that equation do you add the mass of the satellite to the earth's mass?
10. (Original post by CD223)
Makes sense god I prefer this to unit 5 haha. May I ask - are you definitely able to get hold of the EMPA MS soon? Just wondered how you get it so quickly/if you've broken up from college

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me too. To be honest i cannot remember studying unit 5 lool.
In theory definitely

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11. Ay yo
Any1 out there plz help me with this
Q 14 June 10

I dont know y the answer is A
can some1 plz explain y its A, it would b much appreciated

Thanks
Attached Images

12. (Original post by CD223)
In a past paper there was a question where you doubled f and m and were asked the effect on the total energy of the system. In that case it multiplied it by 8. That equation explains why.

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We do need to know the first one but the second one is useful to know and it's a short cut

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Thanks

For June 2014 Section B question 3bii would this be correct? (see attached)

13. (Original post by Paul Dirac)
Thanks

For June 2014 Section B question 3bii would this be correct? (see attached)

Yeah that should be alright

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14. (Original post by Kiki05)
Ay yo
Any1 out there plz help me with this
Q 14 June 10

I dont know y the answer is A
can some1 plz explain y its A, it would b much appreciated

Thanks
CC223 provided an explanation to that question in the form of a picture a few pages earlier

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15. (Original post by Mehrdad jafari)
Yeah that should be alright

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Thanks, I was just a bit confused by the mark scheme: "cosine curve"
Cosine curve starts at the top and moves down but my one starts at the bottom...
I assume they meant "negative cosine curve"
16. (Original post by Paul Dirac)
Thanks

For June 2014 Section B question 3bii would this be correct? (see attached)

By the way the way the question papers and the mark schemes of june 14 paper are on the first page of this thread

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17. (Original post by Paul Dirac)
Thanks, I was just a bit confused by the mark scheme: "cosine curve"
Cosine curve starts at the top and moves down but my one starts at the bottom...
I assume they meant "negative cosine curve"
Oh really? Yeah, cosine curve starts at the bottom. Let me think

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18. (Original post by Mehrdad jafari)
Oh really? Yeah, cosine curve starts at the bottom. Let me think

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Cosine starts at top

My sketch should be right because they said downwards acceleration is negative. So if the mass starts at the highest point of the oscillation, SHM means that the acceleration points downwards to the equilibrium position, so acceleration is negative at t=0.

I think they just weren't being very clear on the mark scheme.
19. How many times do you guys do each past paper?
Spoiler:
Show
Edit: Also why am I so bad at 6 mark questions? Do people just memorize the mark schemes for those? Because it seems everyone is better at it than I am (and I got A*A* in English GCSE )
20. (Original post by Paul Dirac)
Cosine starts at top

My sketch should be right because they said downwards acceleration is negative. So if the mass starts at the highest point of the oscillation, SHM means that the acceleration points downwards to the equilibrium position, so acceleration is negative at t=0.

I think they just weren't being very clear on the mark scheme.
Yeah sorry i meant starts at the top my mind is switched off lol.
Yeah you should be right

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