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# OCR Physics G484 - June 2013 Unit 4 (OFFICIAL RETAKE THREAD) watch

1. I did not like the G485 paper so need to do well in this one!

Any ideas on what is gonna come up based on previous years?
2. When using formulae involving Temperature, do we always have to use kelvin or are there cases where we are supposed to use celcius.
Formulae such as: pV = NkT
pV = nRT
E = 3/2 kT
E = mcΔѲ
3. (Original post by I'm Batman...)
When using formulae involving Temperature, do we always have to use kelvin or are there cases where we are supposed to use celcius.
Formulae such as: pV = NkT
pV = nRT
E = 3/2 kT
E = mcΔѲ
For pV=NkT, pV=nRT and E=3/2 kT, you use the thermodynamic temperature (Kelvin).
For E=mcΔѲ, it depends on the unit of the specific heat capacity; it the unit of c is Jkg^(-1)K^(-1), than your temperature would be in Kelvin, but if the unit of c is Jkg^(-1)°C^(-1), than the temperature would be measured in celcius.
4. After doing all the practise papers, what else is there to do?
5. (Original post by I'm Batman...)
When using formulae involving Temperature, do we always have to use kelvin or are there cases where we are supposed to use celcius.
Formulae such as: pV = NkT
pV = nRT
E = 3/2 kT
E = mcΔѲ
Remember that a temperature change will be the same whether you use kelvin or celsius (they use the same scale, just start at a different point) Therefore you can use either when using E=mcΔѲ, allthough if they ask you to state temperature then check if they want it in kelvin or farenheit.

For the other 3 formulae, kelvin should be used
6. What do we need to know about the inverse square law in relation to gravitational fields and orbits?
7. Does anyone have a copy of the Jan 2013 paper?
8. (Original post by Jullith)
Does anyone have a copy of the Jan 2013 paper?
finally got my hands on them
Attached Images
9. G484_MS_Jan13.pdf (146.2 KB, 345 views)
10. G484-01Jan13.pdf (164.5 KB, 315 views)
11. Hey guys, I have some questions.
Can momentum and impulse be negative? I've done some questions (in maths mechanics), I've decided a direction so I have a positive approach velocity and negative separation velocity, making the impulse negative, however they wanted the impulse to be positive so they simply changed the signs of the velocities, which I don't understand.

Newton's second law is; the force acting on an object is proportional to the rate of change of momentum and acts in the direction of the change. When I say force, is it the net force?
12. (Original post by imogen--)
What do we need to know about the inverse square law in relation to gravitational fields and orbits?
I think it just means as your distance from a point mass increases, the attraction decreases exponentially.
13. (Original post by eggfriedrice)
I think it just means as your distance from a point mass decreases, the attraction decreases exponentially.
I think you might have meant to say that as distance decreases, gravitational attraction increases or vice versa, due to the inverse square law.

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14. What do you guys think the experiment question will be on? Looking at past papers I think it's likely to be brownian motion or specific heat capacity.
15. (Original post by eggfriedrice)
Hey guys, I have some questions.
Can momentum and impulse be negative? I've done some questions (in maths mechanics), I've decided a direction so I have a positive approach velocity and negative separation velocity, making the impulse negative, however they wanted the impulse to be positive so they simply changed the signs of the velocities, which I don't understand.

Newton's second law is; the force acting on an object is proportional to the rate of change of momentum and acts in the direction of the change. When I say force, is it the net force?
If you can post the question, I might be able to make sense of it. You typically should remember that impulse will always be in a direction opposite to the motion before the collision, as a general rule, unless body A collides with body B with a greater velocity than body B, when they were both moving initially in the same direction, then in that case the impulse would be in the direction of motion of B before the collision.

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16. (Original post by Rhodopsin94)
I think you might have meant to say that as distance decreases, gravitational attraction increases or vice versa, due to the inverse square law.

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Aha yep, changed it. Thanks for pointing that out.

Ok so, if I threw a ball at a wall (ball on left, wall on right), and I took going right to be positive and left to be negative, the impulse on the wall would still be negative despite going to the left?
17. (Original post by eggfriedrice)
Aha yep, changed it. Thanks for pointing that out.

Ok so, if I threw a ball at a wall (ball on left, wall on right), and I took going right to be positive and left to be negative, the impulse on the wall would still be negative despite going to the left?
Technically no, I don't think so, but I'll try and explain why:

Before the ball collides it has velocity mu. After the collision the ball has velocity -mv according to your positive/negative directions. So I = (-mv) - mu, which leads to a negative answer.

BUT I had the same problem when I did maths mechanics:

You know that the direction of the impulse is to the left. In questions, like this, always determine the direction of the impulse first. So when you are working out the impulse, take your impulse direction as the positive direction.

So I = mv - (-mu) = mv + mu. Hopefully that should work now, as I found that it did for me.

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18. (Original post by Rhodopsin94)
Technically no, I don't think so, but I'll try and explain why:

Before the ball collides it has velocity mu. After the collision the ball has velocity -mv according to your positive/negative directions. So I = (-mv) - mu, which leads to a negative answer.

BUT I had the same problem when I did maths mechanics:

You know that the direction of the impulse is to the left. In questions, like this, always determine the direction of the impulse first. So when you are working out the impulse, take your impulse direction as the positive direction.

So I = mv - (-mu) = mv + mu. Hopefully that should work now, as I found that it did for me.

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Yes! That's exactly what I was confused about. So the direction of impulse is always the direction the object is traveling after the collision? And we change v to pos, and u to neg?

Also does that mean they wouldn't ever want a negative impulse?
19. The calculation questions in G485 were hard, so they might do the same thing in G484.

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20. (Original post by Jullith)
For pV=NkT, pV=nRT and E=3/2 kT, you use the thermodynamic temperature (Kelvin).
For E=mcΔѲ, it depends on the unit of the specific heat capacity; it the unit of c is Jkg^(-1)K^(-1), than your temperature would be in Kelvin, but if the unit of c is Jkg^(-1)°C^(-1), than the temperature would be measured in celcius.
But u must remember that the thermodynamic scale is made in such a way that a change in temp of say 5 degrees celsius is the SAME as the temp change of 5 Kelvin
21. (Original post by eggfriedrice)
Yes! That's exactly what I was confused about. So the direction of impulse is always the direction the object is traveling after the collision? And we change v to pos, and u to neg?

Also does that mean they wouldn't ever want a negative impulse?
Yeah that sounds right, so long as the v is in the same direction as impulse, which I see no reason why not, then yes.

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22. Q) in order to move a satellite into a new smaller orbit, a decelerating force is applied for a brief period of time. Suggest how the decelerating for could be applied.

Now I don't understand the question. We want a smaller orbit so smaller r. Surely by decelerating the orbit or decreasing the force, wouldn't the radius increase? The mark scheme says to apply something like a rocket motor against the direction of travel, thus decelerating it.

Can anyone explain this to me? Thanks.

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Updated: December 13, 2013
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