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# Unit 5 Physics Edexcel A2 and Edexcel IAL

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1. Did anyone else do IAL? How did you find it? Are there any specific threads to discuss the IAL paper?
2. What did you get for the mass of the moon? And also the one where you had to calculate something with the luminosity and with the 50.1 of the luminosity of the sun. Sorry, I can't remember exactly what it was ahhaa.
3. would say this paper was on par with last years. Bear in mind last years grade boundaries were incredibly low (61/80 = A*) so don't worry if, say, you lost marks on the harder questions - simple harmonic motion and specific heat capacity - you've still got plenty to play with
4. What are yall talkin about did you guys do wph05??!!!!!
5. Message samwray for the Unoffical Markscheme
6. For the time of the barrier;
Using x = Acos(wt):
w=2pi/T=2pi/4.5=1.4
Since x is displacement from the point of equilibrium, find the point of equilibrium for the bottom side using 2.3/2= 1.15
x = 0.6-1.15 = -0.55
Insert this into cos formula to get 1.4t = 1.81, 4.47
t = 1.29, 3.19
The difference in t is the time where it is above 0.6m hence t = 1.9

Did anyone else get this? How did you lot get 3s??
For the time of the barrier;
Using x = Acos(wt):
w=2pi/T=2pi/4.5=1.4
Since x is displacement from the point of equilibrium, find the point of equilibrium for the bottom side using 2.3/2= 1.15
x = 0.6-1.15 = -0.55
Insert this into cos formula to get 1.4t = 1.81, 4.47
t = 1.29, 3.19
The difference in t is the time where it is above 0.6m hence t = 1.9

Did anyone else get this? How did you lot get 3s??
Did you use degree mode? I did the same method but in radian mode and got 3 point something. Remember the time period for oscillation was 4.5 seconds and seeing as half an oscillation occurs during the motion it must be greater than 2.25 seconds
8. (Original post by PhysicsIP2016)
Did you use degree mode? I did the same method but in radian mode and got 3 point something. Remember the time period for oscillation was 4.5 seconds and seeing as half an oscillation occurs during the motion it must be greater than 2.25 seconds
Nope, I used radians, I think I realised where I went wrong now - I should've used x = 0.55 instead of minus, giving me an answer of 2.6s which would make more sense. Is that what you did? Hopefully get 3/4 method marks though.
9. I retract my shm answer.
Correct one is 100% 2.96.
1.15-0.6 gives displacemnt to the critical point find t times 2 4.5- the answer gives 2.96.

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Nope, I used radians, I think I realised where I went wrong now - I should've used x = 0.55 instead of minus, giving me an answer of 2.6s which would make more sense. Is that what you did? Hopefully get 3/4 method marks though.
ah okay! That's weird because I would have thought that because its a cosine graph using a minus wouldn't have made any difference? Yeah I reckon you'll get majority method marks
I remember what I did now, I kind of visualised the situation and split it up into three parts: 0.6m above the base to equilibrium position, then half an oscillation, then back to 0.6m above the base again.
I then used x=Acos(wt), using x=0.55, to work out the time it takes for the object to move from the equilibrium point to 0.55m. Since there is symmetry in SHM, I said that this is also the same amount of time it takes for the object to move from 0.55m to the equilibrium position so I multiplied that by 2 then added it to 2.25 s.
11. (Original post by physicsmaths)
Correct one is 100% 2.96.
1.15-0.6 gives displacemnt to the critical point find t times 2 4.5- the answer gives 2.96.

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But that's displacement from the equilibrium position to 0.6 m above the base, why are you subtracting it from 4.5 when that motion is within the path that you need to include?
12. (Original post by PhysicsIP2016)
But that's displacement from the equilibrium position to 0.6 m above the base, why are you subtracting it from 4.5 when that motion is within the path that you need to include?
No.
Yes but using coswt means t=0 is it is starting at amplitude so we want time taken for when the barrier is greater then or equal to 0.6. Time taken to get 0.6 (when the guy can't go under it) is from 0.55=1.15cos(wt) the principle value. Then we can times by to when it comes down. That gives time when he cant cross. So overall time is 4.5 -that time.

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13. (Original post by physicsmaths)
Correct one is 100% 2.96.
1.15-0.6 gives displacemnt to the critical point find t times 2 4.5- the answer gives 2.96.

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I actually think it's 3.7
Using a sine function, the period is equal to 4.5*2 (up down up down).
W = 2pi/9
then the time during which the height of the shutter is less than 0.6 is then given by
0.6 = 2.3 * sin Wt
find t, multiply it by 2 (one for up and one for down), take that time away from 4.5 and you get the required time, which is 3.7 or something similar.
14. (Original post by physicsmaths)
No.
Yes but using coswt means t=0 is it is starting at amplitude so we want time taken for when the barrier is greater then or equal to 0.6. Time taken to get 0.6 (when the guy can't go under it) is from 0.55=1.15cos(wt) the principle value. Then we can times by to when it comes down. That gives time when he cant cross. So overall time is 4.5 -that time.

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Aaah yes I see it now, damn oh well, method marks 😂
15. (Original post by oShahpo)
I actually think it's 3.7
Using a sine function, the period is equal to 4.5*2 (up down up down).
W = 2pi/9
then the time during which the height of the shutter is less than 0.6 is then given by
0.6 = 2.3 * sin Wt
find t, multiply it by 2 (one for up and one for down), take that time away from 4.5 and you get the required time, which is 3.7 or something similar.
My friends did that. But infact what you dont realise is that at that centre there is no SHM. So you can't use SHM formulae. Look at definitions of SHM.
Tbh everyones mehtods will get 3/5 anyway so everyones done good anyway.

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16. (Original post by physicsmaths)
My friends did that. But infact what you dont realise is that at that centre there is no SHM. So you can't use SHM formulae. Look at definitions of SHM.
Tbh everyones mehtods will get 3/5 anyway so everyones done good anyway.

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But why isn't there SHM in the centre?
17. (Original post by physicsmaths)
My friends did that. But infact what you dont realise is that at that centre there is no SHM. So you can't use SHM formulae. Look at definitions of SHM.
Tbh everyones mehtods will get 3/5 anyway so everyones done good anyway.

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Oh I see, because it rebounds right?
18. (Original post by oShahpo)
Oh I see, because it rebounds right?
The bottom isn't an equilibrium position since accltn is max magnitude at the bottom. So it is centre, at 1.15.

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19. (Original post by physicsmaths)
The bottom isn't an equilibrium position since accltn is max magnitude at the bottom. So it is centre, at 1.15.

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Oh I see, **** in a wheel-barrel it is then.
20. What did people get for the MCQ asking about Mean square speed in relation to temperature

it was either 2, or root 2 i think

I put root 2 then crossed it out as it asked for the mean SQUARE speed ... but im not confident about it haha

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