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# 19/05/10 Edexcel Phy 6(6PH08) watch

1. (Original post by Xtian)
Hmmm are you sure it was >0.5? I think the range was 0.52, and half of that is 0.26.
yes it was 0.26. i just used this because all the old papers suggested the reaction time as 0.1s while it was only the sample paper that suggested 0.5
2. OH MY GOD GUYS.

Did they really give us BG count rate in Q4?

I swear I didn't read that so I just found ln A and used one column. Hopefully I'm only penalised once for it though. I just need a B in this paper.
3. (Original post by wooper)
Can you remember what the masses were? Were they 0.930 kg and 0.030 kg? The height was 885 mm. I can't remember which mass was dropping. If it was the lighter one, using mgh gives energy = 0.03 * 9.81 * 0.885 = 0.26 J. If it was the heavier one, mgh = 0.93 * 9.81 * 0.885 = 8.07 J.

I got 0.584 m/s for the velocity, giving energy = 0.5 *(0.93+0.03)* (0.584)^2 = 0.16 J.

Where is the energy going?

There cannot be friction or air resistance, because using s=0.5(u+v)t where u = 0, gives v = 2s/t , i.e. 2h/t, which is the formula given.
wooper the height was 685 mm.i am sure
so no energy lost
so confirms the formula they gave us
4. (Original post by Xtian)
Hmmm are you sure it was >0.5? I think the range was 0.52, and half of that is 0.26.
hm...i dont know.maybe i dont remember well
but i think the biggest value for time was 3.3 and the smallest about 2 and something?
Does anyone have the paper?

no wait!i think i found 0.26 too.but my %uncertainty for k.e was not that because i did it differently.

what i did is:TOTAL K.E=K.E of mass1+K.E of mass 2
so the % uncertainty of K.E=4 * % uncertainty of u

You guys calculated the K.E differently right?
5. (Original post by LetoKynes)
One would be the corrected Activity, A (i.e. Recorded Count - Background count), and the other would be ln A.

Silly me only realised the background count value when I had a minute left didn't have time to correct my graph

I doubt they are fussy about that
Same (ish)! I finished about half an hour early, so I was able to change it, the only problem is I did it in pen so it's messy. But, the gradient turns out to be the same! :O

[oh and I can't remember any of the numbers, so don't bother asking! ]
6. (Original post by unamed)
Same (ish)! I finished about half an hour early, so I was able to change it, the only problem is I did it in pen so it's messy. But, the gradient turns out to be the same! :O
Yea, but I'm still pissed
7. (Original post by Doughboy)
Yea, but I'm still pissed
oh dont worry.i heard lots of people who did that,so probanly there will be some upgrade in that question.
By the way do you have the paper?
8. (Original post by Doughboy)
Yea, but I'm still pissed
It's all right, they'll deduct a maximum of 2 marks because of that!

[oh and for the material question, I got iron. ]
9. (Original post by unamed)
It's all right, they'll deduct a maximum of 2 marks because of that!

[oh and for the material question, I got iron. ]

i got nichrome.(8400)
and 32swg i think
10. Yea me too ^ Nicrhome and 32 seg.

I'm unsubscribing now.

And no, I don't have the papers from June 2010 [unlike in Jan '10 lol].

Bye.
11. (Original post by fredbraty)

i got nichrome.(8400)
and 32swg i think
I got 8000 <
either way, it's only 1 mark (follow through marks! )
12. (Original post by Doughboy)
Yea me too ^ Nicrhome and 32 seg.

I'm unsubscribing now.

And no, I don't have the papers from June 2010 [unlike in Jan '10 lol].

Bye.
good luck with the rest!
13. (Original post by unamed)
I got 8000 <
either way, it's only 1 mark (follow through marks! )
14. Well lucky you, unamed!

What the heck was that swg stuff anyway? I don't remember reading anything about it o.o (I know it's just matching values, but still Shocked me a lil!)
15. (Original post by Xtian)
I did that too, but got 20%, probably because I took the uncertainty in the time to be 0.3 and you did 0.26?

In the percentage uncertainty of the diameter, I used half the range (0.02mm) rather than the precision of the micrometer (0.01mm) and got 7%. Anyone else?

And in the last questions I got around 0.05mm^-1 for mu, but were we not supposed to include units, since it asked for the value?
Yep, I used 0.26. And used half the range for micrometer, but to 3 dp (0.015). And put mm^-1.
16. (Original post by fredbraty)
K.E=0.5*m1*v^2+0.5*m2*v^2
so the %uncertainty of K.E=4*%uncertainty of u.
yess for %uncertainty of u=%uncertainty of h + %uncertainty of t
how did you find the K.E?Did you do 0.5*(m+M)*v^2?or 0.5*m1*v^2+0.5*m2*v^2 ?
maybe that's why you found it differently.
Yeah I added the masses then found the KE.. I assume thats wrong then? How many marks would I lose?
17. (Original post by dancing.barefoot)
Yeah I added the masses then found the KE.. I assume thats wrong then? How many marks would I lose?
no its not wrong.you got the same result.it is the same thing but factorised.see below.

0.5*m*(v^2)+0.5*M*(v^2) is the same expression as
0.5*(m+M)*(v^2)

the only difference is in the %uncertainty calculation.e.g:If someone used the L.H.S expression then the %uncertainty for K.E=4*%UNCERTAINTY OF U
BUT if someone used the R.H.S expression then the %uncertainty=2*%uncertainty of u.

I guess they should accept both?
18. (Original post by fredbraty)
no its not wrong.you got the same result.it is the same thing but factorised.see below.

0.5*m*(v^2)+0.5*M*(v^2) is the same expression as
0.5*(m+M)*(v^2)

the only difference is in the %uncertainty calculation.e.g:If someone used the L.H.S expression then the %uncertainty for K.E=4*%UNCERTAINTY OF U
BUT if someone used the R.H.S expression then the %uncertainty=2*%uncertainty of u.

I guess they should accept both?
Yeah I see what you mean, and I've thought through it. I think the other method gives a wrong % uncertainty though..

When you're doing the separate KEs, you have to find the uncertainty of KE1 + KE2. So, you find the uncertainty in each KE. To find the total percentage uncertainty, you have to find the absolute uncertainties in each value, and add them together. Then find the percentage uncertainty from that. Which is way too much to ask for in a 2-mark question..
19. (Original post by dancing.barefoot)
Yeah I see what you mean, and I've thought through it. I think the other method gives a wrong % uncertainty though..

When you're doing the separate KEs, you have to find the uncertainty of KE1 + KE2. So, you find the uncertainty in each KE. To find the total percentage uncertainty, you have to find the absolute uncertainties in each value, and add them together. Then find the percentage uncertainty from that. Which is way too much to ask for in a 2-mark question..
what do you mean by the absolute uncertainties?
what i see is that we have two kinetic energies and we want the %uncertainty of the total K.E.So,it is the %uncertainty of K.E1 +%uncertainty of K.E2
Each K.E has %uncertainty=2*the one of u.
If you do it with the factorised formula then you find half the %uncertainty.
But i cant see a reason for not accepting both.
Both are logical.
And as it seems most people didnt do well in that question.
20. (Original post by fredbraty)
what do you mean by the absolute uncertainties?
what i see is that we have two kinetic energies and we want the %uncertainty of the total K.E.So,it is the %uncertainty of K.E1 +%uncertainty of K.E2
Each K.E has %uncertainty=2*the one of u.
If you do it with the factorised formula then you find half the %uncertainty.
But i cant see a reason for not accepting both.
Both are logical.
And as it seems most people didnt do well in that question.
When you're finding the percentage uncertainty of two quantities added together, it isn't as simple as just adding each individual percentage uncertainty.. From what I understand anyway - but I suck at this stuff.

And surely, if both methods were correct, they would give the same percentage uncertainty..

Absolute uncertainty is, for example, 0.26 for reaction time.

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