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Aqa a level physics paper 1 2018 unofficial mark scheme

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Reply 60
Avatar for tomBch
tomBch
10
did anyone have to do anything to do with circumference for the violin question about finding the new frequency? I did and I can't remember what answer I got but I swear it wasn't 440Hz. Was I right in assuming the tension in the string was caused by a reduction in the length of the violin string when it was turned through an angle or what?

TL:biggrin:R: How EXACTLY did people do that part of the violin question and with what values
Original post by BenHayward00
For he first one I got A=2.98
F=ma
F= 1.8x10^4 x 2.98
F=5.36x10^4

Initial velocity = 17.5ms-1
S=(u+v)/2 *T
S=17.5/2 * 12
S= 105m


suvat only applies when acceleration is constant, which it is not here
the length is (pi*diameter)*5 i messed it up i did length as pi*(radius)^2*50
Reply 63
Avatar for BatesWW3
BatesWW3
Anyone want to take a stab in the dark with rough grade boundaries?
Original post by tomBch
did anyone have to do anything to do with circumference for the violin question about finding the new frequency? I did and I can't remember what answer I got but I swear it wasn't 440Hz. Was I right in assuming the tension in the string was caused by a reduction in the length of the violin string when it was turned through an angle or what?

TL:biggrin:R: How EXACTLY did people do that part of the violin question and with what values


they gave the diameter of the peg, so you use pi x d to give the circumference then multiply to 75/360 to give the arc length, or the extension. THEN, find the extension that corresponded to 25N, and add the new extension onto it, then found the new value for tension. it was 34 for me. then put all of it into the first harmonic equation.

but I got 430 anyways sooo
Original post by tomBch
did anyone have to do anything to do with circumference for the violin question about finding the new frequency? I did and I can't remember what answer I got but I swear it wasn't 440Hz. Was I right in assuming the tension in the string was caused by a reduction in the length of the violin string when it was turned through an angle or what?

TL:biggrin:R: How EXACTLY did people do that part of the violin question and with what values


work out the length it extends by when you rotate it, so (2pi*75)/360). multiply that by your diameter. then add it to the original length which was 11mm. then read off the graph. plug in your new value of t, which was around 34N i think.
Reply 66
Avatar for tomBch
tomBch
10
Original post by sdfsdd
work out the length it extends by when you rotate it, so (2pi*75)/360). multiply that by your diameter. then add it to the original length which was 11mm. then read off the graph. plug in your new value of t, which was around 34N i think.


aah thanks I see, I presumed that the length of the string was actually being shortened because the tension increased and so did the frequency. And wasn't the graph extension against tension or not? I guess my memory is fuzzy about the whole thing. Most of it went alright though I think :smile:
Original post by Adxmp
for the 0.36W one did anyone get that it wasnt acceptable as PD across the resistor was like 9.5 which was more than the cell provided?


Yes that's exactly what I got aswell, also for the resistor I'm pretty sure I got 540 ohms what about you?
Original post by tomBch
aah thanks I see, I presumed that the length of the string was actually being shortened because the tension increased and so did the frequency. And wasn't the graph extension against tension or not? I guess my memory is fuzzy about the whole thing. Most of it went alright though I think :smile:


it was a graph of tension against extension. they only said the tension increased, so you know which way the peg is turning
Im not sure if anyone has asked this yet but as far as I can see there were too methods for the acrobat tension questions. You could do the vertical component (which I believe was 85g) divided by cos28.5 or work out the horizontal component/centripetal force and divide this by sin28.8. When I did this I ended up getting difference answers (949N and 960N I think) would some please explain what I'm doing wrong. Thanks.
Reply 70
Avatar for xanax7
xanax7
4
Original post by willfarrar
What did people get for the value of R? The resistor question?

I got 445 ohms

Consider the circuit initially without the resistor, you find the current to be 9 x 10^-3 A
Then add the resistor so its in parallel with the thermistor, substitute 5 ohms into the value for V and work out the required resistance.
Subtract resistor A from value and then use resistors in parallel equation to work out the resistance.

Anyone else do something similar?


Got 535 ohms or something like that. If you want the output to be 5V, you split the 9 V into 5 and 4. I did it as a ratio. If the resistor (in series) is 250 ohms and gets 4 Volts, the resistors in parallel which get 5 volts have a resistance of 250*(4/5). And then u just use 1/Rtotal = 1/R1 + 1/R2
Reply 71
Avatar for xanax7
xanax7
4
Original post by gcsemusicsucks
yeah I guessed the diameter because it's the only value squared in the equation but literally no idea :colondollar:


Wasn’t clear if they were talking about real uncertainty or % wise..
Original post by BatesWW3
Anyone want to take a stab in the dark with rough grade boundaries?


I hope they're lower than last year's but doubt it tbh
Reply 73
Avatar for tomBch
tomBch
10
Original post by sdfsdd
it was a graph of tension against extension. they only said the tension increased, so you know which way the peg is turning


Thanks I think I know where I went wrong now, silly mistake! I got the extension and added tension and everything but when putting it back into the equation I must of used the additional tension instead of overall tension because I remember getting a value of f about 20hz greater than the first i.e I got 260Hz instead of the correct value of 440Hz. Provided I did most of it right apart from the last stupid substitution of the tension, I think I must've therefore still picked up some decent marks for the question despite that mistake :smile:
Reply 74
Avatar for Adxmp
Adxmp
4
Original post by MaDxRussian
Yes that's exactly what I got aswell, also for the resistor I'm pretty sure I got 540 ohms what about you?


In that question instead of doing the the ratio of 4:5 i did 5:4 so i got 450 which means I think what you got is good
Original post by TruthfulHoax
40,000 N ?
244 ms^-1
440 Hz
2.11 x 10^-9 m^2
540 ohms
962 N
40,000 Ns

got exactly those
Original post by tomBch
Thanks I think I know where I went wrong now, silly mistake! I got the extension and added tension and everything but when putting it back into the equation I must of used the additional tension instead of overall tension because I remember getting a value of f about 20hz greater than the first i.e I got 260Hz instead of the correct value of 440Hz. Provided I did most of it right apart from the last stupid substitution of the tension, I think I must've therefore still picked up some decent marks for the question despite that mistake :smile:


you would've probably got 2 or 3 marks. but that paper went really **** for me. literally first time i ever cried after an exam
Original post by Adxmp
for the 0.36W one did anyone get that it wasnt acceptable as PD across the resistor was like 9.5 which was more than the cell provided?


i got this
Original post by DarthRoar
For Q1 I drew a tangent and got A as 2.5ms-2

The ramp was long enough because the initial KE was about 2 million joules but ramp could supply about 6 million joules.

I got 536 ohms for the resistor in Q2


I think I got that for the ohms too, because (1/((1/536)+(1/750)))/5*4=250 (the one we're given).
Original post by TheSwyn
Why was the tension not 1400. Why do you not have to add the centripetal force
I remember adding centripetal force, you definitely needed to as it was rotating. I did this and got 1900 so one of us got it right 😂

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