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AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread]

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i swear to conserve momentum
mv=NV
but the velocity of v was in opposite direction so
-mv=NV
V=-mv/N

i wrote the negative in brackets, so if the negative is wrong will i lose the marks?
Original post by Jordenwilder1998
how many marks was for RC and for resistance. if i did rc wrong because of ln 0.8 i got 1/3???? does that mean i get 1/2 for resistance for follow thru?


It's likely you'll gain some ECF marks if you did the subsequent question correctly.
Original post by ombtom
Wasn't it V = -mv/N?


Arrows on diagram were in opposite directions, so nope.
Reply 1423
Original post by Ben :)
Yeah -mv because it was a velocity so had to be going in the opposite direction to conserve momentum!

It's in the opposite direction so mv-NV=0 but that means V=mv/N not -mv/N.
Original post by Jordenwilder1998
how many marks was for RC and for resistance. if i did rc wrong because of ln 0.8 i got 1/3???? does that mean i get 1/2 for resistance for follow thru?


3 marks, it was 27.9 something so 28s, might get 1 mark for method
for RC i got 202 what did everyone else get
and for R i got some big number 10^5 i think

someone confirm if this is right pls i used v=vo e^-t/rc
where v = 0.8vo

45/(-ln(0.8) = RC
My mistake, thought the alpha particle was classed as daughter nucleus.
Original post by Mango Milkshake
It clearly said recoiling nucleus.

if its recoiling it mustve already emitted the alpha :smug:
What do you think the grade boundaries will be like? In comparison to lasts years paper? :p:D:colone:
DBBDDDDBBCBACCCDBACAADCBD

Can anyone confirm with me?

Posted from TSR Mobile
Original post by DesignPredator
Here are my answers to Written, the question parts might not be correct.

Written answers
1ai) 8.26 (will be a range)
ii) 2.6x10^-4

bi)28
ii)5.9x10^4
iii) bottom box ticked

2ai) Derivation
ii) 6.06x10^6 (3SF only)

b) Downward Curve staring at 9.81
Minimum at g=0 at a point slightly offset from halfway closer to Venus
Rises up to 8.87

3a) V=mv/N

b) Derivation

ci) 1.15x10^-19 kgms-1 OR Ns

cii) Anti electron neutrino released also so Beta minus particles have range of Kinetic Energies upon release.

4a) Small amplitude oscillation as only valid for small angular amplitudes.

b) -Set up light string with Bob on the end,
attach top of string to clamp stand boss.
-Measure length from top to centre of Bob.
-Release the Bob so it oscillates with small amplitude.
-Use stop clock to measure time taken for 20 oscillations.
-Divide this by 20 for mean T
-Plot T^2 against L
-Calc Gradient
g=(4pi^2)/gradient
-Repeat procedure

c) -Student value 4x true value
-Time period half so all values for T^2 will be 1/4 the true
-Gradient is 4x lower
-State equation or show that g is inversely proportional to gradient.

5a) Lenz's Law: Direction of induced EMF/current is always in a direction that opposes the change that caused it.
Faraday's Law: Induced EMF is proportional to rate of change of flux linkage.

b) -Change in flux linkage when current flows.
-EMF induced in wheel (or whatever it was called)
-Current induced in the wheel as good conductor.
-Current carrying conductor (wheel) opposes the field due to coil, therefore there is a force against the direction of motion.

c) Wheel not in contact with electromagnet so no wear, but in brake pads there is wear due to friction.

More energy used or less effective or heating.


Im not sure about your momentum i got the same but i think were wrong because i remember using 220 instead of 216 in the calculations which isnt right...
Original post by johnapplebottom
i swear to conserve momentum
mv=NV
but the velocity of v was in opposite direction so
-mv=NV
V=-mv/N

i wrote the negative in brackets, so if the negative is wrong will i lose the marks?


That's not quite it. The momentum before was zero, and the momentum after was (N)(V)+(m)(-v) as the alpha particle is going in to opposite direction to the daughter nucleus (momentum is a vector so . . .).

As momentum is conserved, 0=NV-mv
Taking mv over you get NV=mv
Judging from reading though answers on here think i may have got around 48-50ish, any ideas of what this might be? Hopefully a B because thats what i need :/
216 was the nuclean number
Original post by johnapplebottom
216 was the nuclean number


220


Posted from TSR Mobile
Original post by shuu00
That's not quite it. The momentum before was zero, and the momentum after was (N)(V)+(m)(-v) as the alpha particle is going in to opposite direction to the daughter nucleus (momentum is a vector so . . .).

As momentum is conserved, 0=NV-mv
Taking mv over you get NV=mv


ah **** yeah your right i see, i put the negative in brackets will i lose the marks?
Original post by Questioner1234
Yeah thanks I was just remembering from the y axis (will edit in a second) - did it give the units or did we have to write them?


They gave you the units (microfarads)



no wat, 220 was the nuclean number before it decayed by alpha particle thing
so after it was 220-4 = 216
Original post by Questioner1234
Was the capacitance in the first question 8.2 microfarads????


I got that I think
Reply 1438
120 UMS boundary predictions?
what did people say about the B- emission any why the equations wouldnt work? I said that B emission is an electron which isnt a nucleon, which is what the equations represent?

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