AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread]

Around what ums will 60 marks most likely be?

Posted from TSR Mobile

for this i didnt know if to calculate the corrected count rate ? as it didnt specify corrected on the count rate at the source ?

That was basically me in unit 4 xD. I can't remember them all no. first part was electron diffraction stuff then alpha scattering. 2nd part was the binding energy part I think. 6 marker on rutherford experiment. Uhh Pv = nRT part with the piston stuff where you calc the new volume, moles and the density of the gas. Did anyone get 4.4x10-4 and 1.2 for that part? Last part was some basic heat capacity stuff. Although I got like 1700 for the heat capacity of ice which I am not sure is correct.

Did we get given accurate atomic values for the protons and neutrons? I just used the same method as you above and got 125MeV assuming the extra 3 missing MeV were just down to the different methods used.

was the answer of 1.2 for density dependant on getting 4.4x10-4 right? because I know I got 1.2 for density but can't remember if I got 4.4... for the volume for the first bit of the question! Also, how did you work out number of moles? I did volume of piston / volume of one mole .. is that right?

No it was more complicated because that only works at constant temperature. They told you that the pressure, volume and temperature all changed.

So it was P1V1/T1 = P2V2/T2 - From pV = nRT your constant becomes nR so that's what you need to equate.

Oh yeah the flywheel design has come up a couple of times so I just said higher density greater mass, concentrate mass at the edge. Also low resistance material, but that was just some bs I came up with lol.

Yeah the lever thing looked like it was gonna reverse the rotation when released, but that wouldn't make much sense so I just ignored that. I think the spring was there just to return the lever though, it wasn't there for the rotation.
For mechanics it said the gears keep rotating so I said it would waste energy stored by the flywheel so you should design it such that gear and flywheel have no contact afterwards? And then I also said the small gear on the flywheel should be made smaller if possible so that a rotation of the big gear will produce more rotations of the small gear and hence flywheel will have a higher angular speed. But i am not sure at all.

Where did you put your X? Am I right in thinking it was right before minimum volume at the end of compression right?

didn't correct for count rate, 1 or 2 gone.
Unit for latent wrong - 1 gone.
Detecting beta - 1 gone.
6 marker - default 2 gone.

Used wrong method for specific charge - 3 gone.
Current - 2 gone.
6 marker - default 2 gone.

So 13 lost. Then presumably losing however many more for mark schemes that are stupidly specific or contradictory.

Probably got around 50, which should hopefully be around an a.

It asked for the count rate that would be recorded by the student. So that means that you need to include the background radiation.

yes but the background will be included in the reading made? its only when the background is taken away does it give the corrected one? hence why add it on ?

Hey, what did you get for efficiency on last question of applied?

I put a similar answer but my change in design was to change the radius of the gear and its cogs to maximise the rotational velocity of the fly wheel.

Depends what option you did but it'll be around the A* boundary. Probably 105-110

Because it's not asking for a corrected count rate. It's asking for the raw count rate seen.

0.41

Yeah that's pretty much what I meant but didn't know how to word

I got 35% for the efficiency? I'm pretty sure i did as I remember checking that question twice.

Also it said in the flywheel question that the size is limited by the torch so the radius can be increased, I put:
Stronger magnet- greater flux density
More turns in the coil
Higher density flywheel
Position magnet far away from centre of flywheel to increase moment of inertia
High tensile stress(dunno what this means)


Posted from TSR Mobile

Pretty sure there'd be a range for that efficiency as the area of graph is so difficult to determine due to round corners