2016 | OCR A2 Advancing Physics B | G494 & G495 | 20th & 28th June

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Thought it was a fairly good paper overall, although my class was split, with about 2/3 saying it was good/alright and the other 1/3 saying it was hard. For the pressure question I got something like 83000 Pa, any thoughts?

Reply 61

smartalan73

Original post by NamelessPersona

I got about 170000 Pa??

Reply 62

NamelessPersona

Original post by smartalan73

I got about 170000 Pa??

The pressure in the vacuum had to be lower than normal, and the normal pressure was 100000 Pa.

Reply 63

M12345689

Original post by smartalan73

I got about 170000 Pa??

i think you had to take that away from the 100kpa atmospheric pressure

Reply 64

M12345689

Original post by M12345689

i think you had to take that away from the 100kpa atmospheric pressure

did anyone get 172 for the bullet velocity?

Reply 65

smartalan73

Original post by NamelessPersona

The pressure in the vacuum had to be lower than normal, and the normal pressure was 100000 Pa.

Original post by M12345689

i think you had to take that away from the 100kpa atmospheric pressure

Whoops i probably added instead of taking away then, didnt think about it having to be lower

Reply 66

username1424648

Original post by M12345689

did anyone get 172 for the bullet velocity?

Yes, but I put 170 because 2 sf and stuff

Reply 67

NamelessPersona

Original post by M12345689

did anyone get 172 for the bullet velocity?

I got 170 to 2 s.f.

Reply 68

BongoBuddha

Original post by M12345689

did anyone get 172 for the bullet velocity?

Yes i got something like that :smile:

Reply 69

BongoBuddha

I'm kicking myself because i ran out of time and just needed 30 more seconds to write my answer to the 2nd last question where you had to show that intial KE was not equal to final KE for 2 marks (all i have to do was work out 1/2mu^2 and 1/2mv^2 to show they were different and KE was not coserved, I also skipped the whole Boltzmann factor Q :frown:

How many marks do you think that double spread on the boltmann factor was?

Reply 70

NamelessPersona

Original post by smartalan73

Whoops i probably added instead of taking away then, didnt think about it having to be lower

I think it was something like the weight of the masses = 1.96 N
Area of nozzle = 1.13*10^(-4)
Pressure = F/A
Pressure ~= 17000
100000 - 17000 = 83000 Pa

Reply 71

Simonium1010

anyone get something like (200*9.8)/0.000113... = ~1.7x10^7 for the pressure??

And then for the precision bit after, (second area)/(first area) = 6.25???

*edit* oh no it was 100g not 100kg.... of course... okay so like 17,345Pa??
Then what are you supposed to do with the 100kPa atmospheric pressure?

(edited 7 years ago)

Reply 72

NamelessPersona

Original post by Simonium1010

anyone get something like (200*9.8)/0.000113... = ~1.7x10^7 for the pressure??

The masses were in grams so that would give a weight of 0.2*9.8 = 1.96
See my post above for my reasoning, not sure if correct though.

Reply 73

BongoBuddha

Original post by Simonium1010

anyone get something like (200*9.8)/0.000113... = ~1.7x10^7 for the pressure??

And then for the precision bit after, (second area)/(first area) = 6.25???

I didn't know you had to do a calculation for the precicion bit, I just tried to explain in words why a larger area readuced percentage error :colondollar:

Reply 74

Simonium1010

That one asking about the kinetic energy not being conserved in the collision... did everyone else get really big energy losses, like 14J before and 0.2J after?

Reply 75

NamelessPersona

Original post by Simonium1010

That one asking about the kinetic energy not being conserved in the collision... did everyone else get really big energy losses, like 14J before and 0.2J after?