OCR Physics A Depth Unofficial Mark Scheme

1. a) In longitudinal waves, the vibrations are along the same direction as the direction of travel. In transverse waves, the vibrations are at right angles to the direction of travel. (2)
b) (i) 40mV (1)
(ii) 667 Hz (1-2)
(iii) 0.495m (1-2)
c) I is proportional to A^2 A is halved to 20mV. (2)

2. Horizontal v = 30, vertical = 6.3 so resultant is 31.
KE = 75J
PE = 3.1J

3.

4. Percentage uncertainty = 4%
Gradient = 5800
Resistivity = 0.11 ohm m

5. KEmax = 1.24x10-19

6. v = 1.03x10^7
de Broglie wavelength = 7.1x10-^11m
Pattern size decreases because v increases.

I can't remember all of the questions and answers so please let me know of any you can remember.

(edited 7 years ago)

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Reply 1

M0nkey Thunder

For 1a I would have expected to also need to mention that longitudinal waves have rarefactions and compressions whereas transverse waves have peaks and troughs

Reply 2

ronnydandam

Original post by M0nkey Thunder

For 1a I would have expected to also need to mention that longitudinal waves have rarefactions and compressions whereas transverse waves have peaks and troughs

I doubt it - there was only 2 lines but then again it was two marks so who knows

Reply 3

Parhomus

I'm assuming the OP used 30ms^-1 to find the values for the original velocity and so the kinetic energy etc etc. I used the 28ms^-1 do you think that's fine.

Reply 4

mightyned

are error marks carried over
for 1b i forgot to times 0.5ms by 3 so got an answer 3 times bigger this means my answer for c is different will i get one mark for using correct equation

Reply 5

dylan1016

Original post by Parhomus

I'm assuming the OP used 30ms^-1 to find the values for the original velocity and so the kinetic energy etc etc. I used the 28ms^-1 do you think that's fine.

Was it asking for the kinetic energy at max height or just after being hit? I can't remember correctly but I think if it was just after being hit you would need to use 31ms^-1. Correct me if I'm wrong.

Reply 6

JN17

Original post by dylan1016

I believe he means that when you were calculating the magnitude of initial velocity you could either use the calculated values, or the ones given to you (6.3 and 30), using calculated values gives 28, otherwise 31

Reply 7

Parhomus

Original post by dylan1016

It asked for both of them; and for how u got 31ms^-1 you used s=ut+1/2at^2 whilst I used v=u+at; the only downside of this exam is how there can be so many different answers because of this. For t i got 0.642 because I used v=u+at whilst if u do s=ut+1/2at^2 you'd get two answers with one being 0.57. So im not sure :C.

Reply 8

dylan1016

Original post by Parhomus

I used 0.64 as well and I calculated it without suvat.

(edited 7 years ago)

Reply 9

samendrag

i used the values that they gave so i calculated it was 28.3 but used 30, is that ok?

Reply 10

Harrysomers1

I also got 28.1 and then 28.9 I think for the overall initial velocity. In most previous papers I've done this has been okay, so hoping it will be fine

Reply 11

Parhomus

Original post by Harrysomers1

I also got 28.1 and then 28.9 I think for the overall initial velocity. In most previous papers I've done this has been okay, so hoping it will be fine

Yeah; I think I should've used 28.1 but instead I used 28 ;P so I got a value of 28.7 for the initial velocity. Pretty ridiculous how OCR did that question since it's so subject to using different answers etc.

Reply 12

zak_121

Original post by dylan1016

I used 0.64 as well and I calculated it without suvat.

Yh I did that too but I used 28...ms^1 afterwards not 31?
Where did 31 come from

Reply 13

chemari1

what were the experiment questions

Reply 14

zak_121

Original post by Parhomus

Dw I used all the values in my calculator and I still got 28.7 to 3sf

Reply 15

dylan1016

Original post by Harrysomers1

I also got 28.1 and then 28.9 I think for the overall initial velocity. In most previous papers I've done this has been okay, so hoping it will be fine