OCR Physics A Depth Unofficial Mark Scheme

Gold leaf electroscope and a part Young's modulus; only asked for stress

How can the resultant initial velocity be less the the horizontal component of 30ms^-1?

Because if you use the exact values it will give you a horizontal component of 28.1 or something like that. Then we got given that vertical was 6.3 if I remember rightly. Then obviously just apply pythag and you get around 28.9. This is the more precise answer.

Because if you use the exact values it will give you a horizontal component of 28.1 or something like that. Then we got given that vertical was 6.3 if I remember rightly. Then obviously just apply pythag and you get around 28.9. This is the more precise answer.

Yeah sorry I've confused myself.

No offence; I think mine is even more accurate ayyy lmao. But yours is also extremely precise; only reason I say so is because I think you did 18/0.64 which gives you 28.1 to 3sf but i did 18/0.6422 ;P so when i did pythag I got 28.7ms^-1. You think I still got full marks, all the working is correct so surely even if they wanna be pricks about it then I should get the rest of the marks.

If I drew Searle's apparatus then said about a control wire, finding the cross sectional area with the micrometer when the spirit level is levelled with both, then find the weight by multiplying the masses of the hanger , weights added to break the wire and control weight then divide this by the area,
said about eye-ware protection and putting a cardboard box absorbing the weight of a possibly bouncing weight will I get full marks?

You'd have to use 28.7ms^-1 and not 30, was the mass 0.160 or 0.180?

I think so; I also wrote about repeating these measurements and taking an average; I forgot about putting cardboard box/reducing distance to the ground to avoid the weights causing damage.

Hahahah yeah you are more accurate, I do think there should be a range of answers allowed

If I drew Searle's apparatus then said about a control wire, finding the cross sectional area with the micrometer when the spirit level is levelled with both, then find the weight by multiplying the masses of the hanger , weights added to break the wire and control weight then divide this by the area,
said about eye-ware protection and putting a cardboard box absorbing the weight of a possibly bouncing weight will I get full marks?

Yeah, that's the funniest thing, me and my friend; we usually get good grades and the first thing we said when we came out was; I'm pretty sure there should be a range of values for the answers to those question. :P

I did pretty much same as you, bar the cardboard box

what did you put for the one about the photoelectric effect?

For all the confusion about the velocity question. The question specifically told us to work out the velocity using the answer that they give you for time which is 0.6s not what you get. That should give you 30ms. I am thinking due to the peculiar wording on that question, ocr will not mind if you used your own answer for time.

For that I think I wrote about how the fact that UV did allow for the leaf to fall but not the white light showed the threshold frequency which the wave like nature couldnt explain because if it was due to the wavelike nature then the continuous flow of energy would allow emission of electrons. Since this didn't happen I said that it showed the wavelike nature; also for that question I wrote about how the increase in intensity which would apparently speed up the process if it was only wavelike in nature didn't make a difference supporting the threshold/particle like nature. Also I wrote about how increasing the intensity when it was above threshold freq showed that there was particulate nature for EM radiation because the process of the leaf falling back into place sped up; this shows the wave particle duality as it meant the number of photons per m^2 increased and so due to the one to one ratio if it was a particle would mean the number of electrons released per second would increase >>>>falls faster as shown in observations.