AQA A Level Physics Paper 3 7408/3 - 16 Jun 2022 [Exam Chat]

Doing alevel aqa paper 3, the previous past papers focus on different practicals. How can I get more practice with the practicla on the advanced information? Do I still have enough time to secure at least an A on this paper?

For those doing turning points the advanced information for AQA says mass and energy, could length and time dilation come up then?

Cosmic rays detected are protons with a total energy of 3.7 x 10^9 eV.Calculate the velocity of the protons as a fraction of the speed of light.Could anyone explain why it's 0.97c? How do you use the formula to get to it?AQA 2017 Paper 3b D Q5

I am assuming you do turning points as they ask these type of questions. All you have to do is use this formula of 𝐸 = 𝑚 𝑐2 = 𝑚0𝑐2/square root(1 − 𝑣2/
𝑐2). the equation is in the data sheet. after you have done that you get the value of v and then divide that by the speed of light to get a decimal of 0.96711 which rounds to 0.97. Hope that helped

Thank you! Though I don't know how to use the formula to get v, I've tried rearranging I don't know what I'm doing wrong, I'm not getting the right answer :/

No problem we have the energy in eV and we want the energy to be in Joules so we multiply by 1.6x10^-19. Then we equate that to the equation that is in the data sheet and bearing in mind that c is the speed of light and m0 is the initial mass of the proton which is 1.673x10^-27. We rearrange the equation to make v the subject by dividing m0c^2/E(energy we calculated)=square root (1-v^2/c^2) bearing in mind that v is the value we are looking for. so after plugging in the values we get a value that is around 290134363.9 and then you divide that by the speed of light to get 0.96711 which rounds to 0.97. Hope that made sense if you still dont get it I will write the whole working out so you can understand it

Thank you again! I'm doing 1.673x10^27 x (3x10^8)^2 divide the energy 5.92x10^-10 J and I'm getting 0.25 what do I do from here?

Then you square it and then take it away from and then multiply it by (3x10^8)^2 and you will get your value

Thank you I'm so sorry I squared it and got 0.064. Took that away from 1 and I got 0.9355. And multiplying that by 3x10^8^2 get's me 8.42 x 10^16?
or
Speed of light subtract 0.064 then multiplying the answer by Speed of light squared = 2.699 x 10^25

Sorry i forgot to tell you the last step you have to square root the answer after you multiplied by the speed of light

Thank you so much literally when I tried doing it I had the same idea for rearranging and basically did the same thing but I don't know if I was missing a step or something, I couldn't see what I was doing wrong but yeah I've got it now, thanks again for the help. I'm resitting this year and like didn't have anyone to ask really so I thought I'd try here, there's no video's covering the paper online, really do appreciate it.

Does anyone have access to the 2021 paper 3 physics turning points?

i have it

Can u send it thru privately pls

Does anyone know here do astrophysics