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# Official OCR A2 G484 June 2016 [Module 1] [90UMS] watch

1. (Original post by Zonothene)
Workings to get the mass of S2 for those wondering.
Well **** me sideways. Goodbye 100%ums. ****ing bull-****.
2. (Original post by Danny.L)
Well, they told you the period was equal for both of them, 4 days.
I've no idea if that is right.
I just used keplers third law with T^2 = (4Pi^2 * R2^3)/(G*M2)
That got me 1.73*10^33.
Was it out of 2 or 3?
i did this very differently

Got

F=mv^2/r equal to the first eqn we wrote F=Gm1m2/(r1+r2)^2

I had V for planet 1 and R1 so put equal to each other M1 cancels in both equations

Leaving

V^2/r1 = Gm2/(r1+r2)^2 leaving only M2 as a variable,

Not sure if anyone else did it it like this.

Got 1x10^30 something like that anyway

Starting to forget answers but SHC of the metal was 128.something ???
3. (Original post by Tabatha Gregg)
i did this very differently

Got

F=mv^2/r equal to the first eqn we wrote F=Gm1m2/(r1+r2)^2

I had V for planet 1 and R1 so put equal to each other M1 cancels in both equations

Leaving

V^2/r1 = Gm2/(r1+r2)^2 leaving only M2 as a variable,

Not sure if anyone else did it it like this.

Got 1x10^30 something like that anyway

Starting to forget answers but SHC of the metal was 128.something ???
I got 130.8 for it
4. (Original post by Danny.L)
Yeah a couple people in my class got that too. I've done it like 10 times and I consistently get 10^33, perhaps you mis-entered something on your calculator?
I get 1.73*10^33 using R2 and period 4 years but get a different answer using R1, 4 years and ratio M1/M2 = 3.
5. (Original post by Zonothene)
Workings to get the mass of S2 for those wondering.
Are you sure this is right I don't really follow what you did
6. (Original post by Zonothene)
Workings to get the mass of S2 for those wondering.
Thanks dawg, glad I know I got the working right but not the final answer
7. i don't think you can use kepler's third law (the normal equation) to calculate S2.
8. Gg g484 no RE
First exam in physics has always been my weakest, no difference this year.
G481 last year I dropped 3/4 of my entire ums i dropped in total on it.
g484, guess I'm gonna end up dropping most ums on this too :/
9. (Original post by tombrunt45)
Are you sure this is right I don't really follow what you did
You know that the gravitational force exerted on the star is equal to the resultant force. Therefore the gravitational force exerted on the star = centripetal force. You know S1 is 3 x greater mass than S2 so you can sub in S1 = 3S2 into both sides of the equation. From this point you can solve for S2
10. Why the f*** is everyone ignoring me? Also, what do you think 45/60 would convert to out of 90?
Fun fact: i thought c was a planet with it's own gravity.
11. (Original post by Zonothene)
You know that the gravitational force exerted on the star is equal to the resultant force. Therefore the gravitational force exerted on the star = centripetal force. You know S1 is 3 x greater mass than S2 so you can sub in S1 = 3S2 into both sides of the equation. From this point you can solve for S2
makes sense its just that I was one of the people who got 1.73x10^33, any ideas what those people(me) did wrong? or you might have done wrong? I honestly have no clue, besides I thought the period was years?
12. (Original post by Danny.L)
I got 130.8 for it
Did we have to give s unit for it...?

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13. GUYS TEACHERCOL HAS DONE IT, THE UNOFFICIAL MARK SCHEME IS HERE
http://www.thestudentroom.co.uk/show....php?t=4176952
14. just need to say one thing, the period was 4 years not days as everyone seems to be saying, I worked it out and if it was days both stars would have to be travelling above the speed of light (clearly impossible)
15. (Original post by tombrunt45)
makes sense its just that I was one of the people who got 1.73x10^33, any ideas what those people(me) did wrong? or you might have done wrong? I honestly have no clue, besides I thought the period was years?
Those who got that answer used keplers third law without accounting for the mass of another star. It is a binary system so both stars rotate around the centre of combined massed rather than 1 star just rotating around the other. If you wanted to approach the problem using keplers third law you would have to use this formula

T^2 = (4pi^2/G(M1 + M2))r^3
16. (Original post by Zonothene)
You know that the gravitational force exerted on the star is equal to the resultant force. Therefore the gravitational force exerted on the star = centripetal force. You know S1 is 3 x greater mass than S2 so you can sub in S1 = 3S2 into both sides of the equation. From this point you can solve for S2
I shall not believe you until I see TeacherCol's solutions. You're method is FAR too complicated for 3 marks and is based on so many assumptions and what you can imply from the data. This is not astrophysics, it's plain simple clear cut Newtonian World Physics. So many people here got answers in the order of 10^33 and not one has got your answer in the order of 10^35. Clearly something is wrong right?
17. (Original post by speed1✈️✈️)
I shall not believe you until I see TeacherCol's solutions. You're method is FAR too complicated for 3 marks and is based on so many assumptions and what you can imply from the data. This is not astrophysics, it's plain simple clear cut Newtonian World Physics. So many people here got answers in the order of 10^33 and not one has got your answer in the order of 10^35. Clearly something is wrong right?
My answer is to the power of 33, not 35. I guess my writing isn't too clear
18. (Original post by tombrunt45)
just need to say one thing, the period was 4 years not days as everyone seems to be saying, I worked it out and if it was days both stars would have to be travelling above the speed of light (clearly impossible)
Omg LMAO. I know I said days but I was just working it out right now from years to seconds (4 *365 *86400). Something wrong with me today
19. (Original post by speed1✈️✈️)
I shall not believe you until I see TeacherCol's solutions. You're method is FAR too complicated for 3 marks and is based on so many assumptions and what you can imply from the data. This is not astrophysics, it's plain simple clear cut Newtonian World Physics. So many people here got answers in the order of 10^33 and not one has got your answer in the order of 10^35. Clearly something is wrong right?
Oi Teachercol has already done the solutions look at my post above
20. (Original post by UnknownAnon)
There was also 1 mark for letters question, and 2 for the one below that, and 3 for the one below that (just before the radius ratio question)

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Could you elaborate pls?

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