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# M1 OCR (Not MEI) Exam - 9/06/2015 watch

See previous post (did this so you get notification I've replied)
2. mm but I thought it says the angle between the two forces is 60 degrees?
3. (Original post by Super199)
Can someone explain how you work out change in momentum

For 3ii. Jan 2010.

Thanks
4. (Original post by Lilli1997)
See previous post (did this so you get notification I've replied)
Thank You!
5. Oops it's upside down!

And I've just realised I think you meant ib, sorry, got ahead of myself!
6. (Original post by Lilli1997)
Pic is upside down
7. (Original post by Lilli1997)
Can you explain 5i. I have no idea how to do it
8. (Original post by Super199)
I don't understand question 4a on the same paper

I just got 0.4gsin60 and 0.4gcos60. Where does the other 0.3gsin60 and 0.3gcos60 come from? I thought it was only considered particle p.
Typed half of it out, decided it was impossible to explain without drawing the diagram. Give me two minutes to draw it out, but I am working on it
9. (Original post by Super199)
mm but I thought it says the angle between the two forces is 60 degrees?
Yes. But when you are finding the resultant force you put the two forces you have been given head to tail. You know when forces are in equilibrium they make a closed triangle, well it's a similar idea. When the forces are head to tail connecting the ends for the shortest root will give you the resultant.

It's hard to explain but I can draw a diagram?
10. (Original post by Super199)
Can you explain 5i. I have no idea how to do it
Yep, I did this paper, workings bit messy but will write it out.
11. (Original post by Lilli1997)
Yes. But when you are finding the resultant force you put the two forces you have been given head to tail. You know when forces are in equilibrium they make a closed triangle, well it's a similar idea. When the forces are head to tail connecting the ends for the shortest root will give you the resultant.

It's hard to explain but I can draw a diagram?
mm I think I get it
12. (Original post by Super199)
Can you explain 5i. I have no idea how to do it
Here you go

Probably upside down, something keeps going wrong with the camera!
13. (Original post by Super199)
I don't understand question 4a on the same paper

I just got 0.4gsin60 and 0.4gcos60. Where does the other 0.3gsin60 and 0.3gcos60 come from? I thought it was only considered particle p.

Sorry it's sideways. I think you probably forgot to count the 0.3g as P is attached to Q. For the diagram, blue is what you're told and orange is what I added in. Also sorry my writing is so messy
14. (Original post by Shakzey)
Lets all hope theres no tractive force questions and we should all be ok tomorrow lmao (hopefully)
Probably sound completely clueless but what are they?!
15. (Original post by chloe-jessica)

Sorry it's sideways. I think you probably forgot to count the 0.3g as P is attached to Q. For the diagram, blue is what you're told and orange is what I added in. Also sorry my writing is so messy
Right I am a bit confused
I still don't see where you get the 0.3g part from like I get it from Q somehow. But how is that affecting p?
16. (Original post by lauren_2)
Probably sound completely clueless but what are they?!
If it makes you feel any better, I have no clue what those are either
17. (Original post by Lilli1997)
Here you go

Probably upside down, something keeps going wrong with the camera!
Thanks

Do you mind helping me with the next part?
18. (Original post by Super199)
Right I am a bit confused
I still don't see where you get the 0.3g part from like I get it from Q somehow. But how is that affecting p?
It's from the weight of particle Q. If we do F=ma for particle Q (Q is a lot simpler than P, there's no angles and no friction) the resultant force will be the tension acting upwards minus the weight acting down (P is heavier than Q so the system would move down on the side of P and up on the side of Q). As it's not moving, a=0. Therefore, T-0.3g=0.3*0 which gives us T=0.3g. T must be acting constantly throughout the string, therefore there's also a T on the side of P (which is the one I marked on). You resolve this along with resolving the weight, then balance.
19. (Original post by chloe-jessica)
If it makes you feel any better, I have no clue what those are either
20. (Original post by chloe-jessica)
It's from the weight of particle Q. If we do F=ma for particle Q (Q is a lot simpler than P, there's no angles and no friction) the resultant force will be the tension acting upwards minus the weight acting down (P is heavier than Q so the system would move down on the side of P and up on the side of Q). As it's not moving, a=0. Therefore, T-0.3g=0.3*0 which gives us T=0.3g. T must be acting constantly throughout the string, therefore there's also a T on the side of P (which is the one I marked on). You resolve this along with resolving the weight, then balance.
Ah I see thank you

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