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# Edexcel - M3 - 18th May 2016 watch

1. Am I thinking about this right, theta goes to 0 but then starts increasing as it goes to other side of circle?
2. (Original post by BBeyond)
Am I thinking about this right, theta goes to 0 but then starts increasing as it goes to other side of circle?
Yeah mate, that's how I did it - theta was defined as the angle that OP makes with the downwards vertical.

3. (Original post by Ayman!)
Yeah mate, that's how I did it - theta was defined as the angle that OP makes with the downwards vertical.
So like this? Didn't even read that part of the question tbh lol

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4. (Original post by BBeyond)

So like this? Didn't even read that part of the question tbh lol

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Looks right!
5. (Original post by Ayman!)
Looks right!
Sound cheers man
6. Who is willing to share any exam TIPS or TRICKS for this M3 exam.......
7. (Original post by justthatboy)
Who is willing to share any exam TIPS or TRICKS for this M3 exam.......
Just have good working out and the 100s is there for grabs. M3 is just mistake land.
****ing retarded it is.

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8. If a question tells me to calculate the magnitude of the resultant force on a particle, why do I have to calculate the magnitude of acceleration and sub it into F=ma. It kind of makes sense but I don't fully understand the reason why
9. (Original post by JustDynamite)
If a question tells me to calculate the magnitude of the resultant force on a particle, why do I have to calculate the magnitude of acceleration and sub it into F=ma. It kind of makes sense but I don't fully understand the reason why
|F|=|ma|=m|a| since m>0.

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10. (Original post by physicsmaths)
|F|=|ma|=m|a| since m>0.

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Ah, Thanks man.
11. (Original post by JustDynamite)
If a question tells me to calculate the magnitude of the resultant force on a particle, why do I have to calculate the magnitude of acceleration and sub it into F=ma. It kind of makes sense but I don't fully understand the reason why
Or you could do it in the other order. Substitute your acceleration into F = ma and then find the magnitude of the force.
12. (Original post by tiny hobbit)
Or you could do it in the other order. Substitute your acceleration into F = ma and then find the magnitude of the force.
If a circular motion question has 'prove the time is less then 2pi...'
This would mean T>0 and I neglected the equality at 0 since it was 'less then 2pi'.

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13. (Original post by physicsmaths)
If a circular motion question has 'prove the time is less then 2pi...'
This would mean T>0 and I neglected the equality at 0 since it was 'less then 2pi'.

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That's the right approach. There have been inconsistencies in questions involving strings in M3, sometimes using T > 0 and sometimes using T >= 0. If you are asked to "show ...", start with whichever version of the inequality will lead to the result you are asked to show.
14. (Original post by tiny hobbit)
That's the right approach. There have been inconsistencies in questions involving strings in M3, sometimes using T > 0 and sometimes using T >= 0. If you are asked to "show ...", start with whichever version of the inequality will lead to the result you are asked to show.
Thanks, that is what I do but i was just woriried since the Mark scheme said otherwise and implied the time is achievable. And used t>=0 but didnt address t>0 in the MS at all.

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15. (Original post by physicsmaths)
Thanks, that is what I do but i was just woriried since the Mark scheme said otherwise and implied the time is achievable. And used t>=0 but didnt address t>0 in the MS at all.

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I see what you mean. To my mind, the version in the mark scheme doesn't tie in with what the question asked. Mark schemes for papers since then, IAL and home, have been consistent with the question asked.

If there is a rod rather than a string, the speed has to be > 0 at the top, otherwise it will just stay there.
16. Bring it on Edexcel
17. (Original post by tiny hobbit)
I see what you mean. To my mind, the version in the mark scheme doesn't tie in with what the question asked. Mark schemes for papers since then, IAL and home, have been consistent with the question asked.

If there is a rod rather than a string, the speed has to be > 0 at the top, otherwise it will just stay there.
Yes it find the critical value.
So from now it is T>=0 and then v^2>0 for a rod right?

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18. (Original post by physicsmaths)
Yes it find the critical value.
So from now it is T>=0 and then v^2>0 for a rod right?

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I honestly don't know for the tension. As I said, there has been inconsistency in the past.
19. Just did Jan 2014 IAL and found it a fair bit harder than any of the non IAL ones
20. (Original post by Messier31)
Just did Jan 2014 IAL and found it a fair bit harder than any of the non IAL ones
Yeh it was. The moments was the one I found difficult. I did overocmplicate it though.

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