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Edexcel Mathematics: Mechanics M3 6679 01 - 17 May 2017 [Exam Discussion]

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MS

1. 3/4
2. theta = 33.6 (degrees) l = 0.2m
3. (a) 4.52ms^-1 (b) 5.42m
4. (a) 25/11a (b) theta =23.7 (degrees)
5. (b) 5mg>mg (c) v(max) = 3sqrt(ga)
6. (b) F=38.4N (c) v=6.5ms^-1
7. (a) AO =2.2m (b) a = -625/9x (c) T=1.02 seconds
Hi, if I did part 6c wrong where I put in mgh as well when finding the speed, i wrote EPE=KE+mgh+EPE. So i got a lower speed. How many marks would i lose ?
70 90 ums
63 80 ums
56 70 ums

Anyone have any predictions??

Forgot to take away base for the COM question (4)... how many marks do you think I've lost (got the answer as 27/11 a not 25/11 a).
Original post by Akkiakki98
Wait, it's not 1/5???

Original post by manoobsmatthew
That's right.


I seen a few people agree they got 2 as the answer, and just assumed I was wrong. Maybe not though, my bad!

Edit: oops they agreed on .2 - need my eyes testing!
(edited 6 years ago)
Original post by WhiteScythe
MS

1. 3/4
2. theta = 33.6 (degrees) l = 0.2m
3. (a) 4.52ms^-1 (b) 5.42m
4. (a) 25/11a (b) theta =23.7 (degrees)
5. (b) 5mg>mg (c) v(max) = 3sqrt(ga)
6. (b) F=38.4N (c) v=6.5ms^-1
7. (a) AO =2.2m (b) a = -625/9x (c) T=1.02 seconds

If you did velocity on question 5b, shouldn't it be 5ag>ag
For question on vertical circles part (b) I worked out the velocity to be 5ag but didn't prove that with that velocity the particle will stay in contact. How many marks lost?

For question 6c, why was the change in gravitational potential energy not calculated? I calculated the GPE during the exam, now I'm worried I made a silly mistake.
My answers with a bit of working - all from memory so there could be mistakes!

https://goo.gl/photos/VJvGEqLUQjb7t4S59
Reply 348
Original post by Sapereaude45
For question 6c, why was the change in gravitational potential energy not calculated? I calculated the GPE during the exam, now I'm worried I made a silly mistake.


the table was turned sideways in a way, causing the force of weight to act perpendicular, no change in gpe occured. I wuldnt stressed too much as i dont see how u could be penalised more than 3 marks for this
Original post by reza.1
the table was turned sideways in a way, causing the force of weight to act perpendicular, no change in gpe occured. I wuldnt stressed too much as i dont see how u could be penalised more than 3 marks for this

Of course!Thanks for your help.
Original post by reza.1
the table was turned sideways in a way, causing the force of weight to act perpendicular, no change in gpe occured. I wuldnt stressed too much as i dont see how u could be penalised more than 3 marks for this

So maximum 3 marks penalise then, man i put GPE in aswell :/. I think i will lose 5-6 marks in this paper
Reply 351
Found it easier than last year but last year's boundaries were 72 and 69...
For question 5b (the vertical motion question where you had to 'verify' that the particle moves in a complete circle). I did something different to what they wanted I think, would appreciate it if anyone could tell if I'm wrong on this one.

So yeah I wasn't sure what to do so I resolved towards the center and got an equation of the normal reaction RR on the particle:

R=mg(7+3sinθ)R=mg(7+3sin\theta)

I then said that since 1sinθ1-1\leqslant sin\theta \leqslant1 then 4mgR10mg4mg\leqslant R \leqslant10mg so R>0R>0 for all values of θ\theta.

I then showed that the speed vv at the top when θ=270\theta=270^{\circ} was greater than zero.

So R>0 R>0 for all θ\theta and V>0V>0 at top of circle is what I showed. Would this be enough to verify that the particle moves in a complete circle? Or I have not really done what they asked?
Reply 353
In question 3 why could you not conserve energy?
Guys for question 3, I was silly enough to do energy conversions to work out V.

Mgxsin60= 0.5mv^2+ work done against resistance. Any marks?
Original post by johnlannister8
For question 5b (the vertical motion question where you had to 'verify' that the particle moves in a complete circle). I did something different to what they wanted I think, would appreciate it if anyone could tell if I'm wrong on this one.

So yeah I wasn't sure what to do so I resolved towards the center and got an equation of the normal reaction RR on the particle:

R=mg(7+3sinθ)R=mg(7+3sin\theta)

I then said that since 1sinθ1-1\leqslant sin\theta \leqslant1 then 4mgR10mg4mg\leqslant R \leqslant10mg so R>0R>0 for all values of θ\theta.

I then showed that the speed vv at the top when θ=270\theta=270^{\circ} was greater than zero.

So R>0 R>0 for all θ\theta and V>0V>0 at top of circle is what I showed. Would this be enough to verify that the particle moves in a complete circle? Or I have not really done what they asked?


Yeah. You find the smallest value of v^2. Then if R>0 for this value, the particle stays in contact.
It was 7+ 2sin (theta) though, not 3.
Original post by JJSinha
Yeah. You find the smallest value of v^2. Then if R>0 for this value, the particle stays in contact.
It was 7+ 2sin (theta) though, not 3.


I think its 3 cause he moves gravitational force component to the other side. What you meant is the speed
How many marks was the question 6? (I messed up and thought it was vertical 😭😭)
Original post by AAls
Found it easier than last year but last year's boundaries were 72 and 69...


completely agree, although needing 73 or 74 for 90UMS seems crazy. Id say it will be same boundaries as last years
Original post by 0192837465
How many marks was the question 6? (I messed up and thought it was vertical 😭😭)


It was a 6 marker, you would lose 3 mark maximum i believe. If you write out everything and you put in mgh by mistake.

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