Hey there! Sign in to join this conversationNew here? Join for free

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

  • View Poll Results: How did you find this exam?
    Amazing! :woo:
    48
    27.43%
    Good
    64
    36.57%
    OKAY. :bebored:
    44
    25.14%
    Bad... :cry2:
    13
    7.43%
    HORRIBLE PAPER!!
    6
    3.43%

    Offline

    11
    ReputationRep:
    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
    Offline

    11
    ReputationRep:
    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 ?
    Offline

    0
    ReputationRep:
    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).
    Offline

    8
    ReputationRep:
    (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!
    Offline

    0
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    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?
    Offline

    2
    ReputationRep:
    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.
    Offline

    1
    ReputationRep:
    My answers with a bit of working - all from memory so there could be mistakes!

    https://goo.gl/photos/VJvGEqLUQjb7t4S59
    Offline

    2
    ReputationRep:
    (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
    Offline

    2
    ReputationRep:
    (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.
    Offline

    11
    ReputationRep:
    (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
    Offline

    14
    ReputationRep:
    Found it easier than last year but last year's boundaries were 72 and 69...
    Offline

    1
    ReputationRep:
    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 R on the particle:

    R=mg(7+3sin\theta)

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

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

    So  R>0 for all \theta and V>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?
    Offline

    0
    ReputationRep:
    In question 3 why could you not conserve energy?
    Offline

    1
    ReputationRep:
    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?
    Offline

    11
    ReputationRep:
    (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 R on the particle:

    R=mg(7+3sin\theta)

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

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

    So  R>0 for all \theta and V>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.
    Offline

    11
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    How many marks was the question 6? (I messed up and thought it was vertical 😭😭)
    Offline

    17
    ReputationRep:
    (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
    Offline

    11
    ReputationRep:
    (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.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.