Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    17
    ReputationRep:
    Can someone explain 4ii to me I just don't get it.

    Like there is no point giving me hints, I don't get what is going on.
    Attached Images
     
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by Super199)
    Can someone explain 4ii to me I just don't get it.

    Like there is no point giving me hints, I don't get what is going on.
    Since you've done the first part, you know the general set up.

    For the second part there is an additional mass added to the string at its midpoint.

    Then tension in the AP part of the string will be what's necessary for the two masses to move in a circle.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by ghostwalker)
    Since you've done the first part, you know the general set up.

    For the second part there is an additional mass added to the string at its midpoint.

    Then tension in the AP part of the string will be what's necessary for the two masses to move in a circle.
    Is the tension in the string the same even though it is only half the length now?
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by Super199)
    Is the tension in the string the same even though it is only half the length now?
    Same as what? It's different to the first part.

    The length of the portion of the string does not effect the tension. It's only the masses and their position (and the rate of revolution) that effects the tension. And there are now two masses.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by ghostwalker)
    Same as what? It's different to the first part.

    The length of the portion of the string does not effect the tension. It's only the masses and their position (and the rate of revolution) that effects the tension. And there are now two masses.
    Hmm I think I've got it cheers.

    Do you mind helping with 6ii.

    I think I have the time for p, but I don't understand how to get the time for q.

    http://www.ocr.org.uk/Images/57767-q...echanics-2.pdf
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by Super199)
    Hmm I think I've got it cheers.

    Do you mind helping with 6ii.

    I think I have the time for p, but I don't understand how to get the time for q.

    http://www.ocr.org.uk/Images/57767-q...echanics-2.pdf
    Q was launched when P was above A.

    So, how long did it take for P to get to the point above A? Knock that off the total flight time fro P, and you have the flight time for Q.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by ghostwalker)
    Q was launched when P was above A.

    So, how long did it take for P to get to the point above A? Knock that off the total flight time fro P, and you have the flight time for Q.
    Time for P to get to A.

    S = 250
    u = 86sin62.3
    v = ?
    a = -9.8
    t = ?
    S= ut -4.9t^2
    250 = 86sin62.2t - 4.9t^2
    t = 10.8 s

    Time from P to the ground
    s = -250
    u = 30
    v =?
    a = -9.8
    t = ?
    -250 = 30t - 9.8t^2
    t = 6.8s

    Total time = 17.6
    17.6 - 16.8 = 10.8s
    But where does the time for p - 15.5s come from?
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by Super199)
    Time for P to get to A.

    S = 250
    u = 86sin62.3
    v = ?
    a = -9.8
    t = ?
    S= ut -4.9t^2
    250 = 86sin62.2t - 4.9t^2
    t = 10.8 s
    IF the vertical speed initially was 86ms^-1 (which it isn't - it's less), then in 10 seconds it's vertical velocity would have become negative i.e 86-10g. But it isn't, it's 30 upwards, so t=10.8 can't be right.

    Time from P to the ground
    s = -250
    u = 30
    v =?
    a = -9.8
    t = ?
    -250 = 30t - 9.8t^2
    t = 6.8s
    If you're going to work out the time from P to the ground, then this is the actual time of flight of Q.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by ghostwalker)
    IF the vertical speed initially was 86ms^-1 (which it isn't - it's less), then in 10 seconds it's vertical velocity would have become negative i.e 86-10g. But it isn't, it's 30 upwards, so t=10.8 can't be right.



    If you're going to work out the time from P to the ground, then this is the actual time of flight of Q.
    Hmm Im not too sure. I've attached the mark scheme if you want to have a look.
    Attached Images
     
 
 
 
  • 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
    Brexit voters: Do you stand by your vote?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • 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

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