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    Hi

    Can someone tell me what am I doing wrong?
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    :giggle: I just got the same answer. Probably because I imputed the same numbers....
    Attachment 615586615588



    Also how would you answer question 5c?
    Because wouldn't the answer by two time should the tension. I calculated the tension at 27.5625 which is correct. By then at the back of my textbook it says these magnitude of the force exerted on the pulley is 39N (why is it not 2T(56.125))?

    Thank you
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    (Original post by Onthedancefloor)
    Hi

    Can someone tell me what am I doing wrong?


    :giggle: I just got the same answer. Probably because I imputed the same numbers....


    Also how would you answer question 5c?
    Because wouldn't the answer by two time should the tension. I calculated the tension at 27.5625 which is correct. By then at the back of my textbook it says these magnitude of the force exerted on the pulley is 39N (why is it not 2T(56.125))?

    Thank you
    When A hits the floor, A won't be pulling B anymore so the string will become slack. When this happens, the only force acting on B will be due to gravity i.e. it will be 3g acting downwards.

    You have assumed incorrectly that the acceleration will remain at g/7 upwards even after A hits the floor.
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    (Original post by notnek)
    When A hits the floor, A won't be pulling B anymore so the string will become slack. When this happens, the only force acting on B will be due to gravity i.e. it will be 3g acting downwards.

    You have assumed incorrectly that the acceleration will remain at g/7 upwards even after A hits the floor.
    S=ut +1/2at^2
    When ut=0 as u=0
    a=3g and t^2= 20/7

    Therefore, S=1/2(3x9.8)x20/7
    S=42m

    This is wrong. What have I done wrong?
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    (Original post by Onthedancefloor)
    S=ut +1/2at^2
    When ut=0 as u=0
    a=3g and t^2= 20/7

    Therefore, S=1/2(3x9.8)x20/7
    S=42m

    This is wrong. What have I done wrong?
    When A hits the floor, B won't be at rest. It would have picked up some speed during the 2m motion.

    So the first thing you need to do is find the final speed of B during the first 2m of motion. Then you can use this as your initial speed for the motion after A hits the floor.

    Does this make sense? This kind of question is very common in M1 exams and often trips students up. You need to get used to splitting up the motion into different stages.
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    (Original post by notnek)
    When A hits the floor, B won't be at rest. It would have picked up some speed during the 2m motion.

    So the first thing you need to do is find the final speed of B during the first 2m of motion. Then you can use this as your initial speed for the motion after A hits the floor.

    Does this make sense? This kind of question is very common in M1 exams and often trips students up. You need to get used to splitting up the motion into different stages.
    Makes sense, but I still can't get the answer. I've filled up like 4 papers trying to get the answers. Could you show me please?
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    (Original post by Onthedancefloor)
    Makes sense, but I still can't get the answer. I've filled up like 4 papers trying to get the answers. Could you show me please?
    Sorry we don't give solutions here and it's much more beneficial to do it yourself.

    Can you please post your working for what you got as the speed of B when A hits the floor?
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    I got v= 2.37
    As at the points u=0 a=g/7 and t= root 20/7
    Therefore, v=u+at
    V= 0+ (g/7)(root 20/7)

    I already know the solution, it's at the back of my textbook. I just would like to see the method
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    (Original post by Onthedancefloor)
    I got v= 2.37
    As at the points u=0 a=g/7 and t= root 20/7
    Therefore, v=u+at
    V= 0+ (g/7)(root 20/7)

    I already know the solution, it's at the back of my textbook. I just would like to see the method
    I don't know how you calculated your 't' but you don't need to use it. Just use s = 2 and v^2=u^2+2as.

    You should get v=\sqrt{\frac{4g}{7}}. If you don't, please post your working.

    Then use SUVAT again where u is the v above but be careful all your directions are correct.
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    (Original post by notnek)
    I don't know how you calculated your 't' but you don't need to use it. Just use s = 2 and v^2=u^2+2as.

    You should get v=\sqrt{\frac{4g}{7}}. If you don't, please post your working.

    Then use SUVAT again where u is the v above but be careful all your directions are correct.
    Okay, your 'u' value isn't the same number, I had just rounded it for convenience.

    I know have a and u, to work out S. You need a third value, how do I find this?
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    (Original post by Onthedancefloor)
    Okay, your 'u' value isn't the same number, I had just rounded it for convenience.

    I know have a and u, to work out S. You need a third value, how do I find this?
    The greatest height occurs when v = 0.

    So taking upwards as positive, you can use v =0, a = -g and u = v (from previous SUVAT) to find s.
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    (Original post by notnek)
    The greatest height occurs when v = 0.

    So taking upwards as positive, you can use v =0, a = -g and u = v (from previous SUVAT) to find s.
    Ahhh okay thank you soooo much Then you add it to 2m as that's how much it travelled previously.

    I probably over complicated this more than I should have. Thank you for taking your time to explain it to me. I get it now
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    What was the answer for 4b
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    (Original post by JickDee)
    What was the answer for 4b
    16/7
 
 
 
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