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    I did the a) part and the answer was 4.66ms-1. I don't know how to do the b) part so anyone please help.
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    Ik this is so not helpful but girl why are you doing that at 5am?!
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    help
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    (Original post by Rashina)
    What do you mean by 5am? Are you talking about the time? Why r u awake at 5am then?
    No I'm talking about the algebraic expression 5am

    Because I am talking w/ someone
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    Using suvat equations.. and the acc u found in part a find the velocity of B when it reaches the ground(in other words its final velocity)
    That final velocity will be the same as the speed of A initially then use suvat again to find v as A reaches the pully with the same acceleration
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    The key thing to understand is that the speed of the object on the right side of the pully as it hits the ground will be the same as the speed of a at that instant
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    (Original post by Rashina)
    Name:  Mechanics doubt.jpg
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    I did the a) part and the answer was 4.66ms-1. I don't know how to do the b) part so anyone please help.
    Do you want any links for good tutorials on this? its about 1 hour (maybe) but its worth watching
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    (Original post by me2*)
    Using suvat equations.. and the acc u found in part a find the velocity of B when it reaches the ground(in other words its final velocity)
    That final velocity will be the same as the speed of A initially then use suvat again to find v as A reaches the pully with the same acceleration
    I did that, but the answer is different from the book's answer.
    My working:
    B:u=0ms-1
    s= 0.6m
    a= 4.66ms-2
    v=?

    v2=u2+2as
    v2=5.592
    v= 2.36 (to 3s.f)

    A:u=2.36ms-1
    s=1m
    a=4.66ms-2
    v=?

    v2=u2+2as
    v2=14.912
    v= 3.86 (to 3s.f)

    But the answer in the book is 1.57ms-1.
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    (Original post by me2*)
    Do you want any links for good tutorials on this? its about 1 hour (maybe) but its worth watching
    Ok that would be useful. Thanks for your help.
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    (Original post by Rashina)
    Ok that would be useful. Thanks for your help.
    https://www.youtube.com/watch?v=r73j4prNAug
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    (Original post by me2*)
    The key thing to understand is that the speed of the object on the right side of the pully as it hits the ground will be the same as the speed of a at that instant
    Is my working correct? Why don't the two answers match(mine and the one in the book)?
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    (Original post by Rashina)
    I did that, but the answer is different from the book's answer.
    My working:
    B:u=0ms-1
    s= 0.6m
    a= 4.66ms-2
    v=?

    v2=u2+2as
    v2=5.592
    v= 2.36 (to 3s.f)

    A:u=2.36ms-1
    s=1m
    a=4.66ms-2
    v=?

    v2=u2+2as
    v2=14.912
    v= 3.86 (to 3s.f)

    But the answer in the book is 1.57ms-1.
    Your working for particle B is correct.

    The surface is rough. Once particle B reaches the ground, it stops moving, and the string stops being taught. As a result, A no longer experiences any tension force from the string, as it is no longer taught.

    A will have moved 0.6m and be travelling at 2.36m/s at this point (same magnitude of motion as B). Then A will no longer experience tension force, but will still experience friction force opposing its motion, as the table is rough. And A will have a further 0.4m to move before reaching P.

    Find the friction force acting on A, then find the acceleration of A, and then apply that over a distance of 0.4m.

    With this method I got an answer of 1.56m/s which is 0.01m/s off the textbook answer, probably due to a rounding error earlier on.
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    (Original post by K-Man_PhysCheM)
    Your working for particle B is correct.

    The surface is rough. Once particle B reaches the ground, it stops moving, and the string stops being taught. As a result, A no longer experiences any tension force from the string, as it is no longer taught.

    A will have moved 0.6m and be travelling at 2.36m/s at this point (same magnitude of motion as B). Then A will no longer experience tension force, but will still experience friction force opposing its motion, as the table is rough. And A will have a further 0.4m to move before reaching P.

    Find the friction force acting on A, then find the acceleration of A, and then apply that over a distance of 0.4m.

    With this method I got an answer of 1.56m/s which is 0.01m/s off the textbook answer, probably due to a rounding error earlier on.
    Check my working for friction force:
    FMAX = µR
    FMAX=0.4 x (ax 0.3)
    =1.2a
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    (Original post by Rashina)
    can I apply the value of v (particle B) for u(particle A) as I did earlier?
    Yes, as particle A will be moving at the same velocity as particle B just before B hits the ground.

    But a and s for particle A are different to what you used earlier, as at this point A will accelerate due to friction (opposing motion), not tension, and will have already travelled 0.6m (same distance B travelled) so will only have a further 0.4m to travel.
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    (Original post by K-Man_PhysCheM)
    Yes, as particle A will be moving at the same velocity as particle B just before B hits the ground.

    But a and s for particle A are different to what you used earlier, as at this point A will accelerate due to friction (opposing motion), not tension, and will have already travelled 0.6m (same distance B travelled) so will only have a further 0.4m to travel.
    I posted my working. Is it correct?
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    (Original post by K-Man_PhysCheM)
    Yes, as particle A will be moving at the same velocity as particle B just before B hits the ground.

    But a and s for particle A are different to what you used earlier, as at this point A will accelerate due to friction (opposing motion), not tension, and will have already travelled 0.6m (same distance B travelled) so will only have a further 0.4m to travel.
    Can you show me your working, please? I couldn't do it.
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    (Original post by Rashina)
    Can you show me your working, please? I couldn't do it.
    Ok, sorry, I was out.

    FULL SOLUTION:

    Spoiler:
    Show



    Friction = 0.4 \times 0.3g = 0.12g

    a = \dfrac{F}{m} = \dfrac{-0.12g}{0.3} = -0.4g

    a = -0.4 \times 9.8 = -3.92ms^{-2}



    u = 2.36 ms^{-1}

    a = -3.92 ms^{-2}

    s = 0.4m


    v^2 = u^2 + 2as

    v = \sqrt{u^2 + 2as} = \sqrt{2.36^2 + 2 \times -3.92 \times 0.4}

    v = \sqrt{2.4336} = 1.56 ms^{-1}



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    (Original post by K-Man_PhysCheM)
    Ok, sorry, I was out.

    FULL SOLUTION:

    Spoiler:
    Show




    Friction = 0.4 \times 0.3g = 0.12g

    a = \dfrac{F}{m} = \dfrac{-0.12g}{0.3} = -0.4g

    a = -0.4 \times 9.8 = -3.92ms^{-2}



    u = 2.36 ms^{-1}

    a = -3.92 ms^{-2}

    s = 0.4m


    v^2 = u^2 + 2as

    v = \sqrt{u^2 + 2as} = \sqrt{2.36^2 + 2 \times -3.92 \times 0.4}

    v = \sqrt{2.4336} = 1.56 ms^{-1}




    Thanks a lot. If I have any doubt I'll ask you.
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    (Original post by K-Man_PhysCheM)
    Ok, sorry, I was out.

    FULL SOLUTION:

    Spoiler:
    Show




    Friction = 0.4 \times 0.3g = 0.12g

    a = \dfrac{F}{m} = \dfrac{-0.12g}{0.3} = -0.4g

    a = -0.4 \times 9.8 = -3.92ms^{-2}



    u = 2.36 ms^{-1}

    a = -3.92 ms^{-2}

    s = 0.4m


    v^2 = u^2 + 2as

    v = \sqrt{u^2 + 2as} = \sqrt{2.36^2 + 2 \times -3.92 \times 0.4}

    v = \sqrt{2.4336} = 1.56 ms^{-1}




    Just a small help. Finished most of it but couldn't do the last one.
    http://www.thestudentroom.co.uk/show....php?t=4400184
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    (Original post by me2*)
    Using suvat equations.. and the acc u found in part a find the velocity of B when it reaches the ground(in other words its final velocity)
    That final velocity will be the same as the speed of A initially then use suvat again to find v as A reaches the pully with the same acceleration
    (Original post by Rashina)
    I did that, but the answer is different from the book's answer.
    My working:
    B:u=0ms-1
    s= 0.6m
    a= 4.66ms-2
    v=?

    v2=u2+2as
    v2=5.592
    v= 2.36 (to 3s.f)

    A:u=2.36ms-1
    s=1m
    a=4.66ms-2
    v=?

    v2=u2+2as
    v2=14.912
    v= 3.86 (to 3s.f)

    But the answer in the book is 1.57ms-1.
    Doing suvat the first time gives the velocity when displacement equals 0.6m. You would suvat for the last 0.4m to find the velocity at A, just remembe to calculate acceleration due to friction
 
 
 
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