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    Two particles A and B, of masses 0.4kg and 0.3kg respectively, are connected by a light inextensible string.
    The particle A is placed near the bottom of a smooth plane inclined at 30 degrees to the horizontal.
    The string passes over a small smooth light pulley which is fixed at the top of the inclined plane and B hangs freely.
    The system is released from rest, with each portion of the string taut and in the same vertical plane as a lane of greatest slope of the inclined plane.

    Calculate:

    a) the common acceleration, in ms^-2, of the 2 particles

    b) the tension, in N, in the string

    Given that A has not reached the pulley, find:

    c) the time taken for B to fall 6.3m from rest

    d) the speed that B then acquireid
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    a) 0.3g - 0.4g sin30 = 0.7 a => a = g/7 m/s²

    b) 0.3g - T = 0.3g / 7 => T = 2.52 N

    c) 6.3 = 1/2 (g/7) t² => t = 3 s

    d) v = 3 * g/7 = 4.2 m/s

    edit: oops, forgot about the g!
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    (Original post by elpaw)
    a) 0.3g - 0.4g sin30 = 0.7 a => a = 1/7 m/s²
    ...
    Err, shouldn't that be,

    a) 0.3g - 0.4g*sin30 = 0.7*a
    0.3g - 0.2g = 0.7*a
    0.7*a = 0.1g
    a=g/7
    a=1.4 m/s
    ======

    For the rest I get,

    b)
    0.3g - T = 0.3*a
    T = 0.3*9.81 - 0.3*1.4
    T = 2.52 N
    ======

    c)
    s=ut + ½at²
    6.3 = 0 + ½*1.4*t²
    6.3 = 0.7*t²
    t² = 6.3/0.7 = 9
    t = 3 s
    ====

    d)
    v² = 2as
    v² = 2*1.4*6.3
    v² = 17.64
    v = 4.2 m/s
    =======
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    yeah, a = 1.4m/s^2
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    yeah, sorry, i dropped a g in the acceleration (e.c.f.)
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    (Original post by Fermat)
    Err, shouldn't that be,

    a) 0.3g - 0.4g*sin30 = 0.7*a
    0.3g - 0.2g = 0.7*a
    0.7*a = 0.1g
    a=g/7
    a=1.4 m/s
    OK Thanks. But shouldn't that 'sin' be 'cos'? And if not, why?

    Also, I would've written:

    Newton II of A: T - 0.4g cos 30 = 0.4a

    Newton II of B: 2.94 - T = 0.3a

    Is that wrong? If it's right, where would I go from there please?

    Thanks
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    Looking at the diagram, the weight of A is resolved into "horizontal" and perpindicular forces, Where "horizontal" means parallel to the slope.
    The "horizontal component of A is 0.4gsinø.

    Doing NewtonII at A and B.

    T - 0.4gsinø = 0.4.a (at A)
    0.3g - T = 0.3.a (at B)

    Adding these eqns,

    0.3g - 0.4gsinø = 0.7.a

    etc.

    So, the first of the two eqns you wrote should have sin30° instead of cos30°.
    0.4gcos30 is the normal reaction of the plane against the mass A. (See diagram)
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