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Math Help - Kinematics

  1. #1
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    Kinematics

    P and Q are 4m apart at opposite ends of a smooth plane inclined at 30 degree to the horizontal. Initially a particle S of mass m is projected up the plane from P with speed 8m/s and a particle T of mass 2m is released from rest at Q.

    Find the distance of the particles from P at the instant when they collide.

    Thanks!
    Last edited by BabyMilo; November 27th 2009 at 05:10 AM.
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  2. #2
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    Quote Originally Posted by BabyMilo View Post
    P and Q are 4m apart at opposite ends of a smooth plane inclined at 30 degree to the horizontal. Initially a particle S of mass m is projected up the plane from P with speed 8m/s and a particle T of mass 2m is released from rest at Q.

    Find the distance of the particles from P at the instant when they collide.

    Thanks!
    magnitude of acceleration a smooth incline is a = g\sin{\theta}

    basic kinematics equation ...

    x = x_0 + v_0 t + \frac{1}{2}at^2

    let point P be x_0 , and up the incline be the (+) direction.

    motion of particle S up the incline ...

    x = 8t - 2.45t^2

    motion of particle T down the incline ...

    x = 4 - 2.45t^2

    solve the system for x , the position relative to point P where the collision occurs.
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  3. #3
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    thanks for the respond.

    can i just ask where did the 2.45 come from?

    oh i see. \frac{1}{2} * g * sin30

    ok thanks!
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  4. #4
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    d) Find the length of the time that elapses from the beginning of the motion until the combined particle reaches P.

    for a,b,c I found out:

    the distance of the particles from P = 3.3875m

    Vs = 5.55ms^-1

    Vt = 2.45ms^-1

    Vs+t(after collision) = \frac{13}{60}


    So for d) i did:
    <br />
3.3875=\frac{13}{60}t+\frac{1}{2}g sin 30*t^2

    which t=1.13s (3.s.f)

    the answer therefore is 1.13+0.5 = 1.63s

    but the actual answer is 1.816 s

    where have i gone wrong?

    thanks for your help!
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