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Math Help - gravity force

  1. #1
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    gravity force

    1>A system consists of three particles, each of mass 5.00 g, located at the corners of an equilateral triangle with sides of 30.0 cm.
    (a) Calculate the potential energy of the system.
    (b) If the particles are released simultaneously, where will they collide?

    2>At the Earth’s surface a projectile is launched straight up at a speed of 10.0 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by gracy View Post
    1>A system consists of three particles, each of mass 5.00 g, located at the corners of an equilateral triangle with sides of 30.0 cm.
    (a) Calculate the potential energy of the system.
    (all quantities will be converted to their SI equivalents in what follows)

    The potential energy of this system is equal to the work required to remove
    the points to infinity. This can be done in a number of ways, but the
    result is always the same, and equal to:

    <br />
E=3 \int_{r=0.3}^{\infty} \frac{G\,m_1\,m_2}{r^2}\, dr=3\,\frac{G\,m_1\,m_2}{0.3}<br />

    but m_1=m_2=0.005 kg and G\approx 6.67\ 10^{-11} \rm{Nm^3/kg^2} , so:

    E\approx 1.67\ 10^{-14}\mbox{joules}


    (b) If the particles are released simultaneously, where will they collide?
    The centre of mass of a free system like this remains at rest as long as there are
    no external forces, so the particles must collide at their centre of mass, which is
    at the centre of the equilateral triangle.

    RonL
    Last edited by CaptainBlack; December 22nd 2006 at 10:38 PM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by gracy View Post
    2>At the Earth’s surface a projectile is launched straight up at a speed of 10.0 km/s. To what height will it rise? Ignore air resistance and the rotation of the Earth
    We have the projectiles acceleration:

    a=-g

    so integrating once gives its velocity:

    v=-gt +c,

    but at t=0\ v=10\ \rm{m/s} so c=10.

    Integrate again to get its displacement:

    d=-gt^2/2+10t +k,

    but at t=0\ d=0\ \rm{m} so k=0.

    Now the maximum height is reached when v=0 which is at t=10/g\ \rm{s}, so the maximum height reached is:

    d_{max}=-50/g+100/g=150/g\ \rm{m}.

    and as g \approx 9.81 \ \rm{m/s^2},

    d_{max}\approx 15.3 \ \rm{m}

    RonL
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