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Total Kinetic Energy and some other problems

  1. Nov 8, 2006 #1
    So, the initial setup
    an assembly of two small dense spheres, each of mass, m, whose centers are connected by a rigid rod of length , l and negligible mass. An astronaut has a device that will shoot a lump of clay mass, m, speed, [tex]v_o[/tex] express your answers in terms of m, [tex]v_o[/tex], l, and fundamental constants.

    for the first problem, the clay ball is shot directly perpindicular to the assembly and sticks to the midpoint of the rod.

    1) Determine Total Kinetic Energy of the system (assembly adn clay lump) after the collision.

    So this would be Translational Kinetic energy plus rotational kinetic energy right?

    [tex] K_t + K_r = (1/2)mv_o^2 + (1/2)Iw^2 [/tex]

    w = omega

    and the initial and final rotational kinetic energy values are zero right, b
    ecause the clay ball strikes directly perpendicular at the midpoint
    The answer I got was
    [tex] K = mv_o^2 [/tex]

    2)Determine the change in kinetic energy as a result of the collision

    so since there is no rotational kinetic energy

    [tex] (1/2)3mv_o^2 - (1/2)mv_o^2 [/tex]

    I don't know if its all right or not....

    and then the next problem

    the assembly is reset, and this time, the clay ball is lanched perpendicularly at one of the clay balls in the assembly. The launched ball sticks to the one already on the assembly (right side)

    1) Determine the distance from the left end of teh rod to the center of mass os the system (assembly and clay lump) immeadiately after the collision. (assume that the radii of the spheres and clay lump are much smaller than the seperation of the spheres.

    i got
    [tex] (L/2) [/tex]

    as the center of mass, but I know thats not right, but don't know how to fix it...

    2) Indicate the direction and motion of the center of mass immeadiately after the collision

    I have no idea how to do this, mainly becasue I don't have the center of mass right.

    3) determine the speed of the center of mass immeadiately after the collision

    again, no idea...

    4) determine the angular speed of the system (assembly adn clay lump) immeadiately after the collision

    ....

    5) Determine the change in kinetic energy as a result of the collision

    no idea...

    please help..
     
    Last edited: Nov 8, 2006
  2. jcsd
  3. Nov 9, 2006 #2

    PhanthomJay

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    First problem first. Kinetic energy is not conserved in a collision. Momentum is. When the clay hits dead center of the rod, there can be no rotation of the rod. So rotational energy is not a factor here. Apply the conservatiion of momentum principle to determine the speed of the clay-sphere system after the collision. Then compare the initial KE of the system just prior to impact (which is not 0) with the final KE after impact, using v_0 in detrermining the initial energy, and the value of v determined above for use in the final energy.
     
  4. Nov 9, 2006 #3
    ok, so you would have
    [tex] m_1v_1 + m_2v_2 = (m_1 + m_2)v [/tex]
    [tex] v_2 = 0 [/tex]
    [tex] v = (m_1v_1)/(m_1 + m_2) [/tex]

    so then...
    [tex] (1/2)3m_1((m_1v_1)/(m_1 + 3m_1))^2 - (1/2)m_1v_1^2 [/tex]

    adn from there I get a really weird answer...... what did I do wrong?
     
  5. Nov 9, 2006 #4

    PhanthomJay

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    Well, you've 'sort of' got the right equation, but I don't know what you did after that.. ..It is given that each sphere and the clay all have the same mass, m, so your equation will simplify. Be sure to include the sum of ALL the masses after the collision, and solve for v. Then determine the total KE after the collision. What was the KE of the system before the collision?
     
  6. Nov 9, 2006 #5
    before the collision, it should've been

    [tex] (1/2)mv_o^2 [/tex]

    and that should be it because nothing else was moving or rotating
     
  7. Nov 9, 2006 #6

    PhanthomJay

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    Yes, and after the collision, KE =?
     
  8. Nov 15, 2006 #7
    Finally, my internets working again. I guess theres no need for this thread anymore because I had to turn it in already. Thanks for the help though.
     
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