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Conservation of Mechanical Energy

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data

    A 50 g ball of clay traveling at speed [tex]v_0[/tex] hits and sticks to a 1.0 kg block sitting at rest on a frictionless surface.

    a. What is the speed of the block after the collision?
    b. Show that the mechanical energy is not conserved in this collision. What percentage of the ball's initial energy is "lost?"

    3. The attempt at a solution

    For part (a), I just said

    m_1*v_0 = (m_1 + m_2)*v_1

    0.05*v_0 = (0.05+1)*v_1

    v_0 = 21*v_1

    I'm hoping this is correct! Now, part (b) I'm not entirely sure of. I guess there is no PE in this case so I should show that

    (.5)(m_1)(v_0)^2 [tex]\neq[/tex] (.5)(m_1)(v_0)^2 + (.5)(m_1)(v_1)^2

    Is this it? How do I go on about finding the percentage "lost"?
  2. jcsd
  3. Nov 14, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    This is fine.
    Not sure what happened here. Find v_1 in terms of v_0.

    This expression seems a bit mixed up. Find the initial KE (the left hand side). Find the final KE (you'll need the speed of the combined masses from part a). (Express everything in terms of v_0.)

    Compare the two KEs and find their difference (the loss of KE). Then you can figure out what percentage of the original KE that loss is.
  4. Nov 14, 2007 #3
    Is the right side of the equation your after? If it is, v_0 shouldn't figure into it, I don't believe.
  5. Nov 14, 2007 #4


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    Homework Helper

    Initial KE = Ei = 0.5*m1*Vo^2
    Final KE = Ef = 0.5*(m1 + m2 )*V1^2
    Show that Ef is not equal to Ei. And percentage loss = [(Ei -Ef)/Ei]*100
  6. Nov 14, 2007 #5
    Thanks everyone, got it! :)

    Oops! I didn't mean to include v_0 there. Sorry for the confusion.
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