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Tough Energy problem Law of conservation of energy

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Bill throws a 10.0g (0.0100kg) straight down froma height of 2.0m. The ball strikes the floor at a speed of 7.5 m/s.
    a) what was the original speed of the ball?
    b) if 30% of the balls energy is transformed in thermal energy during the collision with the floor, find the new height reached by the ball.

    2. Relevant equations
    [tex]\frac{1}{2}mv^2 + mgh = \frac{1}{2}mv'^2 + mgh'[/tex]
    [tex] \sum{E} = \sum{E'} [/tex]

    3. The attempt at a solution
    A) i solved this fairly easily and got the right answer.

    [tex] v = \sqrt{\frac{2\left(\frac{1}{2}mv'^2 - mgh\right)}{m}} [/tex]

    v initial = 4.13 m/s

    B) this is the toughy...

    The answer is supposed to be 2.0 m. i attempted to rearrange the question including the initial kinetic energy + initial potential = end potential and solve for h. but i didnt get it. A friend of mine told me i dont need to include work done by NC forces in this equation. i think i got 1.7m, i know that if i just have initial potential energy = to kinetic energy @ the point of collision.. when losing 30% of energy, it isnt supposed to reach the same height correct? Which leads be to believe that the person throwing the ball supplied the additional 30% energy to return it to the 2.0m height. I don't understand how to mathmatically show how to solve to get h' = 2.0m..

  2. jcsd
  3. Oct 12, 2009 #2


    User Avatar
    Homework Helper

    I get 2.01 m for (b). I just did
    E = .7*.5*m*v^2 = .7*.5*.01*7.5^2 = .1969 J
    .28125 = mgh
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