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Homework Help. Motion.

  1. Nov 10, 2004 #1
    Hi again.
    I have 3 sets of questions which I spose are fairly easy for the ones here.
    1) A tennis ball of mass 0.20kg is released from rest at a height of 2.0m above a concrete floor.
    It rebounds to a height of 1.5m

    a) Calculate its kinetic energy and speed just before impact
    b) calculate its kinetic energy and speed just after impact
    c) calculate its loss of energy between release and its maximum height after rebounding

    I have managed to solve a) I did, V^2 = U^2 + 2as
    So, 0^2 + 2*10*2.0 = 6.3 So the Velocity is 6.3m/s
    Therefore, Kinetic energy = 1/2 mv^2
    So, 0.20 * (6.3*6.3 = 40) / 2 = 4J

    But I dont understand how I can do b and c, can someone please help me :) Thanks in advance.
  2. jcsd
  3. Nov 10, 2004 #2
    Think about all of this in terms of kinetic energy. PE is all it has before it is released, and

    PE = mgh = .2*10*2 = 4.

    Just before impact, PE = 0, and so all energy is kinetic, i.e. KE = 4

    .5*m*v^2 = 4, m*v^2 = 8, v^2 = 40, v = 6.32

    So far so good.

    Now since it came back up to 1.5 m, PE = mgh = .2*10*1.5 = 3. Did the ball lose any energy between right after hitting the floor and going back up to 1.5? Nope. It was in the air. So KE = 3. Do you see this? It's important. Because PE is 3 after the bounce, KE right after the bounce must equal this, because, once again, PE = 0 at that point.


    3 = .5*m*v^2, 6 = m*v^2, 30 = v^2, v = 5.48 or something.

    And so what is the loss of energy? Why, it's simply the change in potential energy. Which is 1.
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