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Momentum Problem

  1. Nov 8, 2004 #1
    A ball with a mass m is attached to the end of a string with a length 50.0 cm and is released from a horizontal position as shown in the diagram. At the bottom of its swing the ball strikes a block of mass M = 2m resting on a frictionless table. Assume that the collision is perfectly elastic. A) What are the velocities of the ball and block after impact. B) To what height does the ball rebound?

    I didn't understand how to find the separate veolicties for each object, but i understood that the equation had something to do with the equation
    mgh = 1/2mv^2
    I found that solving out that equation got me 2.21 m/s for the total velocity, but i didn't know how to find the inidvidual velocities or how to find the height of the rebound, can I get some help please? Thanks
  2. jcsd
  3. Nov 8, 2004 #2
    well you are somewhat correct with your energy equation, moreso for the heigth portion of your question.

    the only energy initially is the potential energy due to gravity. considering the moment both ball and block are sent into motion, you have two objects with kinetic energy.

    break this down into stages and look at the energy/momentum at each stage. if you want to know to what heigth the ball rebounds, think about the system once the ball is not moving again (e.g. when it's only energy is the potential energy associated with gravity).
  4. Nov 8, 2004 #3
    Well I don't really understand the stages that you're talking about, i think i understand that now the mgh = the KE of the ball + the KE of the box, but I don't know the steps to solve for it, and I still don't get how i would be able to know where the ball stops moving and the maximum height it rebounds to... Could you just point me out some of the equations I'm going to need to use or what I need to set equal? Thanks.
  5. Nov 8, 2004 #4
    well you'll need to use both conservation of energy and momentum.

    kinetic energy is 1/2 mv^2, gravitational potential energy is mgh, and momentum is mv

    i would look at this problem in 3 stages.

    1. the block not moving and the ball not moving 50 cm above the block.

    2. the moment the ball collides with the block and both are sent into motion.

    3. the moment the ball stops moving after its gone back into the air.

    look at the first two, and once you've got the velocity for the ball after the collision, you are ready to tackle number 3. the block is out of the equation, so you can solve the final problem with just conservation of energy (1/2mv(postcollision) = mgh).

    i'm kind of drunk and tired, hopefully that helps.
  6. Nov 8, 2004 #5
    I don't get how to solve for part two though, I understand that when the ball collides mgh = 1/2mv^2 and then when solved out v = 2.21 m/s for the total system, but how do you find what part of the velocity causes the ball to rebound back and which part causes the block to move forward?, Sorry if i sound a little dumb, I'm just really confused as to how to solve each of the velocities. Thanks again Teclo.
  7. Nov 8, 2004 #6
    well situation 1 and 2 are what you want to solve for first. 1 is the intial energy and momentum, 2 is the final energy an momentum. try writing the equations out for these two situations.

    the ball will be moving at a velocity you can determine(when the two collide) with conservation of energy -- choose the bottom of the arc for potential energy to e zero.

    mgh = (1/2)mv^2

    now you know the velocity of the ball when it is hitting the block. use conservation of momentum and energy to solve for the velocities of each after the collision.

    once you know the velocity of the ball after the collision, you are solving the same problem you did the first time, except mgh will be the final, and (1/2)mv^2 will be the initial.

    i have to head out now, hopefully that helps or someone else can lend some wisdom.
    Last edited: Nov 8, 2004
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