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Center of mass and momentum

  1. Mar 30, 2008 #1
    A 4.5 kg ball initially rest at the edge of a 2.5 m long, 1.5 m high frictionless table. A hard plastic cube of mass 0.65 kg slides across the table at a speed of 20m/s and strikes the ball to leave the table in the direction in which the cube was moving.
    At t= 1 sec , F = .5 x10^3
    At t= 2 sec , F = 1 x10^3
    At t= 3 sec , F = 1.5 x10^3
    At t= 4 sec , F = 2 x10^3
    At t= 5 sec , F = 2x10^3
    At t= 6 sec , F = 2 x10^3
    At t= 7 sec , F = 1.5 x10^3
    At t= 8 sec , F = 1 x10^3
    At t= 9 sec , F = .5 x10^3
    At t= 10 sec , F = 0 x10^3

    a) Use the time and force to find the total impulse given to the ball
    b) Determine the horizontal velocity of the ball immediately after the collision (you can’t assume the collision is elastic)
    c) Determine the following for the cube immediately after the collision:
    i. Its speed
    ii. Its direction of travel (right or left) , if moving
    d.) determine the distance between the two points of impact of the objects with the floor



    for part a.) i did impulse = momentum = F*t, but i dont think thats right
    for part b.) the equation i used was Vif= (m1-m2)/(m1+m2) but this wasn't correct either
    for part c.) i used the equation v2f= (2m1)/(m1+m2), which again was wrong
    for part d.)to find the distance i used the equation y = v1t + 1/2gt^2 for both distances which again was wrong
     
  2. jcsd
  3. Mar 30, 2008 #2
    a) Regarding your solution, the problem is what F do you use? Instead, plot F vs. t and find the area under the curve. This will be the total impulse and also represents how much momentum was 'given' to the ball by the cube.

    b) Now that you have part (a), you can find this and (c) pretty easily using conservation of momentum. Don't try to plug into a formula like you did above - start over from the law of momentum conservation.

    Considering that (d) asks for the range of both objects, you already know they will probably both be moving to the right.

    d) Both objects will be horizontal projectiles, so the equation you mention will need to be used separately for horizontal and vertical. I'm sure you've solved problems like that before; if you have the correct velocities you should get the answer.
     
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