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A Momentum Problem

  1. Jan 12, 2005 #1
    Someone throws his .20-kg football in the the living room and knocks over his mother's .8 kg antique vase. Ater the collision, the football bounces straight back with a speed of 3.9 m/s, while the vase is moving at 2.6 m/s in the opposite direction. a) How fast did Tyrrell throw the football? b) If the football continued to travel at 3.9 m/s in the same direction it was thrown would the vase have to be more or less massive than .8 kg?

    I'm not sure on how to get the Velocity of the football prior of hitting the vase.

    I did figure out the vase's momentum is
    .8 * 2.6 = 2.08

    If the momentum of the football before it hit the vase was given I know it could be answer with

    p/m=v


    But I have no idea on how to get the velocity of the football prior of hitting the vase.

    I also dont understand the second question at all.
     
  2. jcsd
  3. Jan 12, 2005 #2

    Curious3141

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    It's all about conservation of linear momentum.

    [tex]\vec{p_{initial}} = \vec{p_{final}}[/tex]

    Let the initial velocity of the football be [itex]v[/itex]. Consider both the football and the vase as one entire system. The momentum of this system is conserved.

    Remember that because momentum and velocity are vector quantities, you have to be careful with the signs. You can take the initial direction of travel of the ball as the positive direction, and anything moving in the opposite direction will have a negative velocity and momentum.

    Get an expression for the initial momentum of the system in terms of [itex]v[/itex]. This is the sum of momentums of football and vase. Since the vase didn't move to begin with, it has zero momentum.

    Then get an expression for the final momentum, taking care to get the signs right. The football will have a negative final momentum, the vase will have a positive final momentum.

    Then solve for [itex]v[/itex].

    For the second part, the initial momentum of the system is the same as in the first part. So the left hand side of the equation is the same, meaning the right hand side will also have to be the same. But now you have 2 positive quantities being added (since football and vase are travelling in the same direction). Can you figure out if the vase should be more or less massive in this case ?
     
  4. Jan 12, 2005 #3
    I'm not sure if I did this correctly, but here is what I did. (From what I understood from your explanation)

    Football Vase
    m .2 .8
    v -3.9 2.6

    I used m=.1
    v= -6.5

    Then got the momentum to equal .65

    Then I did p/m=v .65/.2=3.25 m/s

    So I'm wondering if th answer is 3.25 m/s?
     
  5. Jan 12, 2005 #4
    Im assuming they are not interested in friction, etc.

    Does not seem your awnser is right, because how could he throw it, and bounce back at a faster rate? hmmm, remember what Curious3141 said.

    Momentum initial = Momentum final
     
  6. Jan 12, 2005 #5

    Curious3141

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    The answer should be 6.5 m/s.

    I don't understand what you're doing : what do you mean by "I used m=.1", etc ?

    It's quite simple. You should have an equation like this :

    mass of football*initial velocity of football = mass of football*final velocity of football + mass of vase*final velocity of vase.

    You seem to have gotten the right answer, then you did something wrong by multiplying by 0.1. Why ? What's your rationale for this ?
     
  7. Jan 12, 2005 #6
    Ya should be 6.5 m/s

    Momentum Initial = Momentum Final

    (MVi)of ball + (MVi)of vase = (MVf) ball + (MVf) vase
     
  8. Jan 12, 2005 #7
    The .1 is not rational and I know how I got it but that's completly wrong, since .8 + .2 doesn't equal .1... I got confused with another problem I did earlier I guess.

    So the initial velocity of an object thrown is the sum of the two final velocities of the vase and and the returning velocity of the ball? Because that's how it seems, just adding 3.9 + 2.6 = 6.5. Or it just worked out like that for this particular question?

    With the formula you gave me I got 6.45 so I'm guessing your answers were rounded up and that chance of both final velocities been added up to 6.5 wont work on all problems?

    .2 * Vi = .2 * (-3.9) + .8 * 2.6
    .2 * Vi= 1.29
    Vi= 1.29/.2
    Vi= 6.45
     
  9. Jan 12, 2005 #8

    Curious3141

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    It's just that way in this particular case because of the particular masses given. Don't assume it will work out this way in another question.

    The answer is exactly 6.5. You miscalculated.
     
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