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Momentum - Height

  1. Feb 6, 2007 #1
    1. The problem statement, all variables and given/known data
    A stone of mass 'm' is dropped from rest at a height of 1.98 m. From what height would a stone of mass 'm/2' have to be dropped to have the same momentum upon striking the ground?



    2. Relevant equations

    ?

    3. The attempt at a solution

    i don't understand 'm/2' doesn't this mean the first mass divided by 2 and so could i not just simple divide1.98 by 2?
     
  2. jcsd
  3. Feb 6, 2007 #2
    this is probably the simplest question on this assignment and i don't get it:cry:
     
  4. Feb 6, 2007 #3

    PhanthomJay

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    No. The second mass has half the mass of the first mass. What is the momentum of the first ? You'll need to find its speed.Then proceed to work on the momentum of the second mass.
     
  5. Feb 6, 2007 #4
    how do i do that though if i don't have any number for mass the only number i have is the height of 1.98 m i wasnot given any other numbers?
     
  6. Feb 6, 2007 #5

    PhanthomJay

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    use m for the mass of mass1, and use m/2 for the mass of mass 2. That will work. Now you have a distance but not a speed. Use one of your motion eqautions from kinematics that solves for the speed as a function of the height. Are you familiar with these equations?
     
  7. Feb 6, 2007 #6
    ok can i use the formula v=square root of 2(9.8)(1.98) ?
    and then i need to solve for the mass but i don't recall an equation i can use to find mass since i don't have momentum
     
  8. Feb 6, 2007 #7
    by the way, if one rock has half the mass of the other, how should its speed relate to the other to have the same momentum?
     
  9. Feb 6, 2007 #8
    i was thinking that its speed would be less because it is lighter?
     
  10. Feb 6, 2007 #9

    PhanthomJay

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    Yes for the velocity of the first mass. You don't have to solve for the mass. It is given. the mass is 'm'.
    Its momentum is m(sq rt of (2)(9.8)(1.98)). Now what speed must the second mass have (which has a mass of m/2) to have the same momentum? Then work backwards to find its height from which it must be dropped to attain that speed.
     
  11. Feb 6, 2007 #10
    ok i got is 7.920000002m

    THANK YOU!!!!!
     
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