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Momentum Help please

  1. Nov 10, 2005 #1
    Momentum Help!!! please

    Can anyone help me with this problem?

    A 0.19 kg ball of dough is thrown straight up into the air with an initial velocity of 13 m/s. What is the momentum of the ball of dough halfway to its maximum height on the way up.

    I tried (.19)(-13) and (.19)(13) but that was the wrong answer. Thank-you
     
  2. jcsd
  3. Nov 10, 2005 #2

    Pengwuino

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    Well your momentum is your velocity * mass. While the mass does stay the same, the velocity will not. You are looking for the momentum half way through the trip so what do you think the momentum will be halfway through the trip up if its original momentum is 13m/s?
     
  4. Nov 10, 2005 #3
    Oh okay so it will just be half?
     
  5. Nov 10, 2005 #4
    initial velocity is 13 m/s. mass is .19 kg. acceleration of gravity is -9.81 m/s^2. use those to find the velocity at a given time when it is halfway up. when you get the velocity there, use p=mv and you're done. so no, it's not half.
     
  6. Nov 10, 2005 #5
    so it would be the v=(vi)(m)(g) and then plug that into the momentum formula. I got v=(-13)(.19)(-9.8)=24.21 and then 24.21(.19)=4.6 m/2 but that answer didn't work
     
  7. Nov 10, 2005 #6
    You have to use kinematics to determine what the velocity is at half the height, THEN you can use the momentum formula. HINT: the system has constant acceleration!
     
  8. Nov 10, 2005 #7

    Pengwuino

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    No it wouldn't.

    You have to use the kinematic equation for velocity to determine how long it will take to get to the top. Using this number and the equation for position, determine the maximum height. Then use this equation again to determine what at what time value was the ball at 1/2 of it's maximum height. Using that time value, you can determine what the velocity was at that particular time.
     
  9. Nov 10, 2005 #8
    Yes, but to make it even shorter, you don't need to solve for the toal time. Just solve for the max height with the kinematic equation: [tex]v_{f}^2=v_{0}^2+2ad[/tex].
     
  10. Nov 10, 2005 #9

    Pengwuino

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    You can't simplify it like that. its velocity 1/2 the way up isnt the same as its velocity when the velocity is 1/2 the initial.
     
  11. Nov 10, 2005 #10

    daniel_i_l

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    You can use [tex]v_{f}^2=v_{0}^2+2ad[/tex]
    and find the final hight [tex]d[/tex] where [tex]v_{f} = 0[/tex]
    then you can find the speed you want by useing the same equation were for
    [tex]d[/tex] you put half of the final hight.
     
  12. Nov 10, 2005 #11
    once i find the final height (8.6)what equation would I use before I solve for momentum, the same 1? if so would it be 13^2= v^2 + 2(9.8)(8.6/2)?
     
    Last edited: Nov 10, 2005
  13. Nov 10, 2005 #12

    daniel_i_l

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    You can use the same equation, but it is the final speed (vf) that you are looking for, not the initial one. You know that the initial speed (v0)is 13m/s. Now you want to find the final speed, half way up. so the 13^2 has to be on the other side.
     
  14. Nov 10, 2005 #13
    I keep getting the wrong answer I'm not sure where I'm making my mistake.

    So it would be 0^2=13^2+2(9.8)x and x=8.6 after i divide that by half i would do v^2=13^2+2(9.8)4.3 and I would get v^2=253.3 and i would get 15.9 but that is the wrong answer. This is an easy problem in comparison to the others I did with this assignment and I feel dumb because I can't seem to get it.
     
  15. Nov 10, 2005 #14

    daniel_i_l

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    the acceleation in this case is negative because the mass is going up against gravity. That would be: v^2=13^2-2(9.8)4.3
     
  16. Nov 10, 2005 #15
    Remember!
    The vector force of weight (due to gravity) is opposite the direction of initial velocity.. Recalculate your equation... (you are missing a minus sign).

    Sam
     
  17. Nov 10, 2005 #16
    Thats the second time today I've replied at the same time as someone - I'm getting lucky :smile:

    Sam
     
  18. Nov 10, 2005 #17
    even when I do the work with the negative sign it still comes up as incorrect I don't know why
     
  19. Nov 10, 2005 #18

    daniel_i_l

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    the speed should be right, maybe the problem is in the next part.
     
  20. Nov 11, 2005 #19
    Have you considered that the answer your trying to match may be wrong?

    (It happens)! ;-)
     
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