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Momentum of ball

  1. Jan 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A 0.14 kg ball of dough is thrown straight up into the air with an initial velocity of 24 m/s.
    Find the momentum of the ball of dough halfway to its maximum height on the way up.


    2. Relevant equations

    p=mv

    3. The attempt at a solution

    okay so i was trying to figure out the height and then cut it halfway. Then find the velocity and plug it in.But i have no idea wat i'm doing and my answer keeps turning out wrong.
     
  2. jcsd
  3. Jan 3, 2008 #2

    dynamicsolo

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    The dough is being thrown straight up with a given velocity. What force is acting on it? What will its acceleration be on the way up? How do you find the maximum height the dough will reach? How do you find the speed it has when it is halfway there?

    It will help readers here to guide you if you show how you were making your calculations...
     
  4. Jan 3, 2008 #3
    Ok so 9.81m/s2 will be playing a role so the Fnet is going to be 1.37 N but i'm trying to figure out time so i can find distance or find some way i can find distance.
     
  5. Jan 3, 2008 #4

    dynamicsolo

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    How fast is the dough moving vertically when it reaches maximum height? How long after it is tossed will that happen? How do you find the vertical distance the dough will have traveled?
     
  6. Jan 3, 2008 #5
    You know two possibilities about when this object could be at rest (Velocity=0). Intially, before you toss the ball, and when the ball reaches its max hieght of trajectory, right?
    Can you find when [tex] \frac{dy}{dt}=0[/tex]?
     
  7. Jan 3, 2008 #6
    yes i'm trying to figure out Vf but i need time for this formula Vf=(Ft+mu)/m or for Vf=squareroot Vi+2ax then i need x.
     
  8. Jan 3, 2008 #7
    You are fimiliar with the kinematic formula(verticle): [tex] y=y_0+v_ot-\frac{1}{2}gt^2[/tex]
    Differentiating with respect to time, setting equall to 0 and solving for t will give you that time.
     
  9. Jan 3, 2008 #8

    dynamicsolo

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    Well, I see one problem here: if you're using the "velocity-squared" equation from kinematics, that has to be

    Vf = squareroot[ (Vi^2) + 2ax ] .

    You know Vi and also a (what will be the sign of a). The vertical speed at the top of the climb will be what? That will let you find x , the maximum height reached. You could then use this equation again with x set to half the maximum height to find the vertical speed at that point.
     
    Last edited: Jan 3, 2008
  10. Jan 3, 2008 #9

    dynamicsolo

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    Winzer, from some of his other posts, I'm guessing that mortho is not in a calculus-based physics course.
     
  11. Jan 3, 2008 #10
    Right, I just realized that myself too.

    Ok then, use [tex] v_fy=v_i0-gt[/tex], can you find the time when the velocity is zero?
     
  12. Jan 3, 2008 #11
    Ok so i calculated it..is x 1.50 m ? and then i found 5.42 for my almost final velociity so now can i plug it into my momentum equation p=mv or p=(0.14*5.42) or 0.759 kgm/s

    and then for half the height 0.75 m and almost final velocity is 3.84 m/s so momentum is 0.538 kgm/s Right??
     
  13. Jan 3, 2008 #12

    dynamicsolo

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    I think the maximum height is going to be a bit more than 1.5 m, considering that 24 m/sec is about 54 mph. Could you should us your calculation, please?
     
  14. Jan 3, 2008 #13
    o ok i think i see what i did wrong...24squared/(9.81*2) and the answer is 29.4 m right?
    before i squared in the end..
     
  15. Jan 3, 2008 #14

    dynamicsolo

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    That looks better. BTW, what is the speed here? Had you answered the first question in your other post about what the ball's momentum is here?

    Now you can use the velocity-squared equation again to find the dough's speed at half this height. That should get you to the answer for your second question.
     
  16. Jan 3, 2008 #15
    So i did everything all over again and got 3.43 kgm/s for a and for b 2.48 kgm/s
    I used 29.4 m for height. Then did sqaureroot Vi+2ax to find Vf and plugged that in for P=mv for each.
     
  17. Jan 3, 2008 #16

    dynamicsolo

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    I thought part (a) asked for the momentum when the doughball is at the maximum height. The momentum when it is first thrown is indeed (0.14 kg)(24 m/sec) = 3.36 kg·m/sec. At half-maximum height, you would have found the speed at 17.0 m/sec, so the momentum would be 2.38 kg·m/sec. (Notice that this is sqrt(1/2) times the momentum the ball was thrown with. The reason for this will be clearer when you study energy.)
     
  18. Jan 3, 2008 #17
    that's what i thought too for part a and already put in 3.36 for my first answer but it got marked wrong. that 's why i was confused and thought i was supposed to use another velocity or something.
     
  19. Jan 4, 2008 #18

    dynamicsolo

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    That's because you aren't answering the question asked. The 3.36 kg·m/sec is the momentum the doughball has when it is first thrown, when is when it has its maximum speed. The question is asking for the momentum at maximum height.

    So how fast is the ball moving when it is at its maximum height, that is, at the top of the climb?
     
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