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Impulse on a thrown ball

  1. Dec 26, 2004 #1
    A 0.2kg ball is thrown at 15m/s, 45 degrees below the horizontal. The ball is hit by at bat at 40m/s, 30 degrees above the horizontal. What is the impulse delivered to the ball?

    So, impulse is change in momentum. The initial momentum is .2*15=3Ns
    How would I find the final momentum and how does the angles come into play?
  2. jcsd
  3. Dec 26, 2004 #2


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    1.Make a drawing (diagram) in which place the horizontal and vertical directions along with the 2 velocity vectors.
    2.Chose a system of coordinates with 2 axis orthogonal on each other.Project on these 2 axis the 2 velocity vectors.Compute the initial and the final comonents of the momentum vectors.Add/subtract carefully the vectors and find the final answer.

  4. Dec 26, 2004 #3
    how is that possible when one angle is above the horizontal and the other is below? I'm not quite sure how I should place the two diagrams for comparison

    and how would I find final velocity?
    Last edited: Dec 26, 2004
  5. Dec 26, 2004 #4
    http://home.earthlink.net/~urban-xrisis/pic001.jpg [Broken]

    so basically find the x and y componets and subtract right?
    Last edited by a moderator: May 1, 2017
  6. Dec 26, 2004 #5


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    1.The drawing is pretty awkward,as you imagined it like an Arab,from right to left.It's not important.As I said,pick a system of axis and compute this vector:
    [tex] \Delta\vec{p}=\vec{p_{f}}-\vec{p_{i}} [/tex]
    2.If your choise of coordonates is smart,then u should get the result pretty easily.

  7. Dec 26, 2004 #6
    So I do have to split the 40m/s and the 15m/s into it's x and y componets. With that, I have to subtract x-axis momentum of 40m/s from the x-axis momentum of 15m/s. Then subtract the y-axis momentum of 40m/s from the y-axis momentum of 15m/s. The resultant should give me what I'm looking for right?
  8. Dec 26, 2004 #7


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    Yes.To me it looks like those components add instead of subtract,as they correspond to a vector subtraction of vectors which have different dirrections (the support is the same,but point one at another).

  9. Dec 26, 2004 #8
    40 m/s at 30 degrees
    x-componet = 34.6 m/s to the right
    y-componet= 20 m/s upwards

    15 m/s at 45 degrees
    x-componet = -10.6 m/s to the left
    y-componet= -10.6 m/s downwards

    so I add the vectors up?
    x-componet = 24 m/s to the right
    y-componet= 9.4 m/s upwards

    resultant=25.8 m/s
  10. Dec 26, 2004 #9

    Impulse is 8.2Ns

    is this correct?
  11. Dec 26, 2004 #10


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    Here's how i see it.Consider the initial state of the ball being pictured in the left of the screen and having a velocity vector [itex] \vec{v}_{i} [/itex] and moving towards the right part of the screen.In the final state it is hit by a bat and therefore moves to the left again with a velocity [itex] \vec{v}_{f} [/itex].Pick the Oxy system with the direction of 0x from left to right and with direction of 0y upwards.
    U will have
    [tex] \Delta \vec{p}=\vec{p_{f}}-\vec{p_{i}} [/tex]
    [tex] \vec{p_{f}}=m(-v_{f,x}\vec{i}+v_{f,y}\vec{j}) [/tex]
    [tex] \vec{p_{i}}=m(v_{i,x}\vec{i}-v_{i,y}\vec{j}) [/tex]
    [tex] \Delta \vec{p}=m[-(v_{f,x}+v_{i,x})\vec{i}+(v_{f,y}+v_{i,y})\vec{j}] [/tex]

    I'll let u plug in the numbers.

  12. Dec 26, 2004 #11
    so you're saying: Impulese = mass * (resultant force in y + resultant force in x)

    is that correct?
  13. Dec 27, 2004 #12


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    What do you mean resulting force?? :confused: U probably meant "velocity". :wink: I gave you the final vector decomposed on the coordinate axis.Square it and find its modulus and then plug in the numbers.It's not a big deal... :tongue2:

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