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Momentum/impulse question

  1. Jan 22, 2007 #1

    dnt

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    1. The problem statement, all variables and given/known data

    A .30 kg ball with initial velocity of 21i - 1.3j colldes with a wall. The wall exerts a force of -155i N on the ball and its final horizontal component of velocity is -11i m/s.

    (a) Find the vertical velocity component.

    (b) Find the total change in momentum.

    (c) How long was the ball in contact with the wall?

    2. Relevant equations

    change in momentum = impulse = Ft

    3. The attempt at a solution

    I think my main problem is im not sure if this is a conservatoin of momentum problem or not - is momentum conserved after the collision with the wall? Im having a hard time relating the 155 N force applied from the wall to the ball.

    I know that if I can find the vertical component of velocity after hitting the wall, I can easily find the change in momentum (part b) by:

    m (vf - vi) where vf and vi are the overall resultant velocity vectors.

    and i know that if i get that i can set it equal to Ft to find the time in contact with the wall (part c) but i cannot figure out part a. can someone please help. thanks.
     
  2. jcsd
  3. Jan 22, 2007 #2

    AlephZero

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    Strictly speaking momentum is conserved, but that's not very useful to solve the problem, because the change of momentum of the ball equals the change of momentum of the wall plus the earth!

    You said (correctly) change of momentum = impulse = Ft. Apply that equation to the ball.

    Force and momentum are vector quantities. What does the question say about the direction of the force on the ball?
     
  4. Jan 22, 2007 #3

    dnt

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    change in momuntum = m(change in v)

    which is (.30 kg)(vf - 21.04 m/s) = (-155N)t

    {i got the 21.04 m/s by using pythagorean theorem to 21i - 1.3j and found the resultant vector...correct?}

    anyways, now i have two variables. i know if i get vf, then i can solve for t easily. but i cannot get vf! still stuck...

    since it is negative the force on the ball is in the OPPOSITE direction compared to when the ball struck the wall.
     
  5. Jan 22, 2007 #4

    dnt

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    doing some more thinking about it...am i getting closer:

    m(vf-vi)=Ft but perhaps i should consider it in the horizontal only...

    (.03)(-11 - 21) = (.155)(t)

    t = .00619 seconds?

    is that right?

    but i still cannot figure out the final vertical velocity (part a). can i use one dimensional kinematics to find it (using vertical components only)?

    vf = vi + at

    vf = (1.3j) + (9.81)(.00619 sec)

    vf = 1.3608j m/s

    any help is much appreciated. im really trying to get this.
     
  6. Jan 22, 2007 #5

    Dick

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    Your decimal points are sliding around so the answer is not right, but that is essentially it. Now what direction have you assumed the force is acting in? What does that say about vertical change in velocity?
     
  7. Jan 22, 2007 #6

    dnt

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    i assumed the direction was acting in the horizontal direction so it should not affect vertical velocity. however, gravity should make it increase. hence my confusion.
     
  8. Jan 22, 2007 #7

    Dick

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    It would. The usual line is that the collision time is short enough that the correction from gravity is not important. In this case you could make a point that it is (it's a SLOW collision). Not really the point of the exercise though I don't think...
     
  9. Jan 22, 2007 #8

    dnt

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    are you implying that vertical velocity isnt going to change because of such a short time period?
     
    Last edited: Jan 22, 2007
  10. Jan 22, 2007 #9

    Dick

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    No. I'm saying it will change (and in this case by a fair amount compared with vf). I am saying I don't think that correction is the point of the exercise. As for the final horizontal velocity - that was given to you as part of the problem statement. How could you 'figure it out'???
     
  11. Jan 22, 2007 #10

    dnt

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    ok tried to fix it:

    HORIZONTAL: (.03)(-11 - 21) = (-155)(t)

    t = .0619 seconds

    VERTICAL:
    vf = vi + at

    vf = (1.3j) + (9.81)(.0619 sec)

    vf = 1.907j m/s

    so i solve for time BEFORE solving for final vertical velocity? is this correct?
     
  12. Jan 22, 2007 #11

    dnt

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    i know thats why i edited it out - i overthought the problem and i confused myself! i then realized what i typed and took it out. sorry.
     
  13. Jan 22, 2007 #12

    Dick

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    Looks fine. Make it clear that vf has been corrected for gravitational acceleration during the collision. That may not be part of the expected answer.
     
  14. Jan 22, 2007 #13

    dnt

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    you mean for the 1.907 m/s part? how do i make it clear? isnt showing my work sufficient?

    btw thanks for the help!
     
  15. Jan 22, 2007 #14

    Dick

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    Sure. Clearly showing work is sufficient. You're welcome - but I didn't help that much.
     
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