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Momentum and Impulse!

  1. Jul 17, 2004 #1
    This question kinda stumped me :frown: ....

    Suppose that a ball of mass 0.25 Kg is moving downward at a speed of 15 m/s when it lands in your hand.
    (a) If it takes .15 s to bring it to rest in your hand, what is the net impulse your hand exerts on the ball to bring it to rest? Give both size and direction

    What i did for this problem was F = M (V<f> - V<i>) / Time
    .................................OR F = -25N (when u solve it like i did)
    i was thinking if impulse is in units N.S or Kg m/s - then this stuff that i did is incorrect...is the equation im using wrong? because i have different equations but this is the only one that encompasses all information given in the question.

    part b asks: what is the average net force u exerted on the ball to stop it? Give both size and direction...for this i used F = m x a
    .........................................OR F = 3.75 N when i solved...
    this seems to easy, did i do something wrong? or is it correct?


    Thanx
     
  2. jcsd
  3. Jul 17, 2004 #2

    arildno

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    (a) is poorly stated, in the sense that the impulse is independent of the time spent
    (It is simply M(vf-vi))
    (b) Assuming you did the correct algebra, the average F is 25N in the upwards direction (Simply, impulse divided with time)
     
  4. Jul 17, 2004 #3
    I suspect it is something of a trick question, which is why it is stated that way. To test if the reader really knows what is meant by impulse. But in any case, your answer is the correct one.
     
  5. Jul 17, 2004 #4

    arildno

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    Might well be, Tyger.
    However, this is the type of "trick questioning" that I dislike the most:
    It does not necessarily measure what a person knows, but the confidence by which someone knows it

    That is, breezily confident individuals will score high, whereas someone who at times doubt their own competence (which may be higher than the over-confident's) will be thrown in confusion, and enter a wrong answer, against their own, better judgment.
    I.e, these types of questions borders on measuring personalities, rather than competence.
     
  6. Jul 17, 2004 #5
    (a) is poorly stated, in the sense that the impulse is independent of the time spent
    (It is simply M(vf-vi))

    if this is true arildno...then my equation is fine as long as i take away the time right? does anyone else know how to do this or help?

    (b) Assuming you did the correct algebra, the average F is 25N in the upwards direction (Simply, impulse divided with time

    and how can the force be +25 N if the ball is traveling downward - therefore acceleration is -10 (gravity on earth) - therefore force should be proportional and also negative....right?
     
    Last edited: Jul 17, 2004
  7. Jul 17, 2004 #6

    arildno

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    (a)Yes, as I said, the impulse is M(vf-vi)=M(0-vi)=-Mvi, that is, in the opposite direction than the initial velocity of the ball (i.e, upwards)
    (b)Your hand exerts an upwards force on the ball.
     
  8. Jul 17, 2004 #7
    ok, thanx alot for ur help guys, but i still dont understand why time is given but cannot be used.

    but maybe it IS a trick question and it is put there for that reason.

    -Thanks again,
    Arabian Knight (Sari)
     
  9. Jul 17, 2004 #8

    arildno

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    The time is used in (b) and not in (a)
     
  10. Jul 26, 2004 #9
    m(v - u)
    F = .............
    t

    is how you figure out impulse
    or Ft = mv - mu
    so i'm not sure how f= mv-mu can be correct...
     
  11. Jul 26, 2004 #10
    I = MV-MU
    I = 0-(0.25x15)
    I = 3.75 NS acting upwards.
    I = 3.75 = Force x time

    I now presume it means the 'average force'. Net impulse is a pointless term when working in 2D here and they wouldn't give you the time for nothing.

    3.75 = .15 x force.
    3.75/.15 = Force
    Average force exerted = 25N
    Obviously the force is upwards, as it has stopped the downwards motion.
     
    Last edited: Jul 26, 2004
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