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BaseBall Question (Momentum & Impulse in 2 dimensions)

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data
    a baseball with a mass of 0.152 kg is moving horizontally at 32.0 m/s [E] when it is struck by a bat for 0.00200 seconds. The velocity of the ball just after the collision is 52.0 m/s [W 20 N]

    a) Find the impulse experienced by the ball
    b) find the average net force of the ball


    2. Relevant equations
    FnetΔT=M(V2-V1)


    3. The attempt at a solution

    Im confused between impulse and net force first of all.. I think I calculated both but I am unsure, Also I am unsure about which directions I should set at positive and negative,

    I set East and North as positive

    Fnetx(0.002)=(0.152)((-)52cos20-32) ( 52 is negative in this case since movement in x direction is towards W?)

    Fnet= - 6145.36, negative so [W]

    Fnety(0.002)=(0.152)(52sin20-0) ( 52 is positive here since i set north to = positive)

    Fnet=1351.28 N

    I make the triangle with my components and solve for hypotenuse through pythag. = 6292.17 ∅= W 12.4° N

    IS the hypotenuse equal to net force and the components equal to impulse?
    DO my directions and calculations look correct?

    Thank you everyone who takes the time to read this and make an attempt at helping me !
     
  2. jcsd
  3. Dec 13, 2011 #2
    First of all: Impulse is momentum. Momentum is a vector quantity
    Force is rate of change of momentum ie change in momentum (or impulse) per second
     
  4. Dec 14, 2011 #3
    I guess you can't help with regards to math and setting positive and negative directions?

    That seems to be where most of my confusion is coming from.. Thank you for clearing up momentum and force for me though!

    If I change my directions for certain things the numbers can change drastically so its quite confusing..

    thanks again!
     
  5. Jan 23, 2012 #4
    BUMP

    I didn't get full marks for this question,

    Where exactly do I calculate impulse..


    Is it

    FnetΔt=M(V1-V2)
    FnetΔt=answer for impulse ?

    and net force is equal to Fnet= answer for net force?
     
  6. Jan 24, 2012 #5
    Would impulse in this case be equal to

    ΔPx=(Fnetx)(Δt)
    ΔPy=(Fnety)(Δt)

    SO I have an impulse in the x and y dimensions?

    Can someone please shed some light for me!
     
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