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Homework Help: Seriously struggling D: Impulse and Force Question

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Pleaseeee someone help me with this.. its literally hurting my brain

    A baseball with mass 0.152kg is moving horizontally at 32.0m/s[E] when it is struck by a bat for 0.00200 seconds. The velocity of the ball just after the collision is 52.0m/s [W20°N]

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

    3. The attempt at a solution

    someone in my class told me to get the answer we do the following:
    a) I=(m)Δv =0.152(32-(-52))=12.768 (kgm/s) [assume East direct is positive]

    b) I=FΔt, so F=I/Δt=12.768/0.00200=6384 (N)

    part b) I understand..
    where I'm lost is the logic in part a)'s answer..

    can someone please explain to me if I'm doing that right? It's confusing me because of the part in red..
    At least I thought that you cant just add two vectors when they aren't parallel
  2. jcsd
  3. Mar 19, 2012 #2
    You're correct, whoever told you this answer apparently didn't notice that the ball isn't traveling in exactly the opposite direction after the ball is hit, you'd have to take the vector difference of the two velocities.
  4. Mar 19, 2012 #3
    alright cool.. now I just need some advice figuring this thing out.. does this look right?
    I uploaded it to an image hosting site so you can see my FBDs..
    the thing is.. I'm trying to find Net force first and then sub that into
    Impulse=Netforce x Δtime

    but the question asks me for impulse in part a)
    and then netforce in part b)

    Is there a way I can get impulse first with the information I have, that I'm not seeing?
    heres the pic of the work I've done..

    *btw sry just ignore the red highlighter on my work**
  5. Mar 19, 2012 #4
    Your impulse is your change in momentum. In this case, your mass x change in velocity IS your impulse. Therefore, impulse = mΔv = FΔt.
  6. Mar 19, 2012 #5
    can you tell me if my work is right then? Idk I feel like I got some weird values..

    v1x = 32.0
    v2x’ = 52cos(20°) = –48.86
    v2y’ = 52sin(20°) = 17.79

    use the impulse equation for the x-components

    Fnetx ∆t = m(v2x – v1x)
    Fnetx (0.00200) = (0.152)( –48.86 – (+32))
    Fnetx (0.00200) = (0.152)( –48.86 – (+32))
    Fnetx = -6145.36 N

    Now use the impulse equation for the y-components

    Fnety ∆t = m(v2y – v1y)
    Fnety(0.00200) = (0.152)( 17.79 – (0))
    Fnety =1352.04 N

    Find the net force on the baseball using the following triangle:

    |Fnet| = √((–6145)^2+ 1352.04^2)
    |Fnet| =6292.33
    After rounding:
    |Fnet| =6290 N

    To find the direction:

    tan θ = (1352.04N)/( 6145.36N)
    θ = 12.4°
    Therefore, the average net force of the ball is 6290 N [W 12.4° N]

    Solve for impulse:

    Impulse = m(∆t)
    Impulse = (0.152kg)(0.00200s)
    Impulse = 3.04 • 10^-4

    AGHH am I way off base here?!:cry:
  7. Mar 19, 2012 #6
    cuz velocity changes in unparallel directions.. so I was breaking it down into components.. is that even right.. I normally love physics but I hate when I cant understand it:mad:
  8. Mar 19, 2012 #7
    Yeah breaking down components can be tough. I find it usually helps to draw a diagram. It looks like you did it correctly though. You've got the velocity components correct, so assuming you did the rest of your work right, which it looks like you did for finding the force except you messed up finding the impulse. impulse is ∆p (change in momentum) or F∆t, so it would be easiest the way you did your work to take

    I=F∆t=6290N*(0.00200s) = 12.58 N*S. I didn't actually take the time to do the whole problem myself, but it looks like you did all your work right, and that impulse is about the typical impulse of a baseball being hit by a bat, so that should be your answer.
  9. Mar 19, 2012 #8
    Impulse = FΔt. The rest of your calculations looks right to me, although I've been known to get stuff wrong.
  10. Mar 19, 2012 #9
    :surprisedOMG YES! THANK YOU sooo much! I knew my impulse looked funny..
    I accidentally switched Force for mass
    agent_509 I seriously can't thank you enough!:biggrin:

    so just to clarify one thing though,

    the question asks me for impulse in part a)
    and then netforce in part b)

    Is there a way I can get impulse WITHOUT calculating force first, AND considering the fact that my vectors are not parallel?
    is this possible with the information I have and I'm just not seeing it?
  11. Mar 19, 2012 #10
    thanks tal444!
  12. Mar 19, 2012 #11
    Uh, since you calculated the x and y force components separately, couldn't you do the same with impulse? Not sure if you will get the right answer, but that's what I would do.
  13. Mar 19, 2012 #12
    hm I cant really figure out how I would do that could you elaborate a little more.

    v1 = 32.0m/s[E]
    v2 = 52.0m/s[W 20° N]
    m = 0.152kg
    Δt = 0.00200s

    so if we were gonna do impulse separately as you say without force..
    Impulse = mass x Δv

    v1x = 32.0
    v1y = 0

    v2x = 52cos(20°) = –48.86
    v2y = 52sin(20°) = 17.79

    x components:

    Impulse = (0.152)(-48.86 - (+32))

    y components:

    Impulse = (0.152)(17.79 - 0)
    = 2.70408

    and the other way I did it by calculating force first led me to get impulse = 12.58466
    so something seems off either I'm doing something wrong in red or theres no way to get impulse with unparallel vectors without calculating force.
    If someone could give me a definit answer on this that would be great!
  14. Mar 19, 2012 #13
    ***Impulse = mass(Δv)

    ^removed the 'x' representing multiplication to avoid confusion
  15. Mar 19, 2012 #14
    Alright here's how you'd do it without having the force beforehand, you've already calculated the velocity components, so now you just find the difference. The total change in the x direction is just 34-(-48.86) and then in the y direction it's just 17.79, so then you square each of those and then take the square root for:

    √((34+48.84)^2+17.79^2) = 84.73 m/s so that's your Δv

    then just multiply that by the mass to get Δp

    84.73*.152 = 12.88 kg*m/s

    Edit: which is also the same thing as a N*s
    Last edited: Mar 19, 2012
  16. Mar 19, 2012 #15
    mother of god..

    what have you done..

    :D thanks for all your help agent_509 its almost sad how happy this has made me
    your values on the x component part are a bit off but with the right digits it works perfectly to solve for 12. 58466

    just like the answer we got when we solved for force first :D YESSSSSS
  17. Mar 19, 2012 #16
    Ahh put in the 48.84 instead of 48.86 and 34 instead of 32, I can't believe I managed to do that lol. Well glad you caught that mistake, glad I could help!
  18. Mar 19, 2012 #17
    Yes, using the Pythagorean theorem, you find the hypotenuse, which ends up as 12.58 , or the answer you obtained earlier.
  19. Mar 20, 2012 #18
    okay yes the pythagorean theorem of sorts
    ∆v = √[(v1x+ v2x )^2+ (v1y+ v2y )^2]
    Then substitute ‘∆v’ into:
    ∆p = m(∆v)

    can you please explain to me why this method doesn't work for the following sample question, though.. I'm having trouble understanding why not.

    sample Q:

    when I try the pythagorean theorem to solve for ∆v.. to then in turn solve for impulse.. I get
    Δv = 39.99
    (multiplied by the mass which is 0.170kg)
    Impulse = 6.799
    (divided by time which is 0.1)
    Force = 67.99

    and you can see the sample question says that force is 34 N which is almost perfectly half of what the pythagorean theorem method gives me.. what am I doing wrong here or not understanding?

    heres a link to the original thread with the original question that this method did seem to work for:


    the question for parts a) and b) in the thread ^ are worded just like in the text
    (just in case its a wording issue between the two questions and a difference in what we're trying to solve for)
    I would GREATLY appreciate any insight on this.. I've been struggling to figure this out ALL DAY:yuck::cry:
  20. Mar 20, 2012 #19
    In your second question, the board is apply ONLY a horizontal force to the puck. Therefore, there is no need for you to calculate the force at an angle, because you only need the horizontal component.
  21. Mar 20, 2012 #20
    tal444, that is true.

    agent_509 showed me that the equation I wrote in post #18 is off
    the equation we want to use is:

    ∆v = √[(v2x-v1x )^2+ (v2y- v1y )^2]
    Then substitute ‘∆v’ into:

    ∆p = m(∆v)
    Impulse = (mass)(change in velocity)
    Then calculate force using:
    Fnet = ∆p/∆t
    Force = (Impulse)/(change in time)

    this works for both questions.:devil:

    the fact that my wrong equation worked for question 1(the one at the beginning of this thread) was just a fluke
  22. May 13, 2012 #21
    so you just draw the component diagrams and then do what was just said?
    there still seems to be more work done to find net force though, seeing as part
    b is only worth 2 marks it's confusing..
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