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Trajectory and conservation of linear momentum

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    While standing on frictionless ice, you (mass 65.0 kg ) toss a 4.50g rock with initial speed 12.0m/s . The rock is 15.2 m from you when it lands. Take the axis on the ice surface in the horizontal direction of the motion of the rock. Ignore the initial height of the toss.
    At what angle did you toss it?
    How fast are you moving? (cm/s)

    2. Relevant equations

    trajectory equation r=(vo²/g)sin2theta

    m1v1+m2v2=0

    3. The attempt at a solution

    I solved for theta, and got

    theta=(1/2) sin-1(gR/v²) but that's giving me sin-1(1.03)

    obviously something is wrong. please help!

    And for the 2nd part, would you just do m1v1+m2v2cos(theta from part a)=0 ??
     
  2. jcsd
  3. Feb 22, 2009 #2
    Ok, so I was looking at it wrong. The distance 15.2m is the distance from where the rock lands to where the guy slid during the time it was in the air. How do I find that distance without knowing the angle theta?
     
  4. Feb 22, 2009 #3

    Delphi51

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    Amazing how they can come up with new trajectory questions!
    I had to dig out my trig identity book to do this one . . . and it didn't work. Same problem you had with your range equation. There appears to be an inconsistency in the information given.
    I drew 12 at an angle of A and found the initial velocities to be 12cos(A) horizontal and 12*sin(A) vertical. Considering t to be the time to maximum height,
    HORIZONTAL: d = vt -> 15.2/2 = 12*cos(A)*t -> t = 0.633/cos(A)
    VERTICAL: V = Vi + at -> 0 = 12*sin(A) - 9.81*t -> t = 1.22*sin(A)

    Equating the two expressions for t we have 1.22*sin(A) = 0.633/cos(A)
    so sin(A)*cos(A) = 0.5188
    The trig book says sin(A)*cos(A) = 0.5*sin(2A) so
    sin(2A) = 2*0.5188
    which has no solution.
     
  5. Feb 23, 2009 #4
    How come you did 15.2/2 in the horizontal equation?
     
  6. Feb 23, 2009 #5

    Delphi51

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    I was thinking the horiz distance to the point of maximum height is half the range.
    But I didn't have your insight about the distance including the person's slide, so back to the drawing board!

    Perhaps you could begin with the range being 15.2 - x, where x is the slide distance.
    Maybe write x as Vs*t (time of flight) and use conservation of momentum to get another equation with a Vs in it, related to the 12*cos(A).

    Good luck!
     
  7. Feb 23, 2009 #6
    Already did :wink: and........I actually came up with the same answer :mad:

    Spent like 15-20 min working through it, too.

    edit: conceptually, how far would a person slide throwing something that weighs .0045kg? Not very far at all, I'm imagining. Especially when it's not completely horizontal.
     
  8. Feb 23, 2009 #7
  9. Feb 23, 2009 #8

    Delphi51

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    Yes, I see the problem. The slide speed is 0.0008308*cos(A) and that is too small to matter. So, the problem as stated cannot be solved. Perhaps your marker would be okay with you changing the 15.2 to 10, which would certainly make it solvable.
    You could begin with the momentum bit:
    mv = mv
    65Vs = .0045*12*cos(A)
    which shows that you can ignore the distance of the slide (insignificant in 3 digit accuracy).
     
  10. Feb 23, 2009 #9
    When you do find the angle, how would you do the second part?

    m1v1=m2v2, or do you need to multiply it by the angle at all?

    edit: even though my problem says 4.5g, and the other thread has 4.5KG, i solved it the same way and put in the same answer. It turns out the actual answer was 37.7 degrees. Oh well.
     
  11. Feb 23, 2009 #10

    Delphi51

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    Oh, the 4.5 kg instead of grams would make a huge difference! It would make the problem solvable!

    Back to including the slide. The range equation is
    Eq1: 15.2 - Vs*t = 12^2/g*sin(2A) (t is the time of flight in all this)
    Conservation of momentum gives
    Eq2: 65Vs = 4.5*v*cos(A) or Vs = 0.8308*cos(A)
    Vertical velocity at the turnover point is 0 = Vi + at = 12*sin(A) - gt/2 or
    Eq3: t = 2.45*sin(A)
    Sub eq3 into eq1 to get
    15.2 - 2.45*Vs*sin(A) = 14.7*sin(2A)
    Sub eq2 in for Vs:
    15.2 - 2.45*0.8308*cos(A)*sin(A) = 14.7*sin(2A)
    15.2 - 2*sin(A)*cos(A) = 14.7*sin(2A) a bit of a roundoff here to get the 2
    15.2 - sin(2A) = 14.7*sin(2A)
    15.2 = 15.7*sin(2A)
    sin(2A) = 0.97
    A = 37.8 degrees

    To get the speed of the person, you would just use eqn2.
     
  12. Mar 3, 2009 #11
    if you don't mind me asking, how did you get 12^2/g*sin(2A) from .5at^2 for the first equation?
     
  13. Oct 21, 2011 #12
    That is some impressive math Delphi51. To simplify things, you can find the center of the mass of system. This gives where each object was located originally. Then you can find the actual distance the rock went and solve for theta using the range equation. the rest is plug and chug
     
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