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Kinetic Energy, Momentum, Projectile Motion ( A real stumper )

  1. Mar 3, 2007 #1
    1. The problem statement, all variables and given/known data
    From the text:

    "While standing on frictionless ice, you (mass 65.0 kg) toss a 4.50 kg rock with initial speed 12.0 m/s. If the rock is 15.2 m from you when it lands, (a) at what angle did you toss it? (b) How fast are you moving?"

    mass of me = 65.0 kg
    mass of rock = 4.50 kg
    initial velocity = 12.0 m/s
    initial height = ?
    initial angle = ?
    distance = 15.2 m
    speed of my body (recoil) = ?
    acceleration of gravity = 9.8 m/s^2
    time of flight = ?
    xo = 0
    vo = 0

    There is no energy lost to outside forces.


    2. Relevant equations

    These are simply equations i've tried toying with

    Newton's Second Law
    F= Ma

    Projectile Motion Disregarding time
    v^2 = vo^2 +2a(x-xo)

    Definition of work
    Wnet = Integral(F*dx)


    And these are ones i know i'll need later

    Conservation of Momentum
    m1(v1o) + m2(2o) = m1(v1f) + m2(v2f)

    Conservation of kinetic energy
    (1/2)m1*(v1o)^2 + (1/2)m2*(v2o)^2 = (1/2)m1*(v1f)^2 + (1/2)m2*(v2f)^2


    3. The attempt at a solution

    I have ideas, but none of them seem to work because i'm not given an angle or an initial height to work with (i'm actually supposed to find the initial angle). I just don't know where to begin on this problem. I know that in order to make that specific distance that initial velocity will need to be at a specific angle, and I have the notion that this is either going to be simple and i'm going to kick myself for overlooking it or it's going to be overly complicated... although the latter is less likely. Please help by giving me a concept to use, steer me in the right direction.
     
  2. jcsd
  3. Mar 3, 2007 #2

    arildno

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    Remember that the ratio between your horizontal velocity and the rock's horizontal velocity equals the negative reciprocal mass ratio, due to conservation of (horizontal) momentum.

    There being no horizontal forces acting upon either of you, the ratio between your travelled horizontal distance and the rock's travelled horizontal distance equals the same negative reciprocal mass ratio.

    Use this to determine the RANGE of the rock first.
    The range of the rock is the horizontal distance from the point it was thrown, to the point where it landed, relative to the ground.
     
    Last edited: Mar 3, 2007
  4. Mar 4, 2007 #3
    alright, i worked out the reciprocal mass ratio, but the range thing... is that not the distance i already gave? (it was given that it landed at 15.2 m) or are you speaking in terms of it being at the same height it was thrown at after a little fly time?

    i've thought of something - i think this problem may be unsolvable. i'm asked to give an initial angle, but without an initial height or a flight time that's impossible because it can fly to 15.2 m from different heights, which all would require different angles (within a range).
     
    Last edited: Mar 4, 2007
  5. Mar 4, 2007 #4

    Doc Al

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    The problem is quite solvable. Start by figuring out how long the rock was in the air (in terms of its initial speed and angle). Then use that to calculate:
    (1) the horizontal distance the rock traveled from the start point
    (2) the horizontal distance you traveled from the starting point

    Then you can apply the given data to solve for the angle.
     
  6. Mar 4, 2007 #5
    *sigh*

    my point is, the rock can travel 15.2meters being thrown at 12.0 m/s at different heights with a different angle at each height. flight time, launch height, and launch angle are all dependant on eachother, and none are given, so it's impossible to state that for this scenario there is one solveable answer for either part(since part b relies on part a - different launch angles will yield different horizontal momentums, obviously). i think the best i can do is find an equation that works with what i have and what i need, and input different variables to show that without the necessary information there are many answers for this problem.
     
  7. Mar 4, 2007 #6

    arildno

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    Okay, let's continue!

    Due to mass conservation, your velocity wrt. to the ground is:
    [tex]v_{you}=-\frac{4.5}{65.0}*v_{rock}[/tex]
    Your position at the time of impact relative to the origin is therefore:
    [tex]p_{you}=-\frac{4.5}{65.0}p_{rock}[/tex]
    The distance between you and the rock was given to be 15.2m, that is:
    [tex]p_{rock}-p_{you}=15.2\to{p}_{rock}=\frac{65.0}{4.5+65.0}*15.2m[/tex]

    [itex]p_{rock}[/itex] is the sought RANGE of the rock.
    Did you follow this?
     
  8. Mar 4, 2007 #7
    oh my, i feel like such a ditz. i'm sorry if got on your nerves... i'm always overlooking how things are worded. i imagined the distance being the distance the rock travelled, not including the displacement of my body:(

    *yikes*

    i'll rework it and let you guys know how it goes, lol...
     
  9. Mar 4, 2007 #8

    arildno

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    To give you a further hint:
    Remember that the SPEED of the rock contains both the horizontal&vertical velocity components of it!
    Thus, when you are given that the initial speed of the rock is 12m/s, you have to make one of the following interpretations:
    1. This is the speed the rock has relative to YOU. This results in nasty implicit expressions

    2. 12m/s is the initial speed the rock has relative to the GROUND. This makes your problem relatively easy to solve.

    So, choose interpretation 2.
    In this case, when solving the rock problem, you have:
    [tex]v_{0}=12.0,v_{x,0}=v_{0}\cos\theta_{0},v_{y,0}=v_{0}\sin\theta_{0}[/tex]
    where the initial velocity components are determined up to a sine or cosine of the sought launch angle [itex]\theta_{0}[/itex]
     
    Last edited: Mar 4, 2007
  10. Mar 4, 2007 #9
    Alrighty, I'm working on this problem as well. However, I'm running into a good bit of problem.

    I solved time to be [tex]t=\frac{12sin\Theta}{g}[/tex], and [tex]V_{0x}[/tex] to be [tex]12cos\theta [/tex]. But now, I'm running into a bit of a problem.

    When i try to plus these values into the equation [tex]x= x_0 + V_{0x} t[/tex], I wind up with [tex]11.6=cos\theta * sin \theta[/tex], and haven't the slightest clue where I went wrong or what to do to fix it. Any pointers would be greatly appriciated.
     
    Last edited: Mar 4, 2007
  11. Mar 4, 2007 #10
    Luke, i'm having the same problem too. I'm a little lost. I read through this whole post plenty of times and i still can't figure it out.
     
  12. Mar 4, 2007 #11

    Doc Al

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    Is that the total time that it's in the air?

    Hint: Review your trig identities; you can replace [tex]cos\theta * sin \theta[/tex] with a single trig function. (You are almost there! :wink: )
     
  13. Mar 4, 2007 #12
    I do believe it should be the total time in the air. I solved that from [tex]v_y = v_y0 - gt [/tex] , so...Yes, yes it should be the total time in the air.

    As for sin*cos, that can be rewritten as [tex]sin(2\theta)= 2sin\theta*cos\theta[/tex], yes? Maybe my algebra/trig is a bit rusty, but I keep getting domain errors and it is making my TI-89 cry...as well as myself.
     
  14. Mar 4, 2007 #13
    I guess I should write out what i'm doing so you can steer me in the right direction, haha. Because of the aforementioned identity, i multiplied both sides by 2- giving me [tex]23.6=2sin\theta*cos\theta[/tex], then rewrote it as [tex]23.6=sin(2\theta)[/tex]. From here, my first instinct would be to take the inverse sin of 23.6, but that is a domain error in a big way. SO...help? PS, thanks for the help so far.
     
  15. Mar 4, 2007 #14
    Luke, i am completely lost. Could you help me out from what you have so far???
     
  16. Mar 4, 2007 #15

    Doc Al

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    What's the final velocity?

    That's what you'll need.
     
  17. Mar 4, 2007 #16
    Final velocity...oh! because it's on frictionless ice, it should keep it's [tex]V_{0x}[/tex] value for its final velocity, shouldn't it?

    And i'm still hitting a brick wall with the trig.
     
  18. Mar 4, 2007 #17

    Doc Al

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    I'm still talking about [tex]V_y[/tex] that you used in calculating the time. (You made an error that I want you to find.)

    One step at a time.
     
  19. Mar 4, 2007 #18
    can any body help me with this problem. I can't find the angle. I understand how to solve for the distance the rock was thrown from the origin. I then solve for time, and i am stuck. any suggestions?
     
  20. Mar 4, 2007 #19
    Doc, thank you so much for your help. Unfortunatly, I keep staring at this and can't find my error. I sset it up so that [tex]V_y[/tex], the final Y component of the velocity, is zero. As such, [tex]0 = V_{0x} - gt[/tex], or [tex]0=12sin\theta - gt[/tex]. I then add the gt to the opposite side, divide by g to get my t value. I honestly haven't the slightest clue where I'm going wrong. A gentle nudge would be much appriciated.
     
  21. Mar 4, 2007 #20

    Doc Al

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    So, when you toss a ball in the air its speed is zero when it lands back into your hand? :wink:
     
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