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2-dimensional motion question

  1. Jul 23, 2015 #1
    1. The problem statement, all variables and given/known data
    A plane with an initial velocity of (225m/s ŷ + 325m/s Ẑ) intercepts a 20kg piece of luggage with initial velocity of (-75m/s ŷ - 15m/s Ẑ). After, plane is moving at speed 400 m/s. What is the mass of the plane?

    2. Relevant equations
    momentum y: Mplane(225m/s) + 20kg(-75 m/s) = Mplane(Vy) + 20kg(Vy)
    momentum z: Mplane(325m/s) + 20kg(-15m/s) = Mplane(Vz) +20kg(Vz)
    V = Vy^2 + Vz^2 , Tanθ= Vz/Vy

    3. The attempt at a solution
    I know the speed given after the interception is in magnitude form and it needs to be in component form and the speed of the plane and luggage are the same after.
    For the plane I get V=395.3m/s @ 55.3°
    For the luggage I get V=76.5m/s @ 11.3° (assuming this means the Luggage starts in Q1 and moves on the 11.3° angle into Q3?)
    I tried doing conservation of momentum with plugging in 400 in the final velocities but I realized this does not make sense.
     
  2. jcsd
  3. Jul 23, 2015 #2

    Orodruin

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    What is the momentum of the plane+luggage after the collision? How does this relate to the speed of the plane+luggage after the collision?

    Careful with nomenclature here, speed cannot be given on component form as it is a scalar! It is the magnitude of the velocity.
     
  4. Jul 23, 2015 #3
    Express in terms of the magnitudes and angles and solve.
     
  5. Jul 23, 2015 #4
    This is where I seem to be lost. This is for a summer class and I haven't had much time/practice with this sort of problem.
     
  6. Jul 23, 2015 #5
    Well I know the momentum after is Mplane(Vy) + 20kg(Vy) & Mplane(Vz) +20kg(Vz)
    If I knew the angle of the 400m/s I could use sin and cosine to get the components. This is where I am lost
     
  7. Jul 23, 2015 #6

    Orodruin

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    The angle is irrelevant (although you can compute it from the angle of the initial momentum). You only need to find the magnitude of the momentum.
     
  8. Jul 23, 2015 #7
    So : Mplane(395.3m/s) + 20kg(76.5m/s) = Mplane(400m/s) + 20kg(400m/s) ?
     
  9. Jul 23, 2015 #8

    Orodruin

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    No, you need to square the total momentum before the impact! A suggestion is: Do not calculate the direction of the momenta of the separate entities, it has no bearing on the problem - only the total momentum is of relevance.
     
  10. Jul 23, 2015 #9
    We have never squared momentum in any case in class..and isn't calculating the momenta of separate objects crucial to finding the total momentum? I am not grasping your advice, could you dumb it down anymore?
     
  11. Jul 23, 2015 #10

    Orodruin

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    How else do you usually find the magnitude of a vector other than adding the squares of the components, summing, and taking the square root of the result? You do not need to compute the magnitude of the individual momenta, you already have the momenta. What you need to do is to use vector addition to find the total momentum and you already have the vector components so the operation is trivial.
     
  12. Jul 23, 2015 #11
    Could you show me the first step or two? Maybe that will help.
     
  13. Jul 23, 2015 #12

    SammyS

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    If the plane is actually intercepting the luggage, i.e. the plane captures the luggage so that it then becomes part of the plane's cargo, then is problem has no solution. The initial speed of the plane is less than 400 m/s. the initial velocity of the luggage is largely opposite that of the plane. The plane will slow down.

    Maybe there is a language translation problem here.
     
  14. Jul 23, 2015 #13
    Lets say hypothetically it does work. What would the plan of attack be?
     
    Last edited: Jul 23, 2015
  15. Jul 23, 2015 #14

    haruspex

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    Just work with the equations you posted in the OP. Forget the tan theta equation since that is the only one involving the angle and you don't care about that. This leaves you three equations with three unknowns.
    You need to manipulate the equations so as to eliminate vy and vz. Do you understand how to eliminate variables in simultaneous equations?
     
  16. Jul 23, 2015 #15

    SammyS

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    I see that you deleted most of what you posted when you did your Edit..

    It might be helpful for others to know that you altered the values of your homework problem to get the parameters here. Unfortunately, I was not able to find the those values cached by my browser. At any rate, in the homework problem:
    The initial speed of the plane is in excess of 400 m/s.
    As I recall, the final speed was 400 m/s.
    The mass of the luggage was something like 2000 kg.
    Etc.​

    For a problem such as this, you're not free to just pick these values at random.
     
  17. Jul 24, 2015 #16
    A space ship with an initial velocity of (300 m/s y + 400m/s z) captures a 2000kg asteroid moving with an initial velocity of (-100m/s y - 50m/s z.) After, the ship is moving at 498m/s. What is the mass of the spaceship?
     
  18. Jul 24, 2015 #17

    Orodruin

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    Can you think of some alternative way of finding this momentum? Let us say, a conservation law of some sort?
     
  19. Jul 24, 2015 #18
    I was thinking conservation of energy and using kinetic but kinetic energy is dissipated in this problem so I can't set kinetic initial equal to kinetic final
     
  20. Jul 24, 2015 #19

    Orodruin

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    So what other conservation laws do you know?
     
  21. Jul 24, 2015 #20
    I believe we have learned about momentum conservation, energy conservation, and angular momentum conservation.
     
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