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Homework Help: Inelastic Collision Question

  1. Nov 9, 2008 #1
    Hi everyone!

    I've been on the question for a while and I'm not quite sure where I'm going wrong. If someone could help me out, I'd really appreciate it!

    Problem
    Two ice-fishermen are driving trucks across a frozen and frictionless lake. Truck 1 (m1 = 1190kg) is traveling with a speed of 13.2m/s in a direction of 43.0degrees SOUTH of EAST. Truck 2 (m2 = 1000kg) is traveling due NORTH with a speed of 20.1m/s. The trucks collide and lock into a single unit. What is the speed of the joined trucks immediately after the collision?


    2. Relevant equations
    equation for inelastic collision...

    Vf = (m1V1i + m2V2i) / (m1 + m2)



    3. The attempt at a solution
    I have tried it a couple of ways so far, and neither have worked.


    First, I tried just plugging in the velocities without using the angle.

    Vf = (1190kg*13.2m/s + 1000kg*20.1m/s) / (1190kg + 1000kg)
    = 16.4 m/s (wrong answer!!)


    Then I tried incorporating the y-coordinate angle however I'm not sure if I did this properly.

    Vf = (1190kg*13.2m/s(sin43) + 1000kg*20.1m/s) / (1190kg + 1000kg)
    = 14.1 m/s (wrong again!!)




    If you could offer me any help, it would be greatly appreciated!
    :)
     
  2. jcsd
  3. Nov 9, 2008 #2

    LowlyPion

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    Welcome to PF.

    Keep in mind that momentum is a vector. When they join won't the result be a vector addition?

    The equation you are using would be appropriate if the motion was all in 1 direction.
     
  4. Nov 9, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi jessguy! Welcome to PF! :smile:

    Momentum is a vector, so you must use vector addition.

    Add the x-coordinates to get Vx, then the y-coordinates to get Vy (that's easy in this case!), and use Pythagoras' theorem to find the total speed, V. :smile:
     
  5. Nov 9, 2008 #4
    hm.. ok. so is the mass not needed for this part of the question? (there is a second part to this question, you see)

    This is what I've tried after receiving your feedback..

    Vx = 13.2m/s*cos43 = 9.65 m/s

    Vy = 20.1m/s - 13.2m/s*sin43 = 29.1 m/s

    Vf = sqrt((9.65m/s)^2 + (29.1m/s)^2)
    = 14.7 m/s

    Not the correct answer, once again.

    I understand that I need to use the x and y components of velocities. Am I suppose to use the masses and if so, how?


    thanks again
     
  6. Nov 9, 2008 #5

    tiny-tim

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    Yes … use your original equation, Vf = (m1V1i + m2V2i) / (m1 + m2).

    This is a vector equation, but like all vector equations, it also works for coordinates in one direction. :smile:
     
  7. Nov 9, 2008 #6
    ooook! perfect! I got the answer afterwards!


    Thanks so much :)
     
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