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Collisions in two Dimensions

  1. Nov 18, 2006 #1
    Hi, I am having problems with a particular question it says:
    Two 30kg children, each with a speed of 4.0m/s, are sliding on a frictionless frozen pond when they collide and stick together because they have Velcro straps on their jackets. The two children then collide and stick to a 75kg man who is sliding at 2.0m/s. After the collision, the three-person composite is stationary. What is the angle between the initial velocity vectors of the two children?
    So what i did was you have the mass and velocity of both kids, they are both the same, and i found the total velocity of them using V=m1v1+m2v2/m1+m2.
    Then i used the mass and velocity and mass of the man and I used the new velocity I found and the combined mass (doubled the original mass) and this time solved for the angle using Tan(theta)= m2v2/m1v1 and took the inverse of that to find the angle.
    I am not sure if that was the correct angle to find or not, or if I even went about this problem the right way, if anyone could help me out it would be much appreciated. Thanks in advance

    Mike
     
  2. jcsd
  3. Nov 18, 2006 #2

    Doc Al

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    Staff: Mentor

    Careful. The kids have the same speed, but they move in different directions. You can't just add their momenta together, you need to add them like vectors.

    Here's a hint: Since, after the kids collide with the man, the total momentum is zero, what must be the combined momentum of the kids? Use that to figure out the angle between the kid's initial velocities.
     
  4. Nov 19, 2006 #3
    But you don't know the direction they are comming from, how would you figure that out?, because you find the angles at the end of the problem, that is what the question wants....
     
  5. Nov 19, 2006 #4

    Doc Al

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    You'll find that you don't need to know where they are coming from to find the angle the kids make with each other. Try this: Pretend the man is sliding along in the -x direction. What's his momentum? What must be the total momentum of the kids? (Before they collide.)
     
  6. Nov 19, 2006 #5
    So when finding the new velocity when the two kids collide, do I find the center of mass velocities in the x and y components and then find the magnitude of that to find their new velocity?, or do I need to use cosine and sine in my equations and find the velocity that way. If i did use cosine and sine I am not really sure how to go about that.
     
  7. Nov 19, 2006 #6

    radou

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    What center of mass do you mean? Isn't it stated that the two kids stick together when they collide? Or am I missing something?
     
  8. Nov 19, 2006 #7
    No that is right, they do stick together. I just don't really know what to do with this problem and I was just throwing out some ideas I came up with. Im pretty confused at the moment with it though.
     
  9. Nov 20, 2006 #8

    Doc Al

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    First things first: Answer the two (hopefully easy) questions that I asked in post #4. I'll repeat them here:

    (1) Pretend the man is sliding along in the -x direction. What's his momentum?

    (2) Given the above, what must be the momentum of the two kids (together) just before they collide with the man?​

    Once you answer these questions, you'll be in a better position to solve the problem.
     
  10. Nov 20, 2006 #9
    Well
    1. The mans momentum would be his mass(velocity)= -150 because he is moving along the negative x axis.
    2. The momentum of the 2 kids must have a momentum of 150, which will make them all stop once they collide.
     
  11. Nov 20, 2006 #10

    Doc Al

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    Right! Now you can get to work on the real problem:

    (1) What was the momentum of each kid (magnitude only) before they collided?

    (2) Since the total momentum of the two kids together has an x-component of 150, what must be the x-component of the momentum of each kid separately?

    (3) What angle from the x-axis must each kid have had to make their x-component equal to that value? (One kid will be above the x-axis, the other below it.) This requires finding the component using a little trig.

    (4) Now that you know the angle each kid made with the x-axis, what's the angle between them?
     
  12. Nov 20, 2006 #11
    1. Each kid would have a momentum on 120 each ( 30kg*4m/s)
    2. Would each kid have equal momentum?, meaning 75 each?
     
  13. Nov 21, 2006 #12

    Doc Al

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    Right.
    Exactly. Each kid would have an x-component of momentum equal to 75 kg-m/s.

    Keep going.
     
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