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Homework Help: Check answer for momentum prob

  1. Dec 12, 2003 #1
    A 130-KG tackler moving at 2.5 m/s meets head on(and tackles) a 90 k-g halfback moving at 5.0 m/s. What will be their mutual speed immediatley after the collison?

    someone please check my work

    [tex]m_{1}v_{1}+m_{2}v_{2}=v'(m_{1}+m_{2})[/tex]

    so [tex]v'=\frac{325+450}{220}[/tex]

    [tex]v'=3.52 m/s[/tex]
     
  2. jcsd
  3. Dec 12, 2003 #2

    Doc Al

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    Staff: Mentor

    Note that they meet head on. Direction matters!
     
  4. Dec 12, 2003 #3


    so the [tex]m_{2}v_{2)[/tex] becomes negative?

    i dont understand what u mean by "direction matters"
     
  5. Dec 12, 2003 #4

    Doc Al

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    Yes, [tex]m_{2}v_{2}[/tex] would be negative. If the two collide going the same direction (what you had originally plugged in) you get a totally different answer than if they collide going opposite directions (like in this problem). Momentum is a vector.

    In this problem, the motion is along a straight line. So no angles are involved. But the sign sure does matter!
     
  6. Dec 12, 2003 #5
    then i would get a negative v

    since

    v'=325-450/40=-3.125m/s
     
  7. Dec 12, 2003 #6

    Doc Al

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    You divided by the wrong number, but yes the answer will be negative. And what does that mean?
     
  8. Dec 12, 2003 #7
    yah i got the answer as -.57m/s. thnx
     
    Last edited: Dec 12, 2003
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