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Real Problem

  1. Mar 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A 1250 kg car moving with a speed of 28.1 m/s collides head on with a railroad car of mass 2.5 * 10^4 kg moving with a speed of 3.00 m/s The car becomes attached to the railroad car in the collision IF the collison takes 0.952 s what was the average force on the car in N ?

    2. Relevant equations


    F avr. = pf - Pi / t
    3. The attempt at a solution
    I am fine with every thing, but my problem is calculation ??

    I did that and got 211.2 N

    Is that Right ?
     
  2. jcsd
  3. Mar 27, 2009 #2

    alphysicist

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    Hi remaan,

    That force seems rather small for the 1250kg car in this case. Can you show what numbers you used, and how you got that result?
     
  4. Mar 27, 2009 #3

    In fact, I can say that I used the same num. in the ques.

    and the law of cons. of Momentom.

    First, I found the velocity after the collison which equals to 4. 195

    and then I subtract Pf - Pi and / .592

    and then I got 57.16 * 10 ^ 3 What do you think ??
     
  5. Mar 27, 2009 #4
    Or, I am not sure of using the numbers. I mean I used the whole number without rounding ??
     
  6. Mar 27, 2009 #5

    djeitnstine

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    [tex]\Delta p = m \Delta v_f - m \Delta v_i[/tex] then the resulting momentum equation is [tex]m_1v_{i1} + m_2v_{i2} = m_1v_{f1} + m_2v_{f2}[/tex], and since the collision is inelastic the equation reduces to [tex]m_1v_{i1} + m_2v_{i2} = (m_1+ m_2)v[/tex] was this your approach?
     
  7. Mar 28, 2009 #6

    alphysicist

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    Sure, but how did you use those numbers? What equation did you set up, and how did you put the numbers in the equation? It appears to me that you used the numbers in the wrong way, because the velocity after the collision is not 4.195m/s. (The train car here should not speed up due to the collision.) It seems to me that you are perhaps using speed instead of velocity, but without knowing what you did I can't be sure.


    The time given in the original post was 0.952; was this just a typo? (Also, what were you using for Pf and Pi?)

     
  8. Mar 28, 2009 #7
    Yes, I used this equation.
     
  9. Mar 28, 2009 #8
    ok, let me show you what I did.
    m1v1+ m2v2 = V ( m1 + m2)
    I subs. the masses

    (1250 * 28.1 + 2.5 * 10 ^4 * 3 ) = V ( 1250+ 2.5 * 10^4 )

    And V = 4.195 ( after the collision )

    After that, I found the initial and final momentom.

    Pi = ( 1250 * 28.1 + 2.5 * 10^ 4 ) = 60125

    And Pf = (4.195 ) ( 1250 + 3 * 2.5 * 10 ^ 4 )
    And when I subtract them and divide by .592 I got 52.5 * 10 ^ 3

    I don't know why each time I do it I get different answer ??
     
  10. Mar 28, 2009 #9

    Redbelly98

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    Since the collisions is "head on", does that mean the velocities are in the same direction or opposite directions before colliding?

    Can you use + or - to indicate the direction of a velocity?
     
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