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Conservation of momentum problem

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data
    A 1200kg car travelling initially with a speed of 25m/s in an easterly direction crashes into the rear end of a 9000kg truck moving in the same direction at 20m/s. The velocity of the car right after the collision is 18m/s to the east.

    3.4.1) What is the velocity of the truck after the collision
    3.4.2) How much of mechanical energy is lost during the collision


    2. Relevant equations
    p=mv
    ek=1/2mv^2


    3. The attempt at a solution

    The velocity of the truck was worked out to be 21m/s. My problem is with the second question, this is an elastic collision right ?, so their should be no mechanical energy lost ?
     
  2. jcsd
  3. Jan 26, 2013 #2

    gneill

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    Ah, but presumably you used the given final velocity of the car to work out the final velocity of the truck using conservation of momentum, right? You didn't have to assume anything about the nature of the collision. Work out the kinetic energies before and after. What do you find?
     
  4. Jan 26, 2013 #3
    Ok, this is what I get

    ek before:
    ek= 1/2*1200*25^ + 1/2*9000*20^2
    = 2175000 J

    ek after:
    ek = 1/2*1200*18^2 + 1/2*9000*21^2
    = 2178900 J

    They are not the same, what is wrong ?
     
  5. Jan 26, 2013 #4

    gneill

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    First, you should use more digits for intermediate values, particularly if they are going to be squared later on -- precision errors will be greatly multiplied by rounding or truncating too soon; The speed of the truck after the collision is not exactly 21 m/s. Use several more digits for the speed, and only round results for presentation (submission).

    Second, they never said it was a perfectly elastic collision. (Nor, for that matter, did they say that it was a perfectly inelastic collision). So you can't say beforehand whether or not KE will be conserved.
     
  6. Jan 26, 2013 #5
    Ok, thanks for the help. I also have another problem which I completely dont know where to start, here it is:

    A stunt woman sitting on a tree branch wants to drop vertically onto a horse galloping under the tree. The horse gallops at a constant speed of 10m/s and the woman is initially 3m above the level of the saddle.

    2.5.1) Calculate the horizontal distance between the saddle and the tree branch when the woman makes her move.

    2.5.2) How long is she in the air ?

    I do not know where to start with this problem.
     
  7. Jan 26, 2013 #6

    cepheid

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    Next time post a separate thread for each problem and use the homework help template each time.

    Answer these questions:

    How much time is it going to take for the woman to fall 3 m? (Personally I wouldn't want to free fall 3 m if the landing was going to be on my behind).

    How much horizontal distance does the horse travel in the amount of time you calculated above?

    Therefore, how far ahead of the branch does the saddle have to be when she lets go?
     
  8. Jan 26, 2013 #7
    Ok this is what I got:
    I used the equation to find the time:
    delta y = vi*t+1/2*a*t^2 and got my answer to be 0.35s.

    I then used the equation d=st to get the horizontal distance of 3.5m.
    Are my answers correct ?
     
  9. Jan 26, 2013 #8

    cepheid

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    Hmm, your equation looks correct but your answer for the free fall time is wrong. Can you post your work? What values did you use for 'a' and delta y?
     
  10. Jan 27, 2013 #9
    For 'a' I used 9.8m/s and for delta y I used 3m.
     
  11. Jan 27, 2013 #10

    cepheid

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    You used the right values but got the wrong answer. So the problem is with your arithmetic, not with the physics. Unfortunately, you didn't post your calculation steps like I requested, so it's kind of hard for me to see what you did wrong or to correct it...

    So, you should have had

    3 m = vit + (1/2)(9.81 m/s2)t2

    What is vi, and how would you solve this equation for t?
     
  12. Jan 27, 2013 #11
    vi is 0, am I right ?, anyway this is how I solved it:

    3m = 0*t + (1/2)(9.8)t^2
    3m = 4.9t^2
    sqrt 3 = 4.9t
    1.732 = 4.9t
    t = 1.732/4.9
    = 0.35s

    I dont see anything wrong ?
     
  13. Jan 27, 2013 #12

    gneill

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    Why is the 4.9 excluded from the square-rooting operation???
     
  14. Jan 27, 2013 #13
    Sorry my silly mistake, so the answer should have been 0.78s and the distance 7.8m, right ?
     
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