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Free falling momentum

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data
    A ball is falling freely under the action of gravity. When its downward speed is 8m/s, it explodes into two equal parts. One part goes straight up to a height of 12m above the explosion point. What is the velocity of the other part just after the explosion?


    2. Relevant equations
    initial = final
    P1 +P2 = P1 + P2


    3. The attempt at a solution
    i know that the other piece goes with a force downwards
    iam not sure how to start this
    i figure the straight line motion equation s= ut +1/2 at(squared) could be used?
    when the object reaches 12 meters, the velocity would be 0 as gravity would be decelerating it.
     
  2. jcsd
  3. Oct 2, 2007 #2

    Doc Al

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    Staff: Mentor

    You could, but you'll need another step since you don't have the time. Better would be to use another kinematic formula, one that relates velocity, distance, and acceleration.
    Good.
     
  4. Oct 2, 2007 #3
    iam sorry but i dont know of a kinetic formula that relates velocity, distance, and acceleration.
    could you please tell me it?
     
  5. Oct 2, 2007 #4

    Doc Al

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    Staff: Mentor

  6. Oct 2, 2007 #5
    okay so i used the a=x/t equaltion and got a time of 117.6s

    put that into the d=ut +1/2 at(squared) formula saying that acceleration was -9.8 and rearranged for u. i got an answer of 4.9

    i then timesed the time by 9.8 to get 1152.48 and added 4.9 on to it to get the total speed but this didn't work out when the actual answer is 31.3m/s downward
    where did i go wrong?
     
  7. Oct 2, 2007 #6

    Doc Al

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    Staff: Mentor

    That's not a valid equation. Where did you get it? Do a sanity check: how far would an object fall in 117.6 seconds?

    You can either find that formula that relates distance, velocity, and acceleration, or you can do it in two steps: Find the time it takes to fall 12m (which is the same as the time to rise to a max height of 12m), then use that to find the speed you would attain at the end of that 12m fall.

    Once you get the speed of the piece that shot upwards, you'd use conservation of momentum to find the speed of the other piece.
     
  8. Oct 2, 2007 #7
    haha yeah it was morning and i was very tired when i wrote that down :P

    right, so for me to find the velocity of the above object could i use the v(squared)=u(squared) + 2as formula?

    i did this saying that its final speed was zero. i then got an answer of 15.33 m/s

    with the conservation of momentum i divided m by all of the equation as m was in every part of mv + mv = mv + mv
    -then i ruled out the first final velocity because it is zero
    so the equation looked like this: 15.33 + 23.33 = v2f
    i said the second initial velocity was 23.33 as it was already traveling at 8m/s

    obviously this is wrong from the answer of 31.3m/s
    i did think of taking the 8m/s off the 15.33 and then adding it onto the 23.33 but got an answer of 30.66 which is still wrong
     
  9. Oct 3, 2007 #8

    Doc Al

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    Staff: Mentor

    Good. That's the speed of the upward-moving piece immediately after the explosion.

    In general, you'd better write that equation as: m1v1i + m2v2i = m1v1f + m2v2f,
    where initial and final refer to just before and just after the explosion.

    In this case, since they told you the pieces were identical, m1 = m2 and the m cancels:
    v1i + v2i = v1f + v2f

    You are given the initial velocity (v1i = v2i): 8 m/s down, so call it -8 m/s.

    You just figured out the final velocity of one piece (m1 say) as being 15.33 m/s up, so call it v1f = +15.33 m/s. Now you can solve for v2f.

    FYI: the velocity of m1 just after the explosion is 15.33, not 15.33 + 8 (or 15.33 - 8, for that matter!). Remember you figured out the velocity based on how high it reached, so that includes everything already.
     
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