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Momentum Projectile problem

  1. Dec 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A 9.3 kg firework is launched straight up and at its maximum height 45 m it explodes into three parts. Part A (0.5 kg) moves straight down and lands 0.29 seconds after the explosion. Part B (1 kg) moves horizontally to the right and lands 10 meters from Part A. Part C moves to the left at some angle. How far from Part A does Part C land (no direction needed)?

    2. Relevant equations

    3. The attempt at a solution
    can you check my work
    1. Momentum for part A:
    y = (Vo)t - (1/2)(9.8m/s^2)t^2
    Substitute the values for y and t which are given. Solve for Vo, the initial downward velocity. Multiply the value of Vo by 0.5 kg to get the initial momentum down of A.
    momentum A = 78.2967
    2. Momentum for part B:
    y = (1/2)gt^2 = 45m for its downward motion.
    substitute the value of g and solve for t.
    Vx = 10/t = 10/3.03 = 3.3m/s
    momentum B = (1kg) (Vx) = 3.3

    3. Resultant of the two momentums:
    Did a vector addition of the momentums of A and B. The resultant will be a vector downward to the right. Determine both, magnitude and direction.
    4. Momentum of part C:
    That is opposite (equilibrant) of of the vector sum of A and B.
    momentum C =(momA ^2 + momB ^2) ^0.5 = 78.3662
    Divide the momentum of C by its mass to get the initial velocity of C.
    Voc = 10.046m/s
    i am stuck at this point please help thanks
  2. jcsd
  3. Dec 22, 2008 #2


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    Hi myoplex11! :smile:

    Very good technique so far. :biggrin:

    Now the trick is to split C's velocity into horizontal and vertical components :wink:

    (so you find t from the vertical one, and use that to find the horizontal distance)
  4. Dec 22, 2008 #3
    But how can i split C's velocity i dont know the angle?
  5. Dec 22, 2008 #4


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    uhhh? … mCViC = -mAViA - mBViB. :smile:
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