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Exploding Shell Momentum

  1. Nov 20, 2009 #1
    *SOLVED*

    1. The problem statement, all variables and given/known data

    A 75 kg shell is fired with an initial speed of at an angle 55 degrees above horizontal. Air resistance is negligible. At its highest point, the shell explodes into two fragments, one four times more massive than the other. The heavier fragment lands directly below the point of the explosion.

    If the explosion exerts forces only in the horizontal direction, how far from the launch point does the lighter fragment land?

    2. Relevant equations

    To calculate maximum height of a projectile (without air resistance):
    h = (v * sin(x)2) / 19.6

    Kinematics Equation:
    x - xo = vot + 0.5at2

    Momentum:
    P = mv

    3. The attempt at a solution

    This problem took me a long time and a lot of careful writing, and when I got the wrong answer my head almost exploded. Here was my attempt.

    To find the mass of each shell:
    75/5 = 15kg (mass of smaller fragment, m1)
    15*4 = 60kg (mass of larger fragment, m2)

    Then I found the momentum of the larger fragment:
    P1 = (60)(-125) = -7500kg*m/s
    The momentum of the smaller fragment is equal and opposite of this, because of conservation of momentum. Now we can calculate the change in velocity (dv) of the smaller fragment:
    P2 = (15)(dv) = 7500kg*m/s
    7500/15 = dv = 500m/s

    To find the total horizontal velocity of the smaller fragment we add the dv to vox (the initial horizontal velocity):
    500 + 125cos(55) = 571.70m/s

    Now we have to find how long the smaller fragment will free fall before it hits the ground, so we need to calculate its max height (h):
    h = (125sin(55))2/19.6 = 534.93m
    And then we can use this and the kinematics equation to find the time the fragment takes to hit the ground from its max height:
    0 - 534.93 = 0t + 0.5(-9.8)t2
    t = 10.45s

    Now that we know how long the smaller fragment is in the air after the explosion, we can find the horizontal distance it travels during this interval:
    571.70m/s * 10.45s = 5974.27m

    Now we need to find out how far the shell traveled horizontally before the explosion so we can add that to the distance traveled after the explosion to get the total horizontal distance traveled of the smaller shard. Since the time it takes for the shard to fall from the max height is the same as the time it took to reach the max height:
    71.70m/s * 10.45s = 749.23m

    When I add the two intervals together I get:
    5974.27m + 749.23m = 6723.50m

    Where did I go wrong?
     
    Last edited: Nov 20, 2009
  2. jcsd
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