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Bomb explodes at rest, conservation of motion problem

  1. Jul 10, 2012 #1
    1. The problem statement, all variables and given/known data
    A bomb at rest with 1.0x10^4 J explodes into three pieces.
    The first piece is 1.0 kg and travels in the positive y direction with a velocity of 60 m/s.
    The second piece is 4.0 kg and travels in the positive x direction at 40 m/s.
    Find the third piece's mass, velocity, and angle.

    Diagram attached, but link just in case:
    http://www.flickr.com/photos/82417987@N08/7545812162/in/photostream

    2. Relevant equations
    momentum(p) = mass*velocity
    p(before) = p(after)
    Potential Energy = Kinetic Energy
    KE =1/2mv^2

    3. The attempt at a solution
    KE = 1/2*m*v^2
    1.0*10^4 = 1/2*(1)*(60)^2 + 1/2*(4)*(40)^2 + 1/2*(m)*(v)^2 <- where m and v are the unknown mass and velocity

    Apply conservation of momentum in the x-direction:
    The first piece is only in y, so it is not included in the equation. The x-mom eqn becomes:
    4.0kg*40m/s + m(ofx)*v(of x) = 0

    Now the y-mom equation:
    1.0kg*60m/s +m(of y)*v(of y) = 0

    I know: m(of x)=m(of y) because both are piece 3. And v^2 = v(of x)^2 + v(of y)^2.

    Not sure how to find v(of x) and v(of y)
     

    Attached Files:

  2. jcsd
  3. Jul 10, 2012 #2
    Solve the last two equations for vx and vy. Then plug into the energy conservation and solve for m.
     
  4. Jul 10, 2012 #3
    Okay, so
    v(of x) = -160/m
    v(of y) = -60/m
    1.0x10^4 J = (m)*[(160/m)^2+(60/m)^2]
    1.0x10^4 J = m*(25600/m^2 + 3600/m^2) = 29200/m
    1.0x10^4 J*m=29200
    m=2.92kg

    then do I use KE = mv^2 ?
    if so 1.0x10^4 J = (2.92kg)v^2
    therefore v = 58.5 m/s

    then v(of x) = -160/m = -160/(2.92 kg) = -54.8m m/s
    and v(of y) = -60/m = -60/(2.92 kg) = -20.5 m/s
    so theta = inverse tan(-20.5/-54.8) = 20.5 degrees

    So:
    Mass = 2.92 kg
    Velocity = 58.5 m/s
    Theta = 20.5 degrees

    Can someone verify that?
    Thanks
     
    Last edited: Jul 10, 2012
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