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Energy released during explosion

  1. Feb 26, 2007 #1
    A projectile of mass 20.6 kg is fired at an angle of 59.0^\circ above the horizontal and with a speed of 76.0 m/s. At the highest point of its trajectory the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance.

    How much energy is released during the explosion?

    Conservation of momentum and Kinetic Energy Equation

    Change in KE=Kf-Ki

    So. I used conservation of momentum to find that the speed of the top fragment is 152 m/s. I plug it into the kinetic energy equations but my answer is not accepted on my hw site. I plugged in...
    Change in KE= (.5)(10.3)(152)^2-(.5)(20.6)(16)^2=59492.8 J which is wrong.
  2. jcsd
  3. Feb 26, 2007 #2


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    The velocities of 152 m/s and 16 m/s in your energy equation both look wrong.

    How did you get those numbers?
  4. Feb 26, 2007 #3
    [tex]\frac{1}{2} m1 v^2 = Energy Released + \frac{1}{2} m2 v^2 [/tex]
  5. Feb 26, 2007 #4
    m1 and m2 are two the two masses....
  6. Feb 26, 2007 #5


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    Yes that equation is OK. I was asking how you got the two velocities 152 and 16 to plug into the equation.
  7. Feb 26, 2007 #6
    Typo on my part. I wanted initial velocity to be 76 m/s (instead of 16).
    76 was given, and 152 I obtained through the conservation of momentum equation.

    (20.6 kg)(76 m/s)= (10.3 kg) (Velocity of fragment)
    Solving that equation yielded 152 m/s.
    Is something wrong about that?

    Assuming that velocity is correct and I use your equation sidrox, wouldn't I still get 59492.8 J since the velocity of the other particle is zero?
  8. Feb 26, 2007 #7


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    The projectile has a velocity of 76 m/s when it is launched. The explosion occurs when it is at its highest point.

    You need to find the velocity at the highest point first.

    Remember for a projectile the horizontal component of velocity is constant. At the highest point the vertical component of velocity is 0.

    After that mistake, what you did is correct.

    If the whole projectile has velocity V at the highest point, then the two parts will have velocities 0 and 2V. That is what you did, except you used V = 76 m/sec and that is the wrong value.
  9. Feb 26, 2007 #8
    Many thanks. I solved for it correctly and calculated that the speed as the highest point turns out to be 39.14. Then I just plugged it in and it worked. Thank you again.
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