Projectile Trajectory and Fragment Separation: Velocity, Angle, and Distance

  • Thread starter uwrfmike
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In summary, the object falls straight downward as though it had been released from rest. The object's highest point in its path is at the point where the two fragments break apart. The momentum of the object is conserved and it lands about 3/4 of the way from the gun.
  • #1
uwrfmike
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0
a projectile that is fired from a gun has an initial velocity of 90km/h at an angle of 60 degrees above the horizontal. when the projectile is at the top of its trajectory an internal explosion causes it to separate into two fragments of equal mass. one of the fragments falls straight downward as though it had been released from rest. how far from the gun does the other fragment land. I don't even know where to start. Thanks
 
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  • #2
Try to put forth some effort.

The questions employs two types of problems, a projectile motion problem and a concervation of momentum problem.

Try dividing up the initial launch velocity into X and Y components. From that, using the Y component of the velocity, determine the objects highest point in its path. We will assume the X velocity is constant up until the object breaks into. Using the X velocity, determine the objects momentum (you don't have a mass to use, but it really doesn't matter since the object breaks in half).
Using the concept of concervation of momentum, determine how fast half of the object must be going to concerve momentum (in the X direction), when the other half comes to a stop.
Now that you have the objects new X velocity and height, it is back to a projectile motion problem in order to find out where it lands.
 
  • #3
Can explosion be treated as a collision. I don't think so. Anyway some energy is being lost due to explosion. I think this question is meant for a bit more elementar level and such mistakes can be neglected.
 
  • #4
I think the only point of the explosion in this problem is to explain why the object separates in two, especially since we are given no extra information as to any energy added/lost in the explosion.

Look at it as two frames, the initial frame being the instant before the explosion and the final frame being the instant after.
In the intial frame, you have some mass, m, traveling at some velocity (all directed in the X direction), v1.
In the final frame, you now have two masses, each 1/2 m, one of which has no velocity, the other has some new velocity, v2, still direction totally in the X direction.
Momentum must be conserved, it isn't just a good idea, its the Law, so you need to figure out a relation ship between the new and old velocities by taking into account the change in mass of the particle(s) in question.
 

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