Trajectory of an exploding shell

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A 36-kg shell is fired at a muzzle velocity of 185 m/s at an angle of 59 degrees. At the peak of its trajectory, it explodes into two equal mass fragments, with one fragment coming to a stop and falling vertically. To find the horizontal speed of the other fragment, conservation of linear momentum must be applied, considering the masses and velocities involved. The horizontal component of velocity remains constant due to the absence of friction, while the vertical component is zero at the peak. Understanding these principles is crucial for solving the problem effectively.
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A 36-kg shell is fired from a gun with a muzzle velocity 185 m/s at 59 degrees above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. What is the horizontal speed of the other fragment?

Hint: Use conservation of linear momentum at the point of explosion.

My gut approach to this problem was that I could use the angle to find the vertical and horizontal components of velocity. If we assume no friction, the horizontal component of acceleration should remain constant. It is apparently the wrong answer. I don't really understand how to use conservation of linear momentum here? Any help is much appreciated! Thanks!
 
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Perhaps if you posted your working we could help you along. Further, try to use the hint regarding conservation of momentum rather than velocities. An important point here is that the shell splits into to pieces and therefore you have to take into account the masses of the pieces, not just the velocities.
 
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Welcome to PF!

Bizznaatch said:
My gut approach to this problem was that I could use the angle to find the vertical and horizontal components of velocity. If we assume no friction, the horizontal component of acceleration should remain constant. It is apparently the wrong answer. I don't really understand how to use conservation of linear momentum here? Any help is much appreciated! Thanks!

Hi Bizznaatch! Welcome to PF! :smile:

(btw, you mean the horizontal component of velocity should remain constant - the horizontal acceleration, of course, is zero)

Hint: What is the direction of the velocity at the top of the trajectory?

So what is its speed there? :smile:
 
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