Linear Conservation of Momentum

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To solve the problem of a shell exploding into two fragments, the initial horizontal momentum must be conserved. The shell has an initial horizontal velocity component calculated from its muzzle velocity and launch angle. After the explosion, one fragment has zero horizontal speed, allowing for the simplification of the momentum equation. The horizontal speed of the other fragment can be determined by setting the initial horizontal momentum equal to the momentum of the moving fragment. This application of conservation of momentum effectively reveals the horizontal speed of the second fragment.
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A 18-kg shell is fired from a gun with a muzzle velocity 185 m/s at 33o 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?
 
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Well, I believe that the way to solve this problem is by first finding the x-component of the initial velocity.

Since this is an explosion momentum problem:

m_Tv_i_x = m_1v_1_x+m_2v_2_x

And one of the terms on the right will cancel out since v_x=0 for one of them.
 
Use conservation of momentum.
 
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