A 26.0-kg shell is fired from a gun with a muzzle velocity 170.0 m/s at 68.0o...
A 26.0-kg shell is fired from a gun with a muzzle velocity 170.0 m/s at 68.0o 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?
Homework Answers
momentum before firing
Pix = m*V*costheta
after explosion final momentum Pfx = m/2*vx
from conservation of momentum
Pfx = Pix
m/2*vx = m*v*costheta
26/2*vx = 26*170*cos68
vx = 127.4 m/s
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A 26.0-kg shell is fired from a gun with a muzzle velocity 170.0
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