An artillery shell is fired on flat ground at an initial angle of theta above th

Question

An artillery shell is fired on flat ground at an initial angle of theta above the horizontal. If we neglect air resistance then we know that the projectile will follow a parabolic trajectory. Suppose that the projectile is fired in such a way as to hit a target a distance d away. Unfortunately the shell explodes into two pieces of equal mass in midair. The two pieces land at the same time with one piece landing a distance d beyond the target. Where does the other piece land? At the uppermost part of the trajectory the shell bursts into two equal fragments with one piece going directly upward, relative to the ground, with an initial speed v. What is the velocity of the other fragment, relative to the ground?

Explanation / Answer

the trajectory of the center of mass for the system is going to follow the trajectory of the projectile before it exploded. Thus, the CM of the system is at a distance of dm from the origin.

he equation for the center of mass would be d=mx(centre of mass)=(m/2)x1+(m/2)x2

m*d=m/2*(2*d)+(m/2)*x2

x2=0

so second fragment will land at x=0 ( at the origin)

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