Final answer:
The center of mass of a shell exploding in mid-air will continue to follow a parabolic path, in accordance with the conservation of momentum principle. The internal forces of the explosion have no effect on the overall trajectory, which is governed by external forces like gravity.
"The correct option is approximately option A"
Step-by-step explanation:
When a shell moving along a parabolic path explodes, the center of mass of the fragments will follow a parabolic path. This phenomenon is due to the conservation of momentum, which ensures that even after an explosion, the collective mass of the fragments behaves as if it were still a single object. Before the explosion, the shell follows a parabolic trajectory due to the forces applied during the launch phase, but once airborne and no longer propelled, it's in free fall and influenced solely by gravity. Despite breaking apart into pieces, each with its own velocity and trajectory, the fragments' combined center of mass continues along the same parabolic path the un-exploded shell was following.
The trajectory of the center of mass is not affected by the internal forces of the explosion as they are equal and opposite, canceling each other out. It's only the external forces, like gravity, that dictate the center of mass's path. Should air resistance be negligible, the center of mass will continue on a predictable trajectory; however, if air resistance were significant, it would alter the path of the individual pieces but not the center of mass of the system.
Thus, the correct answer to the question is that the center of mass will follow a parabolic path after the explosion, due to the principles of conservation of momentum and projectile motion.