# Motion of center of mass

I was reading about COM and forces and came upon this in my book.

If a projectle explodes in air in different paths,the path of the centre of mass remains unchanged.This is because during explosion no external force (except gravity ) acts on the COM.

My question is, even though the author realises that there is gravity acting on the particle yet he goes on to conclude that the path of COM remains unchanged. But I learned that path will change whenever there is an external unbalanced force.Here gravity acts but why has the author neglected its effect ? (or am I mistaken somewhere?)

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Because gravity was acting on the projectile before it exploded, it was already taken into account. It wasn't turned on at the time of the explosion.

The phrase "the path remains unchanged" is referring to the gravity-induced parabola that the object was on prior to the explosion, not to a straight line that it would have in the absence of gravity.

So the velocity will change due to the force of gravity, but the "path" will not in this case since it assumes the force to be there.

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so the COM being referred to here is the COM of the earth+particle system? – Karan Singh Sep 9 '15 at 20:20
No, it means the center of mass of the projectile mass. If it didn't explode, that path would be a parabola. After it explodes, it is the same parabola. If gravity were turned off, it would be a straight line. – BowlOfRed Sep 9 '15 at 20:21
Wouldn't air friction come into play? When the projectile explodes in air it will potentially create many tiny pieces which may have vastly different drag coefficients for each piece than when compared to the projectile's drag coefficient prior to exploding. Couldn't this change the COM's path? – jake mcgregor Sep 9 '15 at 21:29
If it looks like a Physics 101 question and it doesn't mention air resistance, then we're going to ignore it. Drag becomes a non-linear complication to the problem that is hard to model. In the limit of zero drag and uniform acceleration, the path is identical. Note the problem says "no external force (except gravity)". – BowlOfRed Sep 9 '15 at 21:43

The passage means that the center of mass follows the same parabolic trajectory it would have followed had there been no explosion. This includes the effect of gravity.

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