Applying momentum conservation when gravity acts during explosion

[G.Chandra]

Why do we apply law of conservation of linear momentum on a body that explodes in air when an external force, gravittional force- mg, is acting on it? The law says that the linear momentum is conserved in absence of the external forces.

 

[pardesi]

hmm that's because the force $$mg$$ into the time for which the explaosion happens $$dt=mgdt$$ is aver small qty so
$$P_{f}-P_{i}=mgdt \approx 0$$

 

[tiny-tim]

Welcome to PF!

Why do we apply law of conservation of linear momentum on a body that explodes in air when an external force, gravittional force- mg, is acting on it? The law says that only in absence of external forces the linear momentum remains conserved.

Hi G.Chandra! Welcome to PF!

Because an explosion, like a collision, is taken to be instantaneous.

So the force from gravity (which takes time!) is zero.

 

[pardesi]

So the force from gravity (which takes time!) is zero.

hmmmm i don't get this explanation of urs?

 

[G.Chandra]

This means that the law will not be applicable if duration of explosion is longer?

 

[pardesi]

no see it's an approximatiion!
so don't wry about the "law holding" or not
what's assumed is that the momentum change due to gravity is negligible as comapred to the initial momentum
say if the intial momentum was some 500 untis and the change sone 1 or 2 units then it shud hardly matter u
it's an approximation and works real well

 

[Andy Resnick]

Why do we apply law of conservation of linear momentum on a body that explodes in air when an external force, gravittional force- mg, is acting on it? The law says that the linear momentum is conserved in absence of the external forces.

That's the magic of representing extended objects as mass-points. The center of mass will follow the undisturbed trajectory (within reason- a mid-air collision is different than a 'simple explosion'), even though all the little pieces will tumble hither and yon.

 

[Redbelly98]

So the force from gravity (which takes time!) is zero.

A better way to say it might be that the impulse due to gravity,

$$F \Delta t = m \ g \ \Delta t$$

is very small since delta-t is small. Equating impulse with change in momentum, we can also say this does not affect the momentum during a collision or explosion.

 

[pardesi]

yeah perfect that's hwy i had questioned "tiny-tim"!