Consider a body of mass $m$ moving in space and not interacting with anything else. The body is moving at a constant velocity and now it starts shedding off its parts. Eventually its mass will decrease to $m'$ and in order to conserve its linear momentum, its velocity must increase.

Here, there was a change in velocity, meaning there was an acceleration without any force acting on the object. Is my explanation correct?

  • $\begingroup$ One case would be when describing the system from a non-inertial reference frame, where even without actual forces acting on the particles one has additional contributions due to the transformations laws between inertial vs non-inertial reference frames. $\endgroup$
    – gented
    Dec 28 '16 at 14:15
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    $\begingroup$ What do you mean by "it starts shedding off"? How is it going to "shed" anything without exerting forces? $F=ma$ holds in all inertial frames, so there cannot be acceleration without force. It's not clear to me what exactly you imagine happening here - why does conservation of momentum mean it must speed up - the parts it "sheds" don't vanish from the world, they still have momentum. $\endgroup$
    – ACuriousMind
    Dec 28 '16 at 14:18
  • $\begingroup$ @ACuriousMind The parts can be shed with no net force. For example, two parts "launched" from diametrically opposed positions (diametrically opposed momenta, really). But the important point you made was that the parts still have momentum. $\endgroup$
    – garyp
    Dec 28 '16 at 15:25
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    $\begingroup$ Many people are correctly pointing out that you have to consider all of the mass of the original system. This is because Newton's Laws are valid only for constant mass systems. You cannot focus on the remaining piece of the body and conclude that its velocity must increase. In that case you are considering a system with a changing mass, and Newton's Laws are not valid. $\endgroup$
    – garyp
    Dec 28 '16 at 15:28
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    $\begingroup$ Probably related: physics.stackexchange.com/q/197587 and physics.stackexchange.com/q/153361 (though the latter is backwards to your model: it adds mass to the moving object) $\endgroup$
    – Kyle Kanos
    Dec 29 '16 at 11:07

Conservation of momentum applies to the entire system. So we must consider not only the "body" in your example, but also the "parts" it sheds.

The velocity of the "body" after shedding its parts will depend on the manner in which the parts are shed.

For example, if the parts merely separated from the body, the velocity of the "body" and the "parts" can remain unchanged, and given the total mass is the same, the total momentum remains unchanged also.

If on the other hand the parts are ejected from the body, then this will have been due to an ejection force (that applies to the body and the parts in equal and opposite directions as per Newton's 3rd law) which will influence the velocities of the body and the parts according to Newton's second law, $F=ma$.

  • $\begingroup$ what if a spaceship is collecting interplanetary dust(initially at rest) and its(spaceship's mass) is increasing with time? $\endgroup$
    – Sahil
    Dec 28 '16 at 14:37
  • $\begingroup$ @ManishYadav Likely as the spaceship collides with the dust, this collision will generate a force on the ship, and an equal and opposite force on the dust. This force will change the velocity of the ship and the dust (according to Newton's second law). The total momentum of the ship dust system will be the same before and after the ship collided and "gathered" the dust. $\endgroup$
    – Kenshin
    Dec 28 '16 at 14:39
  • $\begingroup$ Here can we say that no force acts on the system, but there lies an internal contact force? $\endgroup$
    – Sahil
    Dec 28 '16 at 14:42
  • $\begingroup$ Here can we say that no force acts on the system, but there lies a pair of an internal contact force? $\endgroup$
    – Sahil
    Dec 28 '16 at 14:43
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    $\begingroup$ @ManishYadav If you want the system to be defined as the ship, then a scenario where it is sweeping up dust in it's path is not a scenario without external interaction: the dust is external to the system. If you define the system as including the dust then the interaction is internal and the momentum of the system doesn't change but that doesn't preclude the ship (this time only a part of the system) from changing speed. You can't outwit the conservation of momentum (at least without getting very sneaky with general relativity). $\endgroup$ Dec 28 '16 at 15:14

As I don't know how the body starts shedding off its parts,let us consider a bomb moving at a constant velocity in space. Suddenly the bomb explodes and one part of the bomb goes in one direction ,other part goes in a different direction.

Is it against Newton's laws?

Ofcourse not ,the velocities of the parts have changed because of chemical energy (or nuclear energy) of the bomb,in other words there are forces on the individual parts ,thats why velocities of the parts have changed.

But does the velocity of the bomb as a whole change?

No,if you calculate momentums of all the parts and add them vectorially ,you will get the same momentum as before . This is because there are no external forces on the bomb ,the force due to chemical or nuclear energy was internal.


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