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?
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, $F=ma$.
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.
