momentum imparted by decouplers=velocity?

[Andras]

How much momentum is imparted by decouplers when a section is detached?

If the decoupler is rated 25 and the item detached weighs 1.5, how fast will it be going and how can I calculate that?

Thanks

[thegreatpeon]

If it was simply a question of momentum then the answer would be m₁v₁² = m₂v₂²

Where m₁ = initial mass of the ship

m₂ = final mass of the ship

v₁ = initial velocity

v₂ = final velocity

However, moduals do not simply stop existing. They have some momentum of their own that they keep and they have an ejection force of 10.

I would assume that the ejection force pushes the two parts away from each other with equal force in either direction which would lead to 5 units of acceleration for the remaining ship.

So I think that the final speed of the ship would be v₁ + 5x

And the final speed of the decoupler would be v₁ - 5x

Where x is the unknown acceleration from the ejection on the part of the ship it's attached to.

[rdfox]

Decouplers do not, as of yet, apply force to the part of the stack that contains the command pod. This will change eventually, but Harv's said it's fairly low on the priority list.

[Entroper]

It's clear that the decouplers impart an impulse rather than a velocity, because heavy stages separate more slowly than light stages. My best guess would be that the number is an impulse value in kg*m/s (or N*s). Or tonnes*m/s and kN*s, if you consider the base unit of mass in KSP to be tonnes instead of kg.

[TimothyC]

If it was simply a question of momentum then the answer would be m₁v₁² = m₂v₂²

Where m₁ = initial mass of the ship

m₂ = final mass of the ship

v₁ = initial velocity

v₂ = final velocity

It would be a bit more complicated than that from a dynamics point of view as we\'d have to also look at the KE imparted to both objects, but the equation should be m₁v₁= m₂v₂. When you square the velocity you go go from impulse to energy (Force*Time as opposed to Force*Distance).