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So by the Law of Conservation of Momentum, we know that m1v1 = m2v2, which makes sense with momentum remaining constant, and is used for rockets and the like, changing mass by removing parts of itself.

But, this seems to break the Conservation of Energy (KE + PE = C), because, for instance, if you lose half of your mass, you would move double your previous velocity (and to those that are pedantic, yes, I know it refers to a body that is already in motion in a closed system). In the case of Momentum alone this works, but this would cause Kinetic Energy to increase by a multiple of 2, without pulling from Potential Energy as far as I can tell.

Let us give the example of a hypothetical rocket of 20kg, moving at 30 m/s. This would make for 9000 Joules. If removing a stage makes it 10kg, it would move at 60 m/s to conserve momentum, but in doing so would gain 9000 Joules of energy from seemingly nowhere, becoming 18,000 Joules.

So how does Conservation of Momentum not contradict the Conservation of Energy? What am I missing that allows these two seemingly contradictory laws to coexist?

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Your question have two missconceptions:

  1. This is not how the conservation of momentum works. The momentum is conserved for a system in the absence of external forces. If an object losses half his mass it does not mean that its velocity doubles. The other half of the mass is not lost, it does not dissapear. It takes (or may take) with it some of the initial momentum. So, the conservation of momentum will provide an equation like this: $$m\vec{v_0}=\frac{m}{2} \vec{v}_1+ \frac{m}{2} \vec{v}_2$$ where $\vec{v}_1$ and $\vec{v}_2$ are the velocities of the two halves. So the momentum is conserved for the system and not for just one of the halves. The momentum of each piece does not have to be conserved. Why is that? Because an event like "loosing" a piece involves some forces that will separate the pieces. These forces are internal to the system, so the momentum is conserved for the system. But the same forces are external to each individual piece so the momentum does not have to be conserved for them separately. The equation tells you only that $2\vec{v}_0= \vec{v}_1+\vec{v}_2$ but nothing about the individual values of the two final velocities. This is why in collision problems you need at least one more equation in order to determine the final velocities.

  2. The kinetic energy does not have to be conserved. Neither for individual pieces, neither for the system.There is no law of conservation of kinetic energy. Even in the absence of external forces the internal forces of the system can change its kinetic energy. What is conserved is the total energy (in absence of external forces) but not the KE alone. Take the case of a projectile initially at rest and then exploding into parts. There was no KE initially and there is a lot finally. Momentum though is conserved: zero initially and finally. So, there is no contradiction if the kinetic energy is not conserved. And don't forget that momentum is a vector.

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I think that this is due to the fact that the mass in your rocket will not detach and fly away in the opposite way by itself. Therefore what you have introduced in the system is an external force and you have to account for this in terms of work. Your momentum equation would still be correct, but you would have the following for kinetic energy (since there is no potential energy):

$ \frac{1}{2}m_1v_1^2 - \frac{1}{2}m_2v_2^2 = W $

And thus you get a system of decoupled equations, from which you get that $W = 9000J$.

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  • $\begingroup$ That would make sense. $\endgroup$
    – Zoey
    Feb 20 at 0:29

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