# On Conversation of Momentum in Variable-Mass Systems

I am wondering how you would exactly calculate the velocity change in an object that changes mass for whatever reason, such as a rocket leaving behind parts of itself. Is this a simple case of m1v1 = m2v2, thus leaving the momentum the same as mass changes, or is the derivation more complicated?

Like, another example, lets say, hypothetically, an object was flying and suddenly just lost a bunch of mass without deformation, by magic or something. Would the velocity just change to be equal to jv, where j is some factor in which m is decreased (1/j m * jv).

The reason I am confused is that the wikipedia page lists them as separate derivations, it does not explain that well. And also the fact that it means that a change in velocity would change mass, which does not seem reasonable, as we don't see this in reality as far as I am aware. It seems as if my interpretation of the Conservation of Momentum is somehow wrong.

Could you please correct me where I am mistaken and, if it exists, show a more accurate way to calculate the momentum of a system with variable mass?

Conservation of momentum would still hold (as long as there's no magic). Let's call the mass of the rocket $$m_R$$, the mass the ejected fuel $$m_f$$. The rocket and fuel combined is intially moving at speed $$v$$. Then, from conservation of momentum,
$$(m_R + m_f)v = m_f v_f + m_R v_R$$
You can see that the left side of the equation is the rocket and fuel moving together, and the right hand side is after they have split apart (the fuel is ejected). Let's say the rocket just leaves behind the fuel like you stated. This means $$v_f$$, the velocity of the fuel after splitting off, is zero, giving
$$v_R = \frac{(m_R + m_f)}{m_R}v$$
Because $$\frac{m_R + m_f}{m_R}$$ is greater than $$1$$, the rocket's final velocity is greater than what is was when it still had the fuel!