# Change of mass change in velocity [duplicate]

This may seem trivial, so my apologies, I've searched for a quantitative answer and couldn't find anything explicitly.

If a moving object has some velocity, and experiences a decrease in mass will the object lose velocity? My thoughts: Example, a cart is rolling and I take a box off the cart while it's in motion does the cart lose velocity due to the decrease in mass, or maybe it can be shown that the momentum decreases which equates to a decrease in velocity.

• @AndreiGeanta It depends on how the mass is lost. If you just grab boxes off, the speed will stay the same, not increase. – Chris Dec 22 '17 at 13:01
• If the boxes were lifted straight up such that they didn't decrease the velocity from the removal. – user221227 Dec 22 '17 at 13:04
• Can we use some math to solve this. So if I have mass A and B, with A=19units of mass being the cart say, and B=1 being a box. Let's say the sum is C, and it's moving at V= 1unit of speed. Forgetting any friction from removing the box, say it's lifted straight up so there is nothing imparted. Then we have (A+B)V=CV. Then by the conservation of momentum we get (A+B)V-B=CV-B. It's difficult to know how to go about mathematically solving this, I am not a physicist, nor have I taken any college lvl course in mechanics. – user221227 Dec 22 '17 at 13:05
• You need to specify much more carefully whether there is any force acting on the cart or not while it "loses mass" for this to be answerable. "Losing mass" in and of itself has nothing to do with gaining or losing velocity, all that matters is whether you apply a force to the cart in the process. – ACuriousMind Dec 22 '17 at 13:11
• In an ideal system with no forces applied to the cart, the carts velocity will remain unchanged? – user221227 Dec 22 '17 at 13:13

As Chris mentioned in his comment, it depends on how the mass is lost. Specifically, it depends on the velocity of the mass that is lost.

I'm assuming that the entire system, consisting initially of total mass $M$ and moving with a uniform velocity $\vec V$ is in a force-free region. For such a system, I can use Newton's second law extended to a system of particles (assuming all interactions within the system obey Newton's third law)

$$\frac{\mathrm d}{\mathrm dt} \vec P = 0$$

where $\vec P$ is the total linear momentum of the system. This says that the total linear momentum is conserved.

So if a piece of mass $m$ breaks off with velocity $\vec u$, then the new velocity of the remaining system is given by

$$\vec v = \frac{M\vec V -m\vec u}{M - m}$$

Another way to look at the above equation is to consider the system as composed of two parts $A$ and $B$: one part with mass $m_A = M-m$ and the other with mass $m_B = m$. Rearranging terms in the above equation,

$$m_A \vec v = m_A \vec V + \frac{m_B}{m_A}\left( \vec V - \vec u \right)$$

As evident from the above, the impulse on subsystem $A$ is given by

$$\Delta\vec P_A = -\eta \cdot \vec v_{rel}$$

where $\eta$ is the ratio of the mass of $B$ to the mass of $A$, and $\vec v_{rel}$ is the velocity of $B$ relative to $A$.

The above equation generalizes easily to situations where mass is continuously lost, and to situations where the net external force is nonzero; any good book on engineering mechanics (for example the one by Meriam and Kraige) will likely have a section on such variable mass systems.