Newton's second law in momentum form is simplest for a closed material system whose matter does not cross the system boundary. A rocket, snowplow, leaking cart, or similar object is different: the chosen body's mass changes because matter enters or leaves it.

For such a variable-mass system, momentum carried across the boundary must be included explicitly.

Let the body have mass $m$ and velocity $\mathbf v$. During a short interval $dt$, let its mass change by $dm$. Use the sign convention

• $dm>0$ for mass entering the body;
• $dm<0$ for mass leaving the body.

Let the crossing mass have velocity $\mathbf v_c$ in the chosen inertial frame. Its velocity relative to the body is

$$\mathbf u=\mathbf v_c-\mathbf v.$$

Applying momentum conservation to the body together with the small amount of transferred mass and retaining only first-order differential terms gives

$$\boxed{m\frac{d\mathbf v}{dt}=\mathbf F_{\rm ext}+\mathbf u\frac{dm}{dt}}.$$

The term

$$\mathbf u\frac{dm}{dt}$$

is the momentum transferred by the mass flow relative to the body.

Why $d(m\mathbf v)/dt$ is not enough for the changing body

For a fixed set of particles,

$$\mathbf F_{\rm ext}=\frac{d\mathbf P}{dt}.$$

But the selected variable-mass body is not a fixed set of particles. Matter crossing its boundary carries momentum with it. Writing only

$$\mathbf F_{\rm ext}=\frac{d}{dt}(m\mathbf v)$$

for the changing body generally mixes the body's momentum change with the momentum flux across the boundary and gives the wrong equation of motion.

The closed system consisting of the body plus all transferred matter still obeys ordinary momentum conservation.

Example: a snowplow collecting stationary snow

Suppose a snowplow moves horizontally with speed $v$ while continuously collecting snow that was initially at rest relative to the ground. Ignore external horizontal forces.

The incoming snow has ground velocity

$$v_c=0,$$

so relative to the moving plow,

$$u=v_c-v=-v.$$

Because the plow gains mass,

$$\frac{dm}{dt}>0.$$

The variable-mass equation becomes

$$m\frac{dv}{dt}=-v\frac{dm}{dt}.$$

Rearranging,

$$m,dv+v,dm=0,$$

or

$$d(mv)=0.$$

Therefore

$$mv=\text{constant}.$$

If the plow's total mass doubles while no external horizontal impulse acts, its speed falls to half its initial value.

Mass loss and thrust

If a vehicle ejects mass backward, then both $\mathbf u$ and $dm/dt$ are negative along the forward axis. Their product points forward, producing thrust. The force does not arise because momentum is created; the expelled mass carries backward momentum while the vehicle gains forward momentum.

For several independent inflows and outflows, each stream contributes its own relative-velocity momentum-flow term. The essential principle is always the same: once mass crosses the chosen boundary, momentum flux must be included in the balance.