Why Is F = ma Not Always Valid When Mass Changes?

For both SL and HL, $F=ma$ is not generally valid when mass changes because it assumes constant mass. Newton's second law is more generally expressed using momentum: $F_{\text{ext}}=\frac{dp}{dt}$, with careful accounting for momentum carried into or out of the system.

The Reasoning

Linear momentum is defined as

$$p=mv.$$

The resultant external force is the rate of change of momentum:

$$F_{\text{ext}}=\frac{dp}{dt}.$$

If both $m$ and $v$ can change, differentiating $p=mv$ gives

$$F_{\text{ext}}=\frac{d(mv)}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}=ma+v\frac{dm}{dt}.$$

However, for an open system, the product-rule expression alone can be misleading because the entering or leaving material may have a velocity different from the main object's velocity. The safest method is to define a closed system that includes the transferred mass, or use a momentum-flux equation for a clearly defined control volume. The direction of every momentum transfer must be treated vectorially.

Therefore, $F=ma$ follows only when $\frac{dm}{dt}=0$. This is syllabus understanding A.2.10: Forces and momentum.

For example, suppose a cart moves at $4.0\,\text{m s}^{-1}$ while stationary sand falls into it at $0.50\,\text{kg s}^{-1}$. If its speed is kept constant, $a=0$, but a horizontal force is still required:

$$F=v\frac{dm}{dt}=(4.0)(0.50)=2.0\,\text{N}.$$

The force supplies momentum to the incoming sand. Using $F=ma$ would incorrectly give zero. Here the incoming sand initially has zero horizontal momentum. The applied force gives each kilogram of sand horizontal momentum at a rate equal to the mass flow rate multiplied by the cart speed.

A common misconception is that the extra term can always be applied directly to rockets. A rocket is an open system, so the momentum and velocity of the expelled exhaust relative to the rocket must also be included.

Exam Technique

For an Explain question, state that $F=ma$ assumes constant mass, then use $F_{\text{ext}}=\frac{dp}{dt}$ and discuss momentum entering or leaving the system. For calculations, define the system clearly before selecting an equation.