When solving problems on linear momentum, when can external forces be neglected?

Question

I was recently solving a problem in which one end of a massless string (in vertical orientation) was tied to a block of mass $2m$ and the other end to a ring of mass $m$, which was free to move along a horizontal rod. The block is then given a velocity $v$ (consider that this velocity is not caused by application of an external force). To calculate the velocity of the ring, we would have to apply momentum conservation. The problem is, momentum conservation would require net external force on the system to be zero, but in the solution I saw, the normal force exerted by the rod on the ring was neglected and so was the force of gravity. So, when exactly can external forces be neglected in problem-solving?

Answer 1

Remember that momentum and force are both vectors, and when we write Newton's second law to relate force to the rate of change of momentum this is a vector equation:

$$ (F_x, F_y, F_z) = \left( \frac{dp_x}{dt}, \frac{dp_y}{dt}, \frac{dp_z}{dt}\right) $$

which is a set of three equations:

$$\begin{align} F_x &= \frac{dp_x}{dt} \\ F_y &= \frac{dp_y}{dt} \\ F_z &= \frac{dp_z}{dt} \end{align} $$

In the example you give the only external forces present are the normal force between the ring and rod and gravity acting on the mass, and both of these act in the vertical direction. Suppose we call the vertical axis $y$ and the horizontal axis $x$ (we don't need a $z$ axis in this example) then since no external forces act in the $x$ direction we have $F_x = 0$ and therefore $dp_x/dt = 0$ i.e. momentum is conserved in the $x$ direction.

Since there are forces acting in the $y$ direction we cannot just assume that $p_y$ is conserved, though in fact it is since the vertical forces cancel each other out.

Answer 2

momentum conservation would require net external force on the system to be zero

You are correct - moment of a system is conserved when net external force equals zero, because by third Newton's law of motion all internal forces cancel. This means that there can be external forces, it is just that their vector sum must be zero in order for momentum to be conserved.

when exactly can external forces be neglected in problem-solving?

In some special cases when the event happens for (very) short period of time, such as collisions or explosions, the effect of non-impulsive external forces on the system can be neglected during the event. In these cases the momentum is not theoretically conserved just before and just after the event, but for all practical purposes it could be considered as conserved.

This follows directly from the impulse-momentum theorem

$$\Delta \vec{p} = \vec{J} \qquad \text{where} \qquad \vec{J} = \int \vec{F}_\text{ext} dt$$

where $\vec{J}$ is the impulse, $\vec{p}$ is the linear momentum, and $\Delta$ denotes change, i.e. final value minus initial value. If the impulse is (very) small compared to momentum, which happens when either $\vec{F}_\text{ext}$ or $t$ or both are very small, the change of momentum is approximately zero $\Delta \vec{p} \approx \vec{0}$, which means that final momentum equals initial momentum.