Mistake Master
Mistake Master
Pick two objects, draw a boundary around them, and call the inside the system. If the net external force on that system is zero, the system's total momentum is the same before and after, collision or explosion or anything else inside. The boundary is where mistakes happen: move it, and a force that was internal can flip to external. So the procedure is fixed: name the system first, then ask what is pushing on it.
Two carts sit on a frictionless track. Cart 1 rolls toward cart 2. They collide and stick. The carts exert contact forces on each other during the brief collision, large forces, equal and opposite by Newton's third law. Do those forces change the system's total momentum?
The answer depends entirely on where you drew the system boundary.
Same event, same forces. Different system, different answer. That is the whole topic, and it is where almost every conservation-of-momentum mistake starts.
The rule, stated as a procedure:
For a system of $N$ objects, the total linear momentum is the vector sum of the individual momenta:
$$\vec p_\text{system} \;=\; \sum_{k=1}^{N} m_k \vec v_k.$$Whenever the net external force on the system is zero across the interval from $t_i$ to $t_f$:
$$\vec p_\text{system}(t_i) \;=\; \vec p_\text{system}(t_f).$$That is a vector equation. In one dimension it gives a single signed equation; in two dimensions it gives one equation for each component, and they are conserved independently.
The condition is strict: the net external impulse across the interval must be zero or negligible. If gravity, friction, or a wall acts on the system, they deliver external impulse and the system's momentum changes by that amount: $\Delta\vec p_\text{system} = \vec J_\text{ext}$.
A $2$ kg cart moving east at $3$ m/s strikes a $1$ kg cart at rest on a frictionless track. They stick together. Find the velocity of the combined cart.
System: both carts. Internal forces during the collision: the contact pair. External forces: gravity (down) and the normal force (up). They cancel vertically, and there is no friction. Net external is zero horizontally, so horizontal momentum is conserved across the collision.
$$(2)(+3) + (1)(0) = (2 + 1)\,v_f \;\;\Longrightarrow\;\; v_f = +2 \text{ m/s east}.$$
The combined cart moves at $2$ m/s east. Faster than zero, slower than the striker. Total momentum is $6$ kg$\cdot$m/s east, same as before.
These three problem types cover almost every conservation-of-momentum question you will see. The physics is the same; the bookkeeping changes.
Collision. Two objects approach, interact briefly, separate (or stick). The contact force is huge and short, so outside forces like friction barely act during it. Treat the collision as instantaneous, conserve total momentum across it, and handle what happens before or after as a separate problem.
Explosion. A single object at rest breaks into pieces from internal forces, like the chemicals in a firework or a compressed spring. Those forces are internal, so $\vec p_\text{system}$ is the same after as before. If the pieces start at rest, they end with total momentum zero, and the pieces fly apart so their momenta cancel.
Recoil. The two-object version of an explosion. A cannon fires a shell; the cannon recoils. The cannon-on-shell and shell-on-cannon forces are internal, so the system's total momentum stays at its initial value (often zero). The cannon and shell take equal-and-opposite momenta. Same impulse on each, but different velocity changes, because the cannon is much heavier: $\Delta\vec v = \vec J / m$.
Conservation is only as trustworthy as the system you draw it around. Each trap below applies $\vec p_i = \vec p_f$ where it does not hold — an outside force is acting, the boundary is drawn in the wrong place, or equal impulses get mistaken for equal velocities.
The speed is the same before and after, but the direction reversed, and momentum is a vector. With upward positive, the ball's momentum went from $+m(8)$ to $-m(8)$. The change is not zero; it is $-16m$ in units of kg$\cdot$m/s. Where did the change come from? Gravity acted on the ball through the whole flight; that is an external force on the ball-alone system, and it delivered an impulse equal to the change in momentum.
Fix. Before you write $\vec p_i = \vec p_f$, check whether any external force acted on the system during the interval. For a single ball under gravity, gravity is always external. Conservation of momentum applies only to systems with zero net external force, and only across intervals where that holds. If gravity is in play, use $\Delta\vec p = \vec J_\text{ext} = \vec F_\text{net} \,\Delta t$ instead.
True that they push on each other; false that this rules out conservation. The forces between the carts are internal to the two-cart system, and internal forces always come in Newton's-third-law pairs. When you sum the impulses on the system as a whole, the cart-1-on-cart-2 and cart-2-on-cart-1 impulses add to zero. The total momentum of the system does not change.
Fix. When you classify a force, ask whether both ends of the pair are inside your system boundary. If they are, the force is internal and it contributes nothing to the system's total momentum change. If only one end is inside, the force is external and you need its impulse. The contact between two colliding carts is internal whenever both carts are in the system; it is external if you chose "cart 1 only" as your system.
The first half is correct, by Newton's third law. The second half drops the mass. Impulse equals mass times velocity change, so equal impulses on different masses give different speed changes. A $4$ kg rifle and a $0.01$ kg bullet share equal-and-opposite impulses, so the bullet's velocity changes $400$ times more than the rifle's. Same kick, very different speeds.
Fix. Write $\Delta\vec v = \vec J/m$. The symmetry of Newton's third law lives in the impulse pair, not the velocity changes. For each object's speed change, divide the shared impulse by that object's mass. The lighter object moves more, by exactly the mass ratio.