Conservation of linear momentum

Conservation is a statement about a chosen system: when the net external force on it is zero, the net external impulse is zero, and the system's total momentum stays put. Newton's second law for the system reads $F_\text{ext} = dp/dt$, so $F_\text{ext} = 0$ makes $p$ constant: $p_i = p_f$. That is the whole condition.

So isolated is a choice of where you draw the boundary, not a property of the world. A ball in free fall is not isolated, because gravity pulls on it; a sliding block is not isolated, because friction drags on it; a ball bouncing off a wall is not isolated by itself, because the wall pushes back. Each of those outside forces changes the chosen system's momentum.

Redraw the boundary to swallow the other body, ball plus Earth or ball plus wall, and that outside force becomes internal. Internal forces come in equal and opposite pairs that cancel in the total, so the larger system does conserve its momentum even while each part's momentum changes.

Three habits cost points: treating $p_i = p_f$ as automatic, even for a single object or a system that is not isolated (this section); getting recoil and explosions wrong (§3); and confusing the momentum and kinetic-energy ledgers (§4).

In a collision the two objects push hard and briefly on each other. By Newton's third law those forces are equal and opposite, an internal pair that cancels in the system's total. So as long as no outside force acts during the impact, the two-object system is isolated and its total momentum is conserved: $p_i = p_f$.

That one equation finishes a sticking collision. A $3$ kg cart at $4$ m/s that sticks to a $1$ kg cart at rest carries $p = 12$ kg·m/s into a combined $4$ kg, so the pair leaves at $v_f = 12/4 = 3$ m/s. Write the total before, set it equal to the total after, solve for the one unknown.

Signs carry the directions. Two carts closing head-on, $+12$ and $-4$ kg·m/s, have a total of $+8$; after they interact the total is still $+8$, however it is shared. Dropping a sign turns a closing pair into a chasing one and breaks the bookkeeping.

The trap is to think a collision must change the total, or that sticking somehow loses momentum. The total is fixed the instant the system is isolated; the collision only decides how that fixed total is divided.

A system at rest that pushes itself apart, a gun firing, a skater throwing a ball, a spring releasing two carts, keeps its total momentum at zero, because the push is internal. So the pieces fly off with equal and opposite momenta that cancel: $m_1 v_1 = m_2 v_2$ in magnitude, opposite in direction.

Equal momenta are not equal speeds. A $5$ kg rifle firing a $0.05$ kg bullet at $300$ m/s gives the bullet $15$ kg·m/s, so the rifle takes $-15$ kg·m/s and recoils at $15/5 = 3$ m/s. The lighter piece is always the faster one, by the full mass ratio; the heavier piece barely moves.

Energy is not shared like momentum either. With $K = p^2/2m$ and equal momenta, the lighter piece carries the larger share of kinetic energy, and all of that energy was created from the spring, the powder, or a chemical source; the momentum total never left zero.

The traps cluster here: equal speeds, the heavier piece called faster (the mass ratio run backward), or the energy split evenly. Two more are dropped signs, which let opposite momenta add to twice one piece instead of cancelling, and the blast said to add momentum so the total grows. Pin the total first, usually zero, then read the pieces off it.

Momentum and kinetic energy are different quantities with different rules, and a collision keeps two separate books. Momentum is conserved in every collision with no external impulse, because the contact forces are internal. Kinetic energy is conserved only when the collision is elastic.

A perfectly inelastic collision, the objects sticking, loses the most kinetic energy momentum conservation allows. A $1$ kg cart at $4$ m/s sticking to a $1$ kg cart at rest conserves momentum, so the pair moves at $2$ m/s, while its kinetic energy falls from $8$ J to $4$ J. The missing $4$ J became heat, sound, and deformation; no law was broken.

So a kinetic-energy drop is normal, not a violation, and you must never solve a sticking collision by conserving kinetic energy, since that gives the wrong speed. Conserve momentum to find the motion; then, if you need the energy, compute it separately and compare.

The trap is to assume energy is conserved in every collision, to read an inelastic drop as broken conservation, or to think a collision that loses energy must also lose momentum. Keep the books apart: momentum always, kinetic energy only when the collision is elastic.

A conservation question turns on checking the system boundary first, setting the total to zero when everything starts at rest, and keeping the momentum and energy ledgers apart.

Check the boundary first. Before writing $p_i = p_f$, ask what the system is and whether any outside force acts over the interval. Gravity, friction, or a wall on a single object breaks conservation; draw the boundary to make those forces internal, or carry their impulse explicitly.

From rest, set the total to zero. For a recoil or explosion, the pieces carry equal and opposite momenta: $m_1 v_1 = m_2 v_2$. The lighter piece is faster, and the kinetic energy is created, not conserved.

Keep the two ledgers apart. Conserve momentum to find the motion in any collision. Test kinetic energy only against the elastic condition: equal in an elastic collision, lower in an inelastic one. Never use energy conservation to solve a stick.

Setup. A $2$ kg cart moving right at $6$ m/s strikes a $4$ kg cart at rest on a frictionless track, and they stick. (a) Find their common velocity. (b) Find the kinetic energy lost. (c) Later the same combined $6$ kg cart, at rest, is split by an internal spring back into the $2$ kg and $4$ kg carts; the $2$ kg piece leaves at $3$ m/s. Find the $4$ kg piece's velocity.

(a) Conserve momentum. No outside horizontal force acts during the brief collision, so the two-cart total is conserved: $p_i = (2)(6) + (4)(0) = 12$ kg·m/s. The carts stick into $6$ kg, so $12 = (6)v_f$ and $v_f = 2$ m/s, in the original direction.

(b) Compare the energy. Before, $K_i = \tfrac{1}{2}(2)(6^2) = 36$ J; after, $K_f = \tfrac{1}{2}(6)(2^2) = 12$ J. The collision lost $36 - 12 = 24$ J to heat and deformation. Momentum was conserved the whole time; kinetic energy was not, because the carts stuck.

(c) Split from rest. Now the system starts at rest, so its total stays zero: $0 = (2)(3) + (4)v$. The $2$ kg piece carries $+6$ kg·m/s, so the $4$ kg piece must carry $-6$ kg·m/s, giving $v = -1.5$ m/s: the heavy piece moves the other way at half the light piece's speed, equal and opposite in momentum.

The traps to avoid. Do not solve (a) with kinetic-energy conservation, since the stick loses energy and the answer would be wrong; do not call (b) a violation, since the lost energy is real heat; and in (c) do not give the pieces equal speeds, since a two-to-one mass ratio means a two-to-one speed ratio the other way.