Law of Conservation of Linear Momentum and Its Applications

Consider an isolated system of two (or more) bodies on which no net external force acts -- meaning that any forces present are only the internal forces the bodies within the system exert on one another. By Newton's third law, every such internal force comes paired with an equal and opposite reaction force, also internal to the system, so that when the second law is applied to the system as a whole, these internal action-reaction pairs always cancel out of the total. The result is the law of conservation of linear momentum: the total linear momentum of an isolated system remains constant, however the bodies within it interact, so long as no net external force acts on the system as a whole.

For two bodies of masses $m_1$ and $m_2$, with initial velocities $u_1$ and $u_2$ and final velocities $v_1$ and $v_2$ (after any interaction between them -- a collision, an explosion, one pushing off the other),

This single principle, applied as a vector equation, explains a wide range of situations:

Recoil of a gun. A gun and the bullet inside it, taken together just before firing, form a system initially at rest, so their total momentum is zero. After firing, the bullet moves off with some forward momentum $m_{\text{bullet}}\,v_{\text{bullet}}$; since the total momentum of the (isolated) gun-bullet system must still be zero immediately afterward, the gun itself must recoil backward with exactly equal and opposite momentum, $M_{\text{gun}}\,V_{\text{gun}} = -m_{\text{bullet}}\,v_{\text{bullet}}$ -- which is exactly how Example 4 of this chapter is solved.

Rocket propulsion, quantitatively. Building on the qualitative third-law picture of Section 4.5, momentum conservation makes the effect precise: as a rocket continuously ejects a small mass of exhaust gas backward at high relative speed, the rocket's own remaining mass must correspondingly gain forward momentum, so that the total momentum of rocket-plus-ejected-gas at every instant equals the total momentum before any gas was ejected (here, unlike the gun example, the rocket's own mass is steadily decreasing as fuel burns, which is exactly the situation flagged in Section 4.3 as needing the full momentum form of the second law rather than simply $F = ma$). …