conservation of momentum and reactive motion

The momentum of a system of bodies is the vector sum of the momenta of its parts: $\vec p = \sum_i m_i \vec v_i = M \vec v_c,$ where $M$ is the total mass and $\vec v_c$ is the velocity of the centre of mass. Newton's second law in momentum form, $\Delta \vec p = \vec F_{\text{ext}}\,\Delta t$, shows that only external forces change the momentum of a system. If the system is closed — no external forces act, or their vector sum is zero — then $\Delta \vec p = 0$ and $\vec p = \text{const}$: this is the law of conservation of momentum. Internal forces (a shot, a push-off, an explosion, a break-up) redistribute momentum among the bodies but do not change its total, so different parts of the system receive opposite momentum increments: $m_1 \Delta \vec v_1 = -\,m_2 \Delta \vec v_2$. Reactive motion is a direct consequence of this law: a rocket gains forward momentum because gases are ejected backwards. In differential form the momentum balance is $m\,dv = u\,dm_{\text{gas}}$ (for an infinitesimal gas ejection; the approximation $m\,\Delta v \approx u\,\Delta m_{\text{gas}}$ is valid only for very small $\Delta m_{\text{gas}}$), where $u$ is the speed of the gases relative to the rocket. For a finite burn of fuel mass $\Delta m_{\text{gas}}$, this relation must be integrated, giving the Tsiolkovsky formula $\Delta v = u\ln\frac{m_0}{m_f},$ where $m_0$ and $m_f$ are the initial and final masses of the rocket.

A cart of mass $m_1 = 2.0$ kg moves along a smooth horizontal rail with speed $v_1 = 3.0\ \text{m/s}$ and couples with a stationary cart of mass $m_2 = 1.0$ kg. Find the speed of the carts immediately after coupling.

A person of mass $M = 60$ kg stands on smooth ice holding a ball of mass $m = 5.0$ kg. He throws the ball horizontally with speed $v = 8.0\ \text{m/s}$ relative to the ice. With what speed and in what direction will the person slide?

A rocket in deep space has an initial mass $m_0 = 800$ kg, of which $600$ kg is fuel. The exhaust speed of the gases relative to the rocket is $u = 500\ \text{m/s}$. The engine burns $200$ kg of fuel. Neglecting external forces, find the increment in the rocket's speed and state its direction.

Balloon-rocket experiment (safe, at home). Stretch a $2$–$3$ m string between two chairs, thread a straw onto it, tape an inflated balloon to the straw (pinch the neck with a clip) and release the neck. Note the directions of the balloon's motion and the air jet — they are opposite. Then carry out a numerical check in a simulation: open PhET "Collision Lab", set elasticity to $0$ (sticking), and for each run with different masses and initial velocities compute $\sum m_i v_i$ before and after the collision. Record the results in a table and confirm that the sum is conserved to within rounding error. Do not use open flames, pyrotechnics or compressed gases: perform all observations only with a balloon or in a simulation.

To show that in an interaction the individual momenta change while their sum stays constant, and that reactive motion is a symmetric exchange of momentum between a rocket and its exhaust jet.

A two-panel sketch. (a) Collision: two carts before and after coupling, with momentum arrows $m_1 v_1$ and $(m_1+m_2)v$; beside it a bar chart "total momentum before / after" with equal bar heights. (b) Reactive motion: a rocket, an exhaust-jet arrow $u\,\Delta m_{\text{gas}}$ downwards and a rocket arrow $m\,\Delta v$ upwards, together with the relation $m_1 \Delta v_1 = -\,m_2 \Delta v_2$.

A biathlete fires a rifle of mass $m_1 = 3{,}5$ kg with a bullet of mass $m_2 = 10$ g. The bullet leaves horizontally at $v_2 = 700$ m/s, while the rifle is pressed against the shoulder.

The total momentum of the bullet + rifle system along the barrel is zero before firing, because external horizontal forces are negligible. After firing, the momenta of the bullet and rifle are equal in magnitude and opposite: $m_1 v_1 = m_2 v_2$, giving the recoil speed $v_1 = \dfrac{m_2 v_2}{m_1} \approx 2$ m/s. This is a direct consequence of conservation of momentum: the internal forces of the powder gases change the momenta of the bodies separately but not the total momentum.

An inflated balloon of mass $m_1 = 4$ g is released without tying the opening. In a time $t = 0{,}5$ s, air of mass $m_2 = 6$ g leaves it and moves backwards relative to the kitchen at $v_2 = 12$ m/s. The balloon darts in the opposite direction.

Before release, the total momentum of the balloon + air system was zero, and during $0{,}5$ s external forces (gravity and air drag) cannot noticeably change it. Therefore, after the air leaves, $m_1 v_1 = m_2 v_2$, and the balloon acquires the speed $v_1 = \dfrac{m_2 v_2}{m_1} = 18$ m/s. Reactive motion here is the motion of a body due to separation of part of its mass: the escaping air jet carries away momentum, and the balloon receives an equal-magnitude momentum in the opposite direction.

A person of mass $m_1 = 70$ kg jumps horizontally from a stationary boat of mass $m_2 = 140$ kg with a speed $v_1 = 3$ m/s. Water hardly hinders the boat's motion in the first moments.

The momentum of the person + boat system before the jump is zero. After the jump the sum of momenta remains zero: $m_1 v_1 = m_2 v_2$, so the boat moves backwards at $v_2 = \dfrac{m_1 v_1}{m_2} = 1{,}5$ m/s. This is an example of reactive motion: the person pushes off the boat, and the two bodies separate so that their total momentum does not change. At the same time, the centre of mass of the system stays in place while no external horizontal forces act.