Conservation of Linear Momentum

For an extended body or a collection of $N$ interacting particles of masses $m_i$ at positions $\mathbf{r}_i$:

1. Discrete Particle Systems

The center of mass position vector $\mathbf{R}_{\text{cm}}$ is the mass-weighted average position: $$\mathbf{R}_{\text{cm}} = \frac{1}{M} \sum_{i=1}^N m_i \mathbf{r}_i, \quad M = \sum_{i=1}^N m_i$$

2. Continuous Mass Distributions

For a continuous body with mass density $\rho(\mathbf{r})$: $$\mathbf{R}_{\text{cm}} = \frac{1}{M} \iiint_V \mathbf{r} \, \rho(\mathbf{r}) \, dV, \quad M = \iiint_V \rho(\mathbf{r}) \, dV$$

3. Analytical Center of Mass for Standard Geometries

• Uniform Semicircular Wire of Radius $R$: Let wire lie in $xy$-plane ($y \ge 0$). Linear density $\lambda = M/(\pi R)$. $dm = \lambda R d\theta$: $$y_{\text{cm}} = \frac{1}{M} \int_0^\pi (R \sin \theta) (\lambda R d\theta) = \frac{\lambda R^2}{M} [-\cos\theta]_0^\pi = \frac{2 R}{\pi} \approx 0.637 R$$
• Uniform Semicircular Disc of Radius $R$: Surface density $\sigma = \frac{2M}{\pi R^2}$. Slice into concentric rings of radius $r$: $$y_{\text{cm}} = \frac{1}{M} \int_0^R \left(\frac{2r}{\pi}\right) (\pi r \sigma dr) = \frac{2\sigma}{M} \int_0^R r^2 dr = \frac{4 R}{3\pi} \approx 0.424 R$$
• Uniform Solid Hemisphere of Radius $R$: Slice into horizontal discs of height $z$ ($z \in [0, R]$): $$z_{\text{cm}} = \frac{3 R}{8} = 0.375 R$$

The total linear momentum $\mathbf{P}$ of an $N$-particle system is: $$\mathbf{P} = \sum_{i=1}^N \mathbf{p}_i = \sum_{i=1}^N m_i \mathbf{v}_i = M \mathbf{v}_{\text{cm}}$$

1. Equation of Motion for the Center of Mass

Differentiating total momentum with respect to time: $$\frac{d\mathbf{P}}{dt} = M \mathbf{a}_{\text{cm}} = \sum_{i=1}^N \mathbf{F}_i^{\text{ext}} + \sum_{i=1}^N \sum_{j \neq i} \mathbf{F}_{ij}^{\text{int}}$$ By Newton's third law in strong form, mutual internal forces between particles cancel identically in pairs: $$\mathbf{F}_{ij}^{\text{int}} + \mathbf{F}_{ji}^{\text{int}} = \mathbf{0}$$ Therefore: $$\frac{d\mathbf{P}}{dt} = M \mathbf{a}_{\text{cm}} = \mathbf{F}_{\text{net}}^{\text{ext}}$$ Law of Conservation of Linear Momentum: If the net external force on a system vanishes ($\mathbf{F}_{\text{net}}^{\text{ext}} = \mathbf{0}$): $$\mathbf{P} = M \mathbf{v}_{\text{cm}} = \text{constant}$$ The center of mass moves with constant velocity regardless of complex internal explosions or collisions!

Newton's second law $\mathbf{F} = \frac{d\mathbf{p}}{dt}$ must be applied rigorously to open systems where mass continuously enters or leaves the control volume.

1. Derivation of the Rocket Equation of Motion

Consider a rocket of mass $m(t)$ moving with velocity $\mathbf{v}(t)$. During time $dt$, propellant mass $(-dm > 0)$ is ejected with exhaust velocity $\mathbf{u}_{\text{ex}}$ relative to the rocket. Linear momentum at time $t$: $p(t) = m v$. Linear momentum at time $t + dt$: $$p(t+dt) = (m + dm)(v + dv) + (-dm)(v - u_{\text{ex}}) = mv + m\,dv + u_{\text{ex}}\,dm$$ Change in momentum: $$dp = p(t+dt) - p(t) = m\,dv + u_{\text{ex}}\,dm$$ Applying external force $F_{\text{ext}}$: $$F_{\text{ext}} = \frac{dp}{dt} = m \frac{dv}{dt} + u_{\text{ex}} \frac{dm}{dt}$$ Rearranging gives the rocket dynamical equation: $$m \frac{dv}{dt} = -u_{\text{ex}} \frac{dm}{dt} + F_{\text{ext}} = T_{\text{thrust}} + F_{\text{ext}}$$ where $T_{\text{thrust}} = u_{\text{ex}}|\dot{m}|$ is the thrust force.

2. Free Space Tsiolkovsky Equation

In deep space with zero external forces ($F_{\text{ext}} = 0$): $$m dv = -u_{\text{ex}} dm \implies dv = -u_{\text{ex}} \frac{dm}{m}$$ Integrating from initial state $(m_0, v_0)$ to final burnout $(m_f, v_f)$: $$\Delta v = v_f - v_0 = u_{\text{ex}} \ln\left( \frac{m_0}{m_f} \right)$$

3. Vertical Ascent Under Gravity

For vertical launch against uniform gravity $g$: $$v(t) = -gt + u_{\text{ex}} \ln\left( \frac{m_0}{m_0 - \alpha t} \right)$$

Collisions are intense brief interactions where internal contact impulses vastly exceed external forces.

1. Classification by Kinetic Energy and Coefficient of Restitution

The coefficient of restitution $e$ along the line of impact is: $$e = -\frac{v_{2f} - v_{1f}}{v_{2i} - v_{1i}} = \frac{\text{relative separation speed}}{\text{relative approach speed}}$$

• Elastic ($e = 1$): Total mechanical kinetic energy is conserved: $K_f = K_i$.
• Inelastic ($0 < e < 1$): Energy is partially dissipated into deformation and heat: $K_f < K_i$.
• Completely Inelastic ($e = 0$): Bodies coalesce and move together with velocity $\mathbf{v}_f = \mathbf{v}_{\text{cm}}$.

2. Laboratory vs Center-of-Mass (CM) Frame

In the CM frame, total momentum is identically zero: $m_1 \mathbf{u}_1 + m_2 \mathbf{u}_2 = \mathbf{0}$. In an elastic collision in the CM frame, the speeds of the particles are completely unchanged ($u_1' = u_1, u_2' = u_2$); the collision merely rotates the relative velocity vector by scattering angle $\theta^*$. The laboratory scattering angle $\theta$ relates to CM angle $\theta^*$ by: $$\tan \theta = \frac{\sin \theta^*}{\cos \theta^* + \frac{m_1}{m_2}}$$