Page Description Equation
252 SI unit for linear momentum \[ \mathrm{kg}\cdot\frac{\mathrm{m}}{\mathrm{s}} \]
252 linear momentum of a particle of mass $m$ with velocity $\vect{v}$ \[ \vect{p}\equiv m\vect{v} \]
252 linear momentum components in $x, y, \text{ and } z$ directions \[ \begin{align*} p_x&=mv_x\\ p_y&=mv_y\\ p_z&=mv_z \end{align*} \]
253 net force on a particle with linear momentum $\vect{p}$ \[ \sum\vect{F}=\frac{d\vect{p}}{dt} \]
254 law of conservation of linear momentum, two-particle system \[ \begin{align*} \vect{p}_{1i}+\vect{p}_{2i} &=\vect{p}_{1f}+\vect{p}_{2f}\\ \vect{p}_{\mathrm{tot}} &=\sum_{\mathrm{system}}\vect{p}\\ &=\vect{p}_1+\vect{p}_2\\ &=\mathrm{constant} \end{align*} \]
Momentum is independently conserved in the $x,\ y, \text{ and } z$ directions. \[ \begin{align*} \sum_{\mathrm{system}} p_{i_x} &=\sum_{\mathrm{system}} p_{f_x}\\ \sum_{\mathrm{system}} p_{i_y} &=\sum_{\mathrm{system}} p_{f_y}\\ \sum_{\mathrm{system}} p_{i_z} &=\sum_{\mathrm{system}} p_{f_z} \end{align*} \]
impulse. impulse momentum theorem. The impulse of the force $\vect{F}$ acting on a particle equals the change in the momentum of the particle caused by that force. \[ \vect{I}\equiv\int_{t_i}^{t_f}\vect{F}\,dt\equiv\Delta\vect{p} \]
time-averaged force \[ \overline{\vect{F}}\equiv\frac{1}{\Delta t}\int_{t_i}^{t_f}\vect{F}\,dt \]
impulse in terms of time-averaged force \[ \vect{I}\equiv\overline{\vect{F}}\Delta t \]
impulse when impulsive force is constant \[ \vect{I}\equiv\overline{\vect{F}}\Delta t,\ \ \vect{F}=\mathrm{constant} \]
perfectly inelastic collision. total momentum before the collision equals the total momentum of the composite system after the collision. \[ m_1\vect{v}_{1i}+m_2\vect{v}_{2i}=\left(m_1+m_2\right)\vect{v}_f \]
center of mass of a system of particles where the $i\text{th}$ particle has mass $m_i$ and position $\vect{r}_i.$ \[ \vect{r}_{\mathrm{CM}}\equiv\frac{1}{M}\sum_{i}{m_i\vect{r}_i} \]
center of mass of a rigid body \[ \vect{r}_{\mathrm{CM}}=\frac{1}{M}\int\vect{r}\,dm \]
velocity of the center of mass for a system of particles \[ \begin{align*} \vect{v}_{\mathrm{CM}} &=\frac{1}{M}\sum_{i}{m_i\vect{v}_i}\\ &=\frac{d\vect{r}_{\mathrm{CM}}}{dt}\\ &=\frac{1}{M}\sum_{i}{m_i\frac{d\vect{r}_i}{dt}} \end{align*} \]
total linear momentum of a system of particles \[ \vect{p}_{\mathrm{tot}}=M\vect{v}_{\mathrm{CM}}=\sum_{i}{m_i\vect{v}_i}=\sum_{i}\vect{p}_i \]
acceleration of the center of mass of a system of particles \[ \begin{align*} \vect{a}_{\mathrm{CM}} &=\frac{1}{M}\sum_{i}{m_i\vect{a}_i}\\ &=\frac{d\vect{v}_{\mathrm{CM}}}{dt}\\ &=\frac{1}{M}\sum_{i}{m_i\frac{d\vect{v}_i}{dt}} \end{align*} \]
resultant external force on a system of particles \[ \sum\vect{F}_{\mathrm{ext}}=M\vect{a}_{\mathrm{CM}}=\frac{d\vect{p}_{\mathrm{tot}}}{dt} \]
law of conservation of momentum for a system of particles \[ \sum\vect{F}_{\mathrm{ext}}=0\Rightarrow\vect{p}_{\mathrm{tot}}=\mathrm{constant} \]


Page Notes
257 impulsive force
257 When using impulse approximation, as in collision problems, $p_i$ and $p_f$ represent the momenta immediately before and after the collision.
inelastic collision
perfectly inelastic collision
elastic collision
The center of mass of a system of particles of combined mass $M$ moves like an equivalent particle of mass $M$ would move under the influence of the resultant external force on the system.
256 The impulse-momentum theorem is equivalent to Newton's second law.
256 Impulse is a vector quantity having magnitude of the area under the Force-Time curve.
261 Momentum is constant in all collisions, but kinetic energy is constant only in elastic collisions.
Take note of equations 9.15, 9.16, 9.19, 9.20, 9.21, and the special cases mentioned below them on pp. 262-263. Also, note that in these formulas, the $v$'s are signed values according to direction, they are not magnitudes. They are still called speed though in this context. (p. 262)