Chapter 9 Notes

9-1 Center of Mass

• Learning Objectives:

  • Determine the location of the center of mass given the positions of several particles along an axis or a plane.

  • Locate the center of mass of an extended, symmetric object using symmetry.

  • Determine the center of mass of a two or three-dimensional extended object with a uniform distribution of mass.

• Key Idea:

  • The center of mass of a system of n particles is defined as the point with coordinates:
    X{com} = \frac{1}{M} \sum{i=1}^{n} mi xi, Y{com} = \frac{1}{M} \sum{i=1}^{n} mi yi, Z{com} = \frac{1}{M} \sum{i=1}^{n} mi zi
    \vec{r}{com} = \frac{1}{M} \sum{i=1}^{n} mi \vec{r}i

  where M is the total mass of the system.

• What is Physics? Analyzing complicated motion requires simplification with an understanding of physics, such as traffic accident reconstruction or coaching a ballerina.

  • Complex motion of a system can be simplified by determining the center of mass.

  • The center of mass of a system moves in a simple parabolic path, even if the object rotates.

  • Knowing the center of mass is useful for advising long-jumpers or dancers.

The Center of Mass

• Definition:

  • The center of mass (com) of a system of particles is the point that moves as though:

  • All of the system's mass were concentrated there.

  • All external forces were applied there.

• Systems of Particles:

  • Two Particles:

  • The position of the center of mass (com) of this two-particle system is defined to be
    x{com} = \frac{m2}{m1 + m2}d

  • If m2 = 0, then x{com} = 0

  • If m1 = 0, then x{com} = d

  • If m1 = m2, then x_{com} = \frac{1}{2}d

  • The center of mass must lie somewhere between the two particles.

  • A more generalized situation, in which the coordinate system has been shifted leftward, the position of the center of mass is now defined as
    x{com} = \frac{m1x1 + m2x2}{m1 + m_2}

  • We can rewrite Eq. 9-2 as
    x{com} = \frac{m1x1 + m2x_2}{M}

  • Shifting the axis does not change the relative position of the com.

  • Many Particles:

  • For n particles along the x axis:
    x{com} = \frac{m1x1 + m2x2 + m3x3 + \cdots + mnxn}{M} = \frac{1}{M} \sum{i=1}^{n} mixi

  • Three Dimensions:

  • If particles are distributed in three dimensions:
    x{com} = \frac{1}{M} \sum{i=1}^{n} mixi
    
    y{com} = \frac{1}{M} \sum{i=1}^{n} miyi
    
    z{com} = \frac{1}{M} \sum{i=1}^{n} mizi

  • The position of a particle is given by a position vector:\vec{r}i = xi\hat{i} + yi\hat{j} + zi\hat{k}

  • The position of the center of mass of a system of particles is given by a position vector:\vec{r}{com} = x{com}\hat{i} + y{com}\hat{j} + z{com}\hat{k}

  • The three scalar equations can be replaced by:\vec{r}{com} = \frac{1}{M} \sum{i=1}^{n} mi\vec{r}i

• Solid Bodies:

  • An ordinary object can be treated as a continuous distribution of matter.

  • The "particles" become differential mass elements dm, and the sums become integrals.

  • The coordinates of the center of mass are defined as:
    x{com} = \frac{1}{M} \int x dm y{com} = \frac{1}{M} \int y dm
    
    z_{com} = \frac{1}{M} \int z dm

  • For uniform objects:
    \rho = \frac{dm}{dV} = \frac{M}{V}
    
    dm = \frac{M}{V}dV
    
    x{com} = \frac{1}{V} \int x dV y{com} = \frac{1}{V} \int y dV
    
    z_{com} = \frac{1}{V} \int z dV

• Symmetry as a Shortcut:

  • If an object has a point, line, or plane of symmetry, the center of mass lies at that point, on that line, or in that plane.

  • The center of mass of an object need not lie within the object.

9-2 Newton's Second Law for a System of Particles

• Learning Objectives:

  • Apply Newton's second law to a system of particles by relating the net force to the acceleration of the system's center of mass.

  • Apply the constant-acceleration equations to the motion of individual particles and the system's center of mass.

  • Calculate the velocity and acceleration of the system's center of mass.

  • Determine the velocity of the center of mass given its position as a function of time.

  • Determine the acceleration of the center of mass given its velocity as a function of time.

  • Calculate changes in the velocity and displacement of the com by integrating the com's acceleration and velocity functions.

  • Relate displacements and velocities of particles in a two-particle system when the system's com is stationary.

• Key Idea-

  • Governed by Newton's second law for a system of particles, which is\vec{F}{net} = M\vec{a}{com}

  where \vec{F}{net} is the net force of all external forces, M is the total mass, and \vec{a}{com} is the acceleration of the system's center of mass.

• Newton's Second Law for a System of Particles

  • Motion of a System's com\vec{F}{net} = M\vec{a}{com} \text{ (system of particles)}

  • This is Newton's second law for the motion of the center of mass of a system of particles.

  • The three quantities that appear in Eq. 9-14 must be evaluated with some care:

  • \vec{F}_{net} is the net force of all external forces that act on the system. Internal forces are not included.

  • M is the total mass of the system. The system is closed, so M remains constant.

  • \vec{a}_{com} is the acceleration of the system's center of mass. It gives no information about the acceleration of any other point of the system.

  • This equation is equivalent to three equations involving the components of Fnet and acom along the three coordinate axes

  
  F{net,x} = Ma{com,x}
  
  F{net,y} = Ma{com,y}
  
  F{net,z} = Ma{com,z}

  • Solid Body

  • Equation 9-14 applies also to a solid body.

  • M is the mass of the bat and \,F_{net} is the gravitational force on the bat.

  • Equation 9-14 then tells us that \vec{a}_{com} = \vec{g}.

  • Exploding Bodies

  • The acceleration \vec{a}_{com} of the center of mass of the fragments remains equal to \,g.

  • The center of mass of the fragments follows the same parabolic trajectory that the rocket would have followed had it not exploded.

  • Ballet Leap

  • The shifting center of mass faithfully follows a parabolic path across the stage, giving an illusion that the dancer is floating.

• Proof of Equation 9-14\vec{r}{com} = m1\vec{r}1 + m2\vec{r}2 + m3\vec{r}3 + \cdots + mn\vec{r}n M\vec{v}{com} = m1\vec{v}1 + m2\vec{v}2 + m3\vec{v}3 + \cdots + mn\vec{v}n
  
  M\vec{a}{com} = m1\vec{a}1 + m2\vec{a}2 + m3\vec{a}3 + \cdots + mn\vec{a}n M\vec{a}{com} = \vec{F}1 + \vec{F}2 + \vec{F}3 + \cdots + \vec{F}n
  
  M\vec{a}{com} = \vec{F}{net}

9-3 Linear Momentum

• Learning Objectives:

  • Identify that momentum is a vector quantity.

  • Calculate the linear momentum of a particle.

  • Calculate the change in momentum when a particle changes its speed and direction of travel.

  • Apply the relationship between a particle's momentum and the net force acting on it.

  • Calculate the momentum of a system of particles.

  • Apply the relationship between a system's center-of-mass momentum and the net force acting on the system.

• Key Ideas:

  • For a single particle, we define a quantity \,p called its linear momentum as\vec{p} = m\vec{v}

  • Newton's second law in terms of this momentum\vec{F}_{net} = \frac{d\vec{p}}{dt}

  • For a system of particles\vec{P} = M\vec{v}{com} \vec{F}{net} = \frac{d\vec{P}}{dt}

• Linear Momentum

  • Defined as\vec{p} = m\vec{v} \text{ (linear momentum of a particle)}

  • The SI unit for momentum is the kilogram-meter per second (kg m/s).

• Force and Momentum

  • Newton expressed his second law of motion in terms of momentum\vec{F}_{net} = \frac{d\vec{p}}{dt}

  • Manipulating Eq. 9-23 by substituting for p from Eq. 9-22 gives, for constant mass m\vec{F}_{net} = \frac{d\vec{p}}{dt} = \frac{d}{dt}(m\vec{v}) = m\frac{d\vec{v}}{dt} = m\vec{a}

• The Linear Momentum of a System of Particles

  • The system as a whole has a total linear momentum \,P, which is defined to be the vector sum of the individual particles' linear momenta\vec{P} = \vec{p}1 + \vec{p}2 + \vec{p}3 + \cdots + \vec{p}n = m1\vec{v}1 + m2\vec{v}2 + m3\vec{v}3 + \cdots + mn\vec{v}n
    \vec{P} = M\vec{v}_{com} \text{ (linear momentum, system of particles)}

• Force and Momentum\frac{d\vec{P}}{dt} = M\frac{d\vec{v}{com}}{dt} = M\vec{a}{com}
  \vec{F}_{net} = \frac{d\vec{P}}{dt} \text{ (system of particles)}

9-4 Collision and Impulse

• Learning Objectives

  • Identify that impulse is a vector quantity.

  • Apply the relationship between impulse and momentum change.

  • Apply the relationship between impulse, average force, and the time interval taken by the impulse.

  • Apply the constant-acceleration equations to relate impulse to average force.

  • Calculate the impulse given force as a function of time.

  • Calculate the impulse given a graph of force versus time.

  • Calculate the average force on the target.

• Key Ideas

  • Applying Newton's second law in momentum form to a particle-like body involved in a collision leads to the impulse-linear momentum theorem\vec{p}f - \vec{p}i = \Delta \vec{p} = \vec{J}

  where \,Delta \vec{p} is the body's linear momentum, and \,J is the impulse\vec{J} = \int{ti}^{tf} \vec{F}(t) dt If \,F{avg} is the average magnitude of \,F(t) during the collision and \,Delta t is the duration of the collision
  J = F_{avg}\Delta t

  • When a steady stream of bodies collides with a body whose position is fixed, the average force
    F{avg} = -\frac{n}{\Delta t} \Delta p = -\frac{n}{\Delta t} m \Delta v \Delta v = -v \Delta v = -2v F{avg} = \frac{\Delta m}{\Delta t} \Delta v

• Collision and Impulse

  • Single Collision
    d\vec{p} = \vec{F}(t) dt

  \int{ti}^{tf} d \vec{p} = \int{ti}^{tf} \vec{F}(t) dt
  \vec{J} = \int{ti}^{tf} \vec{F}(t) dt \text{ (impulse defined)} \Delta \vec{p} = \vec{J} \text{ (linear momentum-impulse theorem)} \Delta px = Jx = \int{ti}^{tf} Fx dt J = F{avg} \Delta t

  • Series of Collisions
    J = -n \Delta p
    
    F{avg} = \frac{J}{\Delta t} = -\frac{n}{\Delta t} \Delta p = -\frac{n}{\Delta t} m \Delta v F{avg} = \frac{\Delta m}{\Delta t} \Delta v

9-5 Conservation of Linear Momentum

• Learning Objectives

  • Apply the conservation of linear momentum to relate the initial momenta of the particles to their momenta at a later instant(isolated).

  • Apply conservation of linear momentum along an individual axis.

• Key Ideas

  • Closed and isolated system\vec{P} = constant

  • Conservation of linear momentum can be written in terms of the system's initial momentum and its momentum at some later instant\vec{P}i = \vec{P}f

• Conservation of Linear Momentum\vec{P} = constant
  \vec{P}i = \vec{P}f

• If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change.

9-6 Momentum and Kinetic Energy in Collisions

• Learning Objectives:

  • Distinguish between elastic collisions, inelastic collisions, and completely inelastic collisions.

  • Identify a one-dimensional collision.

  • Apply the conservation of momentum for an isolated one-dimensional collision.

  • Identify that momentum and velocity of the center of mass are not changed after objects collide.

• Key Ideas

  • Kinetic energy is not conserved\vec{p}{1i} + \vec{p}{2i} = \vec{p}{1f} + \vec{p}{2f}
    
    m1v{1i} + m2v{2i} = m1v{1f} + m2v{2f}

  • If the bodies stick together, the collision is a completely inelastic collision and the bodies have the same final velocity V.

  • The center of mass is not affected by a collision because\,V_{com} cannot be changed by the collision.

• Momentum and Kinetic Energy in Collisions

  • If that total happens to be unchanged by the collision, then the kinetic energy of the system is conserved.

  • Such a collision is called an elastic collision.

  • If total kinetic energy is not conserved, it is called an inelastic collision.

  • Inelastic : wet putty ball and a bat because the putty sticks to the bat/the bodies stick together, in which case the collision is called a completely inelastic collision.

• Inelastic Collisions in One Dimension\vec{p}{1i} + \vec{p}{2i} = \vec{p}{1f} + \vec{p}{2f}

  • One-Dimensional Inelastic Collision
    m1v{1i} + m2v{2i} = m1v{1f} + m2v{2f}

  • One-Dimensional Completely Inelastic Collision
    m1v{1i} = (m1 + m2)V
    
    V = \frac{m1}{m1 + m2}v{1i}

  • Velocity of the Center of Mass\vec{P} = M\vec{V}{com} = (m1 + m2)\vec{V}{com}
    \vec{V}{com} = \frac{\vec{P}}{m1 + m2} = \frac{\vec{p}{1i} + \vec{p}{2i}}{m1 + m_2}

9-7 Elastic Collisions in One Dimension

• Learning Objectives

  • Apply the conservation laws for both energy and momentum of the colliding bodies

• Key Idea

  • Kinetic Energy is conserved.

  • Linear momentum is conserved\vec{P} = MV_{com}

• Elastic Collisions in One Dimension

  • total kinetic energy before the collision = total kinetic energy after the collision

  • kinetic energy of each colliding body may change, but the total kinetic energy of the system does not change

  • Stationary Target
    m1v{1i} = m1v{1f} + m2v{2f} \text{ (linear momentum)}
    \frac{1}{2}m1v{1i}^2 = \frac{1}{2}m1v{1f}^2 + \frac{1}{2}m2v{2f}^2 \text{ (kinetic energy)}
    
    v{1f} = \frac{m1 - m2}{m1 + m2}v{1i}
    
    v{2f} = \frac{2m1}{m1 + m2}v_{1i}

  • Equal masses If m1 = m2

    • 
      v{1f} = 0 and v{2f} = v_{1i}

    • A massive target

      • m2 >> m1 :
        v{1f} \approx -v{1i} and
        v{2f} \approx (\frac{2m1}{m2})v{1i}

    • A massive projectile

      • m1 >> m2,
        v{1f} \approx v{1i} and
        v{2f} \approx 2v{1i}

  • Moving Target
    m1v{1i} + m2v{2i} = m1v{1f} + m2v{2f}
    \frac{1}{2}m1v{1i}^2 + \frac{1}{2}m2v{2i}^2 = \frac{1}{2}m1v{1f}^2 + \frac{1}{2}m2v{2f}^2
    
    v{1f} = \frac{m1 - m2}{m1 + m2}v{1i} + \frac{2m2}{m1 + m2}v{2i}
    
    v{2f} = \frac{2m1}{m1 + m2}v{1i} + \frac{m2 - m1}{m1 + m2}v{2i}

9-8 Collisions in Two Dimensions

• Learning Objectives:

  • For an isolated system in which a two-dimensional collision occurs

  • apply the conservation of momentum along each axis.

  • relate the momentum components along an axis before the collision to the momentum components along the same axis after the collision.

  • apply the conservation of total kinetic energy to relate the kinetic energies before and after the collision(elastic).

• Key Idea:

  • If two bodies collide and their motion is not along a single axis, the collision is two-dimensional.

  • If the two-body system is closed and isolated\vec{P}{1i} + \vec{P}{2i} = \vec{P}{1f} + \vec{P}{2f}
    
    K{1i} + K{2i} = K{1f} + K{2f}

• Collisions in Two Dimensions\vec{P}{1i} + \vec{P}{2i} = \vec{P}{1f} + \vec{P}{2f} \text{ (Conserved)}
  
  m1v{1i} = m1v{1f}cos\theta1 + m2v{2f}cos\theta2
  
  0 = -m1v{1f}sin\theta1 + m2v{2f}sin\theta2
  \frac{1}{2}m1v{1i} = \frac{1}{2}m1v{1f} + \frac{1}{2}m2v{2f} \text{ (Kinetic energy)}

9-9 Systems with Varying Mass: A Rocket

• Learning Objectives:

  • Apply the first rocket equation to relate the rate at which the rocket loses mass, the speed of the exhaust products relative to the rocket, the mass of the rocket, and the acceleration of the rocket.

  • Apply the second rocket equation to relate change in speed and initial and final mass.

  • For a moving system undergoing a change in mass at a given rate, relate that rate to the change in momentum.

• Key Ideas:

  • In the absence of external forces a rocket accelerates at an instantaneous rate given by
    Rv_{rel} = Ma \text{ (first rocket equation)}

  -

9-1 Center of Mass

• Learning Objectives:

  • Determine the location of the center of mass given the positions of several particles along an axis or a plane.

  • Locate the center of mass of an extended, symmetric object using symmetry.

  • Determine the center of mass of a two or three-dimensional extended object with a uniform distribution of mass.

• Key Idea:

  • The center of mass of a system of n particles is defined as the point with coordinates:
    X{com} = \frac{1}{M} \sum{i=1}^{n} mi xi, Y{com} = \frac{1}{M} \sum{i=1}^{n} mi yi, Z{com} = \frac{1}{M} \sum{i=1}^{n} mi zi
    \vec{r}{com} = \frac{1}{M} \sum{i=1}^{n} mi \vec{r}i

  where M is the total mass of the system.

• What is Physics? Analyzing complicated motion requires simplification with an understanding of physics, such as traffic accident reconstruction or coaching a ballerina.

  • Complex motion of a system can be simplified by determining the center of mass.

  • The center of mass of a system moves in a simple parabolic path, even if the object rotates.

  • Knowing the center of mass is useful for advising long-jumpers or dancers.

The Center of Mass

• Definition:

  • The center of mass (com) of a system of particles is the point that moves as though:

  • All of the system's mass were concentrated there.

  • All external forces were applied there.

• Systems of Particles:

  • Two Particles:

  • The position of the center of mass (com) of this two-particle system is defined to be
    x{com} = \frac{m2}{m1 + m2}d

  • If m2 = 0, then x{com} = 0

  • If m1 = 0, then x{com} = d

  • If m1 = m2, then x_{com} = \frac{1}{2}d

  • The center of mass must lie somewhere between the two particles.

  • A more generalized situation, in which the coordinate system has been shifted leftward, the position of the center of mass is now defined as
    x{com} = \frac{m1x1 + m2x2}{m1 + m_2}

  • We can rewrite Eq. 9-2 as
    x{com} = \frac{m1x1 + m2x_2}{M}

  • Shifting the axis does not change the relative position of the com.

  • Many Particles:

  • For n particles along the x axis:
    x{com} = \frac{m1x1 + m2x2 + m3x3 + \cdots + mnxn}{M} = \frac{1}{M} \sum{i=1}^{n} mixi

  • Three Dimensions:

  • If particles are distributed in three dimensions:
    x{com} = \frac{1}{M} \sum{i=1}^{n} mixi
    
    y{com} = \frac{1}{M} \sum{i=1}^{n} miyi
    
    z{com} = \frac{1}{M} \sum{i=1}^{n} mizi

  • The position of a particle is given by a position vector:\vec{r}i = xi\hat{i} + yi\hat{j} + zi\hat{k}

  • The position of the center of mass of a system of particles is given by a position vector:\vec{r}{com} = x{com}\hat{i} + y{com}\hat{j} + z{com}\hat{k}

  • The three scalar equations can be replaced by:\vec{r}{com} = \frac{1}{M} \sum{i=1}^{n} mi\vec{r}i

• Solid Bodies:

  • An ordinary object can be treated as a continuous distribution of matter.

  • The "particles" become differential mass elements dm, and the sums become integrals.

  • The coordinates of the center of mass are defined as:
    x{com} = \frac{1}{M} \int x dm y{com} = \frac{1}{M} \int y dm
    
    z_{com} = \frac{1}{M} \int z dm

  • For uniform objects:
    \rho = \frac{dm}{dV} = \frac{M}{V}
    
    dm = \frac{M}{V}dV
    
    x{com} = \frac{1}{V} \int x dV y{com} = \frac{1}{V} \int y dV
    
    z_{com} = \frac{1}{V} \int z dV

• Symmetry as a Shortcut:

  • If an object has a point, line, or plane of symmetry, the center of mass lies at that point, on that line, or in that plane.

  • The center of mass of an object need not lie within the object.

9-2 Newton's Second Law for a System of Particles

• Learning Objectives:

  • Apply Newton's second law to a system of particles by relating the net force to the acceleration of the system's center of mass.

  • Apply the constant-acceleration equations to the motion of individual particles and the system's center of mass.

  • Calculate the velocity and acceleration of the system's center of mass.

  • Determine the velocity of the center of mass given its position as a function of time.

  • Determine the acceleration of the center of mass given its velocity as a function of time.

  • Calculate changes in the velocity and displacement of the com by integrating the com's acceleration and velocity functions.

  • Relate displacements and velocities of particles in a two-particle system when the system's com is stationary.

• Key Idea-

  • Governed by Newton's second law for a system of particles, which is\vec{F}{net} = M\vec{a}{com}

  where \vec{F}{net} is the net force of all external forces, M is the total mass, and \vec{a}{com} is the acceleration of the system's center of mass.

• Newton's Second Law for a System of Particles

  • Motion of a System's com\vec{F}{net} = M\vec{a}{com} \text{ (system of particles)}

  • This is Newton's second law for the motion of the center of mass of a system of particles.

  • The three quantities that appear in Eq. 9-14 must be evaluated with some care:

  • \vec{F}_{net} is the net force of all external forces that act on the system. Internal forces are not included.

  • M is the total mass of the system. The system is closed, so M remains constant.

  • \vec{a}_{com} is the acceleration of the system's center of mass. It gives no information about the acceleration of any other point of the system.

  • This equation is equivalent to three equations involving the components of Fnet and acom along the three coordinate axes

  
  F{net,x} = Ma{com,x}
  
  F{net,y} = Ma{com,y}
  
  F{net,z} = Ma{com,z}

  • Solid Body

  • Equation 9-14 applies also to a solid body.

  • M is the mass of the bat and \,F_{net} is the gravitational force on the bat.

  • Equation 9-14 then tells us that \vec{a}_{com} = \vec{g}.

  • Exploding Bodies

  • The acceleration \vec{a}_{com} of the center of mass of the fragments remains equal to \,g.

  • The center of mass of the fragments follows the same parabolic trajectory that the rocket would have followed had it not exploded.

  • Ballet Leap

  • The shifting center of mass faithfully follows a parabolic path across the stage, giving an illusion that the dancer is floating.

• Proof of Equation 9-14\vec{r}{com} = m1\vec{r}1 + m2\vec{r}2 + m3\vec{r}3 + \cdots + mn\vec{r}n M\vec{v}{com} = m1\vec{v}1 + m2\vec{v}2 + m3\vec{v}3 + \cdots + mn\vec{v}n
  
  M\vec{a}{com} = m1\vec{a}1 + m2\vec{a}2 + m3\vec{a}3 + \cdots + mn\vec{a}n M\vec{a}{com} = \vec{F}1 + \vec{F}2 + \vec{F}3 + \cdots + \vec{F}n
  
  M\vec{a}{com} = \vec{F}{net}

9-3 Linear Momentum

• Learning Objectives:

  • Identify that momentum is a vector quantity.

  • Calculate the linear momentum of a particle.

  • Calculate the change in momentum when a particle changes its speed and direction of travel.

  • Apply the relationship between a particle's momentum and the net force acting on it.

  • Calculate the momentum of a system of particles.

  • Apply the relationship between a system's center-of-mass momentum and the net force acting on the system.

• Key Ideas:

  • For a single particle, we define a quantity \,p called its linear momentum as\vec{p} = m\vec{v}

  • Newton's second law in terms of this momentum\vec{F}_{net} = \frac{d\vec{p}}{dt}

  • For a system of particles\vec{P} = M\vec{v}{com} \vec{F}{net} = \frac{d\vec{P}}{dt}

• Linear Momentum

  • Defined as\vec{p} = m\vec{v} \text{ (linear momentum of a particle)}

  • The SI unit for momentum is the kilogram-meter per second (kg m/s).

• Force and Momentum

  • Newton expressed his second law of motion in terms of momentum\vec{F}_{net} = \frac{d\vec{p}}{dt}

  • Manipulating Eq. 9-23 by substituting for p from Eq. 9-22 gives, for constant mass m\vec{F}_{net} = \frac{d\vec{p}}{dt} = \frac{d}{dt}(m\vec{v}) = m\frac{d\vec{v}}{dt} = m\vec{a}

• The Linear Momentum of a System of Particles

  • The system as a whole has a total linear momentum \,P, which is defined to be the vector sum of the individual particles' linear momenta\vec{P} = \vec{p}1 + \vec{p}2 + \vec{p}3 + \cdots + \vec{p}n = m1\vec{v}1 + m2\vec{v}2 + m3\vec{v}3 + \cdots + mn\vec{v}n
    \vec{P} = M\vec{v}_{com} \text{ (linear momentum, system of particles)}

• Force and Momentum\frac{d\vec{P}}{dt} = M\frac{d\vec{v}{com}}{dt} = M\vec{a}{com}
  \vec{F}_{net} = \frac{d\vec{P}}{dt} \text{ (system of particles)}

9-4 Collision and Impulse

• Learning Objectives

  • Identify that impulse is a vector quantity.

  • Apply the relationship between impulse and momentum change.

  • Apply the relationship between impulse, average force, and the time interval taken by the impulse.

  • Apply the constant-acceleration equations to relate impulse to average force.

  • Calculate the impulse given force as a function of time.

  • Calculate the impulse given a graph of force versus time.

  • Calculate the average force on the target.

• Key Ideas

  • Applying Newton's second law in momentum form to a particle-like body involved in a collision leads to the impulse-linear momentum theorem\vec{p}f - \vec{p}i = \Delta \vec{p} = \vec{J}

  where \,Delta \vec{p} is the body's linear momentum, and \,J is the impulse\vec{J} = \int{ti}^{tf} \vec{F}(t) dt If \,F{avg} is the average magnitude of \,F(t) during the collision and \,Delta t is the duration of the collision
  J = F_{avg}\Delta t

  • When a steady stream of bodies collides with a body whose position is fixed, the average force
    F{avg} = -\frac{n}{\Delta t} \Delta p = -\frac{n}{\Delta t} m \Delta v \Delta v = -v \Delta v = -2v F{avg} = \frac{\Delta m}{\Delta t} \Delta v

• Collision and Impulse

  • Single Collision
    d\vec{p} = \vec{F}(t) dt

  \int{ti}^{tf} d \vec{p} = \int{ti}^{tf} \vec{F}(t) dt
  \vec{J} = \int{ti}^{tf} \vec{F}(t) dt \text{ (impulse defined)} \Delta \vec{p} = \vec{J} \text{ (linear momentum-impulse theorem)} \Delta px = Jx = \int{ti}^{tf} Fx dt J = F{avg} \Delta t

  • Series of Collisions
    J = -n \Delta p
    
    F{avg} = \frac{J}{\Delta t} = -\frac{n}{\Delta t} \Delta p = -\frac{n}{\Delta t} m \Delta v F{avg} = \frac{\Delta m}{\Delta t} \Delta v

9-5 Conservation of Linear Momentum

• Learning Objectives

  • Apply the conservation of linear momentum to relate the initial momenta of the particles to their momenta at a later instant(isolated).

  • Apply conservation of linear momentum along an individual axis.

• Key Ideas

  • Closed and isolated system\vec{P} = constant

  • Conservation of linear momentum can be written in terms of the system's initial momentum and its momentum at some later instant\vec{P}i = \vec{P}f

• Conservation of Linear Momentum\vec{P} = constant
  \vec{P}i = \vec{P}f

• If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change.

9-6 Momentum and Kinetic Energy in Collisions

• Learning Objectives:

  • Distinguish between elastic collisions, inelastic collisions, and completely inelastic collisions.

  • Identify a one-dimensional collision.

  • Apply the conservation of momentum for an isolated one-dimensional collision.

  • Identify that momentum and velocity of the center of mass are not changed after objects collide.

• Key Ideas

  • Kinetic energy is not conserved\vec{p}{1i} + \vec{p}{2i} = \vec{p}{1f} + \vec{p}{2f}
    
    m1v{1i} + m2v{2i} = m1v{1f} + m2v{2f}

  • If the bodies stick together, the collision is a completely inelastic collision and the bodies have the same final velocity V.

  • The center of mass is not affected by a collision because\,V_{com} cannot be changed by the collision.

• Momentum and Kinetic Energy in Collisions

  • If that total happens to be unchanged by the collision, then the kinetic energy of the system is conserved.

  • Such a collision is called an elastic collision.

  • If total kinetic energy is not conserved, it is called an inelastic collision.

  • Inelastic : wet putty ball and a bat because the putty sticks to the bat/the bodies stick together, in which case the collision is called a completely inelastic collision.

• Inelastic Collisions in One Dimension\vec{p}{1i} + \vec{p}{2i} = \vec{p}{1f} + \vec{p}{2f}

  • One-Dimensional Inelastic Collision
    m1v{1i} + m2v{2i} = m1v{1f} + m2v{2f}

  • One-Dimensional Completely Inelastic Collision
    m1v{1i} = (m1 + m2)V
    
    V = \frac{m1}{m1 + m2}v{1i}

  • Velocity of the Center of Mass\vec{P} = M\vec{V}{com} = (m1 + m2)\vec{V}{com}
    \vec{V}{com} = \frac{\vec{P}}{m1 + m2} = \frac{\vec{p}{1i} + \vec{p}{2i}}{m1 + m_2}

9-7 Elastic Collisions in One Dimension

• Learning Objectives

  • Apply the conservation laws for both energy and momentum of the colliding bodies

• Key Idea

  • Kinetic Energy is conserved.

  • Linear momentum is conserved\vec{P} = MV_{com}

• Elastic Collisions in One Dimension

  • total kinetic energy before the collision = total kinetic energy after the collision

  • kinetic energy of each colliding body may change, but the total kinetic energy of the system does not change

  • Stationary Target
    m1v{1i} = m1v{1f} + m2v{2f} \text{ (linear momentum)}
    \frac{1}{2}m1v{1i}^2 = \frac{1}{2}m1v{1f}^2 + \frac{1}{2}m2v{2f}^2 \text{ (kinetic energy)}
    
    v{1f} = \frac{m1 - m2}{m1 + m2}v{1i}
    
    v{2f} = \frac{2m1}{m1 + m2}v_{1i}

  • Equal masses If m1 = m2

    • 
      v{1f} = 0 and v{2f} = v_{1i}

    • A massive target

      • m2 >> m1 :
        v{1f} \approx -v{1i} and
        v{2f} \approx (\frac{2m1}{m2})v{1i}

    • A massive projectile

      • m1 >> m2,
        v{1f} \approx v{1i} and
        v{2f} \approx 2v{1i}

  • Moving Target
    m1v{1i} + m2v{2i} = m1v{1f} + m2v{2f}
    \frac{1}{2}m1v{1i}^2 + \frac{1}{2}m2v{2i}^2 = \frac{1}{2}m1v{1f}^2 + \frac{1}{2}m2v{2f}^2
    
    v{1f} = \frac{m1 - m2}{m1 + m2}v{1i} + \frac{2m2}{m1 + m2}v{2i}
    
    v{2f} = \frac{2m1}{m1 + m2}v{1i} + \frac{m2 - m1}{m1 + m2}v{2i}

9-8 Collisions in Two Dimensions

• Learning Objectives:

  • For an isolated system in which a two-dimensional collision occurs

  • apply the conservation of momentum along each axis.

  • relate the momentum components along an axis before the collision to the momentum components along the same axis after the collision.

  • apply the conservation of total kinetic energy to relate the kinetic energies before and after the collision(elastic).

• Key Idea:

  • If two bodies collide and their motion is not along a single axis, the collision is two-dimensional.

  • If the two-body system is closed and isolated\vec{P}{1i} + \vec{P}{2i} = \vec{P}{1f} + \vec{P}{2f}
    
    K{1i} + K{2i} = K{1f} + K{2f}

• Collisions in Two Dimensions\vec{P}{1i} + \vec{P}{2i} = \vec{P}{1f} + \vec{P}{2f} \text{ (Conserved)}
  
  m1v{1i} = m1v{1f}cos\theta1 + m2v{2f}cos\theta2
  
  0 = -m1v{1f}sin\theta1 + m2v{2f}sin\theta2
  \frac{1}{2}m1v{1i} = \frac{1}{2}m1v{1f} + \frac{1}{2}m2v{2f} \text{ (Kinetic energy)}

9-9 Systems with Varying Mass: A Rocket

• Learning Objectives:

  • Apply the first rocket equation to relate the rate at which the rocket loses mass, the speed of the exhaust products relative to the rocket, the mass of the rocket, and the acceleration of the rocket.

  • Apply the second rocket equation to relate change in speed and initial and final mass.

  • For a moving system undergoing a change in mass at a given rate, relate that rate to the change in momentum.

• Key Ideas:

  • In the absence of external forces a rocket accelerates at an instantaneous rate given by
    Rv_{rel} = Ma \text{ (first rocket equation)}

  -

9-1 Center of Mass

• Learning Objectives:

  • Determine the location of the center of mass given the positions of several particles along an axis or a plane.

  • Locate the center of mass of an extended, symmetric object using symmetry.

  • Determine the center of mass of a two or three-dimensional extended object with a uniform distribution of mass.

• Key Idea:

  • The center of mass of a system of n particles is defined as the point with coordinates:
    X{com} = \frac{1}{M} \sum{i=1}^{n} mi xi, Y{com} = \frac{1}{M} \sum{i=1}^{n} mi yi, Z{com} = \frac{1}{M} \sum{i=1}^{n} mi zi
    \vec{r}{com} = \frac{1}{M} \sum{i=1}^{n} mi \vec{r}i

  where M is the total mass of the system.

• What is Physics? Analyzing complicated motion requires simplification with an understanding of physics, such as traffic accident reconstruction or coaching a ballerina.

  • Complex motion of a system can be simplified by determining the center of mass.

  • The center of mass of a system moves in a simple parabolic path, even if the object rotates.

  • Knowing the center of mass is useful for advising long-jumpers or dancers.

The Center of Mass

• Definition:

  • The center of mass (com) of a system of particles is the point that moves as though:

  • All of the system's mass were concentrated there.

  • All external forces were applied there.

• Systems of Particles:

  • Two Particles:

  • The position of the center of mass (com) of this two-particle system is defined to be
    x{com} = \frac{m2}{m1 + m2}d

  • If m2 = 0, then x{com} = 0

  • If m1 = 0, then x{com} = d

  • If m1 = m2, then x_{com} = \frac{1}{2}d

  • The center of mass must lie somewhere between the two particles.

  • A more generalized situation, in which the coordinate system has been shifted leftward, the position of the center of mass is now defined as
    x{com} = \frac{m1x1 + m2x2}{m1 + m_2}

  • We can rewrite Eq. 9-2 as
    x{com} = \frac{m1x1 + m2x_2}{M}

  • Shifting the axis does not change the relative position of the com.

  • Many Particles:

  • For n particles along the x axis:
    x{com} = \frac{m1x1 + m2x2 + m3x3 + \cdots + mnxn}{M} = \frac{1}{M} \sum{i=1}^{n} mixi

  • Three Dimensions:

  • If particles are distributed in three dimensions:
    x{com} = \frac{1}{M} \sum{i=1}^{n} mixi
    
    y{com} = \frac{1}{M} \sum{i=1}^{n} miyi
    
    z{com} = \frac{1}{M} \sum{i=1}^{n} mizi

  • The position of a particle is given by a position vector:\vec{r}i = xi\hat{i} + yi\hat{j} + zi\hat{k}

  • The position of the center of mass of a system of particles is given by a position vector:\vec{r}{com} = x{com}\hat{i} + y{com}\hat{j} + z{com}\hat{k}

  • The three scalar equations can be replaced by:\vec{r}{com} = \frac{1}{M} \sum{i=1}^{n} mi\vec{r}i

• Solid Bodies:

  • An ordinary object can be treated as a continuous distribution of matter.

  • The "particles" become differential mass elements dm, and the sums become integrals.

  • The coordinates of the center of mass are defined as:
    x{com} = \frac{1}{M} \int x dm y{com} = \frac{1}{M} \int y dm
    
    z_{com} = \frac{1}{M} \int z dm

  • For uniform objects:
    \rho = \frac{dm}{dV} = \frac{M}{V}
    
    dm = \frac{M}{V}dV
    
    x{com} = \frac{1}{V} \int x dV y{com} = \frac{1}{V} \int y dV
    
    z_{com} = \frac{1}{V} \int z dV

• Symmetry as a Shortcut:

  • If an object has a point, line, or plane of symmetry, the center of mass lies at that point, on that line, or in that plane.

  • The center of mass of an object need not lie within the object.

9-2 Newton's Second Law for a System of Particles

• Learning Objectives:

  • Apply Newton's second law to a system of particles by relating the net force to the acceleration of the system's center of mass.

  • Apply the constant-acceleration equations to the motion of individual particles and the system's center of mass.

  • Calculate the velocity and acceleration of the system's center of mass.

  • Determine the velocity of the center of mass given its position as a function of time.

  • Determine the acceleration of the center of mass given its velocity as a function of time.

  • Calculate changes in the velocity and displacement of the com by integrating the com's acceleration and velocity functions.

  • Relate displacements and velocities of particles in a two-particle system when the system's com is stationary.

• Key Idea-

  • Governed by Newton's second law for a system of particles, which is\vec{F}{net} = M\vec{a}{com}

  where \vec{F}{net} is the net force of all external forces, M is the total mass, and \vec{a}{com} is the acceleration of the system's center of mass.

• Newton's Second Law for a System of Particles

  • Motion of a System's com\vec{F}{net} = M\vec{a}{com} \text{ (system of particles)}

  • This is Newton's second law for the motion of the center of mass of a system of particles.

  • The three quantities that appear in Eq. 9-14 must be evaluated with some care:

  • \vec{F}_{net} is the net force of all external forces that act on the system. Internal forces are not included.

  • M is the total mass of the system. The system is closed, so M remains constant.

  • \vec{a}_{com} is the acceleration of the system's center of mass. It gives no information about the acceleration of any other point of the system.

  • This equation is equivalent to three equations involving the components of Fnet and acom along the three coordinate axes

  
  F{net,x} = Ma{com,x}
  
  F{net,y} = Ma{com,y}
  
  F{net,z} = Ma{com,z}

  • Solid Body

  • Equation 9-14 applies also to a solid body.

  • M is the mass of the bat and \,F_{net} is the gravitational force on the bat.

  • Equation 9-14 then tells us that \vec{a}_{com} = \vec{g}.

  • Exploding Bodies

  • The acceleration \vec{a}_{com} of the center of mass of the fragments remains equal to \,g.

  • The center of mass of the fragments follows the same parabolic trajectory that the rocket would have followed had it not exploded.

  • Ballet Leap

  • The shifting center of mass faithfully follows a parabolic path across the stage, giving an illusion that the dancer is floating.

• Proof of Equation 9-14\vec{r}{com} = m1\vec{r}1 + m2\vec{r}2 + m3\vec{r}3 + \cdots + mn\vec{r}n M\vec{v}{com} = m1\vec{v}1 + m2\vec{v}2 + m3\vec{v}3 + \cdots + mn\vec{v}n
  
  M\vec{a}{com} = m1\vec{a}1 + m2\vec{a}2 + m3\vec{a}3 + \cdots + mn\vec{a}n M\vec{a}{com} = \vec{F}1 + \vec{F}2 + \vec{F}3 + \cdots + \vec{F}n
  
  M\vec{a}{com} = \vec{F}{net}

9-3 Linear Momentum

• Learning Objectives:

  • Identify that momentum is a vector quantity.

  • Calculate the linear momentum of a particle.

  • Calculate the change in momentum when a particle changes its speed and direction of travel.

  • Apply the relationship between a particle's momentum and the net force acting on it.

  • Calculate the momentum of a system of particles.

  • Apply the relationship between a system's center-of-mass momentum and the net force acting on the system.

• Key Ideas:

  • For a single particle, we define a quantity \,p called its linear momentum as\vec{p} = m\vec{v}

  • Newton's second law in terms of this momentum\vec{F}_{net} = \frac{d\vec{p}}{dt}

  • For a system of particles\vec{P} = M\vec{v}{com} \vec{F}{net} = \frac{d\vec{P}}{dt}

• Linear Momentum

  • Defined as\vec{p} = m\vec{v} \text{ (linear momentum of a particle)}

  • The SI unit for momentum is the kilogram-meter per second (kg m/s).

• Force and Momentum

  • Newton expressed his second law of motion in terms of momentum\vec{F}_{net} = \frac{d\vec{p}}{dt}

  • Manipulating Eq. 9-23 by substituting for p from Eq. 9-22 gives, for constant mass m\vec{F}_{net} = \frac{d\vec{p}}{dt} = \frac{d}{dt}(m\vec{v}) = m\frac{d\vec{v}}{dt} = m\vec{a}

• The Linear Momentum of a System of Particles

  • The system as a whole has a total linear momentum \,P, which is defined to be the vector sum of the individual particles' linear momenta\vec{P} = \vec{p}1 + \vec{p}2 + \vec{p}3 + \cdots + \vec{p}n = m1\vec{v}1 + m2\vec{v}2 + m3\vec{v}3 + \cdots + mn\vec{v}n
    \vec{P} = M\vec{v}_{com} \text{ (linear momentum, system of particles)}

• Force and Momentum\frac{d\vec{P}}{dt} = M\frac{d\vec{v}{com}}{dt} = M\vec{a}{com}
  \vec{F}_{net} = \frac{d\vec{P}}{dt} \text{ (system of particles)}

9-4 Collision and Impulse

• Learning Objectives

  • Identify that impulse is a vector quantity.

  • Apply the relationship between impulse and momentum change.

  • Apply the relationship between impulse, average force, and the time interval taken by the impulse.

  • Apply the constant-acceleration equations to relate impulse to average force.

  • Calculate the impulse given force as a function of time.

  • Calculate the impulse given a graph of force versus time.

  • Calculate the average force on the target.

• Key Ideas

  • Applying Newton's second law in momentum form to a particle-like body involved in a collision leads to the impulse-linear momentum theorem\vec{p}f - \vec{p}i = \Delta \vec{p} = \vec{J}

  where \,Delta \vec{p} is the body's linear momentum, and \,J is the impulse\vec{J} = \int{ti}^{tf} \vec{F}(t) dt If \,F{avg} is the average magnitude of \,F(t) during the collision and \,Delta t is the duration of the collision
  J = F_{avg}\Delta t

  • When a steady stream of bodies collides with a body whose position is fixed, the average force
    F{avg} = -\frac{n}{\Delta t} \Delta p = -\frac{n}{\Delta t} m \Delta v \Delta v = -v \Delta v = -2v F{avg} = \frac{\Delta m}{\Delta t} \Delta v

• Collision and Impulse

  • Single Collision
    d\vec{p} = \vec{F}(t) dt

  \int{ti}^{tf} d \vec{p} = \int{ti}^{tf} \vec{F}(t) dt
  \vec{J} = \int{ti}^{tf} \vec{F}(t) dt \text{ (impulse defined)} \Delta \vec{p} = \vec{J} \text{ (linear momentum-impulse theorem)} \Delta px = Jx = \int{ti}^{tf} Fx dt J = F{avg} \Delta t

  • Series of Collisions
    J = -n \Delta p
    
    F{avg} = \frac{J}{\Delta t} = -\frac{n}{\Delta t} \Delta p = -\frac{n}{\Delta t} m \Delta v F{avg} = \frac{\Delta m}{\Delta t} \Delta v

9-5 Conservation of Linear Momentum

• Learning Objectives

  • Apply the conservation of linear momentum to relate the initial momenta of the particles to their momenta at a later instant(isolated).

  • Apply conservation of linear momentum along an individual axis.

• Key Ideas

  • Closed and isolated system\vec{P} = constant

  • Conservation of linear momentum can be written in terms of the system's initial momentum and its momentum at some later instant\vec{P}i = \vec{P}f

• Conservation of Linear Momentum\vec{P} = constant
  \vec{P}i = \vec{P}f

• If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change.

9-6 Momentum and Kinetic Energy in Collisions

• Learning Objectives:

  • Distinguish between elastic collisions, inelastic collisions, and completely inelastic collisions.

  • Identify a one-dimensional collision.

  • Apply the conservation of momentum for an isolated one-dimensional collision.

  • Identify that momentum and velocity of the center of mass are not changed after objects collide.

• Key Ideas

  • Kinetic energy is not conserved\vec{p}{1i} + \vec{p}{2i} = \vec{p}{1f} + \vec{p}{2f}
    
    m1v{1i} + m2v{2i} = m1v{1f} + m2v{2f}

  • If the bodies stick together, the collision is a completely inelastic collision and the bodies have the same final velocity V.

  • The center of mass is not affected by a collision because\,V_{com} cannot be changed by the collision.

• Momentum and Kinetic Energy in Collisions

  • If that total happens to be unchanged by the collision, then the kinetic energy of the system is conserved.

  • Such a collision is called an elastic collision.

  • If total kinetic energy is not conserved, it is called an inelastic collision.

  • Inelastic : wet putty ball and a bat because the putty sticks to the bat/the bodies stick together, in which case the collision is called a completely inelastic collision.

• Inelastic Collisions in One Dimension\vec{p}{1i} + \vec{p}{2i} = \vec{p}{1f} + \vec{p}{2f}

  • One-Dimensional Inelastic Collision
    m1v{1i} + m2v{2i} = m1v{1f} + m2v{2f}

  • One-Dimensional Completely Inelastic Collision
    m1v{1i} = (m1 + m2)V
    
    V = \frac{m1}{m1 + m2}v{1i}

  • Velocity of the Center of Mass\vec{P} = M\vec{V}{com} = (m1 + m2)\vec{V}{com}
    \vec{V}{com} = \frac{\vec{P}}{m1 + m2} = \frac{\vec{p}{1i} + \vec{p}{2i}}{m1 + m_2}

9-7 Elastic Collisions in One Dimension

• Learning Objectives

  • Apply the conservation laws for both energy and momentum of the colliding bodies

• Key Idea

  • Kinetic Energy is conserved.

  • Linear momentum is conserved\vec{P} = MV_{com}

• Elastic Collisions in One Dimension

  • total kinetic energy before the collision = total kinetic energy after the collision

  • kinetic energy of each colliding body may change, but the total kinetic energy of the system does not change

  • Stationary Target
    m1v{1i} = m1v{1f} + m2v{2f} \text{ (linear momentum)}
    \frac{1}{2}m1v{1i}^2 = \frac{1}{2}m1v{1f}^2 + \frac{1}{2}m2v{2f}^2 \text{ (kinetic energy)}
    
    v{1f} = \frac{m1 - m2}{m1 + m2}v{1i}
    
    v{2f} = \frac{2m1}{m1 + m2}v_{1i}

  • Equal masses If m1 = m2

    • 
      v{1f} = 0 and v{2f} = v_{1i}

    • A massive target

      • m2 >> m1 :
        v{1f} \approx -v{1i} and
        v{2f} \approx (\frac{2m1}{m2})v{1i}

    • A massive projectile

      • m1 >> m2,
        v{1f} \approx v{1i} and
        v{2f} \approx 2v{1i}

  • Moving Target
    m1v{1i} + m2v{2i} = m1v{1f} + m2v{2f}
    \frac{1}{2}m1v{1i}^2 + \frac{1}{2}m2v{2i}^2 = \frac{1}{2}m1v{1f}^2 + \frac{1}{2}m2v{2f}^2
    
    v{1f} = \frac{m1 - m2}{m1 + m2}v{1i} + \frac{2m2}{m1 + m2}v{2i}
    
    v{2f} = \frac{2m1}{m1 + m2}v{1i} + \frac{m2 - m1}{m1 + m2}v{2i}

9-8 Collisions in Two Dimensions

• Learning Objectives:

  • For an isolated system in which a two-dimensional collision occurs

  • apply the conservation of momentum along each axis.

  • relate the momentum components along an axis before the collision to the momentum components along the same axis after the collision.

  • apply the conservation of total kinetic energy to relate the kinetic energies before and after the collision(elastic).

• Key Idea:

  • If two bodies collide and their motion is not along a single axis, the collision is two-dimensional.

  • If the two-body system is closed and isolated\vec{P}{1i} + \vec{P}{2i} = \vec{P}{1f} + \vec{P}{2f}
    
    K{1i} + K{2i} = K{1f} + K{2f}

• Collisions in Two Dimensions\vec{P}{1i} + \vec{P}{2i} = \vec{P}{1f} + \vec{P}{2f} \text{ (Conserved)}
  
  m1v{1i} = m1v{1f}cos\theta1 + m2v{2f}cos\theta2
  
  0 = -m1v{1f}sin\theta1 + m2v{2f}sin\theta2
  \frac{1}{2}m1v{1i} = \frac{1}{2}m1v{1f} + \frac{1}{2}m2v{2f} \text{ (Kinetic energy)}

9-9 Systems with Varying Mass: A Rocket

• Learning Objectives:

  • Apply the first rocket equation to relate the rate at which the rocket loses mass, the speed of the exhaust products relative to the rocket, the mass of the rocket, and the acceleration of the rocket.

  • Apply the second rocket equation to relate change in speed and initial and final mass.

  • For a moving system undergoing a change in mass at a given rate, relate that rate to the change in momentum.

• Key Ideas:

  • In the absence of external forces a rocket accelerates at an instantaneous rate given by
    Rv_{rel} = Ma \text{ (first rocket equation)}

  -