Physics

Chapter 9 Linear Momentum and Collisions

Imron Rosyadi

Chapter 9: Linear Momentum and Collisions

Introduction to Momentum, Impulse, and Collisions

Table of Contents

• 9.1 Linear Momentum
• 9.2 Impulse and Collisions
• 9.3 Conservation of Linear Momentum
• 9.4 Types of Collisions
• 9.5 Collisions in Multiple Dimensions
• 9.6 Center of Mass
• 9.7 Rocket Propulsion

9.1 Linear Momentum

Learning Objectives

• Explain what momentum is, physically.
• Calculate the momentum of a moving object.

Understanding Momentum

• Kinetic energy (\(K = \frac{1}{2}mv^2\)) accounts for mass and speed but lacks directional information.
• Momentum (\(p\)) is a physical quantity that includes both mass and velocity, thus incorporating direction.
• It characterizes the “quantity of motion” of an object.
• From Latin movimentum, meaning “movement.”

Kinetic energy is crucial for understanding motion, but it’s a scalar quantity. It tells us how much “energy of motion” an object has, but not where it’s going. This is where momentum comes in. Momentum is a vector, meaning it has both magnitude and direction, just like velocity. This directional information is critical in analyzing interactions like collisions.

Momentum Formula

• The momentum \(\vec{p}\) of an object is the product of its mass (\(m\)) and its velocity (\(\vec{v}\)):

  \[\vec{p} = m\vec{v}\]

• Momentum is a vector quantity because velocity is a vector.

Caption: The velocity and momentum vectors for the ball are in the same direction. (credit: modification of work by Ben Sutherland)

The arrow over \(p\) and \(v\) indicates that these are vector quantities. This means not only their magnitudes are important, but also their directions. The units of momentum are kg·m/s, which can also be expressed as N·s (Newton-seconds), as we’ll see shortly when discussing impulse.

Momentum in Action

• Momentum helps determine how difficult an object’s motion is to change over a short time interval.

Caption: Supertanker

• High mass means large momentum, difficult to change velocity.

Caption: Gas molecules

• Tiny mass means small momentum, velocities change almost instantaneously upon collision.

Consider the supertanker. Even at a relatively low speed, its enormous mass gives it a huge momentum. This is why it takes a very large force applied for a long time to significantly change its velocity. Conversely, gas molecules have extremely small masses. Even if they’re moving at very high speeds, their momentum is tiny, so their velocities can change dramatically and almost instantly during collisions.

Comparing Momenta

• Air molecule:
  • Average speed: \(\approx 500 \text{ m/s}\)
  • Average mass: \(6 \times 10^{-25} \text{ kg}\)
  • Momentum: \(p_{molecule} = (6 \times 10^{-25} \text{ kg})(500 \text{ m/s}) = 3 \times 10^{-22} \text{ kg} \cdot \text{m/s}\)
• Automobile:
  • Speed: \(\approx 15 \text{ m/s}\)
  • Mass: \(1400 \text{ kg}\)
  • Momentum: \(p_{car} = (1400 \text{ kg})(15 \text{ m/s}) = 21,000 \text{ kg} \cdot \text{m/s}\)

Note

These momenta differ by 27 orders of magnitude! Momentum depends equally on mass and velocity.

This comparison highlights the vast range of momentum values in the universe. It also emphasizes that both mass and velocity are equally important in determining an object’s momentum. A tiny object can have significant momentum if its velocity is extremely high, and a massive object can have large momentum even at low speeds.

9.2 Impulse and Collisions

Learning Objectives

• Explain what an impulse is, physically.
• Describe what an impulse does.
• Relate impulses to collisions.
• Apply the impulse-momentum theorem to solve problems.

Connecting Momentum and Force

• If an object’s velocity changes due to a force, its momentum also changes.
• The change in an object’s motion is proportional to:
  • The magnitude of the force.
  • The time interval over which the force is applied.

Caption: Change in momentum is proportional to the time the force is applied.

This builds on our understanding of Newton’s second law, \(F=ma\). Since acceleration is the rate of change of velocity, a force causes a change in velocity, which in turn causes a change in momentum. The concept of impulse quantifies this relationship.

Defining Impulse

• The product of a force and the time interval over which it acts is called impulse (\(\vec{J}\)).

• For a differential time interval \(dt\): \[d\vec{J} \equiv \vec{F}(t)dt\]

• Total impulse over an interval \(t_f - t_i\): \[\vec{J} \equiv \int_{t_i}^{t_f}\vec{F}(t)dt\]

Caption: A force applied by a tennis racquet to a tennis ball over a time interval generates an impulse acting on the ball.

Impulse is a vector quantity, and its direction is the same as the direction of the average force. The unit of impulse is Newton-seconds (N·s), which is dimensionally equivalent to kg·m/s, the unit of momentum. This equivalence is central to the impulse-momentum theorem.

Average Force and Impulse

• If we don’t know the force function \(F(t)\), we can use the average force \(\vec{F}_{ave}\). \[\vec{F}_{ave} = \frac{1}{\Delta t} \int_{t_i}^{t_f}\vec{F}(t)dt\]

• Therefore, impulse can also be expressed as: \[\vec{J} = \vec{F}_{ave}\Delta t\]

• This means you only need the average force to calculate impulse.

This is a very practical application. In many real-world scenarios, the force during a collision isn’t constant and might be difficult to model as a function of time. However, knowing the initial and final momenta and the duration of the collision allows us to calculate the average force experienced.

Deriving the Impulse-Momentum Theorem

• Starting with \(F(t) = ma(t)\): \[\vec{J} = \int_{t_i}^{t_f}\vec{F}(t)dt = \int_{t_i}^{t_f} m\vec{a}(t)dt\]

• Since \(\vec{a}(t) = \frac{d\vec{v}}{dt}\): \[\vec{J} = m\int_{t_i}^{t_f} \frac{d\vec{v}}{dt}dt = m[\vec{v}(t_f) - \vec{v}(t_i)]\]

• For constant force \(\vec{F}_{ave} = \vec{F} = m\vec{a}\): \[\vec{J} = m\vec{a}\Delta t = m\vec{v}_f - m\vec{v}_i = m(\vec{v}_f - \vec{v}_i)\]

Impulse-Momentum Theorem

An impulse applied to a system changes the system’s momentum, and that change of momentum is exactly equal to the impulse that was applied:

\[\vec{J} = \Delta\vec{p}\]

This theorem is fundamental. It directly links the application of a force over time (impulse) to the change in an object’s momentum. It also implies that the change in momentum is a vector, with the same direction as the impulse.

Impulse-Momentum Theorem Illustrated

• \(\vec{J} = \Delta\vec{p}\)
• Impulse is a vector quantity.
• An impulse causes a change in momentum, not momentum itself.

Caption: Illustration of impulse-momentum theorem.

The diagram visually represents the vector addition of initial momentum and impulse to get the final momentum. It’s crucial to understand that impulse is the cause of the change in momentum. The direction of the impulse is therefore the direction of the change in momentum.

Problem-Solving Strategy: Impulse-Momentum Theorem

1. Express impulse: As force times the relevant time interval (\(\vec{J} = \vec{F}_{ave}\Delta t\)).
2. Express impulse as change of momentum: Usually \(m\Delta\vec{v}\).
3. Equate and solve: Set these two expressions equal to each other and solve for the desired quantity.

Example 9.1: The Arizona Meteor Crater

• Estimate the average and maximum force applied by a meteor to Earth during impact.
  • Meteor radius: \(25 \text{ m}\)
  • Speed: \(1.28 \times 10^4 \text{ m/s}\)
  • Density: \(\rho = 7970 \text{ kg/m}^3\)

Caption: Arizona Meteor Crater.

Strategy:

1. Calculate force Earth applied to meteor using \(\vec{J} = m\Delta\vec{v}\) and \(\vec{J} = \vec{F}_{ave}\Delta t\).
2. Use Newton’s third law for force on Earth.
3. Estimate impact time (\(\Delta t \approx 2 \text{ s}\)).
4. Assume spherical meteor to find mass (\(m = \rho V = \rho \frac{4}{3}\pi R^3\)).
5. Model force function \(F(t) = F_{max}e^{-t^2/(2\tau^2)}\) to find \(F_{max}\).

This example requires us to apply several physics concepts. First, we need to estimate the meteor’s mass. Then, using the impulse-momentum theorem, we can find the average force. Finally, to estimate the maximum force, we need to assume a functional form for how the force varies over time during the impact. The exponential decay model is a reasonable approximation for such brief, intense interactions.

Example 9.1: Solution (Average Force)

• Mass of meteor:

  \(V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (25 \text{ m})^3 = 6.54 \times 10^4 \text{ m}^3\)

  \(m = \rho V = (7970 \text{ kg/m}^3)(6.54 \times 10^4 \text{ m}^3) = 5.21 \times 10^8 \text{ kg}\)

• Average force:

  \(\vec{F}_{ave} = \frac{m\Delta\vec{v}}{\Delta t}\)

  Initial velocity: \(\vec{v}_i = -(1.28 \times 10^4 \text{ m/s})\hat{j}\)

  Final velocity: \(\vec{v}_f = 0\)

  Assumed impact time: \(\Delta t = 2 \text{ s}\)

  \(\vec{F}_{ave} = \frac{(5.21 \times 10^8 \text{ kg})(0 - (-1.28 \times 10^4 \text{ m/s}))\hat{j}}{2 \text{ s}} = +(3.33 \times 10^{12} \text{ N})\hat{j}\)

The average force is directed upward, opposite to the meteor’s initial velocity, as expected. This force is enormous, but as we will see, Earth’s immense mass makes its acceleration negligible.

Example 9.1: Solution (Maximum Force)

• Maximum force (using \(F(t) = F_{max}e^{-t^2/(2\tau^2)}\)):

  We found \(F_{ave} = 3.33 \times 10^{12} \text{ N}\).

  Given \(\Delta t = t_{max} - 0 \text{ s} = 2 \text{ s}\).

  Integral of \(F(t)\) over \(\Delta t\):

  \(\int_{0}^{t_{max}} F_{max}e^{-t^2/(2\tau^2)}dt\)

  Choosing \(\tau = \frac{1}{e} t_{max}\), the integral gives \(F_{ave} = 0.458 F_{max}\).

  \(F_{max} = \frac{F_{ave}}{0.458} = \frac{3.33 \times 10^{12} \text{ N}}{0.458} = 7.27 \times 10^{12} \text{ N}\)

• Force on Earth by meteor:

  \(\vec{F}_{meteor}(t) = -(7.27 \times 10^{12} \text{ N})e^{-t^2/(8\text{s}^2)}\hat{j}\)

The maximum force is significantly higher than the average force, which is typical for collision events where forces ramp up and then decrease. The graph on the next slide illustrates this difference between average and instantaneous forces while maintaining the same total impulse.

Example 9.1: Significance

Caption: Graph of the average force (red) and instantaneous force (blue) during impact.

• The area under both curves is equal to the total impulse (\(\vec{J}\)).
• Earth’s acceleration: \(a = \frac{F_{ave}}{M_{Earth}} = \frac{3.33 \times 10^{12} \text{ N}}{5.97 \times 10^{24} \text{ kg}} = 5.6 \times 10^{-13} \text{ m/s}^2\)
• This acceleration is practically immeasurable.

The key takeaway here is that the total “effect” of the force (the impulse) is the same regardless of whether the force is constant or varying, as long as the area under the force-time curve is the same. Even with such immense forces, Earth’s massive inertia means its overall motion is barely affected. However, the localized effects of such impacts, like crater formation and seismic waves, are very significant.

Example 9.2: The Benefits of Impulse (Airbags)

• A car at \(27 \text{ m/s}\) collides with a building.
• Driver mass: \(m = 87.8 \text{ kg}\) (from \(860 \text{ N}\) weight).
• Case 1: Protected by airbag/seatbelt, stops in \(2.5 \text{ s}\).
  • Average force on driver: \(F = \frac{m\Delta v}{\Delta t} = \frac{(87.8 \text{ kg})(0 - 27 \text{ m/s})}{2.5 \text{ s}} = -948 \text{ N}\)
  • About 1.1 times his weight.
• Case 2: No protection, collision time with steering wheel \(\approx 0.20 \text{ s}\).
  • Average force on driver: \(F = \frac{(87.8 \text{ kg})(0 - 27 \text{ m/s})}{0.20 \text{ s}} = -11,853 \text{ N}\)
  • About 14 times his weight.

Important

Airbags and seatbelts drastically reduce the average force on occupants by increasing the collision time (\(\Delta t\)). The change in momentum (\(\Delta p\)) is the same in both cases.

This example clearly illustrates the practical importance of the impulse-momentum theorem in safety engineering. By extending the duration of the impact, even with the same change in momentum, the force experienced by the driver is significantly reduced, greatly increasing the chances of survival and minimizing injuries. This is a direct application of the relationship \(F_{ave} = \Delta p / \Delta t\).

Newton’s Second Law in Terms of Momentum

• Average force: \(\vec{F}_{ave} = \frac{\Delta\vec{p}}{\Delta t}\)

• For a continuously changing momentum (\(\Delta t \to dt\)):

  \[\vec{F} = \frac{d\vec{p}}{dt}\]

• This is Newton’s second law expressed in terms of momentum.

• If mass is constant, \(\vec{F} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}\).

• This form is particularly useful when the mass of the system changes, like in rocket propulsion.

Newton’s Second Law of Motion in Terms of Momentum

The net external force on a system is equal to the rate of change of the momentum of that system caused by the force:

\[\vec{F}_{net} = \frac{d\vec{p}}{dt}\]

Newton originally formulated his second law in terms of momentum. This form is more general because it accounts for situations where the mass of the system changes over time, which the familiar \(F=ma\) does not explicitly. We’ll explore this further when we discuss rocket propulsion.

Example 9.5: Venus Williams’ Tennis Serve

• Calculate the average force on a tennis ball during Venus Williams’ serve.

  • Final speed: \(58 \text{ m/s}\)
  • Ball mass: \(0.057 \text{ kg}\)
  • Initial horizontal velocity: \(0 \text{ m/s}\)
  • Contact time: \(5.0 \text{ ms} = 5.0 \times 10^{-3} \text{ s}\)

• Strategy:

  • Use \(\vec{F} = \frac{\Delta\vec{p}}{\Delta t}\)
  • \(\Delta p = m(v_f - v_i)\)

• Solution:

  \(\Delta p = (0.057 \text{ kg})(58 \text{ m/s} - 0 \text{ m/s}) = 3.3 \text{ kg} \cdot \text{m/s}\)

  \(F = \frac{3.3 \text{ kg} \cdot \text{m/s}}{5.0 \times 10^{-3} \text{ s}} = 6.6 \times 10^2 \text{ N}\)

This average force is quite large, highlighting the quick, powerful impact a tennis racket has on the ball. Even though gravity also acts on the ball, this force from the racket is far greater during the brief contact time.

9.3 Conservation of Linear Momentum

Learning Objectives

• Explain the meaning of “conservation of momentum.”
• Correctly identify if a system is, or is not, closed.
• Define a system whose momentum is conserved.
• Mathematically express conservation of momentum for a given system.
• Calculate an unknown quantity using conservation of momentum.

Deriving Conservation of Momentum

• Newton’s Third Law: \(\vec{F}_{21} = -\vec{F}_{12}\)
  • Force on \(m_1\) from \(m_2\) equals negative of force on \(m_2\) from \(m_1\).
• In terms of momentum: \(\frac{d\vec{p}_1}{dt} = -\frac{d\vec{p}_2}{dt}\)
• This implies: \(\frac{d\vec{p}_1}{dt} + \frac{d\vec{p}_2}{dt} = 0\)
• Which leads to: \(\frac{d}{dt}(\vec{p}_1 + \vec{p}_2) = 0\)
• Therefore, the sum of momenta is constant: \(\vec{p}_1 + \vec{p}_2 = \text{constant}\)

This derivation shows how the conservation of momentum is a direct consequence of Newton’s third law. The internal forces within a system, by Newton’s third law, always cancel each other out when summed, ensuring that these internal interactions do not change the system’s total momentum.

Total Momentum of a System

• Generalizing to \(N\) objects:

  \[\vec{p}_1 + \vec{p}_2 + \dots + \vec{p}_N = \text{constant}\]

  \[\sum_{j=1}^{N}\vec{p}_j = \text{constant}\]

• If a physical quantity is constant in time, it is conserved.

Caption: Total momentum of the system before and after a collision remains the same.

This image beautifully illustrates conservation of momentum. Even though the individual momenta of the billiard balls change after collision, their vector sum (the total momentum of the system) remains the same. This principle is one of the most powerful in physics.

Requirements for Momentum Conservation

For a system’s momentum to be conserved, it must be a closed system (also called an isolated system), meeting two requirements:

1. Constant System Mass: The total mass of the system must remain constant during the interaction.
   • Mass may be transferred between objects within the system, but the total mass \(M\) is constant: \(\left[\frac{dm}{dt}\right]_{system} = 0\).
2. Zero Net External Force: The net external force on the system must be zero.
   • Internal forces within the system cancel out; they cannot change the total momentum.
   • Any external forces acting on the objects must sum to zero.

Law of Conservation of Momentum

The total momentum of a closed system is conserved:

\[\sum_{j=1}^{N}\vec{p}_j = \text{constant}\]

This law is a cornerstone of physics. It holds true across all scales, from galactic clusters to subatomic particles. It’s crucial to correctly identify a closed system. For example, billiard balls on a table: gravity and normal forces are external, but they cancel out, making the system “effectively” closed in the horizontal plane during collisions.

The Meaning of ‘System’

• A system is the collection of objects whose motion you are interested in.
• The contents (mass) of the system do not change before, during, or after the objects in the system interact.

Caption: The two cars together form the system that is to be analyzed.

Defining your system is the most critical first step in solving any momentum problem. A poor choice of system can make a problem intractable. For instance, in a car crash, if you choose only one car as the system, momentum won’t be conserved because the other car exerts an external force on it. But if both cars are the system, their interaction forces are internal, and total momentum is conserved (assuming other external forces like friction are negligible during the brief impact).

Problem-Solving Strategy: Conservation of Momentum

1. Identify a closed system: Total mass is constant, and no net external force acts on the system.
2. Initial momentum expression: Write an expression for the total momentum of the system before the event (explosion or collision).
3. Final momentum expression: Write an expression for the total momentum of the system after the event.
4. Equate and solve: Set the initial and final momentum expressions equal to each other, and solve for the desired quantity.

Example 9.6: Colliding Carts

• Cart 1: \(m_1 = 675 \text{ g} = 0.675 \text{ kg}\), \(v_{1i} = 0.75 \text{ m/s}\) (right)

• Cart 2: \(m_2 = 500 \text{ g} = 0.500 \text{ kg}\), \(v_{2i} = 1.33 \text{ m/s}\) (right)

• Carts stick together after collision. Find final velocity \(v_f\).

• Strategy:

  • System: Two carts.
  • Closed system: Masses don’t change, friction negligible, internal forces cancel.
  • Apply conservation of momentum: \(\vec{p}_i = \vec{p}_f\).

• Solution: Define right as \(+x\). Initial momentum: \(\vec{p}_i = m_1\vec{v}_{1i} + m_2\vec{v}_{2i} = (0.675)(0.75)\hat{i} + (0.500)(1.33)\hat{i}\) Final momentum: \(\vec{p}_f = (m_1+m_2)\vec{v}_f = (0.675+0.500)\vec{v}_f\)

  \((0.675+0.500)\vec{v}_f = (0.675)(0.75)\hat{i} + (0.500)(1.33)\hat{i}\) \(1.175\vec{v}_f = (0.50625 + 0.665)\hat{i}\) \(\vec{v}_f = \frac{1.17125}{1.175}\hat{i} = (0.997 \text{ m/s})\hat{i}\)

This is a classic perfectly inelastic collision problem, where objects stick together. The total mass increases, but the total momentum remains conserved. The final velocity is a weighted average of the initial velocities, with the weights being the masses.

Example 9.7: A Bouncing Superball

• Superball: \(m = 0.25 \text{ kg}\), dropped from \(h = 1.50 \text{ m}\).
• Bounces with no energy loss, returns to initial height.

1. Ball’s change of momentum (\(\Delta\vec{p}_{ball}\))?
   • System: Ball only (not closed).
   • Use kinematics/conservation of energy to find speed before impact (\(v_1\)).
   • Due to no energy loss, speed after impact (\(v_2\)) is equal in magnitude to \(v_1\).
   • \(v_1 = \sqrt{2gh} = \sqrt{2(9.8 \text{ m/s}^2)(1.50 \text{ m})} = 5.4 \text{ m/s}\).
   • Before impact: \(\vec{p}_1 = m\vec{v}_1 = (0.25 \text{ kg})(-5.4 \text{ m/s})\hat{j} = -(1.4 \text{ kg} \cdot \text{m/s})\hat{j}\).
   • After impact: \(\vec{p}_2 = m\vec{v}_2 = (0.25 \text{ kg})(+5.4 \text{ m/s})\hat{j} = +(1.4 \text{ kg} \cdot \text{m/s})\hat{j}\).
   • \(\Delta\vec{p}_{ball} = \vec{p}_2 - \vec{p}_1 = (1.4 - (-1.4))\hat{j} = +(2.8 \text{ kg} \cdot \text{m/s})\hat{j}\).
2. Earth’s change of momentum (\(\Delta\vec{p}_{Earth}\))?
   • System: Superball + Earth (closed system).
   • Total momentum is conserved, so \(\Delta\vec{p}_{system} = 0\).
   • \(\Delta\vec{p}_{ball} + \Delta\vec{p}_{Earth} = 0 \implies \Delta\vec{p}_{Earth} = -\Delta\vec{p}_{ball} = -(2.8 \text{ kg} \cdot \text{m/s})\hat{j}\).
3. Earth’s change of velocity (\(\Delta\vec{v}_{Earth}\))?
   • \(M_{Earth} = 5.97 \times 10^{24} \text{ kg}\).
   • \(\Delta\vec{v}_{Earth} = \frac{\Delta\vec{p}_{Earth}}{M_{Earth}} = \frac{-(2.8 \text{ kg} \cdot \text{m/s})\hat{j}}{5.97 \times 10^{24} \text{ kg}} = -(4.7 \times 10^{-25} \text{ m/s})\hat{j}\).
   • Utterly negligible!

This example demonstrates the importance of defining the system. When only the ball is the system, momentum is not conserved due to external forces from gravity and the floor. But by including the Earth, the system becomes closed, and total momentum is conserved. Even though Earth’s change in momentum is equal in magnitude to the ball’s, its enormous mass results in an imperceptibly small change in velocity.

9.4 Types of Collisions

Learning Objectives

• Identify the type of collision.
• Correctly label a collision as elastic or inelastic.
• Use kinetic energy along with momentum and impulse to analyze a collision.

Classifying Interactions

• Momentum is conserved in all interactions within a closed system.
• However, kinetic energy may or may not be conserved.
• Interactions are classified based on how kinetic energy changes.

1. Explosions

• A single object breaks into multiple pieces.
• Initial momentum of the system = 0 (if initially motionless).
• Final net momentum of all pieces = 0 (momentum is conserved).
• Kinetic energy increases.
  • Example: Firecracker, bow and arrow, rocket launch.
  • Potential energy is converted into kinetic energy.

Explosions are the reverse of inelastic collisions. Chemical or elastic potential energy stored in the system is converted into kinetic energy of the fragments. Even if the total kinetic energy increases, the total momentum remains conserved, which is why the center of mass continues its original trajectory.

2. Inelastic Collisions

• Two or more objects collide and stick together, forming a single composite object.
• Total mass of composite object = sum of original masses.
• Momentum is conserved.
• Kinetic energy decreases.
  • This is called a perfectly inelastic collision.
  • Energy is lost to heat, sound, deformation.
• General inelastic collision: objects collide but do not stick, and \(0 < K_f < K_i\).
• In the extreme case of perfectly inelastic collision where final velocity is zero, kinetic energy loss is maximum.

Inelastic collisions are very common in the real world. Think of car crashes, a bullet hitting a block of wood, or even a football player tackling another. Some kinetic energy is always converted into other forms, such as heat, sound, and the energy needed to permanently deform the colliding objects.

3. Elastic Collisions

• Two or more objects collide and bounce off each other.
• They move away from each other at the same relative speed as they approached.
• Momentum is conserved.
• Kinetic energy is conserved. (\(K_f = K_i\))
  • Example: Collisions of billiard balls, subatomic particles.
  • This is an idealized model; perfectly elastic collisions are rare in macroscopic objects.

Collision Types and Kinetic Energy

• Inelastic: \(0 < K_f < K_i\)
• Perfectly Inelastic: \(K_f\) is minimum (objects stick together).
• Elastic: \(K_f = K_i\)

Elastic collisions are an important theoretical construct, especially in particle physics, where kinetic energy can be conserved very precisely. For everyday objects, some energy is always lost to friction, deformation, or sound, so “perfectly elastic” is an approximation.

Problem-Solving Strategy: Collisions

1. Define a closed system.
2. Conservation of Momentum: Write down the expression for momentum conservation (\(\sum \vec{p}_i = \sum \vec{p}_f\)).
3. Kinetic Energy:
   • If kinetic energy is conserved (elastic collision), write down the expression for kinetic energy conservation (\(\sum K_i = \sum K_f\)).
   • If kinetic energy is NOT conserved (inelastic/perfectly inelastic), use the expression for the change of kinetic energy (\(\Delta K = K_f - K_i\)).
4. Solve the system: You will typically have two equations (momentum and energy/change in energy) for two unknowns.

Example 9.10: Formation of a Deuteron

• Proton (\(m_p = 1.67 \times 10^{-27} \text{ kg}\)) collides with a neutron (\(m_n \approx m_p\)) to form a deuteron.

• Proton velocity: \(v_p = 7.0 \times 10^6 \text{ m/s}\) (right).

• Neutron velocity: \(v_n = 4.0 \times 10^6 \text{ m/s}\) (left).

• Find the velocity of the deuteron (\(v_d\)).

• Strategy:

  • Collision is perfectly inelastic (particles stick together).
  • Momentum is conserved, kinetic energy is NOT.
  • Define right as \(+x\). Let \(M = m_p = m_n\).

• Solution: Initial momentum: \(p_i = Mv_p - Mv_n\) Final momentum: \(p_f = (M+M)v_d = 2Mv_d\)

  \(Mv_p - Mv_n = 2Mv_d\) \(v_p - v_n = 2v_d\) \((7.0 \times 10^6 \text{ m/s}) - (4.0 \times 10^6 \text{ m/s}) = 2v_d\) \(3.0 \times 10^6 \text{ m/s} = 2v_d\) \(v_d = 1.5 \times 10^6 \text{ m/s}\) (to the right) \(\vec{v}_d = (1.5 \times 10^6 \text{ m/s})\hat{i}\)

This is a good example of how particle collisions are analyzed. By using conservation of momentum, even without knowing the details of the interaction forces, we can predict the outcome of the collision.

Example 9.11: Ice Hockey 2 (Elastic Collision)

• Red puck: \(m_1 = 15 \text{ g} = 0.015 \text{ kg}\), \(v_{1i} = 0 \text{ m/s}\).

• Blue puck: \(m_2 = 12 \text{ g} = 0.012 \text{ kg}\), \(v_{2i} = 2.5 \text{ m/s}\) (left).

• Collision is perfectly elastic. Find final velocities \(v_{1f}\) and \(v_{2f}\).

• Strategy:

  • Closed system: Two pucks, no friction.
  • Momentum is conserved: \(p_i = p_f\).
  • Kinetic energy is conserved: \(K_i = K_f\).
  • Two equations, two unknowns (\(v_{1f}, v_{2f}\)).

• Solution: Define left as \(+x\).

  1. Conservation of Momentum: \(m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\) \((0.012)(2.5) = (0.015)v_{1f} + (0.012)v_{2f}\) \(0.030 = 0.015v_{1f} + 0.012v_{2f}\) (Eq. 1)

  2. Conservation of Kinetic Energy: \(\frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2\) \((0.012)(2.5)^2 = (0.015)v_{1f}^2 + (0.012)v_{2f}^2\) \(0.075 = 0.015v_{1f}^2 + 0.012v_{2f}^2\) (Eq. 2)

  Solving these simultaneous equations (tedious but possible): \(v_{1f} = 2.22 \text{ m/s}\) (left) \(v_{2f} = -0.28 \text{ m/s}\) (right)

Elastic collision problems often involve solving a system of two equations (momentum and kinetic energy) with two unknowns. The algebraic solution can be lengthy but follows a standard procedure. The result shows that the blue puck reverses direction, and the red puck moves to the left.

9.5 Collisions in Multiple Dimensions

Learning Objectives

• Express momentum as a two-dimensional vector.
• Write equations for momentum conservation in component form.
• Calculate momentum in two dimensions, as a vector quantity.

Momentum in Multiple Dimensions

• Momentum is a vector; it can be expressed in components (x, y, z).
• Conservation of momentum applies independently to each dimension.
  • \(\vec{F}_{net} = \frac{d\vec{p}}{dt}\) implies: \(F_x = \frac{dp_x}{dt}\), \(F_y = \frac{dp_y}{dt}\), \(F_z = \frac{dp_z}{dt}\)
• For a collision, we write conservation of momentum for each component: \(p_{f,x} = p_{1,i,x} + p_{2,i,x}\) \(p_{f,y} = p_{1,i,y} + p_{2,i,y}\)

This is a crucial extension from one-dimensional problems. Because momentum is a vector, its conservation must hold true for each of its perpendicular components independently. This means a single vector equation of momentum conservation splits into multiple scalar equations, one for each dimension.

Problem-Solving Strategy: Conservation of Momentum in Two Dimensions

1. Identify a closed system.
2. X-direction: Write down and solve the equation for conservation of momentum in the x-direction. This gives the x-component of the desired vector quantity.
3. Y-direction: Write down and solve the equation for conservation of momentum in the y-direction. This gives the y-component of the desired vector quantity.
4. Magnitude and Direction: Use the Pythagorean theorem (\(v_f = \sqrt{v_{f,x}^2 + v_{f,y}^2}\)) and trigonometry (\(\theta = \tan^{-1}(v_{f,y}/v_{f,x})\)) to calculate the magnitude and direction of the final vector.

This strategy is fundamental for solving any multi-dimensional momentum problem. Breaking the problem into component equations simplifies the vector algebra into scalar algebra for each direction.

Example 9.14: Traffic Collision

• Car: \(m_c = 1200 \text{ kg}\), \(v_c = 60 \text{ km/hr}\) (East)

  • \(60 \text{ km/hr} = 16.67 \text{ m/s}\)

• Truck: \(m_T = 3000 \text{ kg}\), \(v_T = 40 \text{ km/hr}\) (North)

  • \(40 \text{ km/hr} = 11.11 \text{ m/s}\)

• Vehicles lock together (perfectly inelastic collision). Find final velocity of wreckage \(\vec{v}_w\).

• Strategy:

  • System: Car + Truck (closed for instant of collision).
  • Momentum is conserved in x and y components.
  • Define East as \(+x\), North as \(+y\).

• Solution: Initial momentum: \(\vec{p}_i = m_c\vec{v}_c + m_T\vec{v}_T\) Final momentum: \(\vec{p}_f = (m_c+m_T)\vec{v}_w\) \(\vec{p}_i = \vec{p}_f \implies m_c\vec{v}_c + m_T\vec{v}_T = (m_c+m_T)\vec{v}_w\)

  x-direction: \(m_c v_c = (m_c+m_T)v_{w,x}\) \(v_{w,x} = \frac{m_c}{m_c+m_T}v_c = \frac{1200}{1200+3000}(16.67) = \frac{1200}{4200}(16.67) = 4.76 \text{ m/s}\)

  y-direction: \(m_T v_T = (m_c+m_T)v_{w,y}\) \(v_{w,y} = \frac{m_T}{m_c+m_T}v_T = \frac{3000}{1200+3000}(11.11) = \frac{3000}{4200}(11.11) = 7.94 \text{ m/s}\)

• Magnitude of \(\vec{v}_w\): \(|\vec{v}_w| = \sqrt{v_{w,x}^2 + v_{w,y}^2} = \sqrt{(4.76)^2 + (7.94)^2} = \sqrt{22.66 + 63.04} = \sqrt{85.7} \approx 9.26 \text{ m/s}\) (\(9.26 \text{ m/s} \approx 33.3 \text{ km/hr}\))

• Direction of \(\vec{v}_w\): \(\theta = \tan^{-1}\left(\frac{v_{w,y}}{v_{w,x}}\right) = \tan^{-1}\left(\frac{7.94}{4.76}\right) = \tan^{-1}(1.668) \approx 59.1^\circ\) North of East.

This is a realistic scenario that accident investigators might analyze. It shows how the total momentum is conserved in both x and y components separately, leading to a resultant velocity vector for the combined wreckage. The initial kinetic energy is not conserved, as this is a perfectly inelastic collision.

9.6 Center of Mass

Learning Objectives

• Explain the meaning and usefulness of the concept of center of mass.
• Calculate the center of mass of a given system.
• Apply the center of mass concept in two and three dimensions.
• Calculate the velocity and acceleration of the center of mass.

Beyond Single Particles: Extended Objects

• So far, we’ve treated objects as point particles.
• Real objects are made of many particles, distributed non-uniformly.
• Internal forces (particles on each other) and external forces act.
• How do we describe the motion of the object as a whole?

Caption: As the cat falls, its body performs complicated motions so it can land on its feet, but one point in the system moves with the simple uniform acceleration of gravity.

The cat example beautifully illustrates the concept. While different parts of the cat’s body move in complex ways to orient itself for landing, there’s a special point—the center of mass—that follows a simple parabolic trajectory under gravity. This simplification is why the center of mass is so powerful.

Internal and External Forces on a System of Particles

• Consider an extended object of total mass \(M\), made of \(N\) particles (\(m_j\)).
• Net force on \(j^{th}\) particle: \(\vec{f}_j = \vec{f}_{j,int} + \vec{f}_{j,ext}\)
• Net force on the object (system): \(\vec{F}_{net} = \sum_{j=1}^{N} (\vec{f}_{j,int} + \vec{f}_{j,ext})\)
• By Newton’s Third Law, \(\sum_{j=1}^{N} \vec{f}_{j,int} = 0\) (internal forces cancel in pairs).
• So, \(\vec{F}_{net} = \sum_{j=1}^{N} \vec{f}_{j,ext} = \vec{F}_{ext}\) (sum of external forces).

Note

The total change of momentum of the entire object is due only to the external forces. Internal forces do not change the momentum of the object as a whole.

This is a critical point. Even though individual particles within the object experience internal forces and change their momentum, the sum of all internal forces is zero. Therefore, internal interactions cannot change the total momentum of the entire system. This is why you can’t lift yourself by pulling on your shoelaces!

Defining the Center of Mass (\(\vec{r}_{CM}\))

• The total momentum of the system is \(\vec{p}_{CM} = \sum_{j=1}^{N} \vec{p}_j\).

• Newton’s second law for the system: \(\vec{F}_{ext} = \frac{d\vec{p}_{CM}}{dt}\).

• To relate this to acceleration, we define the center of mass: \[\vec{r}_{CM} \equiv \frac{1}{M}\sum_{j=1}^{N}m_j\vec{r}_j\] Where \(M = \sum_{j=1}^{N} m_j\) is the total mass.

• The center of mass is the point in the object that obeys Newton’s second law for the system: \[\vec{F}_{ext} = M\vec{a}_{CM}\] Where \(\vec{a}_{CM} = \frac{d^2\vec{r}_{CM}}{dt^2}\).

Important

There does not have to be any actual mass at the center of mass of an object (e.g., a hollow sphere, a donut).

The center of mass is a mathematical average position, weighted by mass. It’s the unique point where, if all the system’s mass were concentrated there, it would behave exactly as the real system does under the influence of external forces. This simplifies complex motion into a single point’s trajectory.

Components of Center of Mass

• The center of mass can be calculated for each component independently:

  \[x_{CM} = \frac{1}{M}\sum_{j=1}^{N}m_jx_j\]

  \[y_{CM} = \frac{1}{M}\sum_{j=1}^{N}m_jy_j\]

  \[z_{CM} = \frac{1}{M}\sum_{j=1}^{N}m_jz_j\]

• The position vector is \(\vec{r}_{CM} = x_{CM}\hat{i} + y_{CM}\hat{j} + z_{CM}\hat{k}\).

Velocity and Acceleration of Center of Mass

• Instantaneous velocity of the center of mass:

  \[\vec{v}_{CM} = \frac{d\vec{r}_{CM}}{dt} = \frac{1}{M}\sum_{j=1}^{N}m_j\vec{v}_j\]

• Acceleration of the center of mass: \[\vec{a}_{CM} = \frac{d\vec{v}_{CM}}{dt} = \frac{1}{M}\sum_{j=1}^{N}m_j\vec{a}_j = \frac{\vec{F}_{ext}}{M}\]

These equations show that the velocity and acceleration of the center of mass are simply the mass-weighted averages of the individual particle velocities and accelerations, respectively. The second form of \(\vec{a}_{CM}\) is particularly powerful, demonstrating that the center of mass moves as if all external forces acted directly upon it.

Problem-Solving Strategy: Calculating the Center of Mass

1. Define coordinate system: Often placing the origin at one of the particles.
2. Particle coordinates: Determine \((x_j, y_j, z_j)\) for each particle.
3. Particle masses: Determine \(m_j\) for each particle and sum them to get total mass \(M\).
4. Calculate components: Use the component formulas for \(x_{CM}, y_{CM}, z_{CM}\).
5. Magnitude (if needed): Use Pythagorean theorem for \(|\vec{r}_{CM}|\).

Example 9.16: Center of Mass of the Earth-Moon System

• \(m_e = 5.97 \times 10^{24} \text{ kg}\)

• \(m_m = 7.36 \times 10^{22} \text{ kg}\)

• Distance \(r_m = 3.82 \times 10^8 \text{ m}\) (center to center)

• Strategy:

  • Place origin at Earth’s center (\(r_e = 0\)).
  • Use \(\vec{r}_{CM} = \frac{m_e\vec{r}_e + m_m\vec{r}_m}{m_e+m_m}\).

• Solution:

  \(R = \frac{(5.97 \times 10^{24} \text{ kg})(0 \text{ m}) + (7.36 \times 10^{22} \text{ kg})(3.82 \times 10^8 \text{ m})}{5.97 \times 10^{24} \text{ kg} + 7.36 \times 10^{22} \text{ kg}}\)

  \(R = \frac{2.812 \times 10^{31}}{6.0436 \times 10^{24}} = 4.64 \times 10^6 \text{ m}\)

• Significance:

  • Earth’s radius is \(6.37 \times 10^6 \text{ m}\).
  • The center of mass of the Earth-Moon system is \((6.37 - 4.64) \times 10^6 \text{ m} = 1.73 \times 10^6 \text{ m}\) below Earth’s surface.

This surprising result shows that the Earth itself wobbles slightly around this common center of mass. This is why a coordinate system placed at the exact center of the Earth is not truly inertial if you’re looking at the Earth-Moon system as a whole.

Center of Mass of Continuous Objects

• For objects with uniformly distributed mass:
  • Replace discrete masses \(m_j\) with infinitesimal mass elements \(dm\).
  • Replace summation with integration: \[\vec{r}_{CM} = \frac{1}{M}\int\vec{r}dm\]
• To integrate, express \(dm\) in terms of mass density (\(\lambda\), \(\sigma\), or \(\rho\)) and a differential length (\(dL\)), area (\(dA\)), or volume (\(dV\)).

Example 9.18: CM of a Uniform Thin Hoop

• Find the center of mass of a uniform thin hoop of mass \(M\) and radius \(r\).

• Strategy:

  • Symmetry suggests CM is at the geometric center.
  • Place origin at the center of the hoop.
  • Use linear mass density \(\lambda = M/(2\pi r)\).
  • \(dm = \lambda ds = \lambda r d\theta\).

• Solution:

  \(\vec{r}_{CM} = \frac{1}{M}\int_{0}^{2\pi} [(r\cos\theta)\hat{i} + (r\sin\theta)\hat{j}] (\lambda r d\theta)\)

  \(\vec{r}_{CM} = \frac{1}{M}\int_{0}^{2\pi} [(r\cos\theta)\hat{i} + (r\sin\theta)\hat{j}] \left(\frac{M}{2\pi r}\right) r d\theta\)

  \(\vec{r}_{CM} = \frac{1}{2\pi}\int_{0}^{2\pi} [(r\cos\theta)\hat{i} + (r\sin\theta)\hat{j}] d\theta\)

  \(x_{CM} = \frac{r}{2\pi}\int_{0}^{2\pi} \cos\theta d\theta = \frac{r}{2\pi}[\sin\theta]_{0}^{2\pi} = 0\)

  \(y_{CM} = \frac{r}{2\pi}\int_{0}^{2\pi} \sin\theta d\theta = \frac{r}{2\pi}[-\cos\theta]_{0}^{2\pi} = 0\)

  \(\vec{r}_{CM} = 0\hat{i} + 0\hat{j} = \vec{0}\)

• As expected, the center of mass is at the origin (geometric center).

This example confirms our intuition about symmetric objects. For a uniform hoop, the mass is distributed symmetrically around its geometric center, so the center of mass is at that point, which for a hollow object, is a point in space where no physical mass exists.

Center of Mass and Conservation of Momentum

• For a system of \(N\) objects, the velocity of the center of mass is:

  \[\vec{v}_{CM} = \frac{1}{M}\sum_{j=1}^{N}m_j\vec{v}_j\]

• This also means \(M\vec{v}_{CM} = \sum_{j=1}^{N}m_j\vec{v}_j = \vec{P}_{total}\).

• Conservation of momentum implies:

  \(M\vec{v}_{CM,f} = M\vec{v}_{CM,i}\)

  Therefore, \(\vec{v}_{CM,f} = \vec{v}_{CM,i}\)

In the absence of an external force, the velocity of the center of mass never changes.

This applies to a vast collection of interacting particles, even if individual particles’ momenta change!

This is a profound result. It means that the center of mass of an isolated system moves with constant velocity, even if the individual parts of the system are moving chaotically, colliding, or exploding. This simplifies the analysis of complex systems dramatically, allowing us to track the system as a whole.

Example 9.19: Fireworks Display

• A fireworks rocket explodes, sending fragments outward.
• Before explosion: The rocket (shell) is a projectile. Its center of mass follows a parabolic trajectory (after engine cutoff).
• At explosion: Internal forces cause fragments to fly apart. These internal forces sum to zero.
• After explosion: The velocity of the center of mass of the entire system of fragments continues on the exact same parabolic trajectory as the shell would have followed, ignoring air resistance.
• Each fragment then follows its own parabolic trajectory.

Caption: Exploding fireworks are a vivid example of conservation of momentum and the motion of the center of mass.

This visually stunning example reinforces the concept that internal forces do not affect the motion of the center of mass. The explosion is a dramatic internal interaction, but it doesn’t change the overall trajectory of the system’s center of mass.

9.7 Rocket Propulsion

Learning Objectives

• Describe the application of conservation of momentum when the mass changes with time, as well as the velocity.
• Calculate the speed of a rocket in empty space, at some time, given initial conditions.
• Calculate the speed of a rocket in Earth’s gravity field, at some time, given initial conditions.

Rockets: Changing Mass, Changing Momentum

• Rocket engines expel burned fuel gases (mass + velocity = momentum).
• By conservation of momentum, the rocket gains momentum in the opposite direction.
• The rocket’s mass continuously decreases as fuel is ejected.
• Therefore, the acceleration is not constant (force is constant, but mass decreases).

Caption: Rocket accelerating due to expulsion of fuel.

Rocket propulsion is a prime example of a system where mass is continuously changing. We must use the general form of Newton’s second law (\(\vec{F}_{net} = d\vec{p}/dt\)) to analyze it. The key is to consider the system as the rocket plus the fuel being ejected at a given instant, ensuring that the total mass of this instantaneous system is conserved, even though the rocket’s mass changes.

The Rocket Equation (Deep Space)

• Consider a system of rocket + fuel in deep space (no external forces).

• At time \(t\), rocket has mass \(m\) and velocity \(v\).

• During \(dt\), it ejects mass \(dm_g\) (fuel) at velocity \(u\) (relative to rocket) in the \(-x\) direction.

• Rocket’s mass becomes \(m-dm_g\), velocity becomes \(v+dv\).

• Momentum conservation: \(mv = (m-dm_g)(v+dv) + dm_g(v-u)\)

• Simplifying and noting \(dm_g = -dm\) (decrease in rocket mass): \(mdv = -dm \cdot u\) \(dv = -u \frac{dm}{m}\)

• Integrating from initial mass \(m_0\) to final mass \(m\): \[\int_{v_i}^{v_f} dv = -u\int_{m_0}^{m} \frac{dm}{m}\] \[\Delta v = v_f - v_i = u \ln\left(\frac{m_0}{m}\right)\]

Important

This is the Tsiolkovsky rocket equation. It shows that the change in velocity depends on the exhaust speed (\(u\)) and the ratio of initial to final mass.

The rocket equation is incredibly important for space travel. It shows that to achieve a large change in velocity (delta-v), you need a high exhaust velocity and/or a very large mass ratio (meaning a lot of fuel compared to the empty rocket). The logarithmic dependence means that as the mass ratio increases, each additional unit of fuel provides a proportionally smaller increase in delta-v.

Problem-Solving Strategy: Rocket Propulsion

1. Change of Velocity: Use the rocket equation: \(\Delta v = u \ln\left(\frac{m_0}{m}\right)\).
2. Acceleration/Thrust:
   • Thrust force: \(F = u \frac{dm_g}{dt}\) (rate of change of fuel mass).
   • Acceleration: \(a(t) = \frac{F}{m(t)}\), where \(m(t)\) is the instantaneous mass of the rocket.

Rocket in a Gravitational Field

• When launching from Earth, gravity is an external force \(\vec{F}_{ext} = -mg\hat{j}\).

• This adds an impulse \(d\vec{J} = \vec{F}_{ext}dt = -mgdt\hat{j}\).

• Including this in momentum conservation: \(mdv - dm_g u = -mgdt\) \(mdv = -dm u - mgdt\) \(dv = -u \frac{dm}{m} - g dt\)

• Integrating from \(t=0\) to \(t=\Delta t\) (burn time): \[\Delta v = u \ln\left(\frac{m_0}{m}\right) - g\Delta t\]

Note

The longer the burn time (\(\Delta t\)), the smaller the rocket’s change of velocity will be due to gravity’s opposing force. Rockets need to burn fuel as quickly as possible during liftoff.

This modified rocket equation accounts for the constant drag of gravity during launch. The \(-g\Delta t\) term represents the velocity lost due to gravity over the burn time. This term explains why rockets accelerate so powerfully at liftoff: to minimize the time spent fighting gravity and maximize the \(\Delta v\) gained.

Key Takeaways

• Momentum (\(\vec{p} = m\vec{v}\)) is a vector quantity that characterizes an object’s “quantity of motion,” including its direction.
• Impulse (\(\vec{J} = \int \vec{F} dt = \vec{F}_{ave}\Delta t\)) is the product of force and the time it acts, and it equals the change in an object’s momentum (\(\vec{J} = \Delta\vec{p}\)).
• Conservation of Linear Momentum states that the total momentum of a closed (isolated) system remains constant. A closed system has constant mass and zero net external force.
• Collisions are classified by kinetic energy changes:
  • Elastic: Kinetic energy conserved (\(K_f = K_i\)).
  • Inelastic: Kinetic energy not conserved (\(K_f < K_i\)).
  • Perfectly Inelastic: Objects stick together, maximum \(K\) loss.
  • Explosions: Internal energy converts to kinetic energy (\(K_f > K_i\)).
• In multiple dimensions, momentum is conserved independently for each component (\(x, y, z\)).
• The Center of Mass (\(\vec{r}_{CM} = \frac{1}{M}\sum m_j\vec{r}_j\)) is the mass-weighted average position of a system. It moves as if all external forces acted upon it, and its velocity is constant in an isolated system.
• Rocket Propulsion involves changing mass. The rocket equation (\(\Delta v = u \ln(m_0/m)\)) describes velocity change in deep space. In gravity, the effect of gravity subtracts from the \(\Delta v\) gained.

Key Equations

Equation	Description
\(\vec{p} = m\vec{v}\)	Linear momentum
\(\vec{J} \equiv \int_{t_i}^{t_f}\vec{F}(t)dt\)	Impulse (general)
\(\vec{J} = \vec{F}_{ave}\Delta t\)	Impulse (average force)
\(\vec{J} = \Delta\vec{p}\)	Impulse-momentum theorem
\(\vec{F}_{net} = \frac{d\vec{p}}{dt}\)	Newton’s 2nd Law (momentum form)
\(\sum_{j=1}^{N}\vec{p}_j = \text{constant}\)	Conservation of linear momentum

Equation	Description
\(x_{CM} = \frac{1}{M}\sum m_jx_j\)	X-component of center of mass
\(y_{CM} = \frac{1}{M}\sum m_jy_j\)	Y-component of center of mass
\(\vec{r}_{CM} = \frac{1}{M}\int\vec{r}dm\)	Center of mass (continuous object)
\(\vec{v}_{CM} = \frac{1}{M}\sum m_j\vec{v}_j\)	Velocity of center of mass
\(\Delta v = u \ln\left(\frac{m_0}{m}\right)\)	Rocket equation (deep space)
\(\Delta v = u \ln\left(\frac{m_0}{m}\right) - g\Delta t\)	Rocket equation (with gravity)

Key Terms

• Linear Momentum (\(\vec{p}\)): Product of an object’s mass and velocity, a vector quantity.
• Impulse (\(\vec{J}\)): Product of a force and the time interval over which it acts; a vector quantity equal to the change in momentum.
• Impulse-Momentum Theorem: States that the impulse applied to a system equals the change in the system’s momentum.
• Closed System (Isolated System): A system with constant total mass and zero net external force; its total momentum is conserved.
• Conservation of Momentum: Principle stating that the total momentum of a closed system remains constant over time.
• Elastic Collision: A collision in which total kinetic energy is conserved.

• Inelastic Collision: A collision in which kinetic energy is not conserved (some kinetic energy is lost).
• Perfectly Inelastic Collision: An inelastic collision where the colliding objects stick together, resulting in the maximum possible loss of kinetic energy.
• Explosion: An interaction where a single object breaks apart into multiple pieces, and the system’s kinetic energy increases.
• Center of Mass (\(\vec{r}_{CM}\)): The unique point where the entire mass of a system can be considered concentrated for analyzing overall motion; its motion is governed by external forces.
• Rocket Equation: Describes the change in velocity of a rocket due to the expulsion of mass (fuel).
• Thrust: The force exerted on a rocket by the expulsion of its exhaust gases.