Chapter 9 - Linear Momentum

9. 1 Linear Momentum

Chapter 8 discussed situations difficult to analyze with Newton's laws; these problems were solved using the conservation of energy principle.
Consider an archer (60 kg) standing on frictionless ice, firing a 0.030 kg arrow horizontally at 85 m/s. The problem is to find the archer's velocity after firing the arrow.
Newton's third law indicates that the force of the bow on the arrow is paired with an opposite force on the bow and archer, causing the archer to slide backward.
Motion models (e.g., constant acceleration) are not applicable due to lack of information about the archer's acceleration.
Force models (e.g., net force) are not applicable since the forces are unknown.
Energy models are not applicable because the work done in pulling the bowstring and the elastic potential energy are unknown.
Introducing linear momentum simplifies solving this problem.

Isolated System of Two Particles

Consider an isolated system of two particles with masses m1 and m2, moving at velocities v1 and v2.
Since the system is isolated, the only force on one particle is from the other. Action-reaction pairs exist, where F{12} = -F{21}.
From a system perspective, the sum of forces on particles in an isolated system is zero: F{21} + F{12} = 0
Incorporating Newton's second law (\sum F = ma), the equation becomes m1 a1 + m2 a2 = 0
Replacing acceleration with its definition (a = \frac{dv}{dt}):
m1 \frac{dv1}{dt} + m2 \frac{dv2}{dt} = 0
If masses are constant, the equation is rewritten as:
\frac{d(m1 v1)}{dt} + \frac{d(m2 v2)}{dt} = 0
Which simplifies to:
\frac{d}{dt} (m1 v1 + m2 v2) = 0 (9.1)
This implies that the sum m1v1 + m2v2 is constant over time, similar to the conservation of total energy in an isolated system.

Definition of Linear Momentum

The quantity mv is significant because its sum for an isolated system of particles is conserved; this quantity is called linear momentum.
The linear momentum p of a particle or object modeled as a particle with mass m moving at velocity v is defined as:
p = mv (9.2)
Linear momentum is a vector quantity with direction along v, dimensions ML/T, and SI unit kg \cdot m/s.
Component equations:
px = mvx
py = mvy
pz = mvz
Momentum provides a quantitative distinction between heavy and light particles moving at the same velocity.
Newton described mv as "quantity of motion."

Differences Between Kinetic Energy and Momentum

Kinetic energy is a scalar, momentum is a vector.
- For two equal-mass particles moving towards each other with equal speeds, the system has kinetic energy but zero momentum.
Kinetic energy can transform into other forms of energy (potential, internal), while linear momentum has only one type and doesn't transform.
These differences make momentum-based models distinct from and complementary to energy-based models.

Newton's Second Law in Terms of Momentum

Newton's second law relates the linear momentum of a particle to the net force acting on it.
\sum F = ma = m \frac{dv}{dt}
Assuming constant mass, m can be brought inside the derivative:
\sum F = \frac{d(mv)}{dt} = \frac{dp}{dt} (9.3)
This shows that the time rate of change of a particle's linear momentum equals the net force acting on it.
In this form, Newton's second law is more general, applicable even when mass changes (e.g., rocket propulsion).

Quick Quizzes

Quick Quiz 9.1: Two objects have equal kinetic energies. How do the magnitudes of their momenta compare?
- (d) not enough information to tell
Quick Quiz 9.2: Catching a baseball versus a medicine ball (10x mass). Rank from easiest to hardest to catch:
- (a) same speed as the baseball, (b) same momentum, or (c) same kinetic energy.

2 Analysis Model: Isolated System (Momentum)

Equation 9.1 can be written as:
\frac{d}{dt}(p1 + p2) = 0
Since the time derivative of the total momentum P{tot} = p1 + p2 is zero, the total momentum of the isolated system is constant: P{tot} = constant (9.4)
\Delta P_{tot} = 0 (9.5)
Pitfall Prevention 9.1: The momentum of an isolated system is conserved, but not necessarily the momentum of an individual particle within the system.
For a two-particle system:
p{1i} + p{2i} = p{1f} + p{2f}
In component form:
p{1ix} + p{2ix} = p{1fx} + p{2fx}
p{1iy} + p{2iy} = p{1fy} + p{2fy}
p{1iz} + p{2iz} = p{1fz} + p{2fz}
(9.6)
The isolated system (momentum) model applies when no external forces act on the system. This is similar to the energy version but now for momentum.
Whenever two or more particles in an isolated system interact, the total momentum of the system does not change.

Key Points of the Isolated System (Momentum) Model

No restrictions on the type of forces (conservative or nonconservative) acting within the system.
Forces must be internal to the system.

Analysis Model: Isolated System (Momentum)

If no external forces act on the system, the total momentum of the system is constant:
\Delta P_{tot} = 0 (9.5)
Examples:
- A cue ball strikes another ball on a pool table.
- A spacecraft fires its rockets and moves faster through space (Section 9.9).
- Molecules in a gas move and strike each other (Chapter 20).
- An incoming particle strikes a nucleus, creating a new nucleus and a different outgoing particle (Chapter 43).
- An electron and a positron annihilate to form two outgoing photons (Chapter 44).

Example 9.1: The Archer

A 60-kg archer on frictionless ice fires a 0.030-kg arrow horizontally at 85 m/s. Find the archer's recoil velocity.
The system consists of the archer (including the bow) and the arrow.
The system is not isolated vertically due to gravitational and normal forces, but it is isolated horizontally.
The total horizontal momentum before firing is zero.
Using the isolated system (momentum) model:
\Delta p = 0 \implies pi = pf \implies 0 = m1 v{1f} + m2 v{2f}
Solving for v{1f} (archer's velocity): v{1f} = - \frac{m2}{m1} v_{2f} = - \frac{0.030 kg}{60 kg} (85 m/s) = -0.0421 m/s
The negative sign indicates the archer moves to the left.
WHAT IF? If the arrow is fired at an angle \theta with the horizontal, the recoil velocity decreases in magnitude. v{1f} = - \frac{m2}{m1} v{2f} \cos \theta
- For \theta = 0, \cos \theta = 1 and v_{1f} is the same as before.
- For \theta = 90, \cos \theta = 0 and v_{1f} = 0 (no recoil).

Example 9.2: Can We Really Ignore the Kinetic Energy of the Earth?

Verify that the kinetic energy of the Earth can be ignored when considering the energy of a system consisting of the Earth and a dropped ball.
The system is the ball and the Earth (isolated system).
Ratio of kinetic energies:
\frac{KE}{Kb} = \frac{\frac{1}{2} mE vE^2}{\frac{1}{2} mb vb^2} = \frac{mE vE^2}{mb vb^2} (1)
Applying conservation of momentum:
\Delta p = 0 \implies pi = pf \implies 0 = mb vb + mE vE
Solving for the ratio of velocities:
vE = - \frac{mb}{mE} vb
Substituting into Equation (1):
\frac{KE}{Kb} = \frac{mb}{mE} = \frac{1 kg}{10^{25} kg} = 10^{-25}
The Earth's kinetic energy is negligible compared to the ball's.

3 Analysis Model: Nonisolated System (Momentum)

If there is an external force on the system, the momentum of a system changes and the system is nonisolated.
The momentum of a system changes if a net force from the environment acts on the system.
For momentum considerations, a system is nonisolated if a net force acts on the system for a time interval.
Momentum is transferred to the system from the environment by the net force.

Impulse of a Force

For a net force \sum F acting on a single particle in a system:
dp = \sum F dt (9.7)
Integrating this expression to find the change in momentum:
\Delta p = pf - pi = \int{ti}^{t_f} \sum F dt (9.8)
The quantity on the right side of this equation is the impulse of the net force \sum F acting on a particle over the time interval \Delta t = tf - ti:
I = \int{ti}^{t_f} \sum F dt (9.9)
Impulse is a vector quantity with magnitude equal to the area under the force-time curve.
Its direction is the same as the direction of the change in momentum.
Impulse has dimensions of momentum (ML/T).
Impulse isn't a property of a particle; it's a measure of how much an external force changes the particle's momentum.

Time-Averaged Net Force

The time-averaged net force is defined as:
(\sum F){avg} = \frac{1}{\Delta t} \int{ti}^{tf} \sum F dt (9.10)
We can express Equation 9.9 as:
I = (\sum F)_{avg} \Delta t (9.11)
The time-averaged force is the constant force that would give the same impulse over the time interval.
If the net force is constant:
I = \sum F \Delta t (9.12)

Impulse-Momentum Theorem

Combining Equations 9.8 and 9.9 gives us the impulse-momentum theorem:
\Delta p = I (9.13)
The change in momentum of a particle equals the impulse of the net force acting on it, and is equivalent to Newton's second law.
Impulse represents momentum transferred from an external agent to the particle.
Equation 9.13 is the most general statement of conservation of momentum.
This equation is identical in form to the conservation of energy equation, Equation 8.1.

Nonisolated System (Momentum) Model

Equation 9.13 is the mathematical statement of a new analysis model, the nonisolated system (momentum) model.
Equation 9.13 is a vector equation, directions are important.
There is only one type of momentum and therefore only one way to store momentum in a system.
There is only one way to transfer momentum into a system: by the application of a force on the system over a time interval.

Real-World Example

Consider a crash-test dummy in an accident.
The dummy's momentum changes as the car stops.
The same impulse can occur with a large average force over a short time or a small average force over a long time.
Air bags increase the time interval, reducing the average force.

Impulse Approximation

The impulse approximation assumes one force acts for a short time but is much greater than any other force present.
Replace the net force \sum F with a single force F to find the impulse on the particle.
This is useful in treating collisions with short duration.
The single force is referred to as an impulsive force.
pi and pf represent momenta immediately before and after the collision.
The particle moves very little during the collision.

Quick Quizzes

Quick Quiz 9.3: Two objects at rest on a frictionless surface. Object 1 has greater mass than object 2. (i) Constant force on object 1 through distance d, then on object 2 through distance d. Which statements are true? (ii) Constant force on object 1 for time interval \Delta t, then on object 2 for the same time interval. Which statements are true?
Quick Quiz 9.4: Rank automobile dashboard, seat belt, and air bag in terms of (a) impulse and (b) average force delivered to a passenger.

Analysis Model Nonisolated System (Momentum)

If external forces are applied on the system, the system is nonisolated. In that case, the change in the total momentum of the system is equal to the impulse on the system, a statement known as the impulse-momentum theorem:
\Delta p = I (9.13)
Examples:
- a baseball is struck by a bat
- a spool sitting on a table is pulled by a string (Example 10.14 in Chapter 10)
- a gas molecule strikes the wall of the container holding the gas (Chapter 20)
- photons strike an absorbing surface and exert pressure on the surface (Chapter 33)

Example 9.3: How Good Are the Bumpers?

A 1500 kg car collides with a wall. Initial velocity vi = -15.0 m/s, final velocity vf = 2.60 m/s, collision time \Delta t = 0.150 s. Find the impulse and average net force on the car.
Assume the net force exerted on the car by the wall and friction is large compared with other forces.
Apply the impulse approximation in the horizontal direction.
Car's momentum changes due to an impulse from the environment.
I = \Delta p = pf - pi = m(vf - vi) = (1500 kg)[2.60 m/s - (-15.0 m/s)] = 2.64 \times 10^4 kg \cdot m/s
Calculate the average net force exerted on the car:
(\sum F)_{avg} = \frac{I}{\Delta t} = \frac{2.64 \times 10^4 kg \cdot m/s}{0.150 s} = 1.76 \times 10^5 N
WHAT IF? If the car did not rebound (vf = 0), the net force would be smaller because it only has to stop the car. I = 0 - (1500 kg)(-15.0 m/s) = 2.25 \times 10^4 kg \cdot m/s (\sum F){avg} = \frac{2.25 \times 10^4 kg \cdot m/s}{0.150 s} = 1.50 \times 10^5 N

4 Collisions in One Dimension

The term collision represents an event during which two particles come close to each other and interact by means of forces.
The interaction forces are assumed to be much greater than any external forces present, so we can use the impulse approximation
Collisions may involve physical contact between two macroscopic objects but the notion of what is meant by a collision must be generalized.
When two particles of masses m1 and m2 collide the impulsive forces may vary in time in complicated ways
Regardless of the complexity of the time behavior of the impulsive force, however, this force is internal to the system of two particles.
Therefore, the two particles form an isolated system and the momentum of the system must be conserved in any collision.
In contrast, the total kinetic energy of the system of particles may or may not be conserved, depending on the type of collision. In fact, collisions are categorized as being either elastic or inelastic depending on whether or not kinetic energy is conserved.

Elastic and Inelastic Collisions

An elastic collision is one in which the total kinetic energy (as well as total momentum) of the system is the same before and after the collision.
Inelastic collisions are further divided into two types.
- When the objects stick together after they collide, the collision is called perfectly inelastic.
When the colliding objects do not stick together but some kinetic energy is transformed or transferred away, the collision is called inelastic.

Perfectly Inelastic Collisions

Two particles of masses m1 and m2 moving with initial velocities v{1i} and v{2i} along the same straight line collide and stick together, moving with a common velocity v_f after the collision.
Conservation of momentum says:
m1 v{1i} + m2 v{2i} = (m1 + m2) vf \Delta p = 0 \implies pi = p_f
Solving for the final velocity gives:
vf = \frac{m1 v{1i} + m2 v{2i}}{m1 + m_2} (9.15)