Unit 4: Linear Momentum

Momentum

Momentum is a measure of an object’s motion, defined as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction. Momentum is often described as the "quantity of motion" because it reflects how difficult it is to stop or change the motion of an object. The greater the mass or velocity of an object, the greater its momentum. In AP Physics 1, momentum plays a central role in analyzing collisions and interactions between objects.

\[ p = mv \]

• Momentum combines both the inertia (mass) and the motion (velocity) of an object, making it a powerful descriptor of how objects interact. For example, a truck and a car moving at the same speed have very different momenta because of their mass difference.
• Since momentum is a vector, both magnitude and direction must be considered in calculations. When objects collide or interact, momentum components in the x- and y-directions are often analyzed separately.
• The unit of momentum is \( \text{kg·m/s} \), derived directly from multiplying mass in kilograms by velocity in meters per second. This makes momentum consistent with Newton’s second law when expressed in terms of impulse.
• Momentum is conserved in closed and isolated systems where no external forces act. This means that the total momentum before an interaction equals the total momentum after, a principle that applies universally from billiard balls to rocket propulsion.
• Because momentum depends on velocity, an object at rest has zero momentum regardless of its mass. However, even small objects moving at very high velocities can have significant momentum, such as a bullet or meteor.

Impulse

Impulse measures the change in momentum of an object over a certain time interval. It is the product of the net force applied and the time during which the force acts. Impulse is also a vector quantity, pointing in the same direction as the net force applied. The impulse-momentum theorem directly links force and time to momentum changes, providing an alternative way to solve motion problems without explicitly finding acceleration.

\[ J = F \Delta t = \Delta p \]

• Impulse describes how forces that act for a short period of time, like a bat hitting a baseball, can produce large changes in momentum. Even if the force isn’t extremely large, a longer time interval increases the impulse and thus the momentum change.
• Impulse shares the same units as momentum (\( \text{kg·m/s} \)), but it is often written as Newton-seconds (\( \text{N·s} \)), emphasizing its relation to force and time. This dual interpretation highlights the connection between dynamics and kinematics.
• The impulse-momentum theorem states: \( F\Delta t = m\Delta v \). This provides a practical way to solve problems when the force or time of application is known but acceleration is not. It is especially useful in collision problems.
• Impulse can also be visualized as the area under a force–time graph. This is important because in real-world problems, the force is often not constant. The integral of force over time gives the exact impulse.
• Designs in sports and safety engineering rely heavily on impulse concepts. For instance, airbags extend the time of impact in a collision, reducing the average force on passengers while delivering the same impulse to stop them.

Why Momentum Is Conserved

The conservation of momentum is rooted in Newton’s Third Law of Motion, which states that for every action, there is an equal and opposite reaction. When two objects interact, the forces they exert on each other are equal in magnitude and opposite in direction. Because these forces act over the same time interval, the impulses are equal and opposite, meaning one object’s momentum increases by the same amount the other’s decreases. As a result, the total momentum of the system remains unchanged. This principle applies universally, whether in collisions, explosions, or interactions in space where no external forces are present.

• Momentum conservation is a direct consequence of Newton’s Third Law. If two ice skaters push off each other, the forces they exert are equal and opposite, so their momentum changes cancel, keeping total momentum constant.
• In a closed system with no external forces, the internal forces between objects come in equal and opposite pairs. These cancel when summing total momentum, ensuring that the vector sum of momentum remains unchanged.
• This principle is not limited to macroscopic objects; it also governs microscopic interactions such as atomic collisions and particle physics. It is one of the most fundamental and universally applicable conservation laws in nature.
• External forces, if present, can change the system’s total momentum. For example, friction from the ground provides an external force that prevents two colliding pucks from perfectly conserving momentum.
• Because momentum is a vector, conservation applies independently in each direction. This means in two-dimensional problems, momentum in the x-direction and y-direction must be conserved separately.

Conservation of Linear Momentum

The Law of Conservation of Linear Momentum states that if no external forces act on a system, the total momentum of the system remains constant. This law follows directly from Newton’s Third Law: the internal forces between objects cancel, so the system as a whole does not gain or lose momentum. Mathematically, the total momentum before an interaction equals the total momentum after:

\[ \sum p_{\text{initial}} = \sum p_{\text{final}} \]

• This law applies to all types of interactions, including elastic, inelastic, and perfectly inelastic collisions. In each case, momentum is conserved as long as the system is isolated from external forces, such as friction with the ground or air resistance.
• Elastic collisions conserve both momentum and kinetic energy. Inelastic collisions conserve momentum but not kinetic energy, since some energy is converted to heat, sound, or deformation. Recognizing the difference is crucial on the AP exam.
• Momentum conservation works in both one-dimensional and two-dimensional systems. In two-dimensional cases, momentum components must be conserved separately in the x- and y-directions. Vector analysis becomes essential in solving these problems correctly.
• This principle has real-world applications in areas such as rocket propulsion (where expelled gas provides equal and opposite momentum to the rocket), vehicle crash analysis, and even the motion of celestial bodies in space.
• AP Physics 1 problems often involve identifying the system and ensuring it is closed and isolated. If external forces are not negligible, momentum conservation cannot be applied directly, and the effect of those forces must be considered.

Key Idea: Even if kinetic energy is not conserved in a collision, momentum is always conserved in an isolated system.

Types of Collisions

Collisions occur when two or more objects interact through forces over a short time. While the total momentum of the system is always conserved in an isolated system, whether or not kinetic energy is conserved determines the type of collision. There are three primary categories: elastic, inelastic, and perfectly inelastic. Understanding the differences among these is crucial for solving AP Physics 1 problems effectively.

Elastic Collisions

• In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum of the system remains the same before and after the collision, and no kinetic energy is lost to heat, sound, or deformation.
• Elastic collisions are rare in the real world because some energy is almost always transformed into other forms. However, they are a useful idealization for AP Physics 1, and many problems assume elastic collisions when not stated otherwise.
• In one-dimensional elastic collisions, special formulas exist to directly solve for final velocities, but AP Physics 1 usually focuses on using conservation laws instead of memorization.
• Example: Two billiard balls colliding on a frictionless table. After the collision, they move apart with no loss of total kinetic energy.

Inelastic Collisions

• In an inelastic collision, momentum is conserved, but kinetic energy is not. Some kinetic energy is transformed into heat, sound, or internal energy due to deformation.
• The objects usually separate after the collision, but their final speeds are smaller than what they would be in an elastic collision because of the energy loss.
• Inelastic collisions are far more common in real life than elastic ones, since perfect conservation of kinetic energy is nearly impossible.
• Example: Two cars colliding and crumpling during a crash. Although the total momentum is the same before and after the collision, much of the kinetic energy is lost as sound and heat.

Perfectly Inelastic Collisions

• A perfectly inelastic collision is a special type of inelastic collision where the objects stick together after the collision, moving as a single combined mass.
• This type of collision involves the maximum possible loss of kinetic energy while still conserving momentum. The objects cannot separate after impact because they are bound together.
• Even though kinetic energy is not conserved, the principle of momentum conservation still holds, and it is the main tool for solving perfectly inelastic problems on the AP exam.
• Example: A lump of clay thrown at a wall sticks to it. The clay and the wall move together (though the wall’s motion is negligible due to its huge mass), showing a perfectly inelastic collision.

Explosions and Recoil

Explosions and recoil events are the reverse of collisions: a single object breaks into two or more pieces, often at rest initially. These scenarios are governed by the same principle — conservation of linear momentum — even though kinetic energy typically increases during the process due to stored internal energy being released. Because the total momentum before the explosion is often zero (if the object is initially at rest), the vector sum of the final momenta must also be zero. This means the pieces move in opposite directions such that their momenta cancel each other out. Recoil is a specific type of explosion, like a gun firing a bullet, where the object releasing mass (the gun) moves backward in response.

• Momentum is still conserved even if energy is added to the system from internal chemical or explosive processes. This makes explosion problems distinct from inelastic collisions, where energy is lost.
• In many explosion problems, the object starts at rest. This simplifies calculations because the total initial momentum is zero, which means the momenta of the fragments after the explosion must cancel each other out vectorially.
• Recoil problems, like a cannon firing a cannonball or a person jumping off a skateboard, also follow momentum conservation. The object being launched moves forward while the launcher moves backward with equal and opposite momentum.
• Even though kinetic energy increases in an explosion, total energy is still conserved. The extra kinetic energy comes from stored internal energy, such as chemical potential energy in a fuel or spring system.
• Example: A 0.5 kg firework explodes at rest into two pieces: one 0.2 kg piece flies left at 6 m/s. To conserve momentum, the other 0.3 kg piece must fly right at 4 m/s, because their momenta cancel: \( 0.2(-6) + 0.3(4) = 0 \).

Multi-Step Challenge Problems for Unit 4: Momentum and Impulse

Problem 1: Elastic Collision in One Dimension

A 0.20 kg ball moving at 6.0 m/s collides elastically with a 0.30 kg ball initially at rest on a frictionless surface. Find the final velocities of both balls after the collision.

Solution:

Step 1: Apply momentum conservation.
\( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \).
Substituting: \( (0.20)(6.0) + (0.30)(0) = 0.20v_{1f} + 0.30v_{2f} \).
\( 1.20 = 0.20v_{1f} + 0.30v_{2f} \).   (Equation 1)

Step 2: Use the elastic collision relative velocity rule.
For elastic collisions: \( v_{1i} - v_{2i} = -(v_{1f} - v_{2f}) \).
Substituting: \( 6.0 - 0 = -(v_{1f} - v_{2f}) \).
So, \( v_{1f} - v_{2f} = -6.0 \).   (Equation 2)

Step 3: Solve the system of equations.
From Eq. 2: \( v_{1f} = v_{2f} - 6.0 \).
Substitute into Eq. 1: \( 1.20 = 0.20(v_{2f} - 6.0) + 0.30v_{2f} \).
\( 1.20 = 0.20v_{2f} - 1.20 + 0.30v_{2f} \).
\( 2.40 = 0.50v_{2f} \).
\( v_{2f} = 4.8 \, \text{m/s} \).
Then, \( v_{1f} = 4.8 - 6.0 = -1.2 \, \text{m/s} \).

Answer: After the collision, the 0.20 kg ball moves backward at \( -1.2 \, \text{m/s} \), and the 0.30 kg ball moves forward at \( 4.8 \, \text{m/s} \).

Problem 2: Recoil (Explosion Scenario)

A 4.0 kg rifle fires a 0.020 kg bullet horizontally at 300 m/s. The rifle is initially at rest and free to move horizontally. Find (a) the recoil speed of the rifle and (b) the kinetic energies of both bullet and rifle. Then, (c) explain why the total kinetic energy increases even though momentum is conserved.

Solution:

Step 1: Apply momentum conservation.
Initial momentum = 0, so total final momentum must also equal 0.
\( m_b v_b + m_r v_r = 0 \).
\( (0.020)(300) + (4.0)v_r = 0 \).
\( 6.0 + 4.0v_r = 0 \).
\( v_r = -1.5 \, \text{m/s} \) (negative means opposite direction).

Step 2: Find kinetic energies.
Bullet: \( KE_b = \tfrac{1}{2} m_b v_b^2 = \tfrac{1}{2}(0.020)(300^2) = 900 \, J \).
Rifle: \( KE_r = \tfrac{1}{2} m_r v_r^2 = \tfrac{1}{2}(4.0)(1.5^2) = 4.5 \, J \).

Step 3: Explain total energy increase.
Total KE = 900 + 4.5 = 904.5 J.
This increase comes from the chemical potential energy of the gunpowder inside the rifle. Momentum is conserved because the internal forces are equal and opposite, but extra kinetic energy is supplied from stored chemical energy.

Answer: (a) The rifle recoils at \( -1.5 \, \text{m/s} \). (b) The bullet has 900 J of KE, the rifle 4.5 J. (c) Total kinetic energy increases due to conversion of chemical potential energy into motion.

Problem 3: Perfectly Inelastic Collision

A 1.5 kg cart moving at 4.0 m/s collides and sticks to a 2.0 kg cart initially at rest on a frictionless track. Find: (a) the final velocity of the combined carts, (b) the initial and final kinetic energies, and (c) the amount of kinetic energy lost in the collision.

Solution:

Step 1: Apply momentum conservation.
\( m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f \).
Substituting: \( (1.5)(4.0) + (2.0)(0) = (3.5)v_f \).
\( 6.0 = 3.5v_f \).
\( v_f = \tfrac{6.0}{3.5} \approx 1.71 \, \text{m/s} \).

Step 2: Calculate initial and final kinetic energies.
Initial KE: \( KE_i = \tfrac{1}{2}m_1 v_{1i}^2 + \tfrac{1}{2}m_2 v_{2i}^2 \).
\( KE_i = \tfrac{1}{2}(1.5)(4.0^2) + 0 = 12.0 \, J \).
Final KE: \( KE_f = \tfrac{1}{2}(3.5)(1.71^2) \).
\( KE_f \approx \tfrac{1}{2}(3.5)(2.92) \approx 5.11 \, J \).

Step 3: Find kinetic energy lost.
\( \Delta KE = KE_f - KE_i = 5.11 - 12.0 = -6.89 \, J \).
The negative indicates a loss of kinetic energy.

Answer: (a) The combined carts move at about \( 1.71 \, \text{m/s} \). (b) The kinetic energy decreases from 12.0 J to 5.11 J. (c) About 6.89 J of kinetic energy is lost, likely converted to sound, heat, and deformation.

Common Misconceptions in Unit 4: Momentum and Impulse

Momentum

• Misconception: A heavier object always has more momentum than a lighter one.
  Correction: Momentum depends on both mass and velocity (\( p = mv \)). A light object moving very fast can have more momentum than a heavy object moving slowly. For example, a baseball can have more momentum than a truck tire rolling slowly.
• Misconception: Momentum is always conserved, no matter the situation.
  Correction: Momentum is conserved only in a closed and isolated system. If external forces like friction or a push from the ground act, total momentum of the system can change.

Impulse

• Misconception: A larger force always creates a larger impulse.
  Correction: Impulse depends on both force and time (\( J = F\Delta t \)). A smaller force acting for a longer time can deliver the same or more impulse than a short, large force. This is why airbags are effective — they extend the collision time, reducing force while keeping impulse the same.
• Misconception: Impulse is different from change in momentum.
  Correction: Impulse and change in momentum are equal (\( J = \Delta p \)). Impulse is simply the process by which momentum changes.

Collisions

• Misconception: Kinetic energy is always conserved in collisions.
  Correction: Only elastic collisions conserve kinetic energy. Inelastic and perfectly inelastic collisions lose kinetic energy to heat, sound, or deformation. However, momentum is conserved in all collisions (in isolated systems).
• Misconception: In a perfectly inelastic collision, momentum is not conserved since the objects stick together.
  Correction: Momentum is still conserved. The objects share a common final velocity that ensures total momentum before equals total momentum after, even though kinetic energy decreases.

Explosions and Recoil

• Misconception: Explosions violate conservation of momentum because the total kinetic energy increases.
  Correction: Momentum is still conserved. The increase in kinetic energy comes from stored potential or chemical energy inside the system, not from an external source.
• Misconception: The larger object in a recoil (like a rifle) doesn’t move at all.
  Correction: The rifle does move, but because of its much larger mass, its recoil speed is small compared to the bullet’s. Its momentum, however, exactly balances the bullet’s momentum in the opposite direction.