4.3 Conservation of Linear Momentum

Center-of-mass Velocity

The center-of-mass velocity represents how a collection of objects moves as if it were a single entity. When multiple objects interact, we can simplify our analysis by treating the entire system as one object moving with this collective velocity.

• The center-of-mass velocity is calculated using: $\vec{v}_{\mathrm{cm}}=\frac{\sum \vec{p}_{i}}{\sum m_{i}}=\frac{\sum\left(m_{i} \vec{v}_{i}\right)}{\sum m_{i}}$
• This velocity remains constant when no net external forces act on the system
• For example, if a 2 kg object moving at 3 m/s collides with a 1 kg object at rest, the center-of-mass velocity is: $\vec{v}_{\mathrm{cm}}=\frac{(2\text{ kg})(3\text{ m/s})+(1\text{ kg})(0\text{ m/s})}{2\text{ kg}+1\text{ kg}}=2\text{ m/s}$

Sum of Momenta

The total momentum of a system provides crucial information about its motion and how it will interact with other objects.

• Total momentum equals the sum of all individual momenta: $\vec{p}_{\text{total}} = \sum \vec{p}_i$
• For objects moving in different directions, vector addition must be used
• The total momentum connects directly to the center-of-mass motion through: $\vec{p}_{\text{total}} = M\vec{v}_{\text{cm}}$ where $M$ is the total mass

Changes in Momentum

When objects interact within an isolated system, momentum transfers between them while the total remains constant. This principle helps us predict motion after collisions and other interactions.

• In isolated systems, momentum changes follow: $\Delta\vec{p}_1 = -\Delta\vec{p}_2$
• This relationship stems from Newton's third law of equal and opposite forces
• The impulse-momentum relationship governs these changes: $\vec{J}=\Delta \vec{p}$
• For example, when a baseball player hits a ball, the ball gains momentum in one direction while the bat (and player) experience an equal momentum change in the opposite direction

Velocity Before and After Collisions

Conservation of momentum allows us to determine how objects will move after they interact, even when energy might not be conserved.

• For any collision: $m_1\vec{v}_{1i} + m_2\vec{v}_{2i} = m_1\vec{v}_{1f} + m_2\vec{v}_{2f}$
• In elastic collisions, both momentum and kinetic energy are conserved
• In inelastic collisions, momentum is conserved but kinetic energy is not
• In perfectly inelastic collisions, objects stick together after impact, moving with a common final velocity

System Selection for Momentum

Choosing the right system boundaries is critical for effectively applying conservation principles.

Conservation in Interactions

Momentum conservation applies universally to all interactions between objects, making it one of physics' most powerful principles.

• Whether dealing with collisions, explosions, or other interactions, total momentum remains constant in isolated systems
• This principle works for both microscopic (atomic) and macroscopic (everyday) objects
• The conservation holds regardless of whether the interaction is elastic, inelastic, or somewhere in between

Zero Net External Force

When no external forces act on a system, its total momentum remains unchanged over time.

• Examples include collisions on frictionless surfaces or explosions in empty space
• The mathematical expression is: $\vec{p}_{\text{initial}} = \vec{p}_{\text{final}}$
• This principle allows us to analyze complex interactions by focusing only on the objects involved

Nonzero Net External Force

When external forces act on a system, momentum is exchanged between the system and its environment.

• The change in momentum equals the impulse from external forces: $\Delta\vec{p}_{\text{system}} = \vec{J}_{\text{external}}$
• Common external forces include friction, gravity, and contact forces from objects outside the system
• By expanding system boundaries to include all interacting objects, we can often eliminate external forces and apply conservation principles

Practice Problem 1: Inelastic Collision

Solution

Let's apply conservation of momentum. Since no external forces act on our system (the two cars), the total momentum before equals the total momentum after.

Step 1: Find the initial momentum components.

• East component (x-direction): $p_{x,i} = m_1v_{1x} = (2000 \text{ kg})(15 \text{ m/s}) = 30,000 \text{ kg}\cdot\text{m/s}$
• North component (y-direction): $p_{y,i} = m_2v_{2y} = (1500 \text{ kg})(20 \text{ m/s}) = 30,000 \text{ kg}\cdot\text{m/s}$

Step 2: After collision, the total mass is $m_T = 2000 \text{ kg} + 1500 \text{ kg} = 3500 \text{ kg}$

Step 3: Apply conservation of momentum to find final velocity components.

• $v_{x,f} = \frac{p_{x,i}}{m_T} = \frac{30,000 \text{ kg}\cdot\text{m/s}}{3500 \text{ kg}} = 8.57 \text{ m/s}$ (east)
• $v_{y,f} = \frac{p_{y,i}}{m_T} = \frac{30,000 \text{ kg}\cdot\text{m/s}}{3500 \text{ kg}} = 8.57 \text{ m/s}$ (north)

Step 4: Calculate the magnitude and direction of the final velocity.

• Magnitude: $v_f = \sqrt{v_{x,f}^2 + v_{y,f}^2} = \sqrt{(8.57)^2 + (8.57)^2} = 12.1 \text{ m/s}$
• Direction: $\theta = \tan^{-1}\left(\frac{v_{y,f}}{v_{x,f}}\right) = \tan^{-1}\left(\frac{8.57}{8.57}\right) = 45°$ north of east

The combined cars move at 12.1 m/s at 45° north of east immediately after the collision.

Practice Problem 2: Center-of-Mass Velocity

Solution

To find the center-of-mass velocity, we use the formula: $\vec{v}_{\mathrm{cm}}=\frac{\sum\left(m_{i} \vec{v}_{i}\right)}{\sum m_{i}}$

Step 1: Calculate the total mass of the system. $m_{total} = 2 \text{ kg} + 5 \text{ kg} + 3 \text{ kg} = 10 \text{ kg}$

Step 2: Calculate the sum of the products of mass and velocity. $\sum(m_i v_i) = (2 \text{ kg})(3 \text{ m/s}) + (5 \text{ kg})(-2 \text{ m/s}) + (3 \text{ kg})(4 \text{ m/s})$ $\sum(m_i v_i) = 6 \text{ kg}\cdot\text{m/s} + (-10) \text{ kg}\cdot\text{m/s} + 12 \text{ kg}\cdot\text{m/s} = 8 \text{ kg}\cdot\text{m/s}$

Step 3: Calculate the center-of-mass velocity. $v_{cm} = \frac{8 \text{ kg}\cdot\text{m/s}}{10 \text{ kg}} = 0.8 \text{ m/s}$

The center-of-mass velocity of the system is 0.8 m/s in the positive x-direction.