5.4 Conservation of Linear Momentum

Types of Collisions

In physics, if the net external force acting on a system is zero, the total momentum of the system is conserved. However, forces can vary depending on the direction. For example, a ball in free fall might have no net force in the horizontal (x) direction, while a vertical (y) direction net force exists due to gravity.

Understanding the types of collisions—elastic and inelastic—is crucial when analyzing momentum conservation:

• Elastic Collisions: Both momentum and kinetic energy are conserved. Objects rebound after impact.

• Inelastic Collisions: Momentum is conserved, but kinetic energy is not. Some energy is lost as heat, sound, or deformation.

• Completely Inelastic Collisions: Objects stick together after the collision.

Inelastic Collisions

In inelastic collisions, momentum is conserved even though kinetic energy is not. Energy may transform into heat, sound, or other forms. For completely inelastic collisions, the objects stick together after impact, and momentum conservation allows us to solve for the system’s final velocity.

Example: Two carts of the same mass lie on a table. Cart A is moving, while Cart B is stationary. After colliding, they stick together. To find the final velocity:

1. Initial Momentum:

   • pinitial = m1v1 + m2v2

2. Final Momentum:

   • pfinal = (m1 + m2)vfinal

3. Equating Momentum: Solve for vfinal using the conservation law: pinitial = pfinal.

Elastic Collisions

For elastic collisions, both momentum and kinetic energy are conserved. Use these principles to solve problems involving rebound and final velocities. For 2D problems, break vectors into components and solve for momentum and kinetic energy in each direction.

Example Problem:
Two carts, each 2 kg, collide on a frictionless track. Cart A moves right at 3 m/s, and Cart B moves left at 4 m/s. After colliding, both move right at 1 m/s.

Solution:

1. Momentum Before Collision:

   • Cart A: p = m × v = 2 kg × 3 m/s = 6 kg·m/s

   • Cart B: p = 2 kg × (-4 m/s) = -8 kg·m/s

   • Total: p = 6 kg·m/s – 8 kg·m/s = -2 kg·m/s

2. Momentum After Collision:

   • Cart A: p = 2 kg × 1 m/s = 2 kg·m/s

   • Cart B: p = 2 kg × 1 m/s = 2 kg·m/s

   • Total: p = 2 kg·m/s + 2 kg·m/s = 4 kg·m/s

Since momentum is not conserved here, revisit calculations or problem assumptions.

Center of Mass

The center of mass of a system is the point where its mass is evenly distributed in a gravitational field. For irregular objects, the center of mass lies closer to the heaviest region. The formula for the center of mass in one direction (x-axis):

xcm = (m1x1 + m2x2 + …) / (m1 + m2 + …)

Key Notes:

• If the net external force is zero, the center of mass does not accelerate.

• If the system is moving, the center of mass moves at a constant velocity unless acted upon by an external force.

Example Problems

Example 1: Balancing a Stick

A 1-meter stick is balanced on a fulcrum at its midpoint. A 2 kg mass is attached to one end, and a 3 kg mass to the other. Where is the center of mass?

Solution:

Using the formula:

xcm = (m1x1 + m2x2) / (m1 + m2)

• Mass 1: 2 kg at 0.5 m

• Mass 2: 3 kg at 0.5 m

Substitute values:

xcm = (2 × 0.5 + 3 × 0.5) / (2 + 3) = 1.25 m

The center of mass is 1.25 m from the fulcrum.

Example 2: Ladder Against a Wall

A 5-meter ladder leans against a wall at 60°. It has a mass of 10 kg, and the center of mass is 3 meters from the base. Find the horizontal force exerted by the wall.

Solution:

1. Weight of Ladder:

   • W = m × g = 10 kg × 9.8 m/s² = 98 N

2. Horizontal Force:

   • F = W × cos(θ) = 98 N × cos(60°) = 49 N

The horizontal force exerted by the wall is 49 N.

Example 3: Balancing a Bar

A 4-meter bar is balanced on a fulcrum at its midpoint. A 2 kg mass is attached to one end, a 3 kg mass to the other, and the bar itself weighs 5 kg. Where is the center of mass?

Solution:

Using the center of mass formula:

xcm = (m1x1 + m2x2 + m3x3) / (m1 + m2 + m3)

• Mass 1: 2 kg at 0.5 m

• Mass 2: 3 kg at 0.5 m

• Bar (5 kg) at midpoint: 2 m

Substitute values:

xcm = (2 × 0.5 + 3 × 0.5 + 5 × 2) / (2 + 3 + 5) = 1.44 m

The center of mass is 1.44 m from the fulcrum.