apply the principle of conservation of momentum to solve simple problems, including elastic and inelastic interactions between objects in both one and two dimensions (knowledge of the concept of coefficient of restitution is not required)

Linear Momentum and its Conservation

Objective: Apply the principle of conservation of momentum to solve simple problems, including elastic and inelastic interactions between objects in both one and two dimensions (knowledge of the concept of coefficient of restitution is not required).

What is Momentum?

Momentum is a measure of how hard an object is moving. Mathematically it is defined as:

\$p = mv\$

where \$m\$ is mass (kg) and \$v\$ is velocity (m/s). Think of it like a ball rolling on a track – the heavier and faster it rolls, the more momentum it carries.

Calculating Momentum

Use the simple formula above. For example, a 2 kg skateboard moving at 3 m/s has:

\$p = 2\,\text{kg} \times 3\,\text{m/s} = 6\,\text{kg·m/s}\$

Conservation of Momentum

In an isolated system (no external forces), the total momentum before an event equals the total momentum after:

\$p{\text{initial}} = p{\text{final}}\$

Imagine two ice skaters pushing off each other – the total momentum of the pair stays the same even though each skater changes speed.

Elastic vs Inelastic Collisions

• Elastic collision – objects bounce off each other, kinetic energy is conserved. Think of a rubber ball hitting a wall.
• Inelastic collision – objects stick together or deform, kinetic energy is not conserved. Picture a car crash where the cars crumple.

In both cases, momentum is always conserved.

One‑Dimensional Collisions

For two objects A and B moving along a straight line:

\$mA v{Ai} + mB v{Bi} = mA v{Af} + mB v{Bf}\$

1. Write the momentum equation.
2. Write the kinetic energy equation if the collision is elastic.
3. Solve the two equations for the unknown final velocities.

Example problem:

Two carts, one 0.5 kg moving 2 m/s and one 1.5 kg at rest, collide elastically. Find their final speeds.

Solution:

1. Momentum: \$0.5\times2 + 1.5\times0 = 0.5 v{1f} + 1.5 v{2f}\$
2. Kinetic energy: \$0.5\times2^2 + 1.5\times0^2 = 0.5 v{1f}^2 + 1.5 v{2f}^2\$
3. Solving gives \$v{1f}=0\$ m/s and \$v{2f}=2\$ m/s. The lighter cart stops and the heavier cart takes all the speed.

Two‑Dimensional Collisions

Momentum is a vector, so we treat each component separately.

For objects A and B in 2D:

\$mA \vec{v}{Ai} + mB \vec{v}{Bi} = mA \vec{v}{Af} + mB \vec{v}{Bf}\$

Split into x and y components:

• Momentum in x: \$mA v{Ai,x} + mB v{Bi,x} = mA v{Af,x} + mB v{Bf,x}\$
• Momentum in y: \$mA v{Ai,y} + mB v{Bi,y} = mA v{Af,y} + mB v{Bf,y}\$

Example: A 1 kg ball moving east at 4 m/s collides head‑on with a 2 kg ball moving west at 2 m/s. After an elastic collision, find the final velocities.

Solution:

1. Set x‑axis east positive. Initial momenta: \$p_{x,i}=1\times4 + 2\times(-2)=0\$ kg·m/s.
2. Since total momentum is zero, final momenta must also sum to zero. For an elastic collision, speeds reverse: ball 1 goes west at 4 m/s, ball 2 goes east at 2 m/s.

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