Unit 3.5 – Conservation of Momentum

What is the Law of Conservation of Momentum?

• $m_1,m_2$ = Masses of the two objects.
• $u_1,u_2$ = Initial velocities of the two objects.
• $v_1,v_2$ = Final velocities of the two objects.

Key Points About Conservation of Momentum

1. Closed System:
   • A closed system means no external forces (like friction or air resistance) are acting on the objects.
   • Example: Collisions between billiard balls on a smooth table.
2. Direction Matters:
   • Momentum is a vector quantity, so direction must be considered. Velocities in opposite directions are assigned opposite signs (+ and -).
3. Applications:
   • This principle applies to collisions, explosions, and interactions between objects.

Real-Life Examples of Conservation of Momentum

1. Collision Between Two Cars:
   • When two cars collide, the total momentum of the system before the collision equals the total momentum after the collision.
2. Recoil of a Gun:
   • When a bullet is fired from a gun, the forward momentum of the bullet equals the backward momentum of the gun (recoil).
3. Rocket Propulsion:
   • Rockets expel gases downward with high momentum. The upward momentum of the rocket equals the downward momentum of the expelled gases.
4. Skating on Ice:
   • If two skaters push each other on ice, they move apart with equal and opposite momenta.

Types of Collisions

1. Elastic Collision:

• Both momentum and kinetic energy are conserved.
• Example: Collision between two billiard balls.

2. Inelastic Collision:

• Momentum is conserved, but kinetic energy is not conserved (some energy is lost as heat, sound, etc.).
• Example: A car crash where the cars stick together after the collision.

Examples to Understand Conservation of Momentum

Example 1:

• Solution: Using conservation of momentum:
  $$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$$
  Substituting values:
  $$(2)(4)+(3)(-2)=(2)(1)+(3)v_2$$
  $$8-6=2+3v_2$$
  $$2=2+3v_2$$
  $$3v_2=0⟹v_2=0\text{m/s}\mathrm{.}$$
  The $3\text{kg}$ cart comes to rest after the collision.

Example 2:

• Solution: Using conservation of momentum:
  $$m_{\text{gun}}{u}_{\text{gun}}+m_{\text{bullet}}{u}_{\text{bullet}}=m_{\text{gun}}{v}_{\text{gun}}+m_{\text{bullet}}{v}_{\text{bullet}}$$
  Initially, both the gun and bullet are at rest ($u_{\text{gun}}=u_{\text{bullet}}=0$):
  $$0=(5)v_{\text{gun}}+(0.02)(400)$$
  $$0=5v_{\text{gun}}+8$$
  $$5v_{\text{gun}}=-8⟹v_{\text{gun}}=-1.6\text{m/s}\mathrm{.}$$
  The negative sign indicates that the gun recoils in the opposite direction to the bullet.