A2-4. Conservation of Momentum

The Principle of Conservation of Momentum

This is one of the most powerful and universal laws in physics, governing how objects interact during collisions and explosions.

Conservation of Linear Momentum: In a closed, isolated system (where the net external force is zero, $\Sigma F_{ext} = 0$), the total momentum of the system remains constant before, during, and after an interaction.

$$\Sigma \vec{p}_{\text{initial}} = \Sigma \vec{p}_{\text{final}}$$

For a basic two-particle collision, this expands to:

$$m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2$$

2D Collision Conservation: Because momentum is a vector, it is conserved independently in both the X and Y axes. You can split any 2D collision into two separate 1D equations:

  • X-axis: $\Sigma p_{1x} + \Sigma p_{2x} = \Sigma p'_{1x} + \Sigma p'_{2x}$
  • Y-axis: $\Sigma p_{1y} + \Sigma p_{2y} = \Sigma p'_{1y} + \Sigma p'_{2y}$

Center of Mass: Because the total momentum does not change in an isolated system, the velocity of the system's center of mass ($\vec{v}_{CM}$) remains absolutely constant throughout the entire collision process.

Types of Collisions

While momentum is perfectly conserved in all isolated collisions, kinetic energy ($E_k = \dfrac{1}{2}mv^2$) is not always conserved.

Collision Type Momentum Kinetic Energy Description
Elastic Conserved Conserved Objects bounce off each other perfectly without any energy lost to heat or deformation (e.g., ideal gas particles).
Inelastic Conserved Not Conserved Some kinetic energy is transformed into thermal energy, sound, or permanent deformation of the objects.
Totally Inelastic Conserved Max. Energy Lost Objects stick together after the collision and move as a single combined mass ($m_1 + m_2$) at the same final velocity.
Explosion Conserved (Total = 0) Increases A single object splits into multiple pieces. Stored internal energy is converted into kinetic energy, propelling pieces outward.

Example 1: Totally Inelastic Collision

Problem: A member of a skateboard club (mass $m_1$) is running on a perfectly smooth horizontal track. A skateboard (mass $m_2$) is rolling in the same direction. The skateboarder's speed is 3 times the skateboard's speed ($v_1 = 3v_2$). The skateboarder jumps onto the skateboard, and they continue moving together at a new combined speed of $v = \dfrac{31}{11}v_2$. Based on this, determine the mass ratio of the skateboard to the skateboarder ($\dfrac{m_2}{m_1}$).

Solution:

  • a) Set up the Conservation Equation:
    Since the skateboarder and the board move together at the end, this is a totally inelastic collision.
    $$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$$
  • b) Substitute the known velocities:
    Substitute $v_1 = 3v_2$ and $v = \dfrac{31}{11}v_2$:
    $$m_1(3v_2) + m_2(v_2) = (m_1 + m_2)\left(\dfrac{31}{11}v_2\right)$$
  • c) Simplify and solve for the mass ratio:
    Divide the entire equation by $v_2$ to cancel it out:
    $$3m_1 + m_2 = \dfrac{31}{11}m_1 + \dfrac{31}{11}m_2$$
    Multiply everything by 11 to clear the fraction:
    $$33m_1 + 11m_2 = 31m_1 + 31m_2$$
    Group the $m_1$ and $m_2$ terms on opposite sides:
    $$33m_1 - 31m_1 = 31m_2 - 11m_2$$
    $$2m_1 = 20m_2$$
    $$\dfrac{m_2}{m_1} = \dfrac{2}{20} = \mathbf{\dfrac{1}{10}}$$
  • Conclusion: The mass of the skateboard is exactly $\dfrac{1}{10}$ the mass of the skateboarder.