Conservation of momentum Knowledge-Based Study

Study notes

1. Linear Momentum

   The linear momentum of an object of mass m moving with velocity v is defined as:

   $$\mathbf{p} = m\mathbf{v}$$

   Momentum is a vector quantity possessing both magnitude and direction, measured in SI units of kilogram meters per second (kg·m/s). Its direction is identical to the direction of velocity.

   Isolated Systems

   An isolated system is a physical system upon which no net external force acts:

   • Internal forces: Forces exerted by objects within the system on one another (e.g., interaction forces during a collision). According to Newton's third law, these forces always occur in equal and opposite pairs.
   • External forces: Forces exerted on system components by agents outside the system boundary (e.g., gravity, friction, normal forces from external surfaces).

   When the net external force acting on a system is zero, the system is considered mechanically isolated.

   Derivation from Newton's Laws

   Consider two interacting particles, A and B, in an isolated system. By Newton's second law written in terms of momentum, the rate of change of momentum equals the net force:

   $$\mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt}$$

   During their interaction, particle B exerts a force $\mathbf{F}_{\text{AB}}$ on particle A, and particle A exerts a force $\mathbf{F}_{\text{BA}}$ on particle B. By Newton's third law:

   $$\mathbf{F}_{\text{AB}} = -\mathbf{F}_{\text{BA}}$$

   Summing the rates of change of momentum for both particles gives:

   $$\frac{d\mathbf{p}_A}{dt} + \frac{d\mathbf{p}_B}{dt} = \mathbf{F}_{\text{AB}} + \mathbf{F}_{\text{BA}} = \mathbf{0}$$

   $$\frac{d}{dt}(\mathbf{p}_A + \mathbf{p}_B) = \mathbf{0}$$

   Law of Conservation of Momentum

   For any isolated system with no net external force, the total linear momentum remains constant over time:

   $$\mathbf{p}_{\text{total, initial}} = \mathbf{p}_{\text{total, final}}$$

   In component form for one dimension:

   $$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$$

   Classification of Interactions

   Linear momentum is conserved in all collisions and explosions provided external forces are negligible during the brief collision interval. Interactions are categorized based on kinetic energy behavior:

   • Elastic Collisions: Both total linear momentum and total mechanical kinetic energy are conserved. No internal energy dissipation or deformation occurs.
     $$E_{k,i} = E_{k,f}$$
   • Inelastic Collisions: Total linear momentum is conserved, but kinetic energy is not conserved; some kinetic energy transforms into internal thermal energy, sound, or structural deformation.
     $$E_{k,f} < E_{k,i}$$
   • Completely Inelastic Collisions: The colliding bodies stick together after impact and move with a common final velocity $v_f$:
     $$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f$$
   • Explosions / Recoil: An isolated object splits into multiple fragments due to internal forces. Total initial momentum is typically zero or known, and fragments propel in opposite directions such that vector momentum remains balanced.

   Step-by-Step Method

   1. Define the system: Include all interacting bodies so that collision forces are internal.
   2. Verify isolation: Confirm that external forces sum to zero or act over such a short duration that their external impulse is negligible.
   3. Establish a coordinate axis: Assign explicit positive and negative directions for one-dimensional problems.
   4. Express total momentum: Write the initial momentum sum $\sum p_i$ and final momentum sum $\sum p_f$.
   5. Solve algebraically: Compute the unknown mass or velocity component, paying strict attention to signs.

   Illustrative Example

   Problem: A cart of mass 2.0 kg moves east at 3.0 m/s and collides with a stationary cart of mass 1.0 kg. The carts stick together. What is their common final velocity?

   Solution:

   Let east be the positive direction.

   $$p_i = m_1 v_{1i} + m_2 v_{2i} = (2.0\text{ kg})(+3.0\text{ m/s}) + (1.0\text{ kg})(0) = +6.0\text{ kg}\cdot\text{m/s}$$

   $$p_f = (m_1 + m_2)v_f = (2.0\text{ kg} + 1.0\text{ kg})v_f = (3.0\text{ kg})v_f$$

   By conservation of momentum:

   $$6.0 = 3.0 v_f \implies v_f = +2.0\text{ m/s}$$

   The combined carts move east at 2.0 m/s.