Elastic collisions Knowledge-Based Study

Study notes

1. What Is an Elastic Collision?

   An elastic collision is an interaction between two or more bodies in an isolated system where both total linear momentum and total kinetic energy are conserved. In real macroscopic interactions, some kinetic energy is almost always converted into thermal energy, sound, or permanent deformation. However, collisions between fundamental particles (such as electrons or gas molecules) or idealized rigid bodies (such as hard steel ball bearings) serve as excellent models of elastic collisions.

   Key Conservation Laws

   • Conservation of Linear Momentum: Since no external net force acts on an isolated system, the total vector momentum before the collision equals the total vector momentum after the collision:
     $\vec{p}_{\text{total, initial}} = \vec{p}_{\text{total, final}}$
   • Conservation of Kinetic Energy: Unlike inelastic collisions, no mechanical energy is lost to heat or internal deformation:
     $K_{\text{total, initial}} = K_{\text{total, final}}$

   Governing Equations for Direct (Head-On) Collisions

   Consider two masses, $m_1$ and $m_2$, moving along a straight line with initial velocities $u_1$ and $u_2$. After colliding elastically, their velocities become $v_1$ and $v_2$.

   1. Conservation of Linear Momentum

   $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

   Rearranging terms with like masses yields:

   $m_1 (u_1 - v_1) = m_2 (v_2 - u_2)$   (Equation 1)

   2. Conservation of Kinetic Energy

   $\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$

   Multiplying by 2 and rearranging like terms:

   $m_1 (u_1^2 - v_1^2) = m_2 (v_2^2 - u_2^2)$

   Factoring using the difference of squares:

   $m_1 (u_1 - v_1)(u_1 + v_1) = m_2 (v_2 - u_2)(v_2 + u_2)$   (Equation 2)

   3. The Relative Velocity Relation

   Dividing Equation 2 by Equation 1 gives a critical diagnostic rule for all one-dimensional elastic collisions:

   $u_1 + v_1 = v_2 + u_2 \implies u_1 - u_2 = -(v_1 - v_2)$

   In words, the relative velocity of approach equals the relative velocity of separation with opposite sign. The coefficient of restitution $e$ is exactly equal to 1.

   General Velocity Formulas

   Solving the system of linear equations for final velocities yields:

   • $v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2m_2}{m_1 + m_2}\right)u_2$
   • $v_2 = \left(\frac{2m_1}{m_1 + m_2}\right)u_1 + \left(\frac{m_2 - m_1}{m_1 + m_2}\right)u_2$

   Notable Special Cases

   1. Identical Masses ($m_1 = m_2$): Substituting equal masses into the formulas gives $v_1 = u_2$ and $v_2 = u_1$. The two colliding bodies simply exchange their velocities.
   2. Target at Rest with Equal Mass ($m_1 = m_2$, $u_2 = 0$): Mass 1 comes to a complete stop ($v_1 = 0$), while mass 2 continues forward with the initial velocity of mass 1 ($v_2 = u_1$). This effect is directly observed in Newton's cradle.
   3. Colliding with a Much Heavier Target at Rest ($m_1 \ll m_2$, $u_2 = 0$): Here, $m_1 - m_2 \approx -m_2$ and $m_1 + m_2 \approx m_2$, resulting in $v_1 \approx -u_1$ and $v_2 \approx 0$. The light object rebounds with virtually its original speed, while the massive wall or target remains effectively stationary.
   4. Much Heavier Projectile Striking a Light Target at Rest ($m_1 \gg m_2$, $u_2 = 0$): In this limit, $v_1 \approx u_1$ and $v_2 \approx 2u_1$. The heavy mass continues virtually unhindered, while the light target is propelled forward at twice the incoming speed.

   Example Problem

   A glider of mass $m_1 = 0.20\text{ kg}$ moves along a frictionless air track at $u_1 = +3.0\text{ m/s}$. It undergoes a head-on elastic collision with a stationary second glider of mass $m_2 = 0.40\text{ kg}$ ($u_2 = 0$). Determine the final velocities $v_1$ and $v_2$.

   Step-by-Step Solution

   1. Identify Knowns and Assumptions:
      $m_1 = 0.20\text{ kg}$, $m_2 = 0.40\text{ kg}$, $u_1 = +3.0\text{ m/s}$, $u_2 = 0\text{ m/s}$. Friction is negligible, and the collision is purely elastic.
   2. Apply the Velocity Formula for Mass 1:
      $v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 = \left(\frac{0.20 - 0.40}{0.20 + 0.40}\right)(3.0) = \left(\frac{-0.20}{0.60}\right)(3.0) = -1.0\text{ m/s}$
      The negative sign indicates that glider 1 rebounds to the left.
   3. Apply the Velocity Formula for Mass 2:
      $v_2 = \left(\frac{2m_1}{m_1 + m_2}\right)u_1 = \left(\frac{2(0.20)}{0.20 + 0.40}\right)(3.0) = \left(\frac{0.40}{0.60}\right)(3.0) = +2.0\text{ m/s}$
      Glider 2 moves forward to the right at 2.0 m/s.
   4. Verification via Kinetic Energy:
      $K_{\text{initial}} = \frac{1}{2}(0.20)(3.0)^2 = 0.90\text{ J}$
      $K_{\text{final}} = \frac{1}{2}(0.20)(-1.0)^2 + \frac{1}{2}(0.40)(2.0)^2 = 0.10 + 0.80 = 0.90\text{ J}$
      Because $K_{\text{initial}} = K_{\text{final}}$, conservation of kinetic energy is confirmed.