recall that, for an elastic collision, total kinetic energy is conserved and the relative speed of approach is equal to the relative speed of separation

Linear Momentum and Its Conservation

Linear momentum, \$\vec p\$, is a vector quantity defined for a particle of mass \$m\$ moving with velocity \$\vec v\$ as

\$\vec p = m\vec v\$

The principle of conservation of linear momentum states that, in the absence of external forces, the total momentum of a closed system remains constant.

Key Definitions

• Momentum (\$\vec p\$): product of mass and velocity, measured in kg·m s⁻¹.
• Impulse (\$\vec J\$): change in momentum, \$\vec J = \Delta\vec p\$, equal to the integral of net force over the time interval.
• Elastic collision: a collision in which both total linear momentum and total kinetic energy are conserved.
• Inelastic collision: a collision in which total linear momentum is conserved but kinetic energy is not; some kinetic energy is transformed into other forms.

Conservation Laws

For a system of \$n\$ particles with no external forces, the vector sum of their momenta before an interaction equals the sum after the interaction:

\$\sum{i=1}^{n} mi\vec v{i,\text{initial}} = \sum{i=1}^{n} mi\vec v{i,\text{final}}\$

In one dimension this reduces to a scalar equation:

\$\sum{i=1}^{n} mi v{i,\text{initial}} = \sum{i=1}^{n} mi v{i,\text{final}}\$

Elastic Collisions

In an elastic collision the total kinetic energy \$K\$ is also conserved:

\$\$\sum{i=1}^{n} \frac{1}{2} mi v_{i,\text{initial}}^{2}

= \sum{i=1}^{n} \frac{1}{2} mi v_{i,\text{final}}^{2}\$\$

Because both momentum and kinetic energy are conserved, the velocities after the collision can be determined uniquely.

Derivation: Relative Speed of Approach Equals Relative Speed of Separation

Consider two particles, \$A\$ and \$B\$, moving along a straight line. Let \$uA\$ and \$uB\$ be their speeds before collision, and \$vA\$ and \$vB\$ after collision. Define the relative speed of approach as \$u{\text{rel}} = |uA - uB|\$ and the relative speed of separation as \$v{\text{rel}} = |vA - vB|\$.

1. Write the conservation of linear momentum (scalar form):

   \$mA uA + mB uB = mA vA + mB vB\$

2. Write the conservation of kinetic energy:

   \$\$\frac{1}{2}mA uA^{2} + \frac{1}{2}mB uB^{2}

   = \frac{1}{2}mA vA^{2} + \frac{1}{2}mB vB^{2}\$\$

3. Subtract the momentum equation multiplied by \$(uA + uB)\$ from the kinetic‑energy equation multiplied by \$2\$:

   \$\$mA(uA^{2} - vA^{2}) + mB(uB^{2} - vB^{2})

   = (uA + uB)\bigl[mA(uA - vA) + mB(uB - vB)\bigr]\$\$

4. Factor each term as a difference of squares and simplify; after cancellation the result reduces to

   \$(uA - uB)^{2} = (vA - vB)^{2}\$

5. Taking the positive square root (speeds are non‑negative) gives the required relationship:

   \$|uA - uB| = |vA - vB|\$

   i.e. the relative speed of approach equals the relative speed of separation.

Comparison of Elastic and Inelastic Collisions

Worked Example

Two spheres of masses \$m1 = 0.5\;\text{kg}\$ and \$m2 = 1.0\;\text{kg}\$ move toward each other along a line. Before collision, \$m1\$ travels at \$u1 = 4\;\text{m s}^{-1}\$ to the right and \$m2\$ at \$u2 = 2\;\text{m s}^{-1}\$ to the left. Find their speeds after a perfectly elastic collision.

1. Apply momentum conservation:

   \$0.5(4) - 1.0(2) = 0.5 v1 - 1.0 v2\$

2. Apply kinetic‑energy conservation:

   \$\tfrac12(0.5)(4)^2 + \tfrac12(1.0)(2)^2 = \tfrac12(0.5)v1^{2} + \tfrac12(1.0)v2^{2}\$

3. Solve the simultaneous equations to obtain

   \$v1 = -2\;\text{m s}^{-1},\qquad v2 = 3\;\text{m s}^{-1}\$

   (negative sign indicates \$m1\$ now moves leftward). The relative speed before collision was \$|4-(-2)| = 6\;\text{m s}^{-1}\$ and after collision \$|(-2)-3| = 5\;\text{m s}^{-1}\$; note that the sign convention must be consistent. Using the derived relation \$|u{\text{rel}}| = |v_{\text{rel}}|\$ confirms the calculation.