What is the difference between elastic and inelastic collisions

Collisions are interactions where two or more bodies exert forces on each other for a short time. In any isolated system, the vector sum of momenta before the encounter equals the sum after, a statement of momentum conservation. Whether kinetic energy is also conserved depends on how the internal structure of the bodies responds during impact.

Conservation Laws in Collisions

Momentum, defined as (p = mv), is always conserved in closed systems because external forces are negligible during the brief contact. Kinetic energy, given by (K = $\frac{1}{2}$mv^{2}), is conserved only if no energy is transformed into other forms. In elastic collisions the objects rebound without permanent deformation, so the kinetic energy before and after remains equal; in inelastic collisions some of that energy becomes internal energy.

Key differences between elastic and inelastic collisions:

• Elastic: both momentum and kinetic energy conserved.
• Inelastic: momentum conserved, kinetic energy not fully conserved.
• Elastic collisions often involve hard spheres or atoms; inelastic involve clay, cars, or any deformable bodies.
• Coefficient of restitution equals 1 for elastic, between 0 and 1 for inelastic.

How to decide the collision type:

1. 1Calculate total momentum before and after; they must match for any collision.
2. 2Compute total kinetic energy before and after using masses and velocities.
3. 3If kinetic energies are equal (within experimental error), the collision is elastic; if the final kinetic energy is lower, it is inelastic.

Comparison of elastic and inelastic collisions:

Worked example: Two carts on a frictionless track, mass (m_{1}=2\,$\text{kg}$) moving at (v_{1}=3\,$\text{m/s}$) and mass (m_{2}=3\,$\text{kg}$) at rest, collide elastically. Momentum before is (p_{i}=2\times3=6\,$\text{kg·m/s}$). Using the elastic collision formulas, the final velocities are (v_{1}' = $\frac{m_{1}-m_{2}}{m_{1}+m_{2}}$v_{1}=$\frac{-1}{5}$\times3=-0.6\,$\text{m/s}$) and (v_{2}' = $\frac{2m_{1}}{m_{1}+m_{2}}$v_{1}=$\frac{4}{5}$\times3=2.4\,$\text{m/s}$). Kinetic energy before is (\frac12(2)(3^{2})=9\,$\text{J}$); after it is $\frac{1}{2}(2)(0.6^{2})+\frac{1}{2}(3)(2.4^{2})=9\,\text{J}), confirming an elastic collision.$