Problem — 2D elastic collision on a billiard table

Problem

Ball 1 ($m_1 = 4\;\mathrm{kg}$) moves with $\vec{v}_{1A} = (1{,}73;\,-1)\;\mathrm{m/s}$ and collides elastically with ball 2 ($m_2 = 7\;\mathrm{kg}$, at rest). After the collision ball 1 leaves at $20^\circ$ below the $x$ axis with a magnitude $v_{1B}$ to be determined. Write the three conservation equations and state the unknowns.

Solution

Elastic collision in two dimensions: both components of momentum and the kinetic energy are conserved. The three unknowns are the magnitude $v_{1B}$ (direction known, $-20^\circ$) and the two components $v_{2B,x}$, $v_{2B,y}$ of ball 2.

Conservation of $p_x$: $m_1 v_{1A,x} = m_1 v_{1B}\cos(-20^\circ) + m_2 v_{2B,x}$ Conservation of $p_y$: $m_1 v_{1A,y} = m_1 v_{1B}\sin(-20^\circ) + m_2 v_{2B,y}$ Conservation of $E_\text{cin}$ (elastic): $\tfrac{1}{2}m_1 v_{1A}^2 = \tfrac{1}{2}m_1 v_{1B}^2 + \tfrac{1}{2}m_2\left(v_{2B,x}^2 + v_{2B,y}^2\right)$ $\ev{\text{3 equations } (p_x, p_y, E_\text{cin}) \text{ in 3 unknowns } (v_{1B}, v_{2B,x}, v_{2B,y})}$ The system is non-linear (the energy equation contains squares), but can be solved by substitution: from the first two, $v_{2B,x}$ and $v_{2B,y}$ are obtained as functions of $v_{1B}$, which are then substituted into the third.

Links

Topics: Momentum and collisions Concepts: Elastic collision · Conservation of momentum Methods: Decomposition into Cartesian components