Worked example 6.6: 2-dimensional collision

Question: Two objects slide over a frictionless horizontal surface. The first object, mass $m_1=5 {\rm kg}$, is propelled with speed $v_{i1} = 4.5 {\rm m/s}$ toward the second object, mass $m_2 =2.5 {\rm kg}$, which is initially at rest. After the collision, both objects have velocities which are directed $\theta = 30^\circ$ on either side of the original line of motion of the first object. What are the final speeds of the two objects? Is the collision elastic or inelastic?

Answer: Let us adopt the coordinate system shown in the diagram. Conservation of momentum along the $x$-axis yields

$$m_1 v_{i1} = m_1 v_{f1} \cos\theta + m_2 v_{f2} \cos\theta.$$

Likewise, conservation of momentum along the $y$-axis yields

$$m_1 v_{f1} \sin\theta = m_2 v_{f2} \sin\theta.$$

The above pair of equations can be combined to give

$$v_{f1} = \frac{v_{i1}}{2 \cos\theta} = \frac{4.5}{2\times\cos 30^\circ} = 2.5981 {\rm m/s},$$

and

$$v_{f2} = \frac{m_1}{m_2} v_{f1} = \frac{5\times 2.5981}{2.5} = 5.1962 {\rm m/s}.$$

The initial kinetic energy of the system is

$$K_i = \frac{1}{2} m_1 v_{i1}^{ 2} = 0.5\times 5\times 4.5^2 = 50.63 {\rm J}.$$

The final kinetic energy of the system is

$$K_f= \frac{1}{2} m_1 v_{f1}^{ 2}+ \frac{1}{2} m_2 v_{f2... ...imes 2.5981^2 + 0.5\times 2.5\times 5.1962^2 = 50.63 {\rm J}.$$

Since $K_i=K_f$, the collision is elastic.