Worked example 6.5: Elastic collision

Question: An object of mass $m_1=2 {\rm kg}$, moving with velocity $v_{i1}= 12 {\rm m/s}$, collides head-on with a stationary object whose mass is $m_2=6 {\rm kg}$. Given that the collision is elastic, what are the final velocities of the two objects. Neglect friction.

Answer: Momentum conservation yields

$$m_1 v_{i1} = m_1 v_{f1} + m_2 v_{f2},$$

where $v_{f1}$ and $v_{f2}$ are the final velocities of the first and second objects, respectively. Since the collision is elastic, the total kinetic energy must be the same before and after the collision. Hence,

$$\frac{1}{2} m_1 v_{i1}^2 = \frac{1}{2} m_1 v_{f1}^{ 2} + \frac{1}{2} m_2 v_{f2}^{ 2}.$$

Let $x=v_{f1}/v_{i1}$ and $y=v_{f2}/v_{i1}$. Noting that $m_2/m_1=3$, the above two equations reduce to

$$1 = x + 3 y,$$

and

$$1 = x^2 + 3 y^2.$$

Eliminating $x$ between the previous two expressions, we obtain

$$1 = (1-3 y)^2 + 3 y^2,$$

or

$$6 y (2 y-1)=0,$$

which has the non-trivial solution $y=1/2$. The corresponding solution for $x$ is $x= (1-3 y)=-1/2$.

It follows that the final velocity of the first object is

$$v_{f1} = x v_{i1} = -0.5\times12 = -6 {\rm m/s}.$$

The minus sign indicates that this object reverses direction as a result of the collision. Likewise, the final velocity of the second object is

$$v_{f2} = y v_{i1} = 0.5 \times 12 = 6 {\rm m/s}.$$