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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

\begin{displaymath}
m_1 v_{i1} = m_1 v_{f1} + m_2 v_{f2},
\end{displaymath}

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,

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

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

\begin{displaymath}
1 = x + 3 y,
\end{displaymath}

and

\begin{displaymath}
1 = x^2 + 3 y^2.
\end{displaymath}

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

\begin{displaymath}
1 = (1-3 y)^2 + 3 y^2,
\end{displaymath}

or

\begin{displaymath}
6 y (2 y-1)=0,
\end{displaymath}

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

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

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

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


next up previous
Next: Worked example 6.6: 2-dimensional Up: Conservation of momentum Previous: Worked example 6.4: Bullet
Richard Fitzpatrick 2006-02-02