Question

Two objects are involved in a perfectly elastic head-on collision on a frictionless air hockey table. The objects have masses of 6.21kg and 9.4kg respectively. If the first object has a final velocity of 1.9m/s [right]and the second object has a final velocity of 0.80 m/s [left], calculate the initial velocities of each object

Answer 1

Let $m_1=6.21\,\mathrm{kg}$ and $m_2= 9.4\,\mathrm{kg}.$ Let $v_{1} \; \mathrm{and}\; v_2$ be the initial velocities of bodies and $u_{1} = 1.9\,\mathrm{m/s}\; \mathrm{and}\; u_2 = 0.80\,\mathrm{m/s}$ be the velocities after the collision.

We may write the law of conservation of impulses (see https://en.wikipedia.org/wiki/Momentum#Conservation):

$m_1\vec{v_1} + m_2\vec{v_2} = m_1\vec{u_1} + m_2\vec{u_2}.$

$\vec{v_1}$ is directed left because the final velocity of the first body is directed right. In projection

$-m_1v_1+ m_2v_2 = m_1u_1 - m_2u_2.$ Here $v_1 \, \mathrm{and}\; v_2$ are positive values.

Next, we substitute the values of mass and velocity and get

$-m_1v_1 + m_2v_2 = 4.279\,\mathrm{kg\,m/s}.$

Let us write the law of conservation of energy:

$\dfrac{m_1v_1^2}{2} + \dfrac{m_2v_2^2}{2} =\dfrac{m_1u_1^2}{2} + \dfrac{m_2u_2^2}{2}\,.$

We may multiply the expression on 2 and calculate the right hand.

${m_1v_1^2} + {m_2v_2^2} =28.4341\,\mathrm{J}.$

So we have a system of equations, namely

$\begin{cases} -m_1v_1 + m_2v_2 = 4.279\,\mathrm{kg\,m/s}, \\ {m_1v_1^2} + {m_2v_2^2} =28.4341\,\mathrm{J}. \end{cases}$

Therefore,

$\begin{cases} -6.21v_1 + 9.4v_2 = 4.279, \\ {6.21v_1^2} + {9.4v_2^2} =28.4341. \end{cases}$ $\begin{cases} 6.21v_1 = 9.4v_2-4.279, \\ {6.21^{-1}\cdot( 9.4v_2-4.279)^2} + {9.4v_2^2} =28.4341. \end{cases}$

Solving the last equation, we get $v_{2,1} = -0.8\,\mathrm{m/s}\; (u_{1,1} =-1.9\,\mathrm{m/s} )$ and $v_{2,2} = 1.35\,\mathrm{m/s}\; (u_{1,2} =1.35\,\mathrm{m/s} ).$ The first pair of values is negative, so we should choose the second pair.