91Ó°ÊÓ

The first car has twice the mass of the second car.

Let\(m{}_1\)be the mass of the first car and\({m_2}\)be the mass of the second car.

Let the original speed of the first car be \({v_1}\) and that of the second car be \({v_2}\).

For this problem, two equations can be obtained using the car's kinetic energy at different speeds.

By solving those two equations, you can get the values of original speeds.

The relation between the masses of the cars is \({m_1} = 2{m_2}\).

The original kinetic energies of the cars are related as:

\(\begin{array}{c}{K_1} = \frac{1}{2}{K_2}\\\frac{1}{2}{m_1}v_1^2 = \frac{1}{2}\left( {\frac{1}{2}{m_2}v_2^2} \right)\\\frac{1}{2} \times 2{m_2} \times v_1^2 = \frac{1}{4}{m_2}v_2^2\\4v_1^2 = v_2^2\end{array}\)

After taking the square root, you get:

\({v_2} = 2{v_1}\) … (i)

Now, after increasing the speed by \(8.0\;{\rm{m/s}}\), the relation between the kinetic energies is as follows:

\(\begin{array}{c}\frac{1}{2}{m_1}{\left( {{v_1} + 8.0\;{\rm{m/s}}} \right)^2} = \frac{1}{2}{m_2}{\left( {{v_2} + 8.0\;{\rm{m/s}}} \right)^2}\\\frac{1}{2} \times 2{m_2} \times {\left( {{v_1} + 8.0\;{\rm{m/s}}} \right)^2} = \frac{1}{2}{m_2}{\left( {2{v_1} + 8.0\;{\rm{m/s}}} \right)^2}\\\left( {{v_1} + 8.0\;{\rm{m/s}}} \right) = \frac{1}{{\sqrt 2 }}\left( {2{v_1} + 8.0\;{\rm{m/s}}} \right)\end{array}\)

After further calculation, you will get:

\(\begin{array}{c}{v_1} + 8.0\;{\rm{m/s}} = \sqrt 2 {v_1} + \frac{{8.0}}{{\sqrt 2 }}\;{\rm{m/s}}\\\sqrt 2 {v_1} - {v_1} = 8.0\;{\rm{m/s}} - \frac{{8.0}}{{\sqrt 2 }}\;{\rm{m/s}}\\\left( {\sqrt 2 - 1} \right){v_1} = 8.0\left( {1 - \frac{1}{{\sqrt 2 }}} \right)\;{\rm{m/s}}\\{v_1} = 5.66\;{\rm{m/s}}\end{array}\)

Now, from equation (i), you get:

\(\begin{array}{c}{v_2} = 2 \times 5.66\;{\rm{m/s}}\\ = 11.32\;{\rm{m/s}}\end{array}\)

Hence, the original speed of the first car is \(5.66\;{\rm{m/s}}\) and that of the second car is \(11.32\;{\rm{m/s}}\).