Question

(III) One car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 8.0 m/s, they then have the same kinetic energy. What were the original speeds of the two cars?

Short answer

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

Step-by-step solution

Given data

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 \({v_2}\).

Relation between the velocities of both cars

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)

Relation between the kinetic energies of both cars after increasing the speeds by 8 m/s

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 \(11.32\;{\rm{m/s}}\).