Car A hits car B (initially at rest and of equal mass) from behind while going\(38\;{\rm{m/s}}\). Immediately after the collision, car B moves forward at\(15\;{\rm{m/s}}\)and car A is at rest. What fraction of the initial kinetic energy is lost in the collision?

The energy possessed by a body by its motion is called kinetic energy.Mathematically it is given by

\(KE = \frac{1}{2}m{v^2}\), … (i)

Here, m is the mass, and v is the velocity of the body.

The masses of car A and car B are equal, such that\({m_{\rm{A}}} = {m_{\rm{B}}} = m\).

The initial speed of car B is\({v_{\rm{B}}} = 0\).

The initial speed of car A is\({v_{\rm{A}}} = 38\;{\rm{m/s}}\).

The speed of car B after the collision is\({v_{\rm{B}}}^\prime = 15\;{\rm{m/s}}\).

The speed of car A after the collision is \({v_{\rm{A}}}^\prime = 0\).

The initial kinetic energy of the system before the collision is as follows:

\(\begin{array}{c}K{E_{{\rm{initial}}}} = \frac{1}{2}{m_{\rm{A}}}v_{\rm{A}}2 + \frac{1}{2}{m_{\rm{B}}}v_{\rm{B}}2\\ = \frac{1}{2}m{\left( {38\;{\rm{m/s}}} \right)2} + 0\\ = 720m\;{\rm{J}}\end{array}\)

The final kinetic energy of the system after the collision is as follows:

\(\begin{array}{c}K{E_{{\rm{final}}}} = \frac{1}{2}{m_{\rm{A}}}{v_{\rm{A}}}{\prime 2} + \frac{1}{2}{m_{\rm{B}}}{v_{\rm{B}}}{\prime 2}\\ = 0 + \frac{1}{2}m{\left( {15\;{\rm{m/s}}} \right)2}\\ = 110m\;{\rm{J}}\end{array}\)

\(\begin{array}{c}K{E_{{\rm{lost}}}} = \frac{{K{E_{{\rm{initial}}}} - K{E_{{\rm{final}}}}}}{{K{E_{{\rm{initial}}}}}}\\ = \frac{{\left( {720m\;{\rm{J}}} \right) - \left( {110m\;{\rm{J}}} \right)}}{{\left( {720m\;{\rm{J}}} \right)}}\\ \approx 0.85\end{array}\)

Thus, the fraction of kinetic energy lost during the collision process is 0.85.