91Ó°ÊÓ

The mass of the boxcar is\({m_1} = 7700\;{\rm{kg}}\).

The initial speed of the boxcar is \({v_1} = 14\;{\rm{m/s}}\).

The initial speed of the second car is \({v_2} = 0\)

The final speed of the combined system of both cars is\(v = 5.0\;{\rm{m/s}}\).

Let \({m_2}\) be the mass of the second car.

After the collision, the cars stick together; therefore, the collision between the cars is inelastic.

The momentum of the system will remain conserved.

Now the total momentum before the collision is \(\left( {{m_1}{v_1} + {m_2}{v_2}} \right)\).

The total momentum after the collision is \(\left( {{m_1} + {m_2}} \right)v\).

From the concept of momentum conservation,

\(\begin{array}{c}\left( {{m_1} + {m_2}} \right)v = \left( {{m_1}{v_1} + {m_2}{v_2}} \right)\\{m_1}v + {m_2}v = {m_1}{v_1} + {m_2}{v_2}\\{m_2}\left( {v - {v_2}} \right) = {m_1}\left( {{v_1} - v} \right)\\{m_2} = \frac{{{m_1}\left( {{v_1} - v} \right)}}{{\left( {v - {v_2}} \right)}}\end{array}\)

Now substituting the values in the above equation,

\(\begin{array}{c}{m_2} = \frac{{\left( {7700\;{\rm{kg}}} \right) \times \left( {14\;{\rm{m/s}} - 5.0\;{\rm{m/s}}} \right)}}{{\left( {5.0\;{\rm{m/s}} - 0} \right)}}\\ = 13860\;{\rm{kg}}\end{array}\)

Hence, the mass of the second car is \(13860\;{\rm{kg}}\).