91Ó°ÊÓ

The mass of the railroad car is \({m_1} = 7150\;{\rm{kg}}\).

The mass of the load is \({m_2} = 3350\;{\rm{kg}}\).

The initial speed of the railroad car is \({v_1} = 15.0\;{\rm{m/s}}\).

The initial speed of the load is \({v_2} = 0\).

Let \(v\) be the new speed of the car.

The momentum of the car remains constant during the process as no external force acts on the system if you consider load and car as a system and ignore air resistance.

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

After the drop, the load gains a speed equal to the speed of the car, and both move at the same speed.

Therefore, the total momentum after the incident is \(\left( {{m_1} + {m_2}} \right)v\).

From the concept of momentum conservation, you get:

\(\begin{array}{c}\left( {{m_1} + {m_2}} \right)v = \left( {{m_1}{v_1} + {m_2}{v_2}} \right)\\v = \frac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\\v = \frac{{\left[ {\left( {7150\;{\rm{kg}}} \right) \times \left( {15.0\;{\rm{m/s}}} \right)} \right] + \left[ {\left( {3350\;{\rm{kg}}} \right) \times 0} \right]}}{{\left( {7150\;{\rm{kg}}} \right) + \left( {3350\;{\rm{kg}}} \right)}}\\v = 10.21\;{\rm{m/s}}{\rm{.}}\end{array}\)

Hence, the new speed of the car is \(10.2\;{\rm{m/s}}\).