Question

A 11,600 kg railroad car travels along on a level friction less track with a constant speed of 18.0 m/s. A 3,750 kg load, initially at rest is dropped onto the car. What will the car's new speed be?

Answer 1

`\begin{gathered} \text{For railroad car} \\ m_r=11,600kg_{} \\ v_r=18.0\text{ m/s} \\ \text{For load} \\ m_l=3,750kg_{} \\ \text{Question} \\ v_(r+l)=_{}\text{?} \\ \\ Momentum\text{ 1= Momentum 2} \\ P1=P2 \\ P1=m_rv_r,\text{ }P2=(m_r+m_l)v_(r+l) \\ \text{Hence} \\ m_rv_r=(m_r+m_l)v_(r+l) \\ \text{Solving }v_(r+l)\text{ } \\ v_(r+l)=\frac{m_rv_r_{}}{m_r+m_l} \\ \\ v_(r+l)=\frac{(11,600kg_{})(18.0\text{ m/s})_{}}{11,600kg_{}+3,750kg_{}_{}} \\ v_(r+l)=13.6\text{ m/s} \\ \text{The new cars sp}eed\text{ is 13.6 m/s} \end{gathered}`