Chapter 7, Problem 7P

To determine

To Find: Car’s new speed.

Explanation of Solution

Given:

Mass of the railroad car, $m_1=12600\text{kg}$

Mass of the load, $m_2=5350\text{kg}$

Initial velocity of railroad car, $u_1=18.0\text{m/s}$

Initial velocity of load, $u_2=0$

Formula Used:

As per law of conservation of momentum,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=\left(m_1+m_2\right)v\\ v=\frac{m_1{u}_1+m_2{u}_2}{\left(m_1+m_2\right)}\end{array}$

Here, v is the final velocity, u is the initial velocity, and m is the mass.

Calculations:

Let the final velocity of the railroad car + load bev.

  $\begin{array}{l}v=\frac{m_1{u}_1+m_2{u}_2}{\left(m_1+m_2\right)}\\ v=\frac{\left(12600\times 18.0\right)+0}{\left(12600+5350\right)}\\ v=12.63\text{m/s}\end{array}$

Conclusion: 

Thus, the car’s new speedis $12.63\text{m/s}$ .