Question

A 24000- railroad freight car collides with a stationary caboose car. They couple together, and 22 percent of the initial kinetic energy is dissipated as heat, sound, vibrations, and so on. What is the mass of the caboose

Answer 1

Answer:

$m_{c} = 6768\,kg$

Explanation:

According to the Principle of Energy Conservation and the Work-Energy Theorem, the system is modelled as follows:

$K_{o} = K_{f} + W_{loss}$, where $\frac{K_{f}}{K_{o}} = 0.78$.

Then,

$K_{f} = 0.78\cdot K_{o}$

$0.5\cdot (m_{f}+m_{c})\cdot v_{f}^{2} = 0.39\cdot m_{f}\cdot v_{o}^{2}$

Besides, the Principle of Momentum Conservation describes the following model:

$m_{f}\cdot v_{o} = (m_{f}+m_{c})\cdot v_{f}$

The final velocity of the system is:

$v_{f} = \frac{m_{f}}{m_{f}+m_{c}}\cdot v_{o}$

After substituting in the energy expression:

$0.5\cdot \frac{m_{f}^{2}}{m_{f}+m_{c}}\cdot v_{o}^{2} = 0.39\cdot m_{f}\cdot v_{o}^{2}$

$0.5\cdot m_{f} = 0.39\cdot (m_{f}+m_{c})$

The mass of the caboose is:

$0.39\cdot m_{c} = 0.11\cdot m_{f}$

$m_{c} = 0.282\cdot m_{f}$

$m_{c} = 0.282\cdot (24000\,kg)$

$m_{c} = 6768\,kg$