## A 24000- railroad freight car collides with a stationary caboose car. They couple together, and 22 percent of the initial kinetic energy is

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

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1 year 2021-07-27T15:53:03+00:00 1 Answers 6 views 0

$$m_{c} = 6768\,kg$$

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:

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

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

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

$$m_{c} = 6768\,kg$$