Question

A railroad freight car of mass $\mathbf{3}\mathbf{.}\mathbf{18}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{kg}$collides with a stationary caboose car. They couple together, and $\mathbf{27}\mathbf{.}\mathbf{0}\mathbf{\%}$of the initial kinetic energy is transferred to thermal energy, sound, vibrations, and so on. Find the mass of the caboose.

Short answer

The mass of the caboose car is $1.18\times {10}^4\mathrm{kg}$.

Step-by-step solution

Understanding the given information

1. Mass of freight car,$m_F=3.18\times {10}^4\mathrm{kg}$.
2. The initial velocity of the caboose car,$v_C=0.00\mathrm{m}/\mathrm{s}$.
3. The relationship between initial and final kinetic energy is, $K_f=0.73K_i$.

Concept and formula used in the given question

Using the formula of conservation of momentum, you can find the final velocity of the coupled freight car and caboose car. Using this velocity in the given relation between initial kinetic energy and lost kinetic energy, you can find the mass of the caboose car.

Calculation for the mass of caboose

Let us assume that$m_c$is themass of the caboose car,$m_F$is the mass of the freight car, and$V$isfinal velocity, and $v_F$is the velocity of the freight car. Since the collision is an inelastic collision, using the conservation of momentum, we have,

$\left(m_F{v}_F+m_C{v}_C\right)=\left(m_C{m}_F\right)V$

Substitute the values in the above expression, and we get,

$$\begin{gathered} m_F{v}_F=\left(m_C+m_F\right)V \\ V=\frac{m_F{v}_F}{\left(m_C+m_F\right)} \end{gathered}$$

Now, we have total initial kinetic energy,

$K_i=\frac{1}{2}{m}_F{v_F}^2$

And, for total final kinetic energy,

$K_f=\frac{1}{2}\left(m_C+m_F\right)V^2$

Substitute the values in the above expression, and we get,

$$\begin{gathered} K_f=\frac{1}{2}\left(m_C+m_F\right){\left(\frac{m_F{v}_F}{m_C+m_F}\right)}^2 \\ =\frac{1}{2}\frac{{m_F}^2{v_F}^2}{m_C+m_F} \end{gathered}$$

We are given that 27 % of the original kinetic energy is lost, therefore,

$K_f=0.73K_i$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \frac{1}{2}\frac{{m_F}^2{v_F}^2}{m_C+m_F}=\left(0.73\right)\frac{1}{2}{m}_F{v_F}^2 \\ \frac{{\displaystyle \frac{1}{2}\frac{{m_F}^2{v_F}^2}{m_C+m_F}}}{{\displaystyle \frac{1}{2}}{m}_F{v_F}^2}=0.73 \\ \frac{m_F}{m_C+m_F}=0.73 \end{gathered}$$

Solving further as,

$$\begin{gathered} m_F=0.73\left(m_C+m_F\right) \\ 0.23m_F=0.73m_C \\ {m}_C=\frac{0.23}{0.73}{m}_F \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} m_C=\frac{0.23}{0.73}\left(3.18\times {10}^4\mathrm{kg}\right) \\ =1.18\times {10}^4\mathrm{kg} \end{gathered}$$

Therefore, the mass of the caboose car is $1.18\times {10}^4\mathrm{kg}$.