Answer
The mass of the caboose is $~~1.18\times 10^4~kg$
Work Step by Step
Let $m_1$ be the mass of the freight car.
Let $m_2$ be the mass of the caboose car.
Let $v_i$ be the initial speed of the freight car.
We can use conservation of momentum to find an expression for the speed $v_f$ after the collision:
$p_f = p_i$
$(m_1+m_2)~v_f = m_1~v_i$
$v_f = \frac{m_1}{m_1+m_2}~v_i$
We can find an expression for the initial kinetic energy:
$K_i = \frac{1}{2}m_1~v_i^2$
We can find an expression for the kinetic energy after the collision:
$K_f = \frac{1}{2}(m_1+m_2)~v_f^2$
$K_f = \frac{1}{2}(m_1+m_2)~(\frac{m_1}{m_1+m_2}~v_i)^2$
$K_f = \frac{m_1^2}{2(m_1+m_2)}~v_i^2$
It is given in the question that $K_f = 0.73~K_i$, since $0.27~K_i$ is transformed into other kinds of energy.
We can find $m_2$:
$0.73~K_i = K_f$
$(0.73)~(\frac{1}{2}m_1~v_i^2) = \frac{m_1^2}{2(m_1+m_2)}~v_i^2$
$0.73 = \frac{m_1}{m_1+m_2}$
$(0.73)(m_1+m_2) = m_1$
$0.73~m_2 = m_1-0.73~m_1$
$m_2 = \frac{(m_1)(1-0.73)}{0.73}$
$m_2 = \frac{(3.18\times 10^4~kg)(0.27)}{0.73}$
$m_2 = 1.18\times 10^4~kg$
The mass of the caboose is $~~1.18\times 10^4~kg$
