Given data:
* The mass of the first train car is,
![m_1=42000\text{ kg}](https://img.qammunity.org/2023/formulas/physics/college/flp9kv960lldny0zbrplaa93bjulvk6g36.png)
* The initial velocity of the first train car is,
![u_1_{}=10\text{ m/s}](https://img.qammunity.org/2023/formulas/physics/college/5x1ogpqjo0jl3q5ljbigbfvrrr03z9vshp.png)
* The velocity of the couple train car is,
![v=6\text{ m/s}](https://img.qammunity.org/2023/formulas/physics/college/weygoaoid5ydmm5ssw9anhmsing07zcr36.png)
Solution:
By the law of conservation of momentum,
![p_i=p_f](https://img.qammunity.org/2023/formulas/physics/college/e7rj8yeigkhgmt4s6ul64ny08zwwhw91ko.png)
where p_i is the initial momentum of net system, and p_f is the final momentum of the net system,
![m_1u_1+m_2u_2=(m_1+m_2)v](https://img.qammunity.org/2023/formulas/physics/college/6ayyvwrjj5zekw49rg8f4x48rozaguj29n.png)
where m_2 is the mass of the second train car, and u_2 is the initial velocity of the second train car,
![\begin{gathered} 42000*10+0=(42000+m_2)*6 \\ 420000=252000+m_2*6 \\ 6m_2=168000 \\ m_2=(168000)/(6) \\ m_2=28000\text{ kg} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/ip9601leaibqv1dsueip3uc6rw02s0hwk2.png)
Thus, the mass of the second train car is 28000 kg.