Given:
The mass of the first railroad car is,
[tex]m_1=7000\text{ kg}[/tex]The initial speed of the first railroad car is,
[tex]u_1=3.0\text{ m/s}[/tex]The mass of the other railroad car is,
[tex]m_2=4000\text{ kg}[/tex]The initial speed of the second railroad car is,
[tex]u_2=0[/tex]As they collide the cars stick together and move down the track.
To find:
The velocity do the cars travel down the track
Explanation:
The linear momentum before the collision is,
[tex]\begin{gathered} m_1u_1+m_2u_2 \\ =7000\times3.0+4000\times0 \\ =21000\text{ kg.m/s} \end{gathered}[/tex]If the velocity after the collision is 'v', the linear momentum is,
[tex]\begin{gathered} (m_1+m_2)v \\ =(7000+4000)v \\ =11000v \end{gathered}[/tex]According to the linear momentum conservation principle,
[tex]\begin{gathered} 11000v=21000 \\ v=\frac{21000}{11000} \\ v=1.91\text{ m/s} \end{gathered}[/tex]Hence, the required velocity with which the cars will travel is 1.91 m/s.
