Respuesta :
Answer:
[tex]V = (0.129\ m/s)\ i[/tex]
Explanation:
It is given that,
Mass of the car 1, [tex]m_1=1.62\times 10^5\ kg[/tex]
Initial speed of the car 1, [tex]u_1=0.3\ m/s i[/tex]
Mass of the car 2, [tex]m_2=1.11\times 10^5\ kg[/tex]
Initial speed of the car 2, [tex]u_2=-0.12\ m/s i[/tex]
It is mentioned that train cars are coupled together by being bumped into one another. Let V is the final velocity of the train cars after the collision. It can be calculated using the conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
[tex]V=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}[/tex]
[tex]V=\dfrac{1.62\times 10^5\times 0.3+1.11\times 10^5\times (-0.12)}{1.62\times 10^5+1.11\times 10^5}[/tex]
[tex]V = (0.129\ m/s)\ i[/tex]
So, the final speed of the coupled train cars is 0.129 m/s towards x axis. Hence, this is the required solution.
