Answer:
0.13 m/s
Explanation:
[tex]m_1[/tex] = Mass of first car = 140000 kg
[tex]m_2[/tex] = Mass of second car = 95000 kg
[tex]u_1[/tex] = Initial Velocity of first car = 0.3 m/s
[tex]u_2[/tex] = Initial Velocity of second car = -0.12 m/s
[tex]v[/tex] = Velocity of combined mass
For elastic collision
[tex]m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{140000\times 0.3 + 95000\times -0.12}{140000+ 95000}\\\Rightarrow v=0.13\ m/s[/tex]
Their final velocity is 0.13 m/s
Answer:
The final velocity of the two train cars is 0.13 m/s.
Explanation:
Given that,
Mass of the first car, [tex]m_1=14000\ kg[/tex]
Mass of the second car, [tex]m_2=95000\ kg[/tex]
Initial speed of first car, [tex]u_1=0.3\ m/s[/tex]
Initial speed of second car, [tex]u_2=-0.12\ m/s[/tex]
It is mentioned that two train cars are coupled together by being bumped into one another. So, it is a case of inelastic collision. Momentum will remain conserved here. Using the conservation of linear momentum we get :
[tex](m_1u_1+m_2u_2)=(m_1+m_2)V[/tex]
V is the final speed of two cars.
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{140000\times 0.3+95000\times (-0.12)}{(140000+95000)}\\\\V=0.13\ m/s[/tex]
So, the final velocity of the two train cars is 0.13 m/s. Hence, this is the required solution.
