Question

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 140,000 kg and a velocity of 0.300 m/s, and the second having a mass of 95,000 kg and a velocity of −0.120 m/s. (The minus indicates direction of motion.) What is their final velocity (in m/s)?

Answer 1

The final velocity of the two train cars is 0.13 m/s.

Step-by-step explanation:

Given that,

Mass of the first car,
`m_1=14000\ kg`

Mass of the second car,
`m_2=95000\ kg`

Initial speed of first car,
`u_1=0.3\ m/s`

Initial speed of second car,
`u_2=-0.12\ m/s`

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 :

`(m_1u_1+m_2u_2)=(m_1+m_2)V`

V is the final speed of two cars.

`V=(m_1u_1+m_2u_2)/((m_1+m_2))\\\\V=(140000* 0.3+95000* (-0.12))/((140000+95000))\\\\V=0.13\ m/s`

So, the final velocity of the two train cars is 0.13 m/s. Hence, this is the required solution.

Answer 2

0.13 m/s

Step-by-step explanation:

`m_1` = Mass of first car = 140000 kg

`m_2` = Mass of second car = 95000 kg

`u_1` = Initial Velocity of first car = 0.3 m/s

`u_2` = Initial Velocity of second car = -0.12 m/s

`v` = Velocity of combined mass

For elastic collision

`m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=(m_1u_1 + m_2u_2)/(m_1 + m_2)\\\Rightarrow  v=(140000* 0.3 + 95000* -0.12)/(140000+ 95000)\\\Rightarrow v=0.13\ m/s`

Their final velocity is 0.13 m/s