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

Answer:

0.13 m/s

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

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{140000\times 0.3 + 95000\times -0.12}{140000+ 95000}\\\Rightarrow v=0.13\ m/s$

Their final velocity is 0.13 m/s

Answer 2

Answer:

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

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.

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{140000\times 0.3+95000\times (-0.12)}{(140000+95000)}\\\\V=0.13\ m/s$

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