Question

A freight car moves along a frictionless level railroad track at constant speed. The car is open on top. A large load of coal is suddenly dumped into the car. What happens to the velocity of the car?

Answer 1

Velocity of the car decreases.

Step-by-step explanation:

We can understand the situation if we apply the conservation of energy principle to the situation

Let the initial mass of the freight be
`m_(f)`

Initial velocity of the freight be
`v_(fi)`

Thus the initial Kinetic energy of the freight will be
`K.E=(1)/(2)m_(f)v_(if)^(2)`

When a Coal Block of mass M falls into the freight it's energy will become

`K.E=(1)/(2)(m_(f)+M)v_(ff)^(2)`

Equating both the energies we get final velocity as
`v_(ff)`

`(1)/(2)m_(f)v_(if)^(2)=(1)/(2)(M+m_(f))v_(ff)^(2)\\\\v_(ff)=\sqrt{(m_f)/((M+m_(ff)))}\cdot v_(if)`

As we see that
`\sqrt{(m_f)/((M+m_(ff)))}` is less than 1 we can infer that velocity decreases.