Question

An open train car, with a mass of 20102010 kg, coasts along a horizontal track at the speed 2.932.93 m/s. The car passes under a loading chute and, as it does so, gravel falls vertically into it for 2.892.89 s at the rate of 419419 kg/s. What is the car's speed ????fvf after the loading is completed?

Answer 1

The final speed of the train car after being loaded with gravel is determined using conservation of momentum, considering the initial velocity of the car and the mass and duration of the gravel falling into it.

Step-by-step explanation:

The student is asking about the final speed of an open train car after it has been loaded with gravel. To solve this problem, we apply the conservation of momentum, since no external horizontal forces are acting on the train and gravel system.

Using the formula for conservation of momentum:

Since the gravel falls vertically, its initial horizontal velocity is 0. Therefore, the final velocity vf can be calculated using:

vf = (mass of car * initial velocity) / (combined mass)

To find the combined mass, we multiply the rate at which gravel falls (419 kg/s) by the time the gravel is falling (2.89 s) and add it to the mass of the car.

After the calculations, we can determine the car's final speed.

Answer 2

The speed of the train decreases to
`v_(f)=1.828m/s` after the loading is completed.

Step-by-step explanation:

Since the linear momentum of the train is conserved in the direction of motion we can write

`\overrightarrow{p_(i)}=\overrightarrow{p_(f)}\\\\m_(1)v_(1)=m_(2)v_(2)`

When gravel falls at a rate of
`419kg/s` for a duration of
`2.89` seconds the mass of gravel added becomes
`2.89* 419=1210.9kg`

Applying the calculated values in the initial equation we have

`2.93m/s* 2010kg=(2010+1210.9)kg* v_(f)\\\\\therefore v_(f)=(5889.3)/(3220.9)=1.828m/s`