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

Answer:

The speed of the train decreases to $v_{f}=1.828m/s$ after the loading is completed.

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 $2.89\times 419=1210.9kg$

Applying the calculated values in the initial equation we have

$2.93m/s\times 2010kg=(2010+1210.9)kg\times v_{f}\\\\\therefore v_{f}=\frac{5889.3}{3220.9}=1.828m/s$