An open train car, with a mass of $2190 \mathrm{~kg}$, coasts along a horizontal track at the speed $2.79 \mathrm{~m} / \mathrm{s}$. The car passes under a loading chute and, as it does so, gravel falls vertically into it for $2.97 \mathrm{~s}$ at the rate of $457 \mathrm{~kg} / \mathrm{s}$. What is the car's speed $v_{\mathrm{f}}$ after the loading is completed? Ignore rolling friction. \[ v_{\mathrm{f}}= \]

Step 1 ：This problem involves the principle of conservation of momentum. The momentum before the gravel starts falling into the car is equal to the momentum after the gravel has fallen into the car.

Step 2 ：The initial momentum is the mass of the car times its velocity. The final momentum is the total mass (mass of the car plus the mass of the gravel) times the final velocity.

Step 3 ：We can set up the equation as follows: \(m_{car} * v_{initial} = (m_{car} + m_{gravel}) * v_{final}\)

Step 4 ：Given that the mass of the car \(m_{car}\) is 2190 kg, the initial velocity \(v_{initial}\) is 2.79 m/s, the time \(t\) is 2.97 s, and the rate of gravel falling is 457 kg/s, we can calculate the mass of the gravel \(m_{gravel}\) as \(t * rate\), which equals 1357.29 kg.

Step 5 ：We can then substitute these values into the equation and solve for the final velocity \(v_{final}\).

Step 6 ：The final velocity \(v_{final}\) is approximately 1.72 m/s.

Step 7 ：Final Answer: The car's speed after the loading is completed is \(\boxed{1.72 \, \text{m/s}}\).