An open train car, with a mass of 2150 kg, coasts along a horizontal track at the speed 2.83 m/s. The car passes under a loading chute and, as it does so, gravel falls vertically into it for 2.95 s at the rate of 481 kg/s. What is the car's speed v f after the loading is completed? Ignore rolling friction.

The car's final speed after the loading is completed is 1.71 m/s

Explanation:

Given;

mass of the open car, m₁ = 2150 kg

initial speed of the car, u = 2.83 m/s

loading rate of the gravel, m' = 481 kg/s.

loading time of the gravel, t = 2.95 s

Then, mass of the gravel, m₂ = m' x  t  = 481 x 2.95 = 1418.95 kg

To determine the car's speed after the loading is completed, we apply the principle of conservation of linear momentum;

Initial momentum = Final momentum

m₁u = v(m₁ + m₂)

where;

v is the car's final speed after the loading is completed

2150 x 2.83 = v (2150 +  1418.95)

6084.5 = 3568.95v

v = 6084.5 / 3568.95

v = 1.71 m/s

Thus, the car's final speed after the loading is completed is 1.71 m/s

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i don't understand. explain it clearly