An open railroad car of mass 2449 kg is coasting with an initial speed of 17 m/s on a frictionless, horizontal track. It is raining, and water begins to accumulate in the car. After some time, it is found that the speed of the car is only 10 m/s. How much water (in kilograms) has accumulated in the car?
2449*17=(2449+massrain)10
solve for mass of rain.
To determine the amount of water accumulated in the car, we need to use the principle of conservation of momentum.
The initial momentum of the system, consisting of the railroad car and the rainwater, is given by the mass of the car multiplied by its initial velocity:
Initial momentum = (mass of car) × (initial velocity of car)
The final momentum of the system is given by the mass of the car, including the accumulated water, multiplied by its final velocity:
Final momentum = (mass of car + mass of water) × (final velocity of car)
According to the principle of conservation of momentum, the initial momentum of the system is equal to the final momentum of the system. Therefore, we can set up the equation:
(mass of car) × (initial velocity of car) = (mass of car + mass of water) × (final velocity of car)
Substituting the given values into the equation, we have:
2449 kg × 17 m/s = (2449 kg + mass of water) × 10 m/s
To solve for the mass of water, we can rearrange the equation:
mass of water = (2449 kg × 17 m/s - 2449 kg × 10 m/s) / 10 m/s
mass of water = (2449 kg × (17 m/s - 10 m/s)) / 10 m/s
mass of water = (2449 kg × 7 m/s) / 10 m/s
mass of water = 17143 kg·m/s / 10 m/s
mass of water = 1714.3 kg
Therefore, approximately 1714.3 kilograms of water has accumulated in the car.