An open railroad car of mass 2445 kg is coasting with an initial speed of 22 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 5 m/s. How much water (in kilograms) has accumulated in the car?
conservation of momentum applies
2445*22=(2445+M)5
solve for Mass rainwater M in kilograms.
To find the amount of water that has accumulated in the car, we need to use the principle of conservation of momentum. Since the car is initially traveling with a speed of 22 m/s and comes to rest at 5 m/s, we can say that the car has lost momentum.
The momentum of the car before gathering water is given by:
Momentum_initial = mass_car * velocity_initial
The momentum of the car after gathering water is given by:
Momentum_final = (mass_car + mass_water) * velocity_final
Since momentum is conserved, we can equate these two equations:
mass_car * velocity_initial = (mass_car + mass_water) * velocity_final
Now, we can rearrange the equation to solve for the mass of water:
mass_water = (mass_car * (velocity_initial - velocity_final)) / velocity_final
Plugging in the given values:
mass_car = 2445 kg
velocity_initial = 22 m/s
velocity_final = 5 m/s
mass_water = (2445 kg * (22 m/s - 5 m/s)) / 5 m/s
Calculating this expression will give us the mass of water accumulated in the car.