A 14000 kg railroad car travels alone on a level frictionless track with a constant speed of 24.0 m/s. A 6500 kg additional load is dropped onto the car. What then will be the car's speed?
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To find the new speed of the railroad car after the additional load is dropped, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it.
Initially, the railroad car is traveling alone, so its momentum is given by:
Momentum_1 = Mass_car * Velocity_car
Momentum_1 = (14000 kg) * (24.0 m/s)
Now, when the additional load of 6500 kg is dropped onto the car, the total mass of the system becomes:
Total_mass = Mass_car + Additional_load
Total_mass = 14000 kg + 6500 kg
Next, we consider the final momentum of the system after the additional load is dropped. The final momentum is given by:
Momentum_2 = Total_mass * Final_velocity
Since the total momentum is conserved, we can equate the initial and final momenta:
Momentum_1 = Momentum_2
Mass_car * Velocity_car = Total_mass * Final_velocity
Now, we can solve for the final velocity of the car:
Final_velocity = (Mass_car * Velocity_car) / Total_mass
Final_velocity = (14000 kg * 24.0 m/s) / (14000 kg + 6500 kg)
Final_velocity = (336000 kg·m/s) / 20500 kg
Final_velocity ≈ 16.39 m/s
Therefore, after the additional load is dropped, the railroad car's speed will be approximately 16.39 m/s.