A 12100 kg railroad car travels alone on a level frictionless track with a constant speed of 20.5 m/s. A 5850 kg load, initially at rest, is dropped onto the car. What will be the car's new speed? Please help!
Apply the law of conervation of linear momentum before and after the drop, along the direction of motion.
This does not seem like college level physics. In any case, you should be able to take it from there.
m₁v=(m₁+m₂)u
u= m₁v/(m₁+m₂)=
To find the new speed of the railroad car after the load is dropped onto it, we can apply the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of a system remains constant if no external forces act on it. In this scenario, the momentum before the load is dropped is equal to the momentum after the load is dropped.
To calculate momentum, we use the equation:
Momentum = mass × velocity
Before the load is dropped, only the railroad car is in motion with a mass of 12,100 kg and a velocity of 20.5 m/s. So, the momentum of the car before the load is dropped is:
Momentum_before = (mass_car × velocity_car)
Substituting the values, we get:
Momentum_before = (12,100 kg × 20.5 m/s)
Now, after the load is dropped, the system consists of the car and the load combined. So, the total mass of the system is the sum of the car's mass and the load's mass (12,100 kg + 5,850 kg = 17,950 kg).
Let's assume the new velocity of the car after the load is dropped is "v".
The momentum after the load is dropped is:
Momentum_after = (total_mass × new_velocity)
Since momentum is conserved, we can equate the momentum before and after the load is dropped:
Momentum_before = Momentum_after
(12,100 kg × 20.5 m/s) = (17,950 kg × v)
Now, we can solve for v to find the new velocity of the car after the load is dropped:
v = (12,100 kg × 20.5 m/s) / 17,950 kg
Calculating this, we get:
v ≈ 13.96 m/s
Therefore, the car's new speed after the load is dropped will be approximately 13.96 m/s.