A 17000 kg railroad car travels along on a level frictionless track with a constant speed of 23.0 m/s. A 5100 kg additional load is dropped onto the car. What then will be its speed in meters/second?
m₁v=(m₁+m₂)u
u= m₁v/(m₁+m₂)=
=17000•23/(17000+5100)=17.7 m/s
To find the speed of the railroad car after the additional load is dropped onto it, we can use the principle of conservation of momentum.
The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the momentum before the load is dropped is equal to the momentum after the load is dropped.
Before the load is dropped:
Momentum of the railroad car = mass of the car × velocity of the car
After the load is dropped:
Momentum of the car + load = (mass of the car + mass of the load) × velocity of the car
Let's calculate the initial momentum and the final momentum, then set them equal to each other to solve for the final velocity.
Initial momentum = mass of the car × velocity of the car
Initial momentum = 17000 kg × 23.0 m/s
Final momentum = (mass of the car + mass of the load) × velocity of the car
Final momentum = (17000 kg + 5100 kg) × v
Since momentum is conserved, we can set the initial momentum equal to the final momentum:
17000 kg × 23.0 m/s = (17000 kg + 5100 kg) × v
Now, let's solve for v, the final velocity of the railroad car:
(17000 kg × 23.0 m/s) = (17000 kg + 5100 kg) × v
391000 kg·m/s = 22100 kg × v
v = 391000 kg·m/s / 22100 kg
v ≈ 17.68 m/s
Therefore, the final speed of the railroad car, after the additional load is dropped onto it, is approximately 17.68 m/s.