A freight car, open at the top, weighing 2.80 tons, is coasting along a level track with negligible friction at 4.10 ft/s when it begins to rain hard. The raindrops fall vertically with respect to the ground. What is the speed (in feet/second) of the car when it has collected 0.200 tons of rain?
Momentum is conserved so P(i)=P(f). M(1)v(i)=(M(1)+M(2))v(f). Convert the tons to lbs (1 ton= 2000 lbs) and plug them in. (2.8*2000)*(4.10)=((2.8+.2)*2000)*v(f)= 3.8267 ft/s
To solve this problem, we need to consider the conservation of momentum. The initial momentum of the system (freight car and raindrops) is equal to the final momentum when raindrops are collected in the car.
First, let's convert the weights from tons to pounds:
Weight of the car = 2.80 tons * 2000 pounds/ton = 5600 pounds
Weight of rain collected = 0.200 tons * 2000 pounds/ton = 400 pounds
Next, we need to calculate the initial and final momentum. The momentum (p) of an object is given by the product of its mass (m) and velocity (v):
Initial momentum = Final momentum
The initial momentum of the car is given by:
p_car_initial = m_car * v_car_initial
The final momentum of the car and the raindrops is given by:
p_car_final = m_car * v_car_final
p_rain = m_rain * 0 (since the raindrops are initially at rest)
Since momentum is conserved, we can equate the initial and final momenta:
p_car_initial = p_car_final + p_rain
Substituting the momentum equations, we get:
m_car * v_car_initial = m_car * v_car_final + m_rain * 0
Since raindrops have negligible mass, we can ignore the raindrops' momentum term:
m_car * v_car_initial = m_car * v_car_final
Next, we can solve for the final velocity of the car (v_car_final):
v_car_final = (m_car * v_car_initial) / m_car
Plugging in the given values:
v_car_final = (5600 pounds * 4.10 ft/s) / 5600 pounds
Simplifying:
v_car_final = 4.10 ft/s
Therefore, the final speed of the car when it has collected 0.200 tons of rain is 4.10 ft/s.