An automobile with a mass of 1316 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 40,810 kg and is moving to the right at a constant speed v = 8.8 m/s on a frictionless rail. The automobile then accelerates to the left, leaving the railcar at a speed of 22 m/s with respect to the ground. When the automobile lands, what is the distance D between it and the left end of the railcar? See the figure.
IT leaves with respect to the ground. That means it is going 30.8m/s with respect to the rail car.
distance=30.8*time
where time is the time to fall 1.5 m
I tried to solve for the time using:
1.5=-1/2*(9.81)t^2
but I'm not getting the right answer.
t=(1.5*2/g)^1/2
To find the distance D between the automobile and the left end of the railcar when the automobile lands, we can use the principle of conservation of momentum. The total momentum of the system before and after the automobile accelerates will remain the same, assuming no external forces act on the system.
Let's break down the problem step by step:
1. Determine the initial momentum of the system:
The initial momentum of the system is equal to the sum of the momentum of the automobile and the momentum of the railcar. Since both are initially moving to the right at a constant speed, their momenta are positive.
Momentum of the automobile (p1) = mass of the automobile (m1) × velocity of the automobile (v1)
p1 = (1316 kg) × (8.8 m/s)
Momentum of the railcar (p2) = mass of the railcar (m2) × velocity of the railcar (v2)
p2 = (40,810 kg) × ( 8.8 m/s)
2. Determine the final momentum of the system:
The final momentum of the system is equal to the sum of the momentum of the automobile and the momentum of the railcar after the automobile leaves the railcar.
Since the automobile moves to the left with respect to the ground, its momentum is negative. The momentum of the railcar remains positive as it continues moving to the right.
Momentum of the automobile (p3) = mass of the automobile (m1) × velocity of the automobile (v3)
p3 = (1316 kg) × (-22 m/s)
Momentum of the railcar (p4) = mass of the railcar (m2) × velocity of the railcar (v4)
p4 = (40,810 kg) × ( 8.8 m/s)
3. Apply the principle of conservation of momentum:
According to the conservation of momentum, the total momentum before and after the automobile leaves the railcar will be the same.
p1 + p2 = p3 + p4
Now, we can substitute and solve for the distance D:
(m1 × v1) + (m2 × v2) = (m1 × v3) + (m2 × v4)
(1316 kg × 8.8 m/s) + (40,810 kg × 8.8 m/s) = (1316 kg × -22 m/s) + (40,810 kg × v4)
Solve for v4:
(1316 kg × 8.8 m/s) + (40,810 kg × 8.8 m/s) - (1316 kg × -22 m/s) = 40,810 kg × v4
v4 ≈ 1.48 m/s
Now we can calculate the distance D using the equation:
Distance (D) = velocity (v4) × time taken to fall (t)
Since the railcar and automobile were moving at the same speed initially, the time taken for the automobile to fall will be the same as the time taken for the railcar to cover the distance D.
We can use the equation: Distance (D) = velocity (v2) × time taken (t)
D = 8.8 m/s × t
Equate the distance equation and solve for D:
8.8 m/s × t = 1.48 m/s × t
8.8 m/s = 1.48 m/s
t = 8.8 m / 1.48 m/s
t ≈ 5.95 s
Now, substitute the value of time (t) back into the distance equation:
D = 8.8 m/s × 5.95 s
D ≈ 52.36 m
Therefore, the distance D between the automobile and the left end of the railcar when the automobile lands is approximately 52.36 meters.