Question

A boxcar of length 8.00 m and height 2.40 m is at rest on frictionless rails. The empty boxcar has a mass of 3800 kg. Inside the boxcar, located at the left end, is a tank containing 1450 kg of water. The tank is 1.90 m long and 2.40 m tall. The tank starts to leak, and the water fills the floor of the boxcar uniformly. . Assume that all the water stays in the box car. . After all the water has leaked out of the tank, what will be the final velocity of the boxcar?

Answer 1

The approach I think you can take is to find the center of mass of the water when the tank is full and then find the center of mass of the same water volume when the water level is in the box car (i.e. after it has leaked). The center of mass should change because the tank is narrower than the boxcar.

So you can look at the water mass as moving from one height to another (i.e. change is center of mass). Take the entire mass of the water as the object that is moving. We then can use the conservation of energy to find the velocity of the boxcar. If the water were free-falling, then the speed would be along the y-axis. However, because the water stays in the boxcar, the y-forces are transferred along the x-axis and the boxcar is on frictionless rails the entire potential energy, m * g * h is converted to kinetic energy 1/2 * mv^2 Equate these two and solve for v. Note that the final velocity does not depend on mass at all, but you will need h, which will be the distance the center of gravity of the water system has moved.

Answer 2

Answer: The final velocity of the boxcar will be zero meter per seconds .

Explanation: Since tank of water is inside the boxcar so when the water leaks from tank the force on the boxcar due to leaking water is internal . Thus the net external force on the boxcar-tank system is zero .

Therefore momentum of the system of boxcar-tank will be conserved .

final momentum of the system=Initial momentum of system

=>$(M_{boxcar}+M_{tank})\times V_{final}=(M_{boxcar}+M_{tank})\times V_{initial}$

=>$V_{final}=V_{initial}$

Since $V_{initial}=0 \frac{m}{s}$

Therefore , $V_{final}=0 \frac{m}{s}$

Thus final velocity of the boxcar will be zero m/s .