Question

A boxcar of length 10.1 m and height 2.4 m is at rest on frictionless rails. Inside the boxcar (whose mass when empty is 3600 kg) a tank containing 1700 kg of water is located at the left end. The tank is 1.0 m long and 2.4 m tall. At some point the walls of the tank start to leak, and the water fills the floor of the boxcar uniformly. Assume that all the water stays in the boxcar.

A. After all the water has leaked out what will be the final velocity of the boxcar? (Take movement to the right as positive. Assume that the mass of the boxcar is evenly distributed.)

B. What is the displacement of the boxcar 9 s after the water has settled in the bottom. (Take positive displacement as being to the right.)

Answer 1

In this case, the shift is potential energy of the water is the cause for the velocity to the boxcar.

Initially, center of mass of water in the tank is at 1.0/2 = 0.5 m

Volume of water is same ; 1.0 * 2.4 * b = 10.1* b * H

Solve for H

H = 0.2376 m

Center of mass will be now at H/2 = 0.1188 m.

Change in height = 0.5-0.1188 = 0.3811 m.

Conserving energy leads to

Decrease in potential energy = increase in kinetic energy

m g h = 0.5 * ( m +M) V^2

1700 * 9.8 * 0.3811 m = 0.5 * (3600+1700) * V^2

V =1.547 m/s

Round off to two significant digits

V=1.5 m/s

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consider tht the left end of the car to be x = 0, the CM of the combined loads is located at:

Xcmi = (3600 * 5.05 + 1700* 0.5)/(3600+1700) =3.52 m

When the water runs out, Xcmf = 5.05 m (by symmetry)

Due to conservation of momentum, the CM must stay in its original position relative to the outside observer, so the car must move (5.05 - 3.52 m) = 1.52 m to the left of its initial position.

( Here, 9 s is irrelevent )