**Date:** Fri, 16 Feb 2001 16:29:23 -0800

**Author:** Eric Ayars

**Subject:** Re: physics question

**Post:**

>PS I bet that the cart would not move. I might be out 10 bucks.

You're out ten bucks.

Simplest solution:

There is no external force, so momentum is conserved. Initial

velocity of center of mass is zero, so initial momentum is zero, so

the center of mass never moves. Since the water moves from one

location (center of cart) to another location (below off-center

drainpipe) and the CM doesn't move, the cart must move. Once the

water stops draining, the cart stops: final momentum equals initial

momentum. The final position of the cart (measured from x=0) is

x(cart) = - x(water) * m(water) / m(cart)

Complex solution:

Calculate the change in momentum of an infinitesimal volume of water

as it starts moving from some arbitrary location in the cart to the

drain. You may ignore the vertical component, since the cart is on a

track and we're only interested in horizontal motion. F=dP/dt.

Integrate over the entire volume of the cart to find the total force

supplied by the cart on the water during the drain time.

Now calculate the change in momentum of an infinitesimal volume of

water as it stops moving horizontally and moves vertically down the

drain. F=dP/dt again. Integrate over the entire volume of water

drained to find the total force supplied by the water on the cart.

Compare the two total forces. They will have the same magnitude, and

the same duration, so the final momentum of the cart equals the

initial momentum of the cart (which is zero) so you might think the

cart never moves, BUT...

The first of the two forces starts BEFORE the second force, and ends

before the second. The time shift between the two is (on average) the

distance between the center of the tank and the drain, divided by the

flow velocity.

delta t = x(water)/v(water)

The net force is not zero for a brief time at the beginning of flow,

during which the cart starts moving. The net force is zero during the

bulk of the flow, so the cart continues moving at constant velocity.

The net force is not zero (and equal and opposite the initial force)

for a brief time as the flow stops, during which time the cart stops.

I haven't worked it out, but I'd bet considerably more than $10 that

the final position of the cart, as calculated with this method, works

out to

x(cart) = - x(water) * m(water) / m(cart)

Have a good weekend - I'm off to do something useful. Like figure out

how to start water flow from an off-center drain in a tank on an

air-track glider without giving the glider any unwanted momentum.

I've got a couple of students looking for a challenging project lab

anyway...

Eric

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| Dr. Eric Ayars ayarer@wwc.edu |

| Assistant Professor of Physics, Walla Walla College (509) 527-2476 |

| http://homepages.wwc.edu/staff/ayarer |

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