Conservation of Energy. Fountain Question

[mfianist]

Water shoots up from a fountainat a speed of v0 at a rate of dm/dt, and hits a garbage can of mass M suspended in the air. What is the maximum height the fountain can lift the garbage can?

First i approached this problem with K.E = P.E

i assumed that m at time t = (dm/dt)t

so i put {M+(dm/dt)t}gh = 1/2 {(dm/dt)tv^2}

then i solved the equation for h.

i am not so sure if my approach is right...


Is there a better of solving this problem?

 

[Soveraign]

Interesting problem. I was looking at this as an impulse problem. Recall the formula impulse P = Ft? So, assuming constant F, this gives us dP = Fdt for very small impulse bits. Rearrange and you have F = dP/dt, or, force equals the time derivative of momentum. Ah, momentum. So, what is the momentum of an object thrown upward as a function of its height? P=mv, so what is v as a function of h and some initial velocity v0? Find this P as a function of h, then find dP/dt keeping in mind dP = mdv + dmv and v is constant for a particular h. So, the garbage can has mass M, weighs Mg, it's force downward is its weight, where does the equivalent force upward come from?

This would probably be my approach, anyone see if I missed something obvious? I'm assuming inelastic collisions with the water particles, which seems reasonable.

 

[wais]

Your approach is correct in using the conservation of energy principle, but there is a more efficient way to solve this problem. We can use the equation for the conservation of momentum, which states that the initial momentum of the system is equal to the final momentum of the system. In this case, the initial momentum is 0 as the garbage can is initially at rest, and the final momentum is (M + dm/dt)t * v0 as the garbage can is lifted by the water with a velocity of v0. Therefore, we can set these two equal to each other and solve for the maximum height h:

0 = (M + dm/dt)t * v0

h = (M + dm/dt)t * v0 / (M + dm/dt)g

This gives us the maximum height that the fountain can lift the garbage can, taking into account the rate of water flow. This approach is more efficient because it only involves one equation and does not require solving for multiple variables. However, your approach is also valid and will give the same result.