Momentum problem with varying mass

[issacnewton]

Homework Statement

I have posted the attachement.

Homework Equations

$$\vec{F}=\frac{d\vec{p}}{dt}$$

The Attempt at a Solution

let the time taken for the bucket to fill up be T. then
$$\frac{dp}{dt}= m(t)\, \frac{dv}{dt}+v(t)\, \frac{dm}{dt}$$

where m and v are functions of time.

$$\frac{dm}{dt}=\frac{\rho V}{T}$$ since after time T, $$\rho V$$ of mass
of water will fill up in the bucket. so

$$m(t)=\left(\frac{\rho V}{T}\right)t+m$$

and the momentum at time t would be

$$\vec{p}(t)=m(t)\, v(t)$$

but after this I don't have much clue. any hints ?

 

[ehild]

Consider the problem as inelastic collision of the cart with water drop(s). A water drop of mass Δm, falling vertically into the bucket has zero horizontal momentum, and gains Δm*v(t+Δt) momentum in the collision process. At the same time the momentum of the cart with the bucket and the water inside (mass m(t) ) changes to m(t)*v(t+Δt). The net change of the momentum of the cart-water-water drop system is zero, as there is no external force.

Δp=(Δm+m(t))v(t+Δt) -m(t)v(t)=0 --->Δm*v(t+Δt)+m(t)*[v(t+Δt) - v(t)]=0

Turning to derivatives, we arrive that the time derivative of the momentum is zero.

$$\frac{dp}{dt}= m(t)\, \frac{dv}{dt}+v(t)\, \frac{dm}{dt}=0$$.


Substitute the time dependent mass, and solve for v.

ehild

 

[issacnewton]

ah, ehild thanks. I was confused about the force. I forgot to consider the whole system. I now realize that internal forces between the water drops and the bucket don't matter. By the way, the net force is zero only in horizontal direction. I vertical direction, I think, it won't be conserved since the ground will exert some force on the system.

 

[ehild]

Yes, you are right, the vertical component of the momentum is not conserved.

ehild