Solving a Cart Losing Mass Problem

[LondonLady]

Hi I have this problem involving a cart which is losing sand

It says:

A cart with initial mass $$M$$ and a load of sand $$\frac{1}{2}M$$ loses sand at the rate $$k$$ kg/s. The cart is pulled horizontally by a force $$F$$. Find the differential equation for the rate of change of the carts velocity in terms of $$k,M$$ and $$F$$ while there is sand in the cart.

So i said that at $$t = 0$$ the momentum$$= \frac{3}{2}Mv$$. Therefore

$$\displaystyle{dp = \left(\frac{3}{2}Mv\right) - \left[\left(\frac{3}{2}M - dM\right)\left(v + dv\right) - vdM\right]}$$

Simplifying

$$\displaystyle{dp = -\frac{3}{2}Mdv + 2vdM$$

Dividing by $$dt$$

$$\displaystyle{\frac{dp}{dt} = -\frac{3}{2}M\frac{dv}{dt} + 2v\frac{dM}{dt}}$$

As $$\displaystyle{\frac{dM}{dt} = -k}$$

$$\displaystyle{\frac{dv}{dt} = -\frac{4}{3M}vk - \frac{2F}{3M}}$$

Im confused because of the very negative right hand side of the equation. Did i make an error in the set up at the start?

Thankyou in advance

 

[donjennix]

I hesitate to put this forth since I do so poorly at this type of problem; I hope this doesn't confuse the issue but I took a different approach -- it seems too easy even though I can't see the flaw

M(t) = 1.5*M - kt

a(t) = F/M(t)

v(t + dt) = v(t) + a(t)dt

thus

v'(t) = a(t)

integrate and you get a log expression:

[log(1.5M) - log(1.5M - kt)]*F/k

This gives sane answers in Excel.

 

[e(ho0n3]

When you say 'initial mass M' does that include the load of sand? donjennix's approach looks OK to me (but then again, I'm no physics god so...). Using momentum, we have

dp/dt = d/dt (mv)

and since m and v are functions of time, then

dp/dt = mdv/dt + vdm/dt.

Rearranging gives

dv/dt = 1/m(dp/dt - vdm/dt)

Since F is producing the change in momentum, we can substitute it for dp/dt. dm/dt is just k so

dv/dt = 1/m(F - vk)

As a said, I'm not physics guru so let's see what others say.

 

[donjennix]

dv/dt = 1/m(F - vk)

And don't forget that m= 1.5*M - kt or something similar

 

[LondonLady]

No the initial mass M does not include the mass of the sand. Thankyou very much for your comments, I will have a thorough read of them now.