Velocity of a cart moving in the rain as a function of time

  • #1
Hamiltonian
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Homework Statement
A cart of mass ##M_0## is moving with a velocity ##v_0##. At t = 0. water starts pouring on the cart from the container above the cart at the rate ##\lambda##kg/sec. Find the velocity of the cart as a function of time.
Relevant Equations
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At time t = 0, the mass of the cart is ##M_0## and velocity is ##v_0## in a time interval ##dt## let a mass of ##dm## be added to the cart due to the pouring water and let the reduction in speed be ##dv##
##\lambda = dm/dt##

applying conservation of momentum from the ground frame gives $$M_0 v_0 = (M_0 + dm)(v_0 - dv)$$
however on substituting ##dm## in terms of ##\lambda## and ##dt## gives a differential equation that I am unable to solve. :oldconfused:

In class we were given a formula $$ F_{thrust} = v_{relative} \frac{dm}{dt}$$
we derived this formula to give the thrust of a rocket(but it can be used for any variable mass system with a few tweaks)

using this formula here gives $$\lambda (-v) = (M_0 + \lambda t)(dv/dt)$$
solving this equation yields the correct answer(##v = \frac{M_0 v_0}{M_0 + \lambda t}##). Instead of blindly applying the above-stated formula on the given problem I tried to derive it in the process of solving the problem but I am ending up with a rather weird-looking equation.

I also figured out a much simpler method that does not require any calculus its simple directly applying the conservation of momentum $$M_0 v_0 = (M_0 + \lambda t)v$$ which gives the correct answer but I want to know why the first method I am trying to use isn't working.
 
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  • #2
is ##v_0 e^{\frac{-\lambda}{m}t} = \frac {mv_0}{m+\lambda t}##? (or is there a mistake in the answer I have stated above)
also shouldn't the equation for the thrust exeretd by the water be $$v_{rel,x}\frac{dm}{dt} = (m + \lambda t)\frac{dv}{dt}$$
also, your first method makes a lot of sense but the answer isn't matching with the solution?
 
  • #3
Hamiltonian299792458 said:
using this formula here gives $$\lambda (-v) = (M_0 + \lambda t)(dv/dt)$$
solving this equation yields the correct answer(##v = \frac{M_0 v_0}{M_0 + \lambda t}##).

I also figured out a much simpler method that does not require any calculus its simply directly applying the conservation of momentum $$M_0 v_0 = (M_0 + \lambda t)v$$ which gives the correct answer but I want to know why the first method I am trying to use isn't working.
so is ##v = \frac{M_0 v_0}{M_0 + \lambda t}## wrong?(because that's what's given in the solutions)
I took sample values to see if ##v_0 e^{\frac{-\lambda}{m}t} = \frac {mv_0}{m+\lambda t}## and they arent equalo0)
also I understood @etotheipi method but I still don't see what is wrong with my second and third method(mentioned above).
 
  • #4
etotheipi said:
Clear the ##mv## from both sides and you get $$mdv = - \lambda v dt \implies \int \frac{dv}{v} = - \int \frac{\lambda}{m} dt \implies v = v_0e^{-\frac{\lambda}{m} t}$$
Whoa! m=m(t)=M0+λt:
##\int \frac{\lambda}{m} dt=\int \frac{\lambda}{M_0+\lambda t} dt##
##\Delta \ln(v)=-\Delta \ln(M_0+\lambda t)##
##v=\frac{M_0v_0}{M_0+\lambda t}##
 
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  • #5
Arghhh I think my brain switched off... I'm going to delete my stuff now
 

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