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

[Hamiltonian]

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.

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.

 

[Hamiltonian]

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?

 

[Hamiltonian]

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 equal
also I understood @etotheipi method but I still don't see what is wrong with my second and third method(mentioned above).

 

[haruspex]

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:
$\int \frac{\lambda}{m} dt=\int \frac{\lambda}{M_0+\lambda t} dt$
$\Delta \ln(v)=-\Delta \ln(M_0+\lambda t)$
$v=\frac{M_0v_0}{M_0+\lambda t}$

 

[etotheipi]

Arghhh I think my brain switched off... I'm going to delete my stuff now