I am having issue with my working for the problem displayed below.
Here are my working as follows
The questions ask for newtons second law, now as the mass varys with time I use the momentum form and use the product rule like so.
$$F=\frac{d(mv)}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$
the equation what has been giving in the first equation
$$\alpha \frac{t}{T}=v\frac{dm}{dt}+m\frac{dv}{dt}$$
next I made the relevant substitutions for $$\frac{dm}{dt}=-2M_w\frac{t}{T^2}$$ and $$m(t)=M_g+M_w-M_w\frac{t^2}{T^2}=\frac{M_gT^2+M_wT^2-M_wt^2}{T^2}$$
and rearranged and equated in the following way
$$\frac{T\alpha t + 2M_wtV}{T^2}=\big(\frac{M_gT^2+M_wT^2-M_wt^2}{T^2}\big)\frac{dv}{dt}$$
Simplifying through and separating the variable I form the following integrals.
$$\int_0^t \frac{t}{M_gT^2+M_wT^2-M_wt^2}=\int_{v_0}^v \frac{1}{T\alpha 2M_wV}$$
Now I will omit every step I did for the integrals and outline what i did and the final solution.
LHS: I used U substitution to simplify the integral and formed the following solution.
$$-\frac{1}{2M_w} ln\big(\frac{M_gT^2+M_wT^2-M_wt^2}{M_gT^2+M_wT^2}\big)$$
RHS: I also did U substitution and formed the following equation.
$$\frac{1}{2M_w}ln\big( \frac{T\alpha+2M_wV}{T\alpha+2M_wV_0}\big)$$
Now if I equate the RHS with the LHS
$$-\frac{1}{2M_w}\big(\frac{M_gT^2+M_wT^2-M_wt^2}{M_gT^2+M_wT^2}\big)=\frac{1}{2M_w}\big( \frac{T\alpha+2M_wV}{T\alpha+2M_wV_0}\big)$$
simplyfying through
$$-\big(\frac{M_gT^2+M_wT^2-M_wt^2}{M_gT^2+M_wT^2}\big)=\big( \frac{T\alpha+2M_wV}{T\alpha+2M_wV_0}\big)$$
Now this is where my issue lies. Before I sub in the limits given I wanted to check if my equation made sense. So if I set $t=0$ I should get $V=V_0$ but this is not the case.
$$-(T\alpha + 2M_wV_0=T\alpha+2M_wV)$$
as you can see this is not what should be expected at t=0 if there was no '-' then it would be correct however I dont think the integration of the given equations is the issue here. It I believe it something at the beginning I have missed. But what I do not know.
I cant see how not using momentum is incorrect in this case as you have to take into account the varying mass with the water. I seem to not see another way of approaching this problem, could anyone shed light on my issue?
Your issue lies with your integration over $t$
$$\int_0^t \frac{t}{M_gT^2+M_wT^2-M_wt^2}dt$$
Let $u=t^2$ such that $du=2tdt$. Now we have
$$\begin{align} \int_0^t \frac{t}{M_gT^2+M_wT^2-M_wt^2}dt &= \dfrac{1}{2}\int \frac{1}{M_gT^2+M_wT^2-M_wu}du \\ &=-\dfrac{1}{2M_w}\ln|M_gT^2+M_wT^2-M_wu| \\ &=-\dfrac{1}{2M_w}\ln|M_gT^2+M_wT^2-M_wt^2|\Bigg|_0^t \\ &=-\dfrac{1}{2M_w}\ln|M_gT^2+M_wT^2-M_wt^2| + \dfrac{1}{2M_w}\ln|M_gT^2+M_wT^2| \\ &=\dfrac{1}{2M_w}\ln\Bigg|\dfrac{M_gT^2+M_wT^2}{M_gT^2+M_wT^2-M_wt^2 }\Bigg| \end{align}$$
See it now?
Your way would've worked, but when you "got rid of" the logarithms, you were not careful enough about the negative sign. You carried it through although it should've inverted one of the two sides.