There appears to be another question about conveyor belt on this site; however, that question just asks for a correct expression. I actually have asked much more than that question. I provided an additional formula that I derived and asked whether it is correct. This is not included in that question.
A conveyor belt moves at constant velocity $v$. At $t=0$, sand starts to be dropped onto the belt at the rate $\mu \text{kg/s}$. The mass of sand on the belt is, therefore, $m=\mu t$. Please ignore resistive forces in this question.
NB Sand has zero velocity in the direction of the belt's motion before it is dropped onto the belt. Sand does NOT have any kinetic energy before it reaches the belt, and when it reaches the belt, it moves with the belt, with velocity $v$: they are "attached" to each other.
The belt continues to move with constant velocity $v$ throughout.
Now, the problem I am trying to solve is:
What is the additional power needed to keep the speed of conveyor belt constant, after sand starts to drop?
I tried to work it out in two ways, but got different results.
Method 1: $$ P=\frac{dE}{dt}=\frac{d (\frac{1}{2} mv^2)}{dt}=\frac{d (\frac{1}{2} \mu v^2 t)}{dt}=\frac{1}{2} \mu v^2. $$
Method 2: $$ P=\frac{dE}{dt}=\frac{Fdx}{dt}=Fv=v\frac{dp}{dt}=v\frac{d(mv)}{dt}= v^2\frac{dm}{dt}+mv\frac{dv}{dt}=\mu v^2. $$
I have no idea which is right: intuitively, $\frac{1}{2} mv^2$ is the total kinetic energy gained by the sand. Dividing this by $t$ gives the average power $\frac{1}{2} \mu v^2$. So this seems to be the correct value.
However, the mark scheme says the answer is $\mu v^2$. My second method also seems more reliable, in that it just uses the definition of momentum and force.
I doubt that the formula $E=(1/2) mv^2$ may not be correct in this situation: $$ dE=Fdx=vdp=\frac{1}{2m}2p dp=\frac{1}{2m}d(p^2), $$ which is not the same as $E=(1/2) mv^2$, because $m$ is not constant. However, the expression $\frac{1}{2m}d(p^2)$ looks very strange and counterintuitive. (Why $m$ is not differentiated?!)
Which of the above are correct and which are wrong?
Can anyone explain intuitively where the factor of $1/2$ comes from?
Why the statement "intuitively, $\frac{1}{2} mv^2$ is the total kinetic energy gained by the sand. Dividing this by $t$ gives the average power $\frac{1}{2} \mu v^2$" is incorrect?
