Recent content by Su6had1p

Yes you are right, it does simplify the sign consideration.

Yes I realised that I missed the latent heat term in equation. But anyways included it in my calculations. Also thanks for pointing out that the whole ice warms upto 0 deg C and not the 50-x thing. That was the source of my error. Thanks.

Since I don't know what the system will look like in the end I tried to check for the extreme cases. Case 1 if the whole ice melts : $$T_f = 170.5 K$$ Case 2 if the whole water freezes : $$T_f = 291.33 K$$ Both of which are unreasonable. Therefore the mixture contains both water and ice, this...

@kuruman @PeroK @wrobel thanks to all of you

I see, F(ext) = 0

Isn't that the 5th step from my solution above?

Definitely, physics forum is a go to ! Especially when chatGPT doesn't correct my approach instead just gives solution. I expect Physics forum to help me identify what mistake in making so that I can correct my thought process.

Unfortunately the Examiner hasn't provided any solution, neither any standard textbooks have a similar problem as far as i know. so my only go to was ChatGPT. It gave the solution : $$v(t) = \frac{\rho Au^{2}t}{M_{C}+\rho Aut}$$

in my solution there is ##M_{c}## which the the mass of container and ##M## which is shorthand for ##M(t)##. There could be a confusion between those two since the question mentions only ##M##. However now that i have mentioned my notations, i think it would be clear. There is time dependence...

It's not rendered

No, I'm not sure about that. I didn't think about it.

A is the area of the water cannon, forget to mention

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F =...

@TSny @kuruman thanks for your help.

Yeah, I just checked somehow i forgot that $$log(a) + log(b) = log(ab) ≠ log(a+b)$$