Uncertainty in particle position

AI Thread Summary
The discussion revolves around a misunderstanding of symmetry in calculating particle position uncertainty. The original poster incorrectly assumed that certain terms could be disregarded as zero by symmetry, but this assumption is challenged by other participants. They emphasize that the integral in question does not equate to zero due to symmetry, prompting a request for clarification on the symmetry argument. The conversation highlights the importance of accurately applying symmetry principles in physics problems. Overall, the thread underscores the need for careful consideration of mathematical assumptions in particle position uncertainty calculations.
MaxJ
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Homework Statement
below
Relevant Equations
below
For this problem,
1723537877356.png

My solution is
1723537945310.png

1723537992107.png

1723538009333.png

However, the my answer for ##\Delta x## is not correct. The correct answer is ##\Delta x = 0.18a##. Does someone please know what I have done wrong?

Kind wishes
 
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MaxJ said:
what I have done wrong?
You have hand waved "zero by symmetry" where what you have waved away is not zero by symmetry. Why do you think it should be zero and what symmetry would there be?
 
Orodruin said:
You have hand waved "zero by symmetry" where what you hade waved away is not zero by symmetry. Why do you think it should be zero and what symmetry would there be?
Be blessed Sir.

I have doubt what you mean in regards to symmetry. Can you explain some more?
 
MaxJ said:
Be blessed Sir.

I have doubt what you mean in regards to symmetry. Can you explain some more?
You were the one trying to invoke symmetry arguments so you should be able to tell me why you think the integral should be zero "due to symmetry".
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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