12:40 AM
A ring structure could be defined on a proper class right?

6 hours later…
6:42 AM
Hi All..
Is it possible to find trajectory of projectile using work-energy theorem?

2 hours later…
8:51 AM
@JingleBells no
U-235 has a binding energy per nucleon of about 7.6 MeV.
The fission products do not have less binding energy, they have more.
The fission products have binding energies per nucleon of more than 8 MeV.
You might want to check the definition of binding energy again.

1 hour later…
10:19 AM
Hello @JohnRennie

3 hours later…
1:21 PM
@123 If by work-energy theorem you mean the fact that doing $X$ amount of work to an object gives it $X$ additional kinetic energy then no. The trajectory is fixed by the starting position and velocity assuming you know the external forces. If you know the kinetic energy of a particle you know the magnitude of its velocity so that gives you less than half of the required information.

Hello @Charlie

hi

We can find trajectory or position vector using work-energy theorem. How we find trajectory/position vector of projectile motion using W-E Theorem?
You have shared the answer it did not understand it fully.

Unless you're using a different definition from me, unless you're given extra information you can't derive the trajectory of something just by knowing its kinetic energy. You need an initial position and velocity but the kinetic energy only tells you the magnitude of the velocity at that time

We know the launch angle $\theta$ , initial velocity $V_o$, direction of force always acting downward direction with constant magnitude $g = 9.8 \frac{m}{s^2}$

1:34 PM
That is enough to know the full trajectory, assuming you know where you're launching it from

Using K-E theorem. Because the motion is under conservative force. So we also find potential energy function.

Sure, you could find the potential energy as a function of time there if that's what you mean

How do we know the trajectory of projectile using W-E theorem.
Because in W-E theorem we take dot product it means it only care the displacement in the direction of force (Let say y-axis) not in any other component.
How x-axis displacement is handled by W-E theorem.

1 hour later…
3:07 PM
Hello @JohnRennie Sir.

@123 hi :-)

How linear motion with force have torque? If we take moment arm above the motion of particle.
Because there is no rotation why torque is there.
It means torque does not means there is always rotation.

I'm eating lunch at the moment. I'll get back to you in 15 minutes or so.

I have read it in KnK page 266. Example Torque on a sliding block.
@JohnRennie Ooookay...

If you have a geodesic I'm being told you can always reparameterise such that $(ds/d\tau)^2$ along it is $\pm1,0$ for space/time/light-like curves. Is this just a general fact for affine parameters?
We seem to take advantage of this in deriving the Schwarzschild geodesics

3:40 PM
@123 the torque is defined by a vector product τ = r x F where r is the position vector and F is the force vector.
Typically for a rotating system we use the centre of rotation as our origin for the position vector because this makes the calculations simple, but we don't need to do this. We can choose any point as our origin, and the equation τ = r x F still works.
It gives a different value for the torque τ, but this is an important fact to remember in rotational motion. The angular velocity, torque, etc are not uniquely defined. Their values depend on where we choose the origin for our position vectors.
And this equation τ = r x F works even when the system is not rotating.

4:06 PM
@JohnRennie thanks. It means we can define torque for linear motion

1 hour later…
5:11 PM
I am a bit confused thinking about an elevator problem. Suppose if a man (mass m) is inside the elevator and it's going upward with a deacceleration g' (where g'>g). Is this situation possible? Doesn't it make the man's apparent weight negative? Does it mean, he would stick to the ceiling?
Deaccelarating upward = Accelerating downward. But I am confused, it's getting hard for me to intuitively think the situation "Deaccelarating upward". The lift is going upward. But the man is feeling the negative apparent weight when the magnitude of the deacceleration is greater than g!
The net force or apparent weight, F = mg - mg' = -ma ?
So, the man would stick to the ceiling even in the upward* moving elevator?!

2 hours later…
7:42 PM
@raf "Weight" is by definition the force imparted by gravity on a mass $m$. If you're on Earth your weight is $W=mg$. If you have someone in an elevator travelling upwards and you want it to deaccelerate with a force $F>mg$ you need to apply some additional force to achieve this, and since this force is no longer just the gravitational force this force isn't just the weight of the person acting downwards.
If you accelerate the elevator downwards faster than the person inside the elevator is falling under gravity they will be pushed into the ceiling.

2 hours later…
9:30 PM
Can someone help me to understand a concept about symmetry? I think it's the kind of question that is better in chat

If you have a question you don't have to ask if you can ask, just ask @SimoBartz
If someone wants to/can help they will

10:06 PM
-1

I wanted to search for a very nice article that was mentioned in an exchange of comments I had. The discussion wasn't moved to chat. The question was closed though because it was thought to be opinion based which I didn't think it to be, but's another thing (it's this one, for who is interested)....

Exactly! why was it deleted?
are the memory chips too expensive?

10:33 PM
So I'm working through problems in [this textbook](https://www.cambridge.org/core/books/introduction-to-quantum-mechanics/990799CA07A83FC5312402AF6860311E) and am up to problem 1.8.

I need to show that by adding a constant $V_0$ to the potential energy [in the schrodinger equation?] adds a time dependant factor to the wave function.
Any ideas on how to approach this?
So far I have the schrodinger equation in 1 dimension:
$$\bbox[.5em, border:.1em solid gray]{ i\hbar\frac{\partial \Psi}{\partial t} = \frac{-\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi}$$
and this formulation of expectation value of a physical quantity:
$$\bbox[.5em, border:.1em solid gray]{ \left<Q(x, p)\right>=\int\limits_{-\infty}^{+\infty}{\Psi^\ast\left[Q\left(x, −i\hbar \frac{\partial}{\partial x}\right)\right]\Psi}\,\mathrm{d}x }$$