Problem Solving Strategies

12:00 AM

@Abcd Elastic collision of two particles of equal mass : see the following -

2

A: Why does the rule that elastic collisions are at 90 degrees in 2 dimensions not apply?

If the velocity of one particle is zero after the collision then the direction in which it is travelling is ambiguous. The magnitude of velocity is zero, but it can be in any direction. In such situations you need to look at the limiting case. As you reduce the magnitude of velocity of particle ...

harambe

@sammygerbil hi

sammy gerbil

@harambe hello

harambe

@sammygerbil how can I solve this in Com frane? The velocities are at different direction finally

sammy gerbil

@harambe see above -

5 hours ago, by Abcd

harambe

5 hours ago, by sammy gerbil

@Abcd In the COM frame both particles move towards the COM with the same speed in opposite directions along the same straight line. The COM is the point of collision. When they separate they go in opposite directions with the same spee, to conserve momentum and KE in the COM frame.

@sammygerbil why is COMpoint of collision

sammy gerbil

12:13 AM

@harambe In COM frame of reference the COM does not move. By definition. When the (point) particles collide they coincide. The COM must be in the same place as both particles when they collide. Because COM is centre of mass and all the mass is in the same place as both particles.

Perhaps you are thinking of particles which are not point particles. Then the COM can be different from the point at which they make contact.

harambe

12:29 AM

@sammygerbil calculated the velocity and converted it back to com frame

@sammygerbil should I do dot product now

It gives m

Aaryan

5:16 AM

hi

John Rennie

@Aaryan hi

harambe

@JohnRennie good morning

@Aaryan hi. Sup

John Rennie

@harambe morning

harambe

@JohnRennie needed some help in understanding surface tension

John Rennie

@harambe OK ... ?

harambe

5:30 AM

@JohnRennie so If I go under the sea, would surface tension exist or is it just surface property

John Rennie

Surface tension is the energy per unit area associated with an interface.

So for example it is a property of the air/water interface.

harambe

Okay.

@JohnRennie imgur.com/a/89BTavW

How does surface tension prevent the water to flow below textile

Isn't it surface property

John Rennie

Flow through a textile is quite complicted and it's a bit of an oversimplification to just say oh it's surface tension.

Surface tension is involved, but it also involves the contact angle of the water on the textile. If the contact angle was low the water would soak into the textile.

harambe

@JohnRennie okay but this example seems kinda complicated to me

John Rennie

As a (rather rough) comparison, imagine holding an inflated balloon and trying to squeeze it through a small hole.

The balloon wants to stay spherical, so you have to do a lot of work to squeeze it into a long thin shape so you can push it through the hole.

Make sense so far?

harambe

5:42 AM

No

John Rennie

Have you ever played with a party balloon?

harambe

@JohnRennie imgur.com/a/V13GfM1

@JohnRennie yeah lots of time

John Rennie

@harambe the balloon wants to stay spherical because it's the shape that stretches the rubber walls of the balloon least. Yes?

harambe

Yes

John Rennie

And if you wanted to push the balloon through a small hole you'd have to deform it to squeeze it through the hole. That would stretch the rubber walls a lot so it would require considerable force.

harambe

5:45 AM

Okay

John Rennie

So if you put the balloon down on a coarse mesh the balloon is going to just sit on top of the mesh, because the gravitational force isn't enough force to make the balloon squeeze through a hole in the mesh.

And that's pretty much what is happening in that picture when you put the water drop on top of the fabric.

harambe

Okay. Got it

John Rennie

The surface tension makes the water-air interface behave a bit like the rubber walls of a balloon.

But be warned it's actually a lot more complicated than that when you look in detail.

(for now I wouldn't worry about the detail)

harambe

Okay. I just wanted to understand how surface tension works. It's a complicated stuff for me

@JohnRennie Surface tension is responsible for the shape of liquid droplets. Although easily deformed, droplets of water tend to be pulled into a spherical shape by the imbalance in cohesive forces of the surface layer. In the absence of other forces, including gravity, drops of virtually all liquids would be approximately spherical. The spherical shape minimizes the necessary "wall tension" of the surface layer according to Laplace's law.

John Rennie

OK ... ?

harambe

5:51 AM

Imbalance of cohesive forces of the surface layer is due to missing liquid parties due to one side covered by air - water interference

John Rennie

Basically yes. In liquid water the water molecules want to bond to other water molecules. That's why it's a liquid not a vapour.

But at the surface the water molecules at the surface only have water on one side (and air at the other) and that increases their energy.

harambe

Okay.

John Rennie

That means there is an energy associated with the surface. We write this as an energy per unit area, and this is the surface tension.

harambe

Isn't surface tension force per unit length

John Rennie

Work is force times distance, so the dimensions of work are F L (where I'm writing F as shorthand for the dimensiuons of force).

Now, energy per using area is $E/L^2 = $FL/L^2 = F/L$

So energy per unit area has the same dimensions as force per unit length.

harambe

5:57 AM

Okay

John Rennie

If you want I can show you exactly how the energy of the interface and surface tension are related using a diagram ...

Nobody recognizeable

@sammygerbil are you here ?

@JohnRennie whenever you will be free then please ping me.

harambe

@JohnRennie yes please

John Rennie

@harambe we have a wire frame with a film (the blue area) inside it - e.g. this could be a soap film. The right edge of the frame slides so we can pull it out to increase the size of the film.

OK so far?

harambe

Ok

John Rennie

6:04 AM

We pull the right edge a small distance $dx$ and this increases the area of the film by $dA = Ldx$ - this new area is shown in pink.

harambe

Okay

John Rennie

Suppose the energy per unit area of the film is $\gamma$ then when we increase the area by $Ldx$ we increase the energy by $dE = \gamma Ldx$. OK so far?

harambe

Yes.

John Rennie

Now, we moved the slider right by applying a force $F$, and we moved it a distance $dx$, so the work we did was just $dE = Fdx$.

harambe

Okay

John Rennie

6:08 AM

And you can probably now see where I'm going with this. The increase in energy of the film has to be equal to the work we did on it.

That means $\gamma L dx = F dx$

$$ \gamma = \frac{F}{L} $$

harambe

Yeah. Makes sense

John Rennie

The point is that surface tension can seem a bit mysterious the way it's usually taught in books, but actually it's really simple because it's just the energy per unit area of the interface.

That's why the interface wants to reduce its area - because all systems want to reduce their energy.

harambe

Okay. Got it

Abcd

6:30 AM

@JohnRennie hi

John Rennie

@Abcd morning :-)

Abcd

@JohnRennie Do you know Modulation in communication system.

John Rennie

@Abcd modulation as in amplitude modulation and frequency modulation?

Abcd

@JohnRennie ya'

John Rennie

I know the basics

Abcd

6:32 AM

@JohnRennie please explain AM

John Rennie

Where should I start? How much do you already know about it?

Abcd

@JohnRennie can you ask, I will tell you if I know that thing

John Rennie

Suppose we have some high frequency wave $A\sin\omega t$. And suppose we have a circuit that can average out this wave to its root mean square value. Don't worry about exactly how that filter works - just accept that circuits like this exist. Then if we average out our wave we just get a constant $A/\sqrt{2}$. OK so far?

Abcd

@JohnRennie its given completely differently in my book... Can you please see:

@JohnRennie I cant understand $c_m(t)$ and $m(t)$. What exactly are they supposed to represent?

John Rennie

$c(t)$ is the high frequency wave I talked about above.

Abcd

6:41 AM

and m(t) ?

John Rennie

So $c(t) = A\sin\omega_ct$ but the amplitude $A$ can be varied with time i.e. it isn't necessarily a constant.

Suppose we write the amplitude $A$ as $A(t) = 1 + B\sin\omega_m t$ where the constant $B$ is small i.e. the amplitude is approximately one but varies slightly above and below one.

So the root mean square of our carrier wave oscillates with time at a frequency $\omega_m$ where we assume $\omega_m$ is a much lower frequency than the frequency of the carrier wave.

Abcd

@JohnRennie carrier is supposed to be the message we give or the one produced externally??

John Rennie

I'm not sure what you're asking there.

The message is going to be encoded in $A(t)$

Abcd

oh so m(t) is the initial message

John Rennie

Yes

Abcd

6:48 AM

@JohnRennie not getting how c(t) and m(t) are superimposed? Do we just write m(t) inside the Amplitude of c(t)... But how does it work....

John Rennie

Remember that I said we change the amplitude of the carrier with time e.g. something like $A(t) = 1 + B\sin\omega_m t$

Abcd

OK?

John Rennie

That means the wave is $A(t)\sin\omega_c t = (1 + B\sin\omega_m t)\sin\omega_c t$

This is the radio wave we broadcast.

Now remember I said there are circuits that can average out the high frequency carrier?

Abcd

> there are circuits that can average out the high frequency carrier?

I dont understand what this statement means

John Rennie

@Abcd you know what the root mean square amplitude of a wave is?

Abcd

6:53 AM

@JohnRennie No... I just know that to find the root mean square of anything we first find the sum of square then mean and then take root.

@JohnRennie I just know rms current.

John Rennie

RMS current is the same idea. If the current is varying with time like $I_0\sin\omega t$ then the point of the RMS is that it's an average.

We square then integrate along one wave to get the average value of the square.

Abcd

OK ... ?

John Rennie

The point is that there are circuits that can do this i.e. feed in some sinusoidal wave and what you get out is the RMS value of the wave. So feed in $A\sin\omega t$ and what you get out is a constant $A/\sqrt{2}$.

OK so far?

Abcd

yes

John Rennie

So if you feed $(1 + B\sin\omega_m t)\sin\omega_c t$ into a circuit like this it will average the $\sin\omega_c t$ bit to $1/\sqrt{2}$ to get $(1 + B\sin\omega_m t)/\sqrt{2}$

Abcd

6:59 AM

K

John Rennie

So what the circuit does is remove the carrier frequency $\omega_c$ and just leave us with our message $B\sin\omega_m t$.

Abcd

oh

John Rennie

Suppose we're doing a radio broadcast and we want to transmit some music.

Music has frequencies in the range 100Hz to 10kHz (roughly).

But it's really hard to transmit and receive radio waves with these frequencies, so we can't simply convert the music to a radio wave.

Instead we generate a wave $(1 + B\sin\omega_m t)\sin\omega_c t$ where the $B\sin\omega_m t$ bit is our music and the $\sin\omega_c t$ bit is a carrier wave of around 1 to 100MHz. The reason we do this is that 1MHz to 100MHz radio waves are easy to broadcast and receive.

Abcd

ya

John Rennie

Then your radio has to remove the $\sin\omega_c t$ carrier (using the averaging I talked about above) to leave just the music $B\sin\omega_m t$. Then it can amplify the music and play it out of the speaker.

Abcd

7:06 AM

@JohnRennie thats it?

John Rennie

Yes

It's basically pretty simple

Abcd

@JohnRennie I dont understand these graphs:

@JohnRennie how did author make those graphs, how does he identify the missing signal and stuff ??

John Rennie

When we combine the message and the carrier we get a combined wave: $(1 + B\sin\omega_m t)\sin\omega_c t$. Are you happy with this as a starting point?

Abcd

@JohnRennie Getting $(1+B \sin \omega_mt ) $ before $\sin \omega_ct $ is magic right?? I mean its not important to know its details ?

John Rennie

Well suppose $B=0$ then the signal is just $\sin\omega_c t$ i.e. we just have a the carrier.

Abcd

7:12 AM

OK

John Rennie

The $\sin\omega_m t$ term is our message. It's a pretty boring message because it's just a sine wave. Maybe it could be a musical instrument playing just one note at a frequency $\omega_m$.

In practice the message would be a complicated function of time, but it's usual to start with a very simple example to illustrate the principle.

So the constant $B$ determines the amplitude of our message relative to the amplitude of the carrier wave.

OK so far?

Abcd

> So the constant B determines the amplitude of our message relative to the amplitude of the carrier wave.

didnt get

John Rennie

Our wave is $(1 + B\sin\omega_m t)\sin\omega_c t$. Yes?

Abcd

ya

John Rennie

And the $B\sin\omega_m t$ part is the message.

If the constant $B$ is zero then the amplitude of our message is also zero i.e. no message is present. Just the carrier.

OK so far?

Abcd

7:18 AM

hmm = yes

John Rennie

Suppose we now increase $B$ to 0.1. Then our wave is now: $(1 + 0.1\sin\omega_m t)\sin\omega_c t$.

So the carrier is $\sin\omega_c t$ and the message is $0.1\sin\omega_m t$ i.e. the amplitude of the message is one tenth the amplitude of the carrier.

Abcd

hmm

John Rennie

The message $0.1\sin\omega_m t$ oscillates between 0.1 and -0.1 so the total amplitude varies between 1.1 and 0.9. Yes?

Abcd

yes

John Rennie

And if we increase $B$ to 1 we get $(1 + \sin\omega_m t)\sin\omega_c t$. Now the amplitudes of the message and carrier are equal and the total amplitude varies between 0 and 2. Still OK?

Abcd

7:23 AM

yes

John Rennie

If you look at your picture of the five graphs the constant $B$ is called $\mu_a$. So the fourth graph down shows $B (i.e. \mu_a) = 1$. Note that the total amplitude varies between zero and two.

(well the y axis doesn't have a scale, but the minimum amplitude is certainly zero)

The third graph down is for $B=0.5$ so the total amplitude varies between 1.5 and 0.5.

Does that make more sense of the graphs?

Abcd

@JohnRennie yes

@JohnRennie are you going to explain more

John Rennie

24 mins ago, by Abcd

@JohnRennie how did author make those graphs, how does he identify the missing signal and stuff ??

Abcd

@JohnRennie you have just explained amplitudes till now... not the missing signals, frequency stuff

John Rennie

The bottom graph shows what happens if $B > 1$. e.g. for $B = 1.5$ the total amplitude would vary between 2.5 and -0.5

But you can't have a negative amplitude. Zero amplitude means no signal, and you can't have less than no signal.

So if you try to make $B$ greater than one the bits of the wave where the total amplitude would be negative just go to zero and that bit of the wave is lost.

That's what missing signal means. It's the bits of the message that get lost if you make $B > 1$.

Abcd

7:36 AM

@JohnRennie OK, and how did he make the dotted curves? How did he decide their zeros.. that is when they intersect the x axis.

Graph (c) doesnt intersect the x axis in the middle, why?

John Rennie

The dotted lines are the total amplitude of $(1 + B\sin\omega_m t)\sin\omega_c t$ i.e. they show $1 + B\sin\omega_m t$

So the dotted lines touch the $x$ axis when $1 + B\sin\omega_m t <= 0$

i.e. they can only touch the $x$ axis when $B >= 1$

Abcd

@JohnRennie I think it should be $=$ and not $\le$

John Rennie

@Abcd yes, though the point about the missing signal is that the amplitude cannot go less than zero so for $1 + B\sin\omega_m t < 0$ the amplitude is zero.

Abcd

Oh i see. Understood prof, thanks.

John Rennie

I need to work now for about 20 minutes ...

harambe

7:52 AM

@JohnRennie please ping when free

John Rennie

8:04 AM

@harambe free

harambe

@JohnRennie do you know why the pressure inside a drop is 2S/R and that of air bubble is 4S/R

Why is there an excess pressure

John Rennie

@harambe because a soap bubble has two liquid air interfaces, one on the outside and one on the inside, so the total tension is doubled. A water drop has just the one liquid air interface on the outside.

harambe

@JohnRennie more inteface means more surface energy right

John Rennie

8:22 AM

@harambe Yes. A soap film has two sides so it has double the energy of a film with only one side.

harambe

@JohnRennie but is there an intuitive reason to understand why pressure inside a soap bubble is greater than outside

The simple answer to "why is the pressure inside a soap bubble higher than outside," is that a higher pressure than the local atmosphere is required to make the bubble in the first place! This requirement comes from the need to counterbalance the surface tension force.

Can you explain what the author means here

John Rennie

@harambe go back to our example of party balloon. Is it obvious why the pressure inside the balloon is higher than atmospheric?

harambe

Yes. If I poke a hole air comes out

That means it is higher

John Rennie

@harambe OK, but why is the pressure higher? What is making it higher?

harambe

8:44 AM

@JohnRennie no idea

John Rennie

@harambe the walls of the balloon are stretchy. They want to shrink back down to their original size, and it's the pressure of the air inside the balloon that stops them shrinking.

The pressure inside the balloon has to be higher than the pressure outside so the excess pressure can push on the balloon walls and keep them stretched.

Isn't that obvious?

harambe

Yes

John Rennie

It's the same with a bubble.

The surface tension means the soap film around the bubble is under tension. The pressure has to be higher inside the bubble to stop the walls from shrinking under their tension.

harambe

Okay

@JohnRennie physics.aps.org/assets/b4c2e0a3-7437-4592-be71-7a666a298652/…

John Rennie

OK ... ?

harambe

8:59 AM

When the pressure exceeds the surface tension I guess the bubbles are formed just like in the video

John Rennie

In a flat film the surface tension every where in the film acts horizontally. Yes?

(give me a moment and I'll draw a diagram)

But now consider what happens if we make a dimple in the film ...

harambe

Okay

John Rennie

There is now a net force in the direction shown.

So to make a dimple in the film we have to apply an equal and opposite force, and that's what the air stream is doing.

harambe

Okay. Got it

@JohnRennie how did you determine the direction of the surface tension in both diagram. I have studied that surface tension is a pull force which acts normal to the boundry.

So theybact towards the film surface?

Arjun

11:17 AM

Someone pls pls pls help me with this :- reddit.com/r/Physics/comments/abt3o0/…

It would be a tremendous help. Thanks in anticipation.

MartianCactus

guys i have a question

so mass * velocity of center of mass of a system = momentum of the system

mass * acceleration of center of mass of a system = total ezternal force on the system

then why is 1/2 * mass * square of velocity of center of mass of a system smaller than or equal to its velocity?

sorry dont know how to weite in equation form

John Rennie

@harambe the surface tension always acts in the interface.

MartianCactus

12:52 PM

"All the particles of a system are situated at a distance x from origin. The distance of the center of mass of the system from the origin is ___"

when could os be equal to r?

Anonymous

← 7 messages moved from The h Bar

sammy gerbil

1:07 PM

@Arjun The link does not contain a clear question. Please type your question here, or post an image of it, and explain what your difficulty is.

@MartianCactus $\frac12 mv^2$ is kinetic energy, $v$ is velocity. These are different quantities, they cannot be compared because they do not have the same units. One cannot be bigger or smaller than the other. Your question does not make sense.

Dante

@sammygerbil Hi! Perfect timing

sammy gerbil

@Dante hello

Dante

@sammygerbil While solving a oscillations problem in which I was given a function for potential energy, I found that they had differentiated the function twice, to find 'K'

K is of course F/x but I didn't understand the meaning of it

What is K exactly here? It's called a constant but is not a constant here...

sammy gerbil

@Dante I need to see the question and the solution you have been given.

Dante

k'

Trying to transfer pictured from my dad's phone to my PC, it showing 'queued' from last 10 mins. Please hold

sammy gerbil

1:37 PM

@Dante I think what you might be referring to is checking that $U$ is a minimum rather than a maximum or an inflexion point. $d^2U/dx^2>0$ for a minimum.

Dante

Yes, they did that too, I got that point.

Then they substituted the point with minimum energy is the double derivative to get 'K'

I'll send the picture when that problem is sorted

sammy gerbil

@Dante $U(x)=U(0)+x\frac{dU}{dx}|_0+\frac12 x^2 \frac{d^2U}{dx^2}|_0+...$

Dante

What's that?

sammy gerbil

At equilibrium the 1st term is an arbitrary constant, 2nd term is zero, the 3rd term is stored PE and has the form $\frac12 kx^2$.

That is the Taylor Expansion of $U(x)$ about an equilibrium point $x=0$.

Dante

@sammygerbil This is new to me, please link some nice source if you know.

sammy gerbil

1:49 PM

1

Q: How to find an effective spring constant of a quadratic potential

If a potential energy is given like $U(r)=A^3/r^2+2B^3r$, how do I find the effective spring constant using Taylor Expansion? I compared spring constant $k$ to be equal to second derivative of potential energy with respect to $r$. Am I going in the correct direction?

Dante

Thanks

sammy gerbil

@Dante web.physics.ucsb.edu/~fratus/phys103/LN/SHM.pdf

Dante

Thanks again'

I don't get this

Along X axis $E=-2x$ along Y axis, it's $2Y$

I don't understand how first option corresponds with this

sammy gerbil

2:07 PM

@Dante Sketch equipotentials $V=$constant. Then the electric field lines are perpendicular to the equipotentials.

MartianCactus

yes

can anyone amswer my question, the center of mass one?

Dante

@sammygerbil 1st and 3rd option look same to me.

Equipotential surface will be hyperbola

Anonymous

room topic changed to Problem Solving Strategies: General chat for high school physics. For MathJax see meta.stackexchange.com/a/220976. Related rooms: Mathematics (chat.stackexchange.com/rooms/76340) and Chemistry (chat.stackexchange.com/rooms/74807). (no tags)

Dante

If I imagine normal to it, can't differentiate between 1st and 3rd option

Anonymous

room topic changed to Problem Solving Strategies: General chat for high school physics. For MathJax see meta.stackexchange.com/a/220976. Related rooms: Mathematics (chat.stackexchange.com/rooms/76340) & Chemistry (chat.stackexchange.com/rooms/74807). (no tags)

MartianCactus

2:13 PM

electric field lines are gonna be present between 2 points when potential changes between 2 points

hmm..

sammy gerbil

@Dante In 1st option the curves get closer to x and y axes for large values of x and y. They are not closed. In 3rd option the curves are more circular, possibly closed.

MartianCactus

@sammygerbil sorry, I meant to say "then why is 1/2 * mass * square of velocity of center of mass of a system smaller than or equal to the system's actual kinetic energy?"

sammy gerbil

@MartianCactus The system's total kinetic energy is made up of the KE of its centre of mass plus some "internal" energy. A good example is a rolling cylinder. It has translational KE related to the velocity of its COM, and rotational KE related to motion of particles around the COM. The rotational KE is "internal" energy.

Another example is thermal energy. A solid might be at rest, so its COM has zero KE. But individual particles within the solid (atoms, molecules) are moving about in random directions at high speed. The latter is "internal" energy, thermal energy which is reflected in the temperature of the body.

MartianCactus

ohh alright

well, is there anything as internal momentum?

sammy gerbil

@MartianCactus KE is a scalar so the KE of constituent particles adds up to a non-zero amount. Whereas momentum is a vector, and the sum of a random collection of vectors is zero. So the total internal momentum is zero.

MartianCactus

2:31 PM

ohh alright. And samw for force. Cool thanks!

can you answer my other question?

sammy gerbil

2:49 PM

2 hours ago, by MartianCactus

"All the particles of a system are situated at a distance x from origin. The distance of the center of mass of the system from the origin is ___"

@MartianCactus This one ^^^ ?

This question cannot be answered. If the particles are symmetrically distributed about the origin the COM is at the origin. If all the particles are clumped together at the position (x,0) then the COM is at the position (x,0).

2 hours ago, by MartianCactus

when could os be equal to r?

@MartianCactus It is not clear to me what you mean by this ^^^ nor how it relates to the question which precedes it.

Dante

4:03 PM

@sammygerbil You'll be here for few more hours right?

sammy gerbil

@Dante Not continuously, but on and off for the next 9 hours. If you have a question, post it along with a description of what you are having difficulty with. Be as specific as possible.

Dante

OK

MartianCactus

4:35 PM

alright

harambe

11:58 PM

@sammygerbil hi

sammy gerbil

@harambe hello

harambe

Transcript for

Problem Solving Strategies