Master Physics Formulas with Formula Mania: Tips for First-Year Students!

[holly]

Buddy, can you spare a formula?

Physics Expert Marcus has suggested first-year physics students get a list of formulas, memorize when they are used, and learn them.

That is EXACTLY what I started to do earlier today! My list is very lean, however.

Can anyone help us beginning students amass some formulas and hints on when to use them? I have gone thru the boards and looked for formulas, but honestly, I can't make heads nor tails out of them, especially those using calculus.

Thanking you in advance.

So far, I have (from Marcus et al) d=1/2gt^2 and one about when something is raised, it uses Joules, and is mgh where g is 10m/s^2. And I have F=ma, which someone else on the boards says uses N-s, and then a=F/m, which I guess will always have the time unit squared. But I have a=delta v/t interval of change, and which one is the best? And I've got p=mv (for momentum). I need work ones, power ones, impulse, KE, so forth. They are proliferating like a bunch of little bunnie rabbits.

Thank you for any help. Perhaps if the weaker students such as myself get a good list of formulas, we will stop posting easy stuff and leave the boards clear for more intriguing problems! I solemnly promise to go away for at least a month.

 

[marcus]

Originally posted by holly


Physics Expert...

they passed out funny hats at the Christmas party. I get to wear the Fireman's hat. There was a vote of the membership. You get to mention this once and that's it.

Originally posted by holly

Thank you for any help. Perhaps if the weaker students such as myself get a good list of formulas, we will stop posting easy stuff and leave the boards clear for more intriguing problems! I solemnly promise to go away for at least a month.

You have already earned an A+ in physical science this semester by your brilliant adaptation of a Famous Saying of Newton's. We do not want people who ask physics questions to go away. We like questions. Making a list of formulas is only part of it. Once you get the basic formulas down you will still have questions. You may be surprised by this but it has been widely observed that even after learning the
first semester's list of formulas there are still things about nature which one cannot understand.

Originally posted by holly


So far, I have
d=1/2gt^2 and

one about when something is raised, it uses Joules, and is mgh where g is 10m/s^2.

F=ma,

which someone else on the boards says uses N-s,
[attention, N-s is the unit of impulse, not force, a quantity
of impulse is what you get when you multiply a force in Newtons
by the length of time, in seconds, that it is applied to something]

and then a=F/m, which I guess will always have the time unit squared.

But I have a=delta v/t interval of change,

And I've got p=mv (for momentum). I need work ones, power ones, impulse, KE, so forth.

They are proliferating like a bunch of little bunnie rabbits.
[this is an excellent analogy because by putting formulas together one can produce more formulas, they breed so to speak]

easy does it
you are making good progress
rome was not built in a day
when is the midterm and/or final?

 

[HallsofIvy]

What Marcus is saying, in his own inimitable way, is that memorizing formulas is a really, really bad way to learn physics. Physics is not formulas. Physics is ideas and concepts. Learn those instead.

 

[chroot]

I'm trying to decide if I should post my own personal list.

- Warren

 

[marcus]

Originally posted by chroot
I'm trying to decide if I should post my own personal list.

- Warren

Please do Warren, I cannot imagine a better outcome from this thread.

especially if you still have your list from freshman year

 

[chroot]

These are the formulas I put on flashcards to help me study the GRE, in no particular order. I feel they are formulas that all physics students should understand and have quick at hand.

ONE-DIMENSIONAL KINEMATICS:

\begin{equation*}<br /> \begin{split}<br /> \textrm{acceleration} &amp;\equiv a(t) = a\\<br /> \textrm{velocity} &amp;\equiv v(t) = v_0 + at\\<br /> \textrm{position} &amp;\equiv x(t) = x_0 + v_0 t + \frac{1}{2} a t^2<br /> \end{split}<br /> \end{equation*}<br />

SIMPLE HARMONIC MOTION:

\begin{equation*}<br /> \begin{split}<br /> \textrm{frequency} &amp;\equiv \omega = \sqrt{k / m}\\<br /> \textrm{velocity} &amp;\equiv \omega x_{max} \sin (\omega t + \phi)\\<br /> \textrm{acceleration} &amp;\equiv \omega^2 x_{max} \cos (\omega t + \phi)<br /> \end{split}<br /> \end{equation*}

PERIOD OF A PHYSICAL PENDULUM:

T = 2 \pi \sqrt{\frac{I_{support}}{M g L_{cm}}}

CHANGE IN ENTROPY:

\Delta S = \int_i^f \frac{dQ}{T}

BERNOULLI'S EQUATION:

p_1 + \frac{1}{2} \rho v_1^2 + \rho g y_1 = p_2 + \frac{1}{2} \rho v_2^2 + \rho g y_2

FARADAY'S LAW:

\varepsilon = -N \frac{d}{dt} \phi_B = -N \frac{d}{dt} (\vec B \cdot \vec A)

\nabla \times \vec E = -\frac{\partial}{\partial t} \phi_B

\oint_F \vec E \cdot d\vec s = -\frac{\partial}{\partial t} \phi_B

BINOMIAL EXPANSION:

(1 + x)^n \approx 1 + nx

CONSERVATION OF MOMENTUM:

general case:

v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} + \frac{2 m_2}{m_1+m_2} v_{2i}

v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i} + \frac{m_2 <br /> - m_1}{m_1+m_2} v_{2i}

stationary target:

v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i}

v_{2f} = \frac{2 m_1}{m_1 + m_2} v_{1i}

E FIELD DUE TO A POINT CHARGE:

\vec E = \frac{q}{4 \pi \epsilon_0 r^2} \vec e_r

E POTENTIAL DUE TO A POINT CHARGE:

\phi = \frac{q}{4 \pi \epsilon_0 r}

E FIELD DUE TO A DIPOLE:

along dipole axis:

E = \frac{1}{2 \pi \epsilon_0} \frac{qd}{z^3}

E POTENTIAL DUE TO A DIPOLE:

\phi = \frac{zqd}{4 \pi \epsilon_0 r^2}

LORENTZ FORCE LAW:

\vec F = q (\vec E + \vec v \times \vec B )

ENERGY OF A PHOTON:

E = hf = \frac{hc}{\lambda}

blue photon with \lambda = 400 nm has energy E = 3.1 eV.

red photon with \lambda = 700 nm has energy E = 1.77 eV.

CENTRIPETAL ACCELERATION:

a = \frac{v^2}{r}

CENTRIPETAL FORCE:

F = ma = \frac{m v^2}{r}

STOKES' THEOREM:

\oint_\Gamma \vec C \cdot d \vec s = \int_S (\nabla \times \vec C) \cdot \vec n dA

GAUSS' THEOREM:

\int_S \vec C \cdot \vec n dA = \int_V \nabla \cdot \vec C dV

GAUSS' LAW:

\oint_S \vec E \cdot \vec n dA = \frac{Q_{int}}{\epsilon_0}

\nabla \cdot \vec E = \frac{\rho}{\epsilon_0}

SNELL'S LAW:

n_1 \sin \theta_1 = n_2 \sin \theta_2

TOTAL INTERNAL REFLECTION:

\theta_c = \sin^{-1} \left( \frac{n_2}{n_1} \right)

DIFFRACTION MINIMA:

a \sin \theta = m \lambda

SCHWARZSCHILD RADIUS:

r_s = \frac{2 G M}{c^2}

SPECIAL RELATIVISTIC VELOCITY ADDITION:

v = \frac{a + b}{1 + \frac{ab}{c^2}}

PARALLEL-AXIS THEOREM:

I_f = I_i + Mh^2

WORK DONE IN A CONSTANT-PRESSURE PROCESS:

W = p \Delta V

WORK DONE IN AN ISOTHERMAL PROCESS:

W = n R T \ln \frac{V_f}{V_i}

WORK DONE IN A CONSTANT-VOLUME PROCESS:

W = 0

DOPPLER SHIFT APPROXIMATION:

u \cong \frac{\Delta \lambda}{\lambda} c

VELOCITY OF WAVE PROPAGATION IN A MEDIUM:

taut string:

v = \sqrt{\frac{\tau}{\mu}}

sound waves:

v = \sqrt{\frac{B}{\rho}}

- Warren

 

[paul11273]

Holly,
While memorizing formulas is not necessarily a bad thing, it can hurt you if that is what you will rely on to get through.
It is better to concentrate on learning concepts, and that will help you to learn when to apply the formulas you have committed to memory.

This is what I do...
As we cover material in class, I have one sheet (now it is about 3 pages) where I write important formulas and a quick note about it.

I use this sheet as a quick reference sheet to jog my memory while doing practice problems.

Here is another tip...by "practice problems" I do not mean assigned HW. I mean going way beyond the HW and doing a ton of extra problems. By doing this extra drilling, memorization is automatic because you have worked with the formulas so much. I try to do this for every chapter as we cover it. When midterm comes around, I have quite a lot of problems worked out. Then I go back and try to redo them. This helps to solidify them in my memory, and is kind of a confidence builder. The last thing I want one week from a major exam is to get bogged down on some crazy problem, wasting time and killing my momentum.

One last thing. I read in your impulse question that your professor has a less than desirable attitude, and the text you are using kind of lacks. I have found that purchasing the solution manual for a text (if there is one) helps me greatly. It let's me practice problems, and I know that if I get stuck, the solution manual will usually unstick me. This forum has been great too!

As for asking everyone for our formula lists, it would probably be better if maybe once a week you post what topics you have covered, and we can send you some focused notes / formulas which cover that material. Just an idea...

 

[himanshu121]

I agree with Halls

So u have got list for most of the formulas(thnxs Chroot). u should remeber those formulas +++++with Concepts as well

 

[marcus]

tuning the discussion to Holly's particular situation

Paul's advice is good and Warren posted a great list for Geek Freshmen and maybe even sophomores too.
Holly is being given a glimpse of the Culture.

But we really have to tune this to a two-year college in a dry oily part of texas
and to the non-engineers introductory physical science course
where the math is high school algebra---in part already forgotten

we need a list which is not (like Warren's) for techie majors at Stanford but which is fine tuned for a different setting

It is a difference not in kind but in degree. Holly is as smart as anyone here, and Nature is the same for everybody, but she has only high school algebra at best and probably mistrusts that.

so this is like building a bridge across the gulf of mexico. its a big jump.

maybe Holly the truth is that EVERYBODY HAS TO MAKE THEIR OWN LIST
you have to sit in class and figure out by listening to the prof what formulas you have to memorize
it should be be the right list for you and for the class
(Warren's list was the right one for him in the class he took)

Personally I think the one you came up with already was very good.
so maybe we got off on the wrong track when we started giving you suggestions.

Let's post again what you have so far:

 

[marcus]

So far, I have

d=(1/2)gt²
(for when something is falling and you know, or you want to find out, the time it takes)

PE = mgh
(when something is raised, it uses Joules)

F=ma

and then a=F/m
(acceleration always has the time unit squared)

a=delta v/delta t
(acceleration is the change in speed divided by the time interval)

p=mv
(for momentum, how it's defined).

I need work ones, power ones, impulse, KE, so forth.

They are proliferating like a bunch of little bunnie rabbits.

------------------------
my comment is
this is the same kind of thing as Warren's list, the difference is only in difficulty and amount

now we all know what you have to do with such a list
you have to practice and practice so you get to really understand
when and how to use the formulas
(that is, like HallsofIvy suggests, the concepts embodied in them, like the potential energy something has because it is up, become second nature)

the concepts embodied in the formulas must become "operational" in your head---which only happens by working examples---which is why they hand out long homework assignments which appear, and in fact are, rather repetitious. and if the prof does not assign enough then you do them on your own, while you are eating a peanut butter sandwich or going to sleep at night, and if the textbook does not have enough examples then you find another textbook. and so it goes.

i suspect this process is already occurring to some extent with holly which is quite fascinating, given the different cultural setting.
if so, good for you and for us.

 

[marcus]

I need work ones, power ones, impulse, KE, so forth.

Well, write down the KE formula
we are all waiting in suspense

 

[holly]

W O W

Geez, I leave for a few hours and when I come back, horrible calculus-y formulas are all over the place. Sin, sin, everywhere a sin...help! Thank you chroot, but you must have mistaken me for someone with a brain...thank you ALL for the help, thank you singly and grouply for the ideas...for the idea of understanding the concept, etc...for cleaning up my little list of formulas so far...

The professor DOESN'T supply ANY formulas whatsoever. Nope. He says if we understand intuitively, we can discern the formula. Well, let's dig up Galileo and tell him he should have discerned F=ma & saved himself a bunch of trouble. Even big ol' brains can't just intuitively build formulas! I MUST memorize them! I MUST do that though it isn't the optimal thing! And I HAVE bought other texts, I have bought study guides, I have been scouting around for formulas...

So, I will do as suggested and just slog thru problem by problem, and ask youse for help, and write down whatever formulas you use.

And I thank each of you from the bottom of my heart! And the rest of my class will thank youse too when I share the formulas. And somebody in west Texas thinks youse are great.

 

[marcus]

holly,
the prof does not supply any formulas whatever
and says if you understand intuitively you will be able
to discern the formula!

this is not a standard course in college physics
it is not something I am familiar with
I am afraid of giving you advice that does not
fit the actual situation on the ground.

this must be a new approach to teaching physics
maybe someone here has encountered this kind of course
and can give you more appropriate coaching

I want to hold off giving advice because I don't have a good
feel for what's going on. But I am still very curious.

If you wouldn't mind, would you please tell us some
homework problems or some exercises from out of the textbook.
this would give me, and whoever else is interested, some idea
of what its like to be learning "intuitive physics"
or "low-formula physics"
if that is what we are dealing with here


I am trying, just out of imagination, to think of what a problem could be in that book:

"Two stones are dropped out of a tall building---from one from one story above streetlevel and the other from nine stories above streetlevel.

which stone hits first and how much longer does it take to fall?

A. 9 times longer to fall
B. 5 times longer to fall
C. 3 times longer
D. they take the same length of time to reach the ground

(imagine the stones are streamlined enough so air resistance plays no role)

 

[chroot]

I believe good physics is about equal parts quick-recall and derivation.

In other words, you need a handful of quick relationships at your disposal. You know, little gems like <E> = 3/2 kT. Sure, you can derive that from stat. mech., but that would take far too long.

On the other hand, you need to be able to mold your memorized equations to the task at hand, which often requires some manipulation, approximation, or other "editing." Knowing when and how to make those edits determines the difference between a formula-dependent physics student and a truly creative one.

Also, don't forget dimensional analysis -- it's an often-overlooked, but very powerful tool.

- Warren

 

[holly]

Actually, that sort of "dropping stones" scenario is what the teacher talks about! We have heavy and thin skydivers falling, too, and we have melting ice blocks, sliding crates, and steel balls rolling down inclines onto carpeting. I can follow that sort of thing...yes, I agree it can be "intuitive," but the problem comes when the *conceptual* part is emphasized, but the *math* part is tested! Because, how else can you test what the person knows? But it seems silly to me to SAY the course has no math, but have it actually be FULL of math...just today, we were listening to a lecture about a big coal cart and a little coal cart having a collision, and the big one was going faster and one way, and the little one was slower and the other way...and the upshot was, "Okay, dorky non-science majors, what might happen? Might the big one slow down and keep going the same direction?" Duh, yes...but then, wham-o, on the test, we have to figure out the momentum of each, and the velocity once they collide...we can't just say, "Gosh, da big one won but it slowed down!" That's why I am desperate for formulae.

Okay, I am stuck on this problem, it's part of our book's companion volume, Practical Physical Science: We have "Bronco," one of our little characters, pushing a block of ice up an incline. The ice weighs 500 N. The incline is 6m long, it is 3m high at the high end. We ignore friction..."How much force is needed to push it up the incline?" How the heck can I figure that out? I look at my little list of formulas, I try to make the ones with an "F" in them work...but I don't know...Then, we must answer this: "How much work is required to push it up the incline?" Well, I don't know. W = F d, I have that written down, but...I didn't know how to get the F in the first part of the problem...Two pages later, there is a little drawing of a guy with an overbite who is intoning, "Making the distinction between momentum and kinetic engery is high-level physics." And we have a pig-tailed girl saying, "Let equations guide your thinking!" Only, as they say around here, "No hay." It means, There ain't any equations lying around for me to use...

Thank you.

 

[chroot]

Hmmm holly, it sounds like professor is really doing a rather poor job if he tests you with methods you've never seen used in class.

- Warren

 

[marcus]

holly, the situation is strange and unfamiliar with lots of
contradictions---this is not the beginning college physics class we are used to.

about the ice
if you lifted it straight up 3 meters you would do
1500 joules of work.

you going to have to do the same amount of work to push it up the
incline
no free lunch, same 1500 joules either way, just more gradual than heaving it straight up

so what force, if applied for 6 meters, does 1500 joules of work
has to be 250 Newtons of force

half the force, and push it twice as far----same work as twice the force push it half as far

250 x 6 is same as 500 x 3

with Bronco and the guy with the overbite and the peppy girl with pigtails it sounds more and more like the end of civilization as we know it

 

[NateTG]

Another option is to make a list of all of the forces that are acting on the block, and make the total equal to zero.

 

[marcus]

are the exams going to be "short answer"
like choose between answers A, B, C, and D

or are they written out

 

[holly]

*sigh*

Well, I am thinking I ought to start placing my questions over in the K-12 board, tho' it IS a college course, because it is too easy for youse and is messing up the boards? Our poor little college here, they try...but...it's only 25 years old...we don't even get to use the lab for the lab that comes with the course, we are told to do the labs at home, I spent all weekend trying to find a 2-meter long piece of aluminum guttering and a 1/2 steel ball & a meterstick.

Thank you for the help in the block of ice problem, I guess there wasn't an equation for it, just thinking. Either way, I'm doomed.

Thanks again...

BTW, the exams, they are a mix, they have the multiple choice, they have short answer, and they have finishing diagrams, the only test so far, it was 74 questions in 50 minutes, I didn't finish...

 

[NateTG]

Originally posted by holly
Well, I am thinking I ought to start placing my questions over in the K-12 board, tho' it IS a college course, because it is too easy for youse and is messing up the boards?..

No, this stuff is typical of a freshman physics course.
The ice block question can be solved using energy, or using a free-body-diagram. (Energy is probably easier, but you may not have covered it.)

 

[marcus]

the distinction between K-12 help and college help
is artificial
you are fine where you are and fine if you move over there
(Nate is right it does sound like college freshman subject matter tho)

sounds like metric units are part of the problem
(for many of you it is the first encounter with them and there are the frustrating details like not having a meter-stick or
a kilogram scale)

what did they want you to do with the ball rolling down the
gutter?

 

[holly]

The steel ball had to be rolled down the gutter at 30, 60, 90, & 120 cm up from the bottom of the ramp. We were supposed to measure how high the ball was. And note how far it rolled. And roll it three times each cm division. Then, we were supposed to prop up the ramp more, and this time roll it three times from how high the ball was initially, ignoring the 30, 60, etc divisions. Then prop it up more, etc. Then average how far it rolled, then answer what conclusions we drew from our data. And make a graph, too. I didn't know what to conclude. It didn't go as far when the ramp was steeper. Oh, and answer why we didn't bother with K.E. calculations but just P.E. ones, and it said "only physics types know why these intermediate steps can be skipped, and why high-level thinking uses energy to solve these problems" or some such drivel. At least the lab didn't feature Nellie Newton or Bronco this time!

 

[Pesci]

hello,
My name is David and I'm new to the site and to physics.
i just returned to college after a minor set back and I have choosen
the wounderful world of physics as my "main course". since returning to school i am basically starting out with a foundation less than that of a high school student. i am currently taking a general physics course to get me started and i have found a great site that has quite a few basic formulas. the address is http://www.hazelwood.k12.mo.us/~grichert/sciweb/formulas.htm

hope it has a at least a little of what you are looking for.

[?]

 

[holly]

Thank you very much, I will go over to that site right now! Because my homework is just busting my chops.

 

[marcus]

Originally posted by holly
The steel ball had to be rolled down the gutter at 30, 60, 90, & 120 cm up from the bottom of the ramp. We were supposed to measure how high the ball was. And note how far it rolled. And roll it three times each cm division. Then, we were supposed to prop up the ramp more, and this time roll it three times from how high the ball was initially, ignoring the 30, 60, etc divisions. Then prop it up more, etc. Then average how far it rolled, then answer what conclusions we drew from our data. And make a graph, too. I didn't know what to conclude. It didn't go as far when the ramp was steeper. Oh, and answer why we didn't bother with K.E. calculations but just P.E. ones, and it said "only physics types know why these intermediate steps can be skipped, and why high-level thinking uses energy to solve these problems" or some such drivel. At least the lab didn't feature Nellie Newton or Bronco this time!

this sounds like the physics course from hell

I assume you have the option of teaming up with a partner to do the experiments with

in normal physics courses with a lab the students often get to team up---so scrounging equipment (if necessary) is easier and setting up the experiment is less tedious. then you and your partner both write the same observations down in your lab book

so the little girl with pigtails is named Nellie Newton?

enough is unfamiliar or just plain wrong in this situation that I don't know what to say

several people have shown they want to help but I suspect that we just do not know how to help in an effective way

Tell us some written homework problems, maybe we can help you get the answers.
(the prof seems to want you to be able to think in a commonsense practical intuitive way BEFORE you apply the formula, I think he wants you to get a visual or gutfeeling about how it is going to turn out before you do any calculation, picture it in your mind etc.
If this is true then maybe the homework problems are of a special
"intuitive" kind designed to foster this kind of thinking, should be interesting to see some of those problems)

 

[holly]

Well, Marcus, thank you for trying to help me.
Anyway, the problems: One of my ex's did 1-30 on my chapter review already. But I still have 31-50 to go...example: A 2kg mass has 40J of P.E. with respect to the ground. How far is it located above the ground? Also: In raising a 5000-N piano with a pulley system, it is noted that for every 1m of rope pulled down, the piano rises 0.1 m. Ideally, this means the force needed to lift it is...? A rifle of mass 2 kg is suspended by strings. The rifle fires a bullet with mass of 1/100 kg at a speed of 200 m/s. What is the recoil velocity of the rifle?
You don't have to do them, I can keep working on them, you said you wanted to see some of the kinds of problems we are doing. Thanks for any enlightenment. I am losing hope rapidly anyway. That's the only "gut" feeling I get about physics. Physics just makes no sense to me intuitively. Wasn't it Gauss who said that unless he could understand a concept mathematically, he had only a meager understanding? Well, physics is like that, too, I think. Gut feelings just don't cut it.

 

[marcus]

Originally posted by holly
A 2kg mass has 40J of P.E. with respect to the ground. How far is it located above the ground?

well, a 2kg mass weighs 20 Newtons
(thats how you figure the force of weight in metric)

so how high do you have to lift a 20 Newton weight in order to do 40 Newton-meters of work

2 meters


a joule is just another name for a "Newton-meter"
(the work of lifting a one-Newton weight by one meter
or pushing with a one-Newton force for one meter of distance)

the only reason they called it joule instead of "Newton meter"
was to be nice and respectful to a Mister Joule who was a famous Beer-maker in London, whom they wanted to honor. But it really means "Newton-meter"----push with Newton force for meter distance.

I picture this 2 kilo mass as a gallon of beer
and this gallon of beer weighs 20 of the diddly-size force units which Frenchmen call "Newtons" (each one is only about 4 ounce but that doesn't matter its what they like to measure force with)

To me it is obvious to my gut (I have a very perspicacious gut)
that it would take 40 joules to lift that beer 2 meters

 

[marcus]

Originally posted by holly

In raising a 5000-N piano with a pulley system, it is noted that for every 1m of rope pulled down, the piano rises 0.1 m. Ideally, this means the force needed to lift it is...?

these schoolteachers who wrote the textbook probably never
did use a block-and-tackle to hoist a piano

so they are writing their problems for your homework enjoyment out of pure theoretical imagination.

the block-and-tackle system here obviously has a
"ten-to-one advantage"

you have to pull ten foot of rope for the load to raise one foot.
the tradeoff is that the pull force is only one tenth the load.

if you have tackle with a ten-to-one advantage then the force you have to pull with is only 100 pounds, to lift a 1000 pound

oops forget I said pounds, I meant to say it in French:
pull with 500 Newtons in order to lift 5000 Newtons.

Im sure little Nellie Newton and the chinless man know all about this so what's wrong with you? I learned all about pulleys and ropes when I was a pirate and used to sail on those oldtime squarerigger pirate ships.

 

[marcus]

Originally posted by holly
A rifle of mass 2 kg is suspended by strings. The rifle fires a bullet with mass of 1/100 kg at a speed of 200 m/s. What is the recoil velocity of the rifle?

this is how Frenchman schoolteachers like to shoot rifles
they hang it with string from their TV antenna
or from the branch of a tree if they have a tree
and then they wait to see if anything walks in front
like a squirrel or gopher and if it does they pull the trigger

you probably wouldn't know about this because in west texas
they shoot rifles differently

French schoolteachers use these very delicate little rifles that only weigh about 4 pounds just so it won't break the string or bend their TV antenna ( no rifle I know of has a mass of 2 kilogram)

the rifle mass is how much more than the bullet mass?
come on holly, the rifle mass is what times the bullet mass?
cant you see the rifle mass is 200 times the mass of the slug?

if the rifle mass is 200 times the mass of the slug then the
slug has to go 200 times faster so the forwards and backwards momentum balance out!

Im sure Nellie and Bronco understand this very well at an intuitive gut level. Bronco, who pushes blocks of ice up a sloping board to get them into his pickup, has a very good gut feel for rifles and Nellie Newton has been shooting since she was six years old.

 

[holly]

Ow, my head hurts.

Thank you for the help. I have another problem, and it's very bad because it was solved by a PhD and I get a different answer. But I think he may have made an error. Here is the problem:

a 5 kg blob of clay moving at 2 m/s slams into a 4 kg blob of clay at rest. The speed of the two blobs stuck together is...He says 1.5 m/s, I say 1.1 m/s. But what are the chances I could be right? I need ol' Marcus "Big Brained Anti-Texan" to solve this. Where'd I put that rifle? I'll shoot the danged clay and we won't have any more problems. There was a rattler in the garage again, I'm unnerved.

 

[marcus]

Originally posted by holly
There was a rattler in the garage again, I'm unnerved.

they had a lot of rattlers where I used to live in some dry hills north of Los Angeles

we left the cars out in the yard, didnt have a garage
but they got under the front porch in the crawl space

 

[marcus]

sometimes i think you know more physics than you let on
and are just having fun with us
the momentum of that blob of clay is 10 kg m/s
and then it hits the other blob and they merge
so now the combined blob has a mass of 9 kg

yes you are right the combined blob must be moving 1.111 m/s
so it will have the same momentum
and let's round it off, like you did, to 1.1

 

[Moonbear]

Hi Holly,
I don't have much help for you...I'd have to dig out dusty old books with that sheet of paper with all the formulas crammed onto it to do that...but I do have lots of sympathy to offer, if that helps. It sounds like you have a truly evil prof. I had a physics lab once, a long time ago, taught by a TA who did evil things like that...I wonder if he's now your prof. He doesn't happen to have a funny French Canadian accent, does he? His quizzes always had questions completely unrelated to anything in lecture or lab. I have this vivid recollection of a quiz that asked us what color an electron was. I also recall answering "sky blue pink" because that answer seemed as logical as the question.

As for the whole formula issue, yes, you do need formulas to solve physics questions, at least a few basic ones that you already seem to have. Your professor's approach is a good one for a few lectures, just to get your thinking cap on and make sure you do intuitively understand the problems you're solving. It will help you catch mistakes when you use the formulas if you have some sense of what direction to expect something to change. However, to never give you any formulas and then expect you to solve problems that require formulas is completely unreasonable. The problem is, once you get beyond a few very basic formulas, deriving the rest of the formulas you'd need does require calculus. I'm not fond of physics courses that don't require calculus as a prerequisite. That just makes no sense because you really can't understand what you're being taught and do end up needing to rely on a memorized list of formulas you don't really understand.

It's true what others have told you that you need to write your very own list of formulas. Just the process of writing them all down will help you remember a lot of them. Besides, it's an age-old tradition among Freshmen to locate the pencil with the thinnest lead, sit down with your one sheet of paper and cram every formula imaginable onto it in your smallest handwriting , writing upside-down, sideways, along the edges, until there is no white space left on the paper. Every physics course I've ever heard of has allowed one sheet of paper with anything you want on it to be brought to the exam. Then the students all arrive at the exam and spend half their time trying to find the right formula amid that sea of scribble. Here's a trick everyone can benefit from. Don't bother cramming every formula onto it. Get the basics down...standard velocity, acceleration formulas, free energy, entropy, enthalpy, etc. Write those large enough so you can read them, and know how to derive the remaining formulas from them. Then, use the rest of the space to write sample problems with the solutions...put on the toughest homework solutions, the ones that if you can solve those, you can solve anything (assuming you are explicitly told you can write ANYTHING on the paper, and not just formulas...don't want to encourage cheating). Afterall, the formulas are useless if you don't know how to use them.

 

[holly]

Thank you Marcus for answering the question about clay blobs and Moonbear for the advice. Unfortunately, we don't get to bring in a piece of paper. It's really the dummy physics, oops, poet's physics course, so I guess I really ought to be able to do the work without a sheet of paper. But how I want a little sheet of paper with the formulas on it! I know I could do better if only I had my little formulas with me!
BTW, my prof is home-grown, no funny accent, just a dry, disparaging way of addressing us lowly students...
Thx again

 

[marcus]

hello holly
i was enjoying the problems you shared with us about
the rifle hung by string
and the clay blobs
and all
tell us some more if you want

 

[marcus]

does the chinless man have a name
(he's not the same as Bronco is he? I thought not)

 

[holly]

Hmmm. Perhaps I am being baited and mocked by a snide physics expert who enjoys laughing at my tiny brain...on the other hand, maybe I'll get some answers out of it...youse really want more questions? I have some that are stumping me.

Okay, I can get this right, but I can't defend my method. We have a car, going 50 km/h, and it skids 20m with locked brakes. And then, it is going 150 km/h and it has locked brakes, so how far will it skid now?

I say, 150 is to 50 as what is to what? Well, it is 3 to 1. And the first car, it went 20m. So, I am going to square the 3, making it nine. And times this 9 by 20m. Giving me 180m. This is always working for skidding cars where I can find the clear relationship and then square how many times more the speed is, and multiply it by the original skid length. But why did I do that? It seemed right...it seems repeatable, but I don't like it when I don't know what I'm really doing. It seems similar to where the arrow goes twice as fast but four times as far into a hay bale. Is there a formula?

Okay, this is a stumper: The force of gravity acts on some apples up in a tree. Some of the apples are twice as far from the ground as others. These twice-as-high apples, for the same mass, have: a) 1/4 the weight. b) 1/2 the weight or c)practically the same weight. It must be a trick question, otherwise it is entirely too stupid.

And this is a horrible one:
How many kilometers pre liter will a car obtain if its engine is 25 percent efficient and it encounters an average retarding force of 1000N? Assume energy content of the gas to be 40MJ/L.

Thanking you in advance.
An aside: The overbite boy has no name, sadly. The pig-tailed girl is not Nellie Newton. Nellie has curly hair. There is a really, really muscular guy with glasses and a cowlick, a buff physicist, I guess, and he, too, is nameless. But much later in the year, we have Sammie Sodium and Connie Chlorine coming up.

 

[paul11273]

Holly,
For your car question, you are getting the right answer, but kind of for the wrong reason.
You are correct in setting this up as a proportion, but it is not a direct proportion, which is clear because you are having to square the relation (in this example the 3). There may be a simpler way, but here is what I have for you...get a pencil and write it down as you go through.

Use the kinetic energy formula, KE=1/2(mass*velocity^2) This simply means that you square the velocity, multiply by the mass, and divide by 2. I am aware that you are not given the mass of the car, but that's coming up.
Know let KEa and KEb represent the kinetic energy of the car at 50 and 150 km/h, respectively. You can then divide the KEa by the distance it stops in...Da
KEa/Da
And do likewise for the other set, KEb/Db

Now set these two equal and solve:
KEa/Da=KEb/Db
KEa * Db = KEb * Da ( I cross multiplied)
1/2(mass*velocity^2)a * Db = 1/2(mass*velocity^2)b * Da (subbed in the KE formula for a and b)

Now we can see that each side has the mass as a multiplier, so it simply cancels out...
(velocity^2)a/2 * Db = (velocity^2)b/2 * Da

At this point you can plug in your known values and solve...
(50^2)/2 * Db = (150^2)/2 * 20
1250 * Db = 11250 * 20
1250 * Db = 225000
Db = 225000/1250
Db = 180m

When you use the kinetic energy formula, you can see how the velocity is squared, it is not linear. This is what you compensated for by squaring the 3 in your ratio.
So that is how I arrived at 180 meters for the unknown distance. I hope you stuck through that explanation, and see how to apply the formula. It is kind of hard to follow some of this typed text. Print it out, and write every step in you own handwriting and notation, and I think you will realize it is not too bad.

For the apple, I would have to say nearly the same weight. As you increase in altitude, the force of gravity lessens, but for the height of a tree, it is negligible (unless maybe for Jack's Beanstalk).

For your last question, good luck. I will try to work it out tomorrow on the bus.

As for the buff physicist character in your text, would you say he is a cross between Johnny Bravo and Dexter from the Cartoon Channel?
Just trying to get a good visual here...

 

[holly]

Howdy Paul...what a horrid mishmash of equations you gave me to solve that skidding problem! Can't I just keep squaring the number I get for the proportion? It's SO much easier. We get only those easy kinds of numbers, something is almost always half as much, twice as much, 100 times as much, etc. I printed out the real method you did -- thank you -- and will puzzle over it tomorrow. I used to dislike physics, but now I truly loathe it.

I don't watch television, so I am not sure if the Buff Physics Grad looks like those cartoons or not. But his arms are enormous. That's because he has to wrestle with such weighty problems.

Thanks again for the real method, guess I'd better learn it.

 

[marcus]

And this is a horrible one:
How many kilometers per liter will a car obtain if its engine is 25 percent efficient and it encounters an average retarding force of 1000N? Assume energy content of the gas to be 40MJ/L.

a joule is just another name for a Newton-meter
(the work done in pushing with a force of 1 Newton for a distance of one meter)

so put a liter in the car is like
putting in 40 million joule
but engine only delivers 10 million joule (25 percent, the rest is waste like exhaust heat and friction heat and all kinds waste energy)

the car has to push with 1000 Newton to keep moving because that is the retarding force

everytime it advances by one meter it does 1000 Newton meters of pushing work----1000 joules of work to go one meter (because pushing against the damn 1000 Newton retarding force!)

so with a million joules of work it can go 1000 meters (a kilometer)
and with the ten million joules the engine can put out (burning that liter of fuel) it can go 10 kilometers.
-------------------

BTW you were right to see the similarity between the arrow into the strawbale and the car skidding
kinetic energy is proportional to SQUARE of speed
and the energy is shown by the length of the skidmarks times the retarding force of the skidding tires----force x distance.


the apples in top of tree weigh practically the same
because their distance from the center of the Earth is practically the same as that of those at bottom of tree.
variation of gravity depends on distance from Earth center

height of tree makes percentagewise almost no difference at all

 

[marcus]

tell us some more problems holly