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

[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