Author Topic: Pre-transposed formula's (Read 3716 times) Tweet Share

cohen · « **on:** June 05, 2011, 09:15:39 pm »

Projectile Motion
Symmetrical "Flights"
$Range = \frac{U^{2}sin(2\Theta)}{g}$
$time-of-flight = \frac{2Usin(\Theta)}{g}$
$maximum-height = \frac{U^{2}sin(\Theta)^{2}}{2g}$
$time-@-max-height = \frac{Usin(\Theta)}{g}$

Any projectile flight
$height = xtan(\theta))- \frac{g x^{2}}{2U^{2}cos(\theta)^{2})}$ -> This can be annoying to put in the calc!
$height = utsin(\theta)-5t^{2}$

Non-projectile Motion

Gravitational Motion
$Time = (7.693 * 10^{5}) * \sqrt[2]{\frac{R^3}{M}}$
The above equation comes from manipulating keplers 3rd law: and the 7.693 * 10^5 is simply $\sqrt{\frac{4\pi ^2}{G}}$

Potential Energy
- Spring Potential
$PE_{S} = \frac{1}{2}kx^2$
$PE_{S} = \frac{1}{2}Fx$
$PE_{S} = \frac{F^2}{2k}$

Structures and Materials
-For the area under the stress-strain graph: Let E = energy per unit volume
$E = \frac{1}{2}\sigma \epsilon$
$E = \frac{1}{2}\gamma \epsilon^{2}$
$E = \frac{\sigma^{2}}{2\gamma}$

I suggest only using these equations (Besides the primary 2 spring potential energy equations) to check your answers only. Most of these formulas (primarily the projectile motion formulas) are extremely annoying to write on the calculator, and you can easily stuff it up.

- If i've miswrote any of these formulas, could you also please tell me? I have a feeling i may have messed one of them up...
- I have a few more equations relating to the kinetic energy and momentum, but i left my folder at school, so I'll post them tomorrow

- If you have any other formula's that i don't have here, feel free to post them & I'll add them!
- If you wish me to write a "Key" for the above symbols used in the equations, i'll do it after school tomorrow

(6/6/2011)

Lasercookie · « **Reply #1 on:** June 05, 2011, 09:34:23 pm »

$Ek=\frac{p^2}_{2m}$

Design Angle:
$tan \theta = \frac{v^2}_{gr}$

Bozo · « **Reply #2 on:** June 06, 2011, 11:56:44 am »

So just to make it clear, you DO NOT need to show steps as to how the pre transposed formula was obtained?

cohen · « **Reply #3 on:** June 06, 2011, 04:17:21 pm »

Geoff didn't mention anything about showing the steps.
I even asked my teacher today, and he said
"The assessors have advanced degree's, they've either been engineers, physicists or teaching vce physics for years. They know all the formula's."

jinny1 · « **Reply #4 on:** June 06, 2011, 07:07:20 pm »

How did you guys derive:
$ PE_{S} = \frac{1}{2}Fx$
$PE_{S} = \frac{F^2}{2k}$
$Ek=\frac{p^2}_{2m} $
??

moekamo · « **Reply #5 on:** June 06, 2011, 07:27:34 pm »

Quote from: jinny1 on June 06, 2011, 07:07:20 pm

How did you guys derive:
$ PE_{S} = \frac{1}{2}Fx$
$PE_{S} = \frac{F^2}{2k}$
$Ek=\frac{p^2}_{2m} $
??

$PE_S=\frac{1}{2}kx^2=\frac{1}{2}(kx)x=\frac{1}{2}Fx \qquad \mbox{ since }F=kx$

$PE_S=\frac{1}{2}kx^2=\frac{1}{2}\frac{k^2x^2}{k}=\frac{1}{2}\frac{F^2}{k} \qquad \mbox{ since }F=kx$

$E_k=\frac{1}{2}mv^2=\frac{1}{2}\frac{m^2v^2}{m}=\frac{1}{2}\frac{p^2}{m} \qquad \mbox{ since }p=mv$

jinny1 · « **Reply #6 on:** June 06, 2011, 07:32:26 pm »

I didnt know that Force = kx....

moekamo · « **Reply #7 on:** June 06, 2011, 07:41:08 pm »

yea it does, remember for a spring with spring constant k, F=-kx, but this is energy and it cant be negative...

jinny1 · « **Reply #8 on:** June 06, 2011, 07:51:14 pm »

oh rite... hooke's law...

those ones arent really important/necessary right??

i mean i wudnt have a clue exactly when to use them derived ones.,

moekamo · « **Reply #9 on:** June 06, 2011, 08:05:57 pm »

nah they're not really needed, you'll be fine without them, all they do is skip an in between step, and physics has enough formulas anyway

pi · « **Reply #10 on:** June 06, 2011, 08:18:07 pm »

Quote from: Bozo on June 06, 2011, 11:56:44 am

So just to make it clear, you DO NOT need to show steps as to how the pre transposed formula was obtained?

My teacher said you do, to not advantage the spesh kids who actually prove the parabolic equations

FromBaghdadWithLove · « **Reply #11 on:** June 06, 2011, 09:53:03 pm »

for that Ek= p^2 / 2m

im guessing that ek is kentic energy, p is power and m is mass? right?

cohen · « **Reply #12 on:** June 06, 2011, 09:58:08 pm »

p is the momentum
also, Ek can also equal 1/2*p*v

Bozo · « **Reply #13 on:** June 06, 2011, 10:48:55 pm »

Quote from: Rohitpi on June 06, 2011, 08:18:07 pm

Quote from: Bozo on June 06, 2011, 11:56:44 am
So just to make it clear, you DO NOT need to show steps as to how the pre transposed formula was obtained?

My teacher said you do, to not advantage the spesh kids who actually prove the parabolic equations

I'm scared now.

pi · « **Reply #14 on:** June 07, 2011, 05:05:09 pm »

Quote from: Bozo on June 06, 2011, 10:48:55 pm

Quote from: Rohitpi on June 06, 2011, 08:18:07 pm
Quote from: Bozo on June 06, 2011, 11:56:44 am
So just to make it clear, you DO NOT need to show steps as to how the pre transposed formula was obtained?

My teacher said you do, to not advantage the spesh kids who actually prove the parabolic equations

I'm scared now.

Nah, its actually pretty straightforward, especially compared to some of the integration stuff we do

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Author Topic: Pre-transposed formula's (Read 3716 times) Tweet Share

cohen

Pre-transposed formula's

Lasercookie

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Bozo

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cohen

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pi

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