https://www.symbolab.com/
http://onsolver.com/

Original First Post

College Algebra

Calculus :: Early Transcendentals

Advanced Engineering Mathematics

Numerical Methods (a.k.a. Computational Methods for Applied Sciences)

Euler Angles

Operators & Symbols ::
(–x, y) ± + − × ÷ √ ² ³ ^ ∫ Σ Δ ∇ ∂ ∠ ° Ω “” → ← ↑ ↓ ∵ ∴ ½ ∞ ≈ ≠ ≪ ≤ ≥ ≫ • · ∝ † ⊗ ✔ ✘ 2⁄2 x≈-1.25992 ∧ ‖a‖ y≈-1.5874 …

Common Greek alphabets in Rotational dynamics:
ϕ, θ, ψ, ω

PID Controller ::
u = − (Ki*∫ x + Kp*x + Kd*ẋ)

15°	π/12
30°	π/6
45°	π/4
60°	π/3
75°	5π/12
90°	π/2

① ② ③ ④ ⑤ ⑥ ⑦ ⑧ ⑨ ⑩

b² – 4ac

Superscripts and subscripts ::
x⁰ x¹ x² x³ x⁴ x⁵ x⁶ x⁷ x⁸ x⁹ x⁺ x⁻ x⁼ x⁽ ⁾ xⁿ x* x˙ xˣ

x₀ x₁ x₂ x₃ x₄ x₅ x₆ x₇ x₈ x₉ x₊ x₋ x₌ x₍ ₎ xᵢ xᵣ xₑ xₙ

Greek alphabets (lowercase) ::
α, β, γ, δ, ε, ζ, η, θ, κ, λ, μ, ξ, π, ρ, σ, τ, υ, ϕ, φ, χ, ψ, ω

Greek alphabets (uppercase & lowercase) ::

Αα Alpha	Νν Nu
Ββ Beta	Ξξ Xi
Γγ Gamma	Οο Omicron
Δδ Delta	Ππ Pi
Εε Epsilon	Ρρ Rho
Ζζ Zeta	Σσς Sigma
Ηη Eta	Ττ Tau
Θθ Theta	Υυ Upsilon
Ιι Iota	Φϕφ Phi
Κκ Kappa	Χχ Chi
Λλ Lambda	Ψψ Psi
Μμ Mu	Ωω Omega

[attachmentid=4328645]

This post has been edited by Critical_Fallacy: Apr 29 2023, 03:50 PM

Mostly are out of syllabus, like Cauchy-Schwarz Inequality, AM-GM Inequality, but Number Theory(Modular Arithmetic, Diophantine Equation) and Euclidean Plane Geometry do cover on syllabus of STPM further maths...

9th Dec 2013: Updated questions and answers...

OMK 2013 Sulong

Section A

1. A large cube is divided into 99 cubes with integer lengths, 98 of them with side 1. Find the volume of the large cube.

2. Let M be the set of all nine-digit positive integers that contain each digit from 1 to 9 once. Find the highest common factor of all elements of M.

3. What is the remainder when 5^5555 is divided by 10000?

4. Given a trapezium with perpendicular diagonals and height 12. The length of one of its diagonals is 15. Find the area of the trapezium.

5. The digits of 2013 can be rearranged to form an arithmetic progression. Determine the number of four-digit positive integers with this property.
(Note: An arithmetic progression might have common difference 0.)

6. Determine the smallest prime factor of 8051.


Section B

1. Given a triangle ABC. The midpoints of AB, BC, CA are C1, A1, B1 respectively.
Construct another triangle DEF with side lengths equal to the lengths of AA1, BB1, CC1.
(a) Prove that the ratio of the area of triangle DEF to the area of triangle ABC is a constant, regardless of the choice of triangle ABC.
(b) Find the area of triangle DEF if AB = 13, BC = 14, CA = 15.


2. Prove that there exist integers a1, a2, a3, …, a2013, b, all greater than 1, such that
(a1!)(a2!)(a3!)…(a2013!) = b!


3. A sequence x1, x2, x3, … is defined as follows: x1 = 1, x2 = 143, and
xn+1 = 5(x1 + x2 + … + xn)/n for all n ≥ 2.
Prove that all terms of the sequence are integers.


Hints & Answers:


This post has been edited by ystiang: Dec 9 2013, 05:53 PM

This is amazing . This is something big .
I hope people will start to utilise this thread as much as possible . I'll share it to everyone I know .

Thank you so much .

Thanks to maximR for showing me this thread.

I got question.



Show means to prove right?

This this correct? P(Z<500-510/s.d.)= 0.01

P(X < 500) = 0.01

Show basically means prove. Don't forget [shown].

Edited for: wrong info... sorry T_T

This post has been edited by ystiang: Nov 29 2013, 07:56 AM

Critical_Fallacy

I consulted the Bronze Medallist , and he told me that this solution is incorrect , and it'll be worth one point only . He attached a link to an accurate solution :

http://www.artofproblemsolving.com/Forum/v...ce9c11#p3152687

He added : The solution gives a construction of 2^k-1 terms but the problem statement requires exactly k terms

I never knew the questions in IMO would be this challenging !

This post has been edited by maximR: Aug 6 2013, 10:33 PM

Try to participate OMK if you intend to study form 6.
OMK considered as very hard, IMO I can't imagine my brain blank 4 hours...

For Malaysia, Top OMK performers are selected to attend the training camps, and the final IMO representatives are selected based on the students' performance in the camps and race. (I think only 6 in the final stage, and 3~4 of them are "bumiputera".

This post has been edited by ystiang: Aug 6 2013, 10:39 PM

I don't think I'll make the cut for IMO camps ( IPhO , probably . In the near future . ) though . You know the Bronze Medallist I mentioned ? He's as old as me , but he has far more experience and he started very young , at about 7 .

How do you find STPM ? Enjoying it ?

You ask the same question to me several months ago... But I think my answer is different now.

Quiet enjoying... especially maths =)

My target: 3.5 or above.

Quit working this sem, and rapidly done all the PBS, focusing on study now.

The only 'tak syok' thing in this sem so far is my monthly test math paper, originally can get perfect score, but so much careless T_T

edited for

This post has been edited by CallMeBin: Aug 7 2013, 02:31 AM

Malaysia got 37/97 on this year's IMO, singapore on the other hand got 6th. China's 14-year-old got gold ffuuu so pro these guys

But when does the 4.3g come into play?

Since P(Z < z) = 0.01

Since the probability is less than 0.5, z must be negative.

P(Z < -z) = 1 - 0.01 = 0.99

From the normal distribution table, P(Z < 2.326) = 0.99

Thus, P(Z < -2.326) = 0.01

(500-510)/SD = -2.326
-10/SD = -2.326

SD = -10/-2.326 = 4.3g [shown]


This post has been edited by ystiang: Nov 29 2013, 07:55 AM

Impressive! Thanks for sharing the link. Tell your Bronze Medallist friend that he obviously has a good understanding of mathematics that is far superior to mine. Through the composed power of his highly intelligent mind, he could probably solve some of the unsolved problems in mathematics in near future. Perhaps you don’t know; I’m particularly fascinated by the Navier–Stokes existence and smoothness problem.

If you want to claim $1 million Millennium Prize from the Clay Mathematics Institute, you’ll need to either prove or disprove this statement: “In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier–Stokes equations.”

In physics, the Navier–Stokes equations are a system of nonlinear partial differential equations for describing the motion of fluid substances in almost every real situation. In fact, these equations are derived from applying Newton's 2nd law to fluid motion. The animation below shows how a Kármán vortex street develops behind a cylinder moving through a fluid.


(courtesy of Cesareo de la Rosa Siqueira at the University of Sao Paulo, Brazil)

I see !

I've heard about turbulence in fluid motion and how hard it is to model that accurately ( I think it's still on Wikipedia's List of unsolved problems in Physics ) .

If you don't mind me asking , what led you to to become a Mathematical Physicist ? Was is a smooth pathway ?

Are you sure of your last statement?

I see. Thanks for the heads up

GreatFish posted a physics question about Newton’s Law of Universal Gravitation on this thread. Because there is slight inaccuracy about the direction of the gravitational force vector on the Wikipedia, so I decided to provide a salient answer on this matter. The negative-sign equation arises partly due to our interest in researching the kinodynamics behavior of the object being accelerated.



This post has been edited by Critical_Fallacy: Oct 2 2014, 04:02 PM

I see . I've thought about that for a long time . Thanks for clarifying !

Now I've a question , it's regarding the constants that appear in Physics equations .

Example :

1 . E = mc^2
2 . F = kx
3 . F(grav) = G (Mm)/r^2

My question is , will a constant always appear if a physical quantity varies directly as the other ? If it will , why do some constants behave in different ways . Why is the speed of light squared in Einstein's equation ? Since there were no experimental data at first to support his theory , how did he derive c ?

Why can the constant in F = kma be defined so that F = ma , but not with other equations ?

I don't know about others but E=mc^2 isnt the full formula. By using some complex special relativity theories, he came up with the momentum in 4D equation.. And if I remembered correctly by using binomial expansion, he got the equation E = mc^2 + 1/2(mv^2) + ... From this he found the rest mass energy equation E = mc^2. There's a second way that he did to derive it which involved E^2 = (pc)^2 + (mc^2)^2 .. after a few algebraic operation, we can obtain E =mc^2. But of course this is the end product of a long complex mathematical calculation and awesome thought experiments.
I think Critical_Fallacy can explain more to you about Time Dilation, Length contraction and so on, until the part where he got the rest mass energy equation.

For F = ma, that is not actually the real formula. From Newton's 2nd law, force can be expressed as:
--->

Force is said to be exerted when the mass of an object changes, or the velocity of the object changes.

I am more interested in the general relativity of Einstein. Anyone here knows how he got the christoeffel symbol in terms of the metric tensor? Probably Critical_Fallacy knows?

I am aware of the actual equations .
But I've always wondered about the constants . Why must constants appear in every Physics equation ? And why can the constant in Newton's 2nd Law be simplified to 1 ?

Thank you for your reply .

What are you doing , currently ? Are you a Physics major ?

Why must constants appear in every Physics equation ?
Idk, probably that's how the world works.
Let say you conduct an experiment to find the relationship between force and the extension of a spring.
After you have get the data, you plot a graph, you'll get something like this:
(Dependant variable on y axis, independent variable on x axis)

So you'll get the equation, F = kx (y = mx) where k is the gradient. The c-intercept is 0, so thats why we do not get F = kx + c.

Some teachers tend to teach proportionality like this:
F α x
To remove α we must add a constant:
F = kx

And why can the constant in Newton's 2nd Law be simplified to 1 ?
The equation F=ma is not derived from F=kma where k = 1.
If you want to write it in proportionality form you'll write the equation F=ma as:
F α a (By assing that mass is constant)
And you get F = ma.
This means that "m" is actually the constant for the equation since m do not change. (For secondary physics level).

Another way to view proportionality is by intuition.
Lets take the Newton's gravitational equation F = -GMm/r^2

From this we can see that the attraction force will of course increase when the masses of the two body increases, therefore we can conclude that F α Mm. And intuitively, we know that when the distance, r increases, the force will decrease. so F α 1/r^2 (Using inverse-square law that newton found, since the body is sphere we cant use 1/r).
By combining these two equations, we get:
F α Mm/r^2
To remove the α, we must add a constant.
F = GMm/r^2.

Now you might be wondering how the heck they figured out G. Try to look up on a brilliant experiment called Cavendish experiment.



And no, I'm not a physics major

I hope you get a light on how classical physics equations are derived.

As fallacy said to err is human, so please fix if there is anything wrong with my explanation.

can you write the deriavation of gravitational potential energy at here

As a unsuccessful wannabe math olympian who never had any formal training in mathematical olympiads, I participated in this year's OMK Sulong 2013. Here are a few of my crude attempts at solving those questions.
The answers that I gave during the actual competition are:

1) 125. Looks quite obvious. So didn't give much thought to it.

3) 8125. Through last 4 digit analysis. the last 4 digits 0625 , 3125, 5625 and 8125 recur for every increment of 4 powers of 5. Since 5555= 4(1388) + 3, 8125 is the answer.

4) Is it 108? Wait, I think I made a mistake here. Not sure what I was thinking back then.[/URL][/IMG]

5) 279? Forgive me for my crude workings. 4! implies the number of permutation of the 4 digits. Minus 3! at combinations including the digit zero because one does not simply start a number with digit 0.
[/URL][/IMG]

6) 83. It took me forever to do it by trial and error but I got there.

As for section B,

1) b) Heron's Formula!

3) I tried doing it by induction but I am not sure if I have done it correctly. The working is so long that I feel it is almost certainly wrong.

This post has been edited by Intermission: Aug 11 2013, 02:18 PM

You should be able to find a lots of detailed explanations from many physics textbooks and websites.

The derivation of Einstein's most famous equation E = mc² is purely theoretical, which arises as a direct consequence of his Special Theory of Relativity that involves some algebraic manipulations using the Pythagorean theorem (SPM level). To understand how he derived the equation, you must understand a portion of his Special Relativity; Time Dilation (see the embedded Spaceship figure). The following derivation is highly simplified for the mathematically untrained. To further enhance your understanding, you should watch the YouTube video by minutephysics.

Hey Critical_Fallacy! I am so curious of your education background! Are you currently a researcher in a university? You took up a daunting task of explaining time dilation to the masses.

Bravo. Me too haven't had any formal training but surely you're more success than me.

For Section A, I think I only have one question correct, yup that prime factor.
Actually, 8051 = 8100 - 49 = (90^2) - (7^2) = (90-7)(90+7) = 83*97

Q3 should be easy though but I wrote 625 and the answer is 8125 TvT
Q1, Q4, Q5... just randomly put some answers...

Section B, I hate those prove, prove and prove...

Physicists know some things never change and they call them the fundamental physical constants. Such frequently used constants as the speed of light in vacuum, c, magnetic constant, μ0, electric constant, ε0, Newtonian constant of gravitation, G, Planck constant, h, elementary charge, e, and the Proton-to-electron mass ratio, μ, are assumed to be the same at all places and times in the universe. They form the scaffolding around which the theories of physics are built on, and they define the fabric of the cosmos.

Despite that, one of the most fundamental properties of Newton’s Mechanics, Maxwell’s Macroscopic Electrodynamics, and the Laws of Thermodynamics, is the absence of any physical constants in their basic equations. In fact, all necessary fundamental physical constants appear only at the stage of applications of these theories to specific phenomena. Nevertheless, these constants play an important role in physics and metrology because scientists need reference values for measurements in experimental physics and making theoretical prediction on papers.

Other than Newton’s and Maxwell’s equations, centripetal acceleration (a = v²/r), density (Q = m/V), pressure (P = F/A), and electric power (P = V*I) seem to be self-evident counterexamples to your above question. You can also see other frequently used physics equations for yourself in this link.


Although physical constants can appear to be dimensional or dimensionless, the imaginary constant factor k in your F = kma does not exist. That’s because the equation F = ma is self-contained and Newton did NOT require adding a dimensionless constant factor k = 1 to his mechanics to reproduce that “observation of force”. In fact, as guided by VengenZ in Post #20, Newton’s 2nd Law originally states that the net force on an object is equal to the rate of change of its linear momentum p in an inertial reference frame, which can be manipulated algebraically to be stated in terms of an object's acceleration, i.e. F = ma.

--->

Since the dimensionless constant factor k does not exist in F = dp/dt in the first place, then it follows that k is not needed in F = ma either. If your friends or physics teacher insist on the existence of k = 1, be sure to ask them what they think of a = dv/dt, p = mv, v = dx/dt, and other counterexamples as shown, so that other similar fundamental physics equations cannot be ignored, and they will thus be psychologically compelled to consider their merit.

~ HAVE A NICE DAY! ~

This set of frequently used Fundamental Physical Constants is recommended for international use by CODATA (The Committee on Data for Science and Technology). The full 2010 CODATA set of constants may be found at its website (click the image) at the time of this writing.



For your convenience, a PDF document (Extensive Listing) is available which can be read online or printed.
 Fundamental_Physical_Constants_____Extensive_Listing__2010_.pdf ( 124.83k ) Number of downloads: 1

Thank you for your detailed explanation ! I really appreciate it .




Thanks !


This is what I need ! I've been thinking about this since Form Four . You've cleared everything up . Thank you !

As for why I proposed that there is a constant in F = ma , here's what's written in our textbooks , and reference books . One of the errors in our Physics syllabus . My aim is to major in Physics , then one day hopefully , revamp the Physics syllabus in Malaysia .

Here's the explanation provided by textbooks :

From Experiment 1 , a α F
From Experiment 2 , a α 1/m

The two results are combined .

a α F/m

OR

F α ma

Therefore : F = kma

Unit of force is newton , N .

In order to make the formula as simple as possible , we make k = 1 by defining a force of 1 N as

1 N is the force which gives a mass of 1 kg an acceleration of 1 m s^-2 ,

1 N = (k) (1 kg) (1 m s^-2)
k = 1

Therefore , F = ma

This post has been edited by maximR: Aug 12 2013, 04:22 PM

What you see in the following figure is an excerpt from The Principia: Mathematical Principles of Natural Philosophy (2010), the reprinted text which based on Motte’s 1729 translation of the 1726, 3rd edition of Philosophiæ Naturalis Principia Mathematica, that was the final version corrected by Sir Isaac Newton.




As you can see, Newton never explicitly stated the formula F = ma in his Principia. The “motion” and “impressed” in which Newton used his terminology, and how he understood the second law and intended it to be understood, have been extensively discussed by historians of science. It was until James Clerk Maxwell published a treatise in Matter and Motion in 1876 (see excerpt below) that the modern ideas of how Newton was using his terminology is understood.



According to Maxwell, Newton meant by motion “the quantity of matter moved as well as the rate at which it travels” and by impressed force he meant “the time during which the force acts as well as the intensity of the force”. And so in modern terms, motion is Newton’s name for momentum, which allows Newton’s 2nd Law to be formulated as follows:

→

Now, allow me to make a few additional points after my Post #29. Based on what the mathematicians and physicists understood from the experimental observations of Newton’s 2nd Law: the change of momentum of a body is proportional to the impulse impressed on the body, a proportionality constant k, shall be assigned to the formula F = kma. In order to standardize the unit force and to eliminate the proportionality constant k, Conférence Générale des Poids et Mesures (CGPM) in 1948, adopted the name "newton" with symbol N for the SI unit force to define the amount needed to accelerate 1 kg of mass at the rate of 1 m/s² by choosing a proportionality constant of 1.

In a sense, the proportionality constant k can be seen as a conversion factor if we apply F = kma in other system of measurement, such as the Imperial units. For example, 1 pound-force (lbf) is defined as the amount needed to accelerate 1 pound (lb) of mass at the standard gravity of 32.174 ft/s². Consequently, we can deduced from this observation that k = 1/32.174. Thus, to eliminate the proportionality constant k, the preferred unit of mass is the slug because a slug is defined to have a mass of 32.174 lb, so that 1 lbf = 1 slug × 1 ft/s².

To summarize this post and Post #29, naturally, the proportionality constant k = 1 doesn’t appear in momentum (p = m*v), centripetal acceleration (a = v²/r), density (Q = m/V), pressure (P = F/A), electric power (P = V*I), and a bundle of other primary equations of physics, because the physicists relate the basic physical quantities by defining the formulation of physical laws in a way to achieve universality and generality. All in all, to determine whether a significant proportionality constant is needed or not, at the first place, we should at least understand how strong the correlation of physical quantities is by performing the dimensional analysis, and ensure that the experimental data are consistent.

Oh! It is not as daunting as the quantum mechanics. Coincidentally, 12 August marks the birthday of Erwin Schrödinger, the Nobel prize-winning quantum physicist whose eponymous equation lies at the heart of quantum mechanics.

In a shooting competition, the probability that ali hits the target is 0.4. Find the minimum number of trials that ali needs to make such that the probability that he hits the target at least once is 0.8. << add maths question. How do I do it? Which method should I use? Is it part of binomial distribution?

Another great post !
This is the first time I've read that Newton originally stated that Impulse is directly proportional to the Change in momentum . So he actually considered the time in which a force acts on a body . Interesting .

So if I get this right , the reason why G appears in Newton's Law of Universal Gravitation is because the unit of force is defined as amount needed to accelerate 1 kg of mass at the rate of 1 m/s² , therefore G should be added to ensure that the units on both sides of the equation are the same ? And that after careful experiments , experimental physicists discovered that the force on gravity between two bodies does not vary simply as F = Mm/r^2 , so there must be a constant to account for the interaction between the two bodies ?

And that means Newton never really had accurate calculations since he couldn't determine G , and did not include them in his equations ? ( or did he ? if yes , then how did he come to the conclusion that there is a constant just by empirical observation ? or did he derive it mathematically , or did he think that since the force of gravity is usually very small between two bodies , there must be something that accounts for this , therefore , a constant must exist ? )



This post has been edited by maximR: Aug 13 2013, 01:19 PM

Are you planning to go for it again next year? It will be the last time we are allowed to participate in OMK.

This surprised me when I re-learn physics at AS level too. Actually it is not the formula F=ma that leads to F=(mv-mu)/t but rather the other round when newton first formulated his 3 laws of motion.

I was well aware that Newton did not explicitly state that F = ma ( It's derived from his 2nd law ) , and I've read that force is defined as the rate of change of momentum early on when I first started learning dynamics . Did you really not read anything other than your SPM Physics reference book in form four and form five ?

But I didn't know that he said Ft α mv - mu

This post has been edited by maximR: Aug 13 2013, 10:49 PM

I hope I can. I'm upper six this year, should be last year for me to participate =(

In Newtonian Kinetic Energy, as long as the object is not at microscopic level and the speed involved is very much lower than the speed of light, the mass is considered an absolute, or in a state of constant in an inertial frame of reference.



P.S. Thanks sharp-eyed Krevaki for pointing out the typo in Line 4 of the previous image. Correction has been made.

This post has been edited by Critical_Fallacy: Aug 14 2013, 07:28 PM

Though Newton’s Principia (1687) theorized the presence of the gravitational constant G, it was not until 1798 that the constant was determined through observing the density of the Earth in the Cavendish experiment. Because Newton’s calculations could not use the mathematical value of G, he could only calculate a force relative to another force as a ratio using Kepler’s Third Law. The following propositions for Newton’s Law of Universal Gravitation are the excerpts from The Book III of Principia. If you are interested, you probably find this link useful (Read Proposition VIII. Theorem VIII. Corollary 2).





As you probably already know, most modern physics textbooks states Newton’s law of universal gravitation according Maxwell’s Matter and Motion (1876); “Every portion of matter attracts every other portion of matter, and the stress between them is proportional to the product of their masses divided by the square of their distance.” Maxwell also acknowledged that although the idea of inverse square law had been the prior work of others, but in the hands of Newton, the doctrine of gravitation assumed its final form.



This post has been edited by Critical_Fallacy: Oct 24 2014, 12:05 AM

For beginners like me , how do I develop the intuition to manipulate the equation so p exists in that KE equation ?
Also , if KE becomes 2KE , p becomes 2p as well ? How to prove this ?



All right , noted . Thank you .

Here's a question from an SPM Olympiad Fizik book :



This post has been edited by maximR: Aug 14 2013, 06:24 PM

The answer is B?

According to the guy who provided the question , yes .

If so, you may experience it when the car/bus where you sit brakes.

Solution provided is about pressure , artificial gravity , inertia ...

That's complicated.
My reason :
When you are sitting in a moving bus, you are moving as well. When the bus stops, your inertia causes you to continue to move.
Same to the question.
The initial state of the objects is static. When you push the box, the inertia will cause the object to try to stay in the initial state, that is not moving.

You might use Newton's Third Law to think about it.

This post has been edited by kingkingyyk: Aug 14 2013, 06:18 PM

Erratum: Please take note that there was a typo in the last line of the previous image. Last line “√(2p)” should be “√2*p”. Thanks Krevaki for pointing it out. It has been corrected.

Basically, it is very similar to proving Trigonometric Identities in SPM Add Maths. When you have the knowledge of momentum p = m*v, you will think of ways to express KE in terms of momentum and at the same time, conserving the dimensions of the original KE equation.

By the way, care to tell me about the reception of your Science Fair project?

The steel (?) ball and the helium balloon are attached to the box by strings. The connections are not rigid and when you move the box, you would expect their inertia to hold them in their original position.

Alamak, why mention my name? Anyway, you're welcome.

Under normal circumstances, your daily experience, common sense plus Newton’s 1st Law generally would tell you the “rational” answer seems to be B. However, if B were the intended answer, why would the person who set the question bother to add the floating helium-filled balloon in the boxcar? Is it a tricky question?

There are 2 clues provided:
(1) If the helium-filled balloon wasn't tied to the floor of the boxcar, it would float all the way up to the ceiling.
(2) In physics, Impulse of Force is used to refer to a fast-acting force that is applied briefly: I = F*δt = m*δv.

Lastly, logical thinking is not just using our “common sense.” After all, “common sense” tells us the earth is flat. But even when “common sense” is correct, it may tell us, for example, that gravity cannot be pulling a thing down and pushing up the other at the same time. Therefore, critical thinking goes well beyond just using such basic principles.

Perhaps Krevaki and work_tgr can offer some clues as well.

The answer would be A then.
What I meant by "logical thinking" here is think based on law's logic.

Admittedly, no. Not quite. I was paying more attention to the application rather than the origin of the ideas which was fine at SPM level. But now at A level I realize it would be imprecise to define or explain force using F=ma instead of F=(mv-mu)/t. That could pose some problems when answering structure questions at A level.

Of all posts, why quote this one?

Didn't manage to finish before the deadline due to many reasons but mostly because of my incompetence. Might participate again next year if possible.

Hmm....I think A levels would serve a better purpose for this because for AS physics mechanics/Mathematics Mechanics 1 the knowledge from SPM can be extrapolated quite easily to fit the needs of the question. If you do exercises from STPM however, watch out for pre requisite knowledge.

Learn to relate different quantities together by the use of formula you already know. A good exercise would be proving formulas that you know, such as differentiation by 1st principle, proving E=mgh and E=0.5mv^2 from W=FS, proving P=h(rho)g from P=F/A and other formulas and so on. Proving trigonometric identities as mentioned by Critical_Fallacy is good too.

I couldn't stress how important this skill is if you are planning to study Physics/Mathematics/Further mathematics at A level/STPM.

Perhaps watching the YouTube video below would give you new insights into the physics of helium balloon.

P(X >= 1) = 1 - P(X < 1) = 1 - P(X = 0) = 0.8

Yup, binomial distribution.

Technically speaking, the method employed by ystiang is called Complementary Cumulative Distribution Function. The solution is exactly the same as the one beautifully brought by ystiang, with a little extra information of how the principle of Cumulative Distribution Function works.

Alright thanks people!

Helium gas balloon will move forward, pendulum will move behind?
Weird guy.

You're right! As the box moves to the right, the air gets compressed to the left, making the left side of the box denser than the right side. The helium balloon actually gets pushed to the right!

I get it now .

But looking for explanations on the internet , there are some who use General Relativity to explain this . So is the explanation using air compression / buoyant force correct , or General Relativity , or both ?

All right . Thank you ! I got only some of them covered through watching Walter Lewin's lectures , like proving the Work-Energy Theorem , P= h(rho)g , etc . I learned the general proof for derivatives too , but not the SPM way where you just add delta x and delta y to x and y respectively . I prefer the usual method , those that are available online .

It takes the knowledge of Density of Steel, Density of Helium, Newton’s Law, Archimedes’ principle (Buoyancy), Impulse of Force, and a few assumptions to understand the tricky physics problem.

Statics ::

(1) The boxcar is at rest on a frictionless surface. Between the two objects inside the boxcar, a solid steel ball is much more dense than a helium-filled balloon. The steel pendulum and helium balloon are rigid bodies.

(2) Newton’s Law tells us the downward force of gravity pulls the steel pendulum and helium balloon, which results in their respective weights, W↓ = mg.

(3) Archimedes’ principle tells us the upward buoyant forces exerted on the steel pendulum and helium balloon “immersed fully” in the air, are equal to the respective weights of the air that the objects displace, F↑ = ρVg.

(4) For the heavier-than-air steel ball, because W↓ > F↑, the steel ball is pulled downward by the Net force.

(5) For the lighter-than-air helium balloon, because F↑ > W↓, the helium balloon is pushed upward by the Net force.

(6) Because the freedom of movements of the steel pendulum and helium balloon are constrained in the direction of the respective Net forces (vertical) axes, Newton’s Law tells us their respective normal Reaction forces are generated, so that they appear to be held in their places.



Dynamics ::

(7) For the impulse of force, in which we assume the force exerted on boxcar acts for a very short time δt but is much greater than any other force present. Imagine flicking your finger to apply the impulse of force on the outer left side of the boxcar! The duration of the collision is very short and bring the boxcar to an abrupt STOP.



(8) Now, because the steel pendulum and helium balloon are not constrained horizontally, any applied horizontal force will cause the objects to swing left and right.

(9) Since the boxcar comes to an abrupt STOP, when we reapply the Newton’s Laws and Archimedes’ principle from (1)−(5), the most probable positions of the steel pendulum and helium balloon caused by the impulse is depicted in Image C. What do you think?

My first respond to this question is thinking about inertia and my answer goes to C.

Reedit : second thought, answer is B.

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My explanation:

1. An opposite force, f occurs at the string to resist F. The direction of f is to the left.

2. Resultant force of L is pointing south west.

3. Resultant force of B is pointing north west.

This post has been edited by work_tgr: Aug 15 2013, 11:38 PM

Probably could use general relativity for that question. Einstein's famous thought experiment (physicist in a box):

http://www.pitt.edu/~jdnorton/teaching/HPS...hway/index.html
(Principle in equivalence section)

Lets imagine the force exerted causes the box to accelerate with acceleration = g. Assuming that this experiment is not conducted in vacuum, we can say that the condition of the inside of the box will be exactly the same as if we were to rotate the box 90 degree anticlockwise.

Using this principle of equivalence, we know that in normal gravity condition (with air), the steel ball will move downward and the helium balloon will move upward. Now lets rotate this image back again 90 degree clockwise and we will get the answer.

The steel ball should move to the left and the helium to the right. So actually there is no answer?

This post has been edited by VengenZ: Aug 16 2013, 02:32 AM

Would you be so kind as to provide the reference link?

But are we not looking for the moment when the box is "pushed"?

Have you discovered this link?

Possible, if it is an “open” boxcar.

In the physics of general relativity (non SPM), we can certainly describe the motion of the steel pendulum and helium balloon by the equivalence principle.

True, but since the impulse acts for a very short time δt ~ 0 sec, the closest image that describes the positions (during δt) of the steel pendulum and helium balloon caused by the impulse that brings the boxcar to an abrupt STOP would probably be Image C. Just my 2 cents.

Refer to Tensile testing and Stress–strain curve.

The formula to calculate the “apparent” engineering stress, σ, is given by σ = F / Ao, assuming that the original cross-sectional area Ao is UNCHANGED (not reducing) through which the force is applied. In fact, as deformation continues, cross-sectional area decreases.

Similarly, the formula to calculate the engineering strain, ε, is given by ε = ΔL / Lo. Some questions give the elongation measurement L instead, and if that is the case, use ΔL = L − Lo.



Refer to Properties of Water, Continuity equation, Bernoulli's principle and Venturi effect.

Water is treated as an incompressible fluid because the changes in relative volume ΔV / V is negligible as the pressure increases.

In fluid mechanics, the continuity equation states that, in any steady state process, the rate of change of mass within the control volume is equal to the rate at which mass leaves the surface S of volume V.



In other words, mass inflow = mass outflow. However, mass is also given by the product of ρV (density × volume). Because water is essentially incompressible, therefore the density ρ of water is treated as constant throughout the streamline from the hose to the head. For that reason, we can rewrite the continuity equation as follows:

Volumetric inflow rate, Q1 = Volumetric outflow rate, Q2[/B][/COLOR]



Because Volumetric flow rate, Q is defined as Q = dV / dt,

and the volume of a mass, V is given by V = A*l (cross-sectional area × traversed length), therefore

A1 × (dl1 / dt) = A2 × (dl2 / dt)

Since dl / dt is defined as the velocity, υ, we can simplify the equation to

A1 × υ1 = A2 × υ2

I wonder where does we learn this
So deep

If the gas is continuously supplied at a constant pressure, the new level of “A” would probably look like this:

Ok thanks! Im impressed how well you edited the pic. I also want to know that,when an object fully immersed in water and its weight is then lets say, 5N , it will still be 5N when it is immersed further right as the vol of water displaced is still the same. Am I correct?

At first glance, you maybe tempted to solve ODE using Laplace Transforms. But it is not easy to find the Laplace Transform of 1/(1+t²). Since the ODE is a nonhomogeneous equation y'' + p(t)*y' + q(t)*y = g(t), your Lecturer probably has taught you about the Variation of Parameters formula.



This post has been edited by Critical_Fallacy: Nov 15 2013, 01:58 PM

Because the gas undergoes expansion from L to J

W(L,J) = (0.005 − 0.003)m³ × (100)kPa = 200 J

and then the gas undergoes compression from J to L

W(J,L) = ½ × {(0.005 − 0.003)m³ × (600 + 100)kPa} = 700 J

Therefore, the net work of the system is:

W(L,J) − W(J,L) = 200 − 700 = -500 J

Ans (A)...

This post has been edited by Critical_Fallacy: Nov 14 2013, 12:39 AM

Technically NO because you missed out the Matrix B.

Given a system of linear equations in the form of Ax = B:



Transform the linear system into augmented matrix:



Then, perform Gaussian Elimination to put the augmented matrix into the upper triangular form:

hi tanks for ur reply! i can understand finally!!

how to do this???? i need ur help!!!!!


Attached thumbnail(s)

|-2/(x+1)| < 1

-1 < -2/(x+1) < 1

0 < -2/(x+1) + 1 or -2/(x+1) - 1 < 0

(x-1)/(x+1) > 0 or -(x+3)/(x+1) < 0

By number line or sign table, you'll get

{x| x<-3 or x>1}



This post has been edited by ystiang: Nov 14 2013, 06:35 PM

This belongs to the topic called Absolute-Value Inequalities.

To solve this problem, let's understand what is absolute value.

The absolute value of a real number R (denoted “|R|”) may be defined as the distance of R from zero.

For example, both –1 and 1 are one unit from zero, and therefore we have |–1| = |1| = 1.

The absolute value of R can be written as a function of the positive square root: |R| = √(R²).

Come back to the problem: |–2 / (x + 1)| < 1.

Let's define f(x) = –2 / (x + 1), and it follows that |f(x)| < 1.

By the above definition, we can intuitively imagine that f(x) = –1 and f(x) = 1.

To satisfy the condition of the inequality |f(x)| < 1, let's do four possible logic tests:

Test 1: f(x) < –1 :: Let f(x) = –1.1, and |f(x)| = |–1.1| = 1.1 < 1 ~ FALSE

Test 2: f(x) > –1 :: Let f(x) = –0.9, and |f(x)| = |–0.9| = 0.9 < 1 ~ TRUE

Test 3: f(x) < 1 :: Let f(x) = 0.9, and |f(x)| = |0.9| = 0.9 < 1 ~ TRUE

Test 4: f(x) > 1 :: Let f(x) = 1.1, and |f(x)| = |1.1| = 1.1 < 1 ~ FALSE

From the results, because Test 2 and 3 are TRUE, we can conclude that given the inequality |x| < a, the solution is always of the form –a < x < a.

Therefore, we have two inequalities: –1 < |f(x)| < 1.

Solve inequality #1: –1 < –2 / (x + 1) ... → ... we have x > 1.

Solve inequality #2: –2 / (x + 1) < 1 ... → ... we have x < –3.

Once you understand the concept, you can use the same simple approach as ystiang did.

Question: What if you were given the inequality |x| > a?