Collection of Physics Problems

Homework Help from TutorVista

2008-10-27T06:59:00.002+08:00

Today I got a website that could help you to solve your homework or any questions you want to ask. It's an online homework help on TutorVista. There is also math homework help and algebra homework help. Because they have experts as math homework helper.

They also provide free homework help service for demo. You just send your questions to them, and they will give response for you within 48 hours. If you still have any questions, just send to them again. Their experts need to check their answers many time to keep their accuracy and correctness. This service also include free math homework help.

To get unlimited access for all subjects, you just need to pay about $99.99 a month and you can use their service whenever you want, because their service is available 24/7.
To remind again, they also provide free online homework help for their demo.

Guaranteach - A Solution for Math Problems

2008-10-10T08:23:00.002+08:00

Today I found a website that provide online tutoring on math. The website is guaranteach. This website gives math helps using videos from over 120 teachers and tutors. The videos have been screened by expertise to make sure the videos suit and accurate for online learning.

Those videos can help whoever you are, because they have more than 2700 math topics, including algebra help. Duration of the videos are approximately 90 seconds, so you don't have to spend much time only for watching video.

How about pricing?
They provide 5 free videos for trial. And if you want to have free access to all of videos, quizzes, plus progress reports, you just pay $9.95/month.

Online Calculus Tutoring - TutorVista

2008-07-29T09:51:00.000+08:00

Having problems with calculus? Do you need calculus help? Don't worry, online calculus tutoring is available now in TutorVista.

TutorVista is a leading company on tutoring online specially for K-12 and college tutoring. They provide 24/7 service for calculus and precalculus help online. You can ask about your calculus and precalculus homework whenever you want.

They provide an incredible unlimited tutoring package for only $99.99/month for ALL subjects: calculus, precalculus, and other subjects help. You can use their service as much as you want and whenever you need.
And they also provide a free demo for the first time user to use their services for FREE.

Course Hero - Study Guides, Homework Solutions, Papers, Exam Answer Keys and more

2008-07-11T11:12:00.003+08:00

Now learning can be done anywhere, even it's on internet. One example is Course Hero.

This web builds a social learning network which provides an environment where students, professors, and other partners to meet, share and collaborate each other.

They provide immediate access to study materials and academic network of students, professors, and other partners.
For history, they have available HIST Paper, HIST Lecture Note, etc in 24/7 in many file type format, .ppt, .pdf, and .doc.

It's easy to join, if you have a facebook account, you can easily sign up. You can create your account less than one minute.
Once you sign up, you can meet over 300,000 fellow students and get over 200,000 textbook solutions immediately.

So, just join Course hero now for your success!

Gold Coins Gain

2008-07-15T08:09:00.000+08:00

Confused on crisis nowadays? Don't worry www.goldcoinsgain.com offers you a growing and safety asset. It's gold.
Gold Coins Gain is online aurum advisors for gold coins. The team consists of hand-selected world's finest precious asset metal managers.

In this site, you can buy gold, gold coins, gold bullions, and also sell gold.
There are many type of gold bullions offered, from $2.5 Liberty Quarter Eagle to Chinese Gold Panda. They also offer gold coins from many countries.

Not only gold, they also offer silver, platinum, and palladium. They also show "Spot Market Prices" of bullions (gold, silver, platinum, and palladium) on the top-right corner of the website.

So, if you want to buy gold, this is the right place!

Announcement

2009-03-15T07:58:00.004+08:00

I can't provide some new problems from now (~~maybe until the end of April~~) because I'm busy preparing myself for National Exam. (and also other business)
Therefore, if you want to become a contributor of this blog by providing some physics problems with solutions, you can leave comment below.
Thank you for your sincerity.

Leak Charge

2009-03-13T21:39:00.005+08:00

A conductor ball of radius R is charged with charge Q0. The ball is placed in an infinite space filled with a material with conductivity σ and permittivity ε, so the charge leaks away from the ball.
Find the remaining charge on the ball as function of time!

Click here to read the solution.

Solution - Leak Charge

2003-03-13T21:49:00.000+08:00

The ball is a conductor, so all of the charge is on its surface. The electrical field on its surface is

Because it's placed inside a material with conductivity σ, the current density on its surface is

Hence, the total current flow out from the ball is

The current equals to the decreasing rate of charge.

Hence,

Integrate both sides of the equation.

That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

Semi-infinite RLC Circuit

2009-03-11T21:58:00.001+08:00

Consider a semi infinite circuit like shown above. Magnitude of impedance of inductor and capacitor equal to R (i.e. ωL=1/ωC=R). Find the total impedance of the circuit!

Click here to read the solution.

Solution - Semi-infinite RLC Circuit

2003-03-10T21:52:00.006+08:00

See picture above. Because the circuit is semi-infinite, total impedance of circuit in the box is the same with total impedance of the whole circuit. So we can draw the circuit like picture below.
To make it easier, we should work with complex number.
Hence, the impedance of resistor, inductor, and capacitor are respectively

So the total impedance of the circuit is

Substitute the impedance of resistor, capacitor, and inductor at the equation above.

Solve the equation for Ztot, and we get

And now we need to change the form, so we assume that

Then give the power of two for each sides of the equation.

And now we can get

and

Solving these equation, and we get

So the total impedance is

Because the real part of impedance cannot be negative, so the total impedance is

So, the magnitude of total impedance is

That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

Free Downloads IPhO Problems and Solution

2005-02-28T14:08:00.004+08:00

Older IPhO Problems >>

IPhO 2008 (Hanoi, Vietnam)

IPhO 2007 (Isfahan, Iran)

IPhO 2006 (Singapore)

IPhO 2005 (Salamanca, Spain)

IPhO 2004 (Pohang, Korea)

Older IPhO Problems >>

Free Downloads IPhO Problems and Solution (2)

2005-03-10T10:16:00.001+08:00

<< Newer IPhO Problems

IPhO 2003 (Taipei, Taiwan)

IPhO 2002 (Bali, Indonesia)

IPhO 2001 (Antalya, Turkey)

This list will be updated. Wait for the other IPhO problems.

<< Newer IPhO Problems

Turning Scooter

2009-03-09T21:49:00.002+08:00

A scooter of length l has two wheels. The scooter moves with CM's speed v on a rough plane and is going to turn its direction without change its speed. To turn its direction, the front wheels is rotated to an angle θ.
Calculate the minimum coefficient of friction between the plane and the wheels, μ, so that the scooter does not slip to the plane! Assume CM of the scooter is at the middle.

Click here to read the solution.

Winding Spring

2009-03-07T21:37:00.002+08:00

A uniform solid cylinder of mass M and radius R can freely rotate around its axis O. And there is a spring of relaxed length L and stiffness K attached to the cylinder and a static wall. See picture below.Initially, the spring is relaxed. As the cylinder starts rotating, the spring will wind the cylinder. The surface of cylinder is very rough, so that the spring does not slip with the cylinder's surface.
Find the minimum initial angular speed of the cylinder, ω0, so that it can rotate to angle 2π!

Click here to read the solution.

Solution - Turning Scooter

2003-03-06T21:44:00.006+08:00

If the scooter does not slip to the plane, path of both wheels must be parallel with its direction.

From the picture above, we can get

Hence, the radius of CM's path is

So the total friction force must equal to

Hence, the x and y component of the friction is

Now, draw the friction forces work on both wheels. Each friction has perpendicular direction to each wheel. Sum of these two frictions must equal to the total friction (both in x and y direction).

First, for y-direction,

From equation (5), we can get

Then for x-direction,

Substitute fx from equation (4) and f2 from equation (6),

From equation (6) and (7) above, we can see that f2 is greater than f1, so the coefficient of friction must equal to

For the normal forces, because the CM of the scooter is at the middle, we can say that

And the total normal forces must equal to its weight, so

Hence,

Back to coefficient of friction. From equation (6) and (8), we can get the coefficient of friction,

That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

Solution - Winding Spring

2003-03-03T22:36:00.002+08:00

Assume the spring consists of N small springs, where N>>1. Hence, relaxed length of the small spring is .
The large spring equals to the small springs in series, so

From the equation above, we can conclude that multiply between spring constant of a part and its relaxed length is constant. And we can write

differentiate it, and we get

The total length of unwinding spring is L. If the cylinder rotates by small angle , a small part of the spring with total length will wind the cylinder, while the other part deformed to length L.
Relaxed length of the small part is,

Because the small part will wind the spring, the change of relaxed length of unwinding spring is,

Then, from equation (1) and (3), we can get

Integrate the equation, we will get

When the unwinding spring have relaxed length l, it will give force to the cylinder,

and give torque,

Thus, the change of energy when it rotates by small angle is

Substitute kl from equation (1),

Substitute k from equation (4),

Integrate two sides of equation,

And now, substitute and , and we get

Substitute for uniform solid cylinder.

That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

Minimum Friction

2009-03-05T21:17:00.003+08:00

A uniform rod of mass M and length l is placed on a rough floor and a rough wall. The rod makes an angle θ to the floor.What is the minimum coefficient of friction of the floor and the wall in order to prevent the rod from slipping? Assume the floor and the wall made by the same material.

Click here to read the solution.

Solution - Minimum Friction

2003-03-02T22:11:00.005+08:00

First, define .

Then, draw all forces that works on the rod. The total force and torque on the rod must be zero, because it is in equilibrium.

For x-component,

For y-component,

And for torque, we can take an arbitrary point on the rod because it is in equilibrium. In this case, we take the contact point to the floor.

Solving these three equations, and we get the relation between and ,

Because and may have different values, the coefficient of friction of wall and floor must be the greatest value between these two values, and , to keep the value of both not higher than the coefficient of friction.

Then draw a graph with as x-axis and as y-axis (can be or ).

From the graph, we can conclude that it reach its minimum value when .

And the equation (4) will become

Solve for ,

That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

Breaking A Ruler

2009-03-03T22:46:00.000+08:00

Why is it much easier to break a ruler by bending it than by pulling it?

Click here to read the solution.

Trapping Lenses

2009-03-01T16:32:00.001+08:00

Some identical thin convex lenses are arranged at edge of a full circle of radius R, according to the picture below. The lenses have equal angle spacing of θ (θ<<1). Each lens has aperture length of D (D<<R).Find the longest focus distance, f, of the lenses so that it is possible to trap a ray of light inside the circle!

Click here to read the solution.

Swinging Ball on Rough Inclined Plane

2009-01-29T22:17:00.003+08:00

A ball of mass m is hang by a string of length l on a rough inclined plane with coefficient of friction μ and angle θ with the horizontal (μ=tanθ). Initially the ball is at the lowest position.
Calculate the minimum initial velocity, v, should be given to the ball to make a full circle of its path!

Click here to read the solution.

Solution - Swinging Ball on Rough Inclined Plane

2003-01-27T20:50:00.001+08:00

There are two conditions that have to be fulfilled to make one full swing. One is to maintain the string in tension, and another one is to have enough energy to complete one-cycle friction.

For the first condition, consider when the ball has made an angle φ like picture below.
From the picture, we can get tension of the string,

In order to make the string in tension at all time, the minimum value of T must equal to zero.
The condition when T reach its minimum value,

Because , so

Now, consider the forces that are align to ball's velocity.

Substitute for , and we get

Substitute dv'/dt from equation (3) to equation (2),

Insert φ to equation (1)

By using conservation of energy,

Now, we need to find the minimum speed required for second condition (i.e. the ball has to have enough energy to complete one-cycle friction.
By using conservation of energy,

As we see, the speed required for the second condition is larger than the speed required for the first condition. So, we may conclude that the minimum speed required for both conditions is

That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

Spring and Hanging Ball

2009-02-27T21:53:00.002+08:00

A small ball of mass M is hung with a spring of stiffness k and zero relaxed length. The ball is in gravitational field, g. Then the ball is brought down to a point (x0, y0) as shown in picture below, and then released.How will path of the ball? And also find coordinate of the ball when it reaches its highest position!

Click here to read the solution.

Solution - Trapping Lenses

2003-03-03T18:11:00.000+08:00

First, we need to find the angle deflection of light ray if it passes a convex lens.

Hence,

By using lens equation,

we can get

If the lenses are separated by angle of θ, then the total number of lenses is

If we want to trap a ray of light, we just need to have the light forms a (nearly) circle path. On one cycle of the path, the light passes N lenses, so the total deflection is

The total deflection must equal to 2π,

In order to get the longest focus length, variable h must be at its maximum value (i.e. aperture length, D)

That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

Solution - Breaking A Ruler

2003-02-25T23:02:00.003+08:00

Assume the ruler has length of 2l and thickness of d.
Consider only a half part of the ruler. The ruler does not move, so we can say that the total force and torque on the ruler is zero.

First, consider when the ruler is pulled with a force F.If we pull the ruler with a force F, the other part of ruler will exert a same force, F, to this part in order to cancel the pulling force.

In other hand, if we bend the ruler with a force F at edge of the ruler, the other part will exert a torque with the same magnitude of

Therefore, we can calculate roughly the magnitude of force exerted by the other part

Because a ruler is commonly very thin compared to its length, the value of l/d will be very large.
So the force exerted by the other part is much larger if we bend the ruler, and it will break easily.

That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.