Tell us about the high school physics competition

The science competition is not about finishing college physics in advance, but about finishing college physics (and advanced mathematics) in the process of studying. (Physics in general engineering is very simple, depending on the physics department).

Physics competitions are mainly about reading books and doing problems.

I recommend to you a few of the ones I've read.

The first Fan Xiaohui, a beginner's book.

Second, "High School Physics Competition Guidance" or something I can't remember the name Zheng Yongling's "White Paper" is not difficult, but it uses a lot of calculus knowledge.

Third, "Advanced Mathematics" only looks at the differential equations section of the first and second volumes.

Fourth, the highlight of the three books edited by Cheng Jiafu, including "Lectures", "Mechanics", and "Electricity" These three books should be read carefully repeatedly, and if you do these three books thoroughly, you already have the strength to win awards.

Fifth, "Higher and Better Physics" is a rare and good book, and it must be done.

Sixth, the more difficult book "Collection of Difficult Problems" Shu Yousheng There is also a red-skinned book edited by Zhu Hao, and the name is incomprehensible.

You don't have to do all the questions above, it's difficult.

Sixth, extracurricular books Zhao Kaihua's "New Concepts of Mechanics" and "New Concepts of Electromagnetism" are actually textbooks for the Department of Physics, which should be read as extracurricular books, but there are also useful ones, such as Coriolis forces, complex solutions for alternating current.

Participating in the physics competition is not to finish college physics in advance, but to complete college physics (and advanced mathematics) in the process of studying. (Physics in general engineering is very simple, depending on the physics department).

Physics competitions are mainly about reading books and doing problems.

I recommend to you a few of the ones I've read.

The first Fan Xiaohui, a beginner's book.

Second, "High School Physics Competition Guidance" or something I can't remember the name Zheng Yongling's "White Paper" is not difficult, but it uses a lot of calculus knowledge.

Third, "Advanced Mathematics" only looks at the differential equations section of the first and second volumes.

Fourth, the highlight of the three books edited by Cheng Jiafu, including "Lectures", "Mechanics", and "Electricity" These three books should be read carefully repeatedly, and if you do these three books thoroughly, you already have the strength to win awards.

Fifth, "Higher and Better Physics" is a rare and good book, and it must be done.

Sixth, the more difficult book "Collection of Difficult Problems" Shu Yousheng There is also a red-skinned book edited by Zhu Hao, and the name is incomprehensible.

You don't have to do all the questions above, it's difficult.

Sixth, extracurricular books Zhao Kaihua's "New Concepts of Mechanics" and "New Concepts of Electromagnetism" are actually textbooks for the Department of Physics, which should be read as extracurricular books, but there are also useful ones, such as Coriolis forces, complex solutions for alternating current.

It should be noted that the rotation of the drum has no effect on the velocity of the cylinder C, because the friction generated by the rolling is perpendicular to the velocity direction of the cylinder's motion.

sinβ=(a/2)/(r+r)

So cos = the square of [1-(sin)] under the root sign and finally substitute cos into the formula in the diagram to get f, I won't simplify, it's too difficult to express (and I don't want to redraw the diagram.) ):p

A and B cylinders are rotated or reversed.

In the whole process of motion, the mechanical energy of the particle is conserved, obviously, if the particle can return to the original starting point, then its upward motion and downward motion along the inclined plane should be symmetrical with each other, that is, after the particle moves along the inclined plane to the highest point it can reach, it should return along the "original road". In this way, this requirement can be met in both cases. First, when the particle collides with the inclined plane for the last time, its velocity direction is just perpendicular to the inclined plane, then its speed must be equal to the velocity before it touches and the direction is opposite, so that the movement of the particle after that will "invert" the motion of its rising process once, and it can return to the original starting point.

The second is that after the particle point collides with the inclined plane for the last time, its speed is just along the vertical upward direction, then the particle bounces up and makes a vertical upward throwing motion, and when the particle reaches the apex of the vertical upward throwing, it will then "invert" the previous motion and return to the original starting point.

Due to the symmetry of the motion, for the convenience of solving, we can find the velocity of the particle falling along the inclined plane each time in the above two cases, and reverse this velocity, that is, the throwing velocity that meets the requirements of the problem.

Solution: (1) Establish a coordinate system as shown in Figure 3, let the particle point be thrown at a certain initial velocity v0 along the direction perpendicular to the inclined plane (y direction in the figure), and before the particle point collides with the inclined plane again, the two components of the acceleration of the particle when moving in the air are respectively

xy фax＝gsinθ

ay＝－gcosθ

At a certain time t, the velocity of the particle has two components:

vy＝v0－gcosθt

vx＝gsinθt

At time t, the position coordinates of the particle are:

x＝gsinθt2/2

y＝v0t－gcosθt2/2

From the above equation, let y 0 to get the time from the start to the first collision with the inclined plane

t＝2v0/gcosθ

At this point, the two velocity components of the particle are:

vy＝－v0

vx＝gsinθ·t

When the particle is raised from the inclined plane**, its velocity perpendicular to the inclined plane direction is v0, and it can be seen that the time interval between the particle and the inclined plane every two times is t, and the velocity components of the particle when it collides with the inclined plane for the nth time are:

vy＝－v0

vx＝gsinθ·nt

Let the angle between the velocity direction and the inclined plane be , then there is:

tgф＝∣vy/vx∣＝v0/(ngsinθ·2v0/gcosθ)＝ctgθ/2n . n
……1

2) Let the particle fall freely from the air to the inclined plane, and the velocity when it collides with the inclined plane is v0, which is still in the coordinate system shown in the figure

t＝2v0cosθ/gcosθ＝2v0/g

After that, the velocity component of the particle when it collides with the inclined plane for the nth time is:

vx＝v0sinθ＋gsinθ·(n－1)t

vy＝－v0cosθ

When the particle collides with the inclined plane for the nth time when it is solved by the above three formulas, the angle between the velocity direction and the inclined plane should satisfy:

tgф＝∣vy/vx∣＝v0ctgθ/【v0＋2v0(n－1)】＝ctgθ/(2n-1). n
……

3) from the above analysis, it can be seen that if the motion situation is reversed, then the particle point is thrown in the angular direction determined by the formula or formula, then the particle point will be able to return along the original road, and the two cases of (1) and (2) can be known that the angle between the velocity direction of the particle and the inclined plane on the inclined plane can be known as long as it satisfies: tg ctg k (k
……, the particle can return to its original starting point.

The final situation is that the ball must collide vertically with the inclined plane before it can return to the starting point.

The final answer is.

tgфtgθ=2

You can refer to the 1st floor to do it yourself, if your foundation is very good, you can understand it, if it is not very good, it is recommended not to do it, the competition is more difficult in this province.

v1, v2 and v enclose a closed vector triangle, the angle between v1 and v2 in the vector triangle is -a, the opposite side is the velocity v, and the cosine theorem has v1 +v2 -v =2v1v2cos( -a)=-2v1v2cosa

v= (v1 +v2 +2v1v2cosa).

Let the angle between v1 and v be , then there is:

v×cosφ=v1

v×cos(α-=v2

Solution: =acos((v1sin ) (v1 +v2 -2v1v2cos ))).

v=√(v1²+v2²-2v1*v2cosα)/sinα

With such an attitude, it is estimated that no one will solve the problem for you. Be humble, not to mention that you are still begging for help.

In fact, is this a solid geometry? Think of the board as an isosceles right-angled triangle with right-angled vertices resting on the tabletop, the other two vertices are the same height, (the hypotenuse is parallel to the tabletop), the right-angled edge is at an angle to the tabletop, the hypotenuse midpoint is o, the vertex is p, the photograph of o on the desktop is q, the non-right-angled vertex of the triangle is a, and the photograph of a on the desktop is b.

Then now there is the west it = angular apb, we know sinapb, ask sinopqFound sinapb=ab ap sinopq=oq op

There is also ab=oq (parallel relation), in an isosceles right triangle, ap=root number 2 op, and replacing oq and op bands with ab and ap is the b answer.

By the way, I'll ask how to fight the West and the root number.

The direction of this friction force is the AC direction, and it is decomposed along the direction perpendicular to the AB edge and the AD edge, and the magnitude of the two components is MGSIN, so the friction force of the two components is synthesized as 2mgsin

The spring is only in contact with the block, not connected, and only accelerates the two pieces, after which it separates. Momentum is conserved horizontally during block acceleration.

According to the fact that they can pass the orbit to the fixed point, their speed at the top of the orbit can be found in the method of "Personality 10 Class Boss".

Then use the Conservation of Energy to find their velocities at the bottom end of the arc orbit, respectively. (The "Boss of Personality Class 10" didn't think about this, and it was right to add this.) If the top velocity is used, the horizontal momentum is not guaranteed, because the title does not say that they arrive at the same time, nor does it say that the orbit gives them equal horizontal impulses).

Since there is no friction, this velocity is the velocity at which the block and spring are separated.

Conservation of momentum analysis is then used to cut the rope ——— the block to separate from the spring during this process.

Momentum is a vector quantity, for the problem mentioned in the "Problem Supplement", because the direction of the orbital elastic force when the spring is doing work is upward, and the whole system is not affected by external forces in the horizontal direction, so the momentum in the horizontal direction is conserved, so the usable momentum is conserved.