Physics Problem Solving, High School Level, High School Physics Problems, Solving

There are a lot to consider. From a cosmological point of view, I can't help it. But according to the level requirements of high school, it can still be answered: (but this is not quite like a high school question, it is a bit difficult).

I can help you simplify the meaning of this question as: m is at rest, m is moving in a uniform linear motion and colliding with m, and finally m and m are added together to do a uniform linear motion.

Let's start solving the problem:

1) Conservation of momentum: mv1 = (m+m)v2. v2=m/（m+m）v1

2) Friction is a constant force, proportional to m (Earth's equilibrium sphere). The asteroid is small relative to the Earth (the frontal drag is ignored and only the friction is calculated). v1 is a fixed value for the initial velocity of the asteroid.

3) Asteroid momentum: f Mo t = (v1-v2) m = mm (m + m) v1 (f and v1 are normal numbers).

4) To simplify, the constant part is set to.

5) Let m m be initially a, so it is simplified to: t=b (a+1) (b is a constant).

6) How does A need to change if this question becomes 1 (A+1) and increases by 29%.

After reading the original English text, I realized that I had misunderstood.

If only the mass of the asteroid is changed (other distances, initial relative velocity and nothing change), and the premise is that the asteroid is very small relative to the Earth (the asteroid is a mass point), then the asteroid acceleration (a=gm d, m is the mass of the earth, d is the distance)) The law of change is the same as before, then the collision time should not change!

The differential equation of motion is: d r dt = gm r

Does the time it takes for an asteroid to reach the Earth to be related only to the initial velocity of the asteroid, the initial distance from the Earth, and what does it have to do with the mass of the asteroid?

If there is English, is it an International Physics Olympiad question, although I have ideas, but I need to use advanced mathematics.

I didn't understand that this extension 29% is said to be sprinkled.

1.This question should be done at an average speed. Let the instantaneous velocity at point b be v, and since the two phases of motion are uniformly variable linear motions, v 2 is the average velocity of the two segments.

Then t1:t2=2:3 and t1+t2=10s, we get t1=4s and t2=6s to get v=2m s.

So a1=, a2=

2.Accelerate first and then decelerate, so the maximum speed is the change of a. Still do it at average speed.

Let the maximum velocity be v, and the equation (v 2)*(v a1)+(v 2)*(v a2)=s can be obtained, and v = under the root number [(2a1a2s) (a1+a2)].

3.The first 3s displacement is, it is easy to calculate a=1m s and the last 3s displacement is, then we can set the end velocity to v, and we can get the final velocity v=8m s

So s=v (2a)=32m

1.First of all, I only know the size of the distance traveled by the object, 10m, the average speed is good, and 1m s

Therefore, the maximum speed must be 2m s, and it can also be judged that the time in the inclined plane is 4s and the time in the horizontal plane is 6s (the average speed knows, this is easy to find) Isn't it easy to find the acceleration?

Inclined plane a1 = 2 4 = 1 2 horizontal plane a2 = 2 6 = 1 3

2.This one is similar to the one above, but different.

We know that the displacement formula for the uniform acceleration motion at rest is s=(1 2)a*t 2=(1 2)v 2 a

And this v is the maximum speed (or change in speed).

Therefore (1 2)v 2 a1 + (1 2) v 2) a2=s

v^2=2s/(1/a1+1/a2)=2s*a1*a2/(a1+a2)

Just open the root number.

The third question is that the acceleration is easy to find: 1 2*a*t 2=s a=1m s

The average velocity is found as:

3*(0+v1) 2+3*(v2+v3) 2 16=0+v1+v2+v3=2v4 (v4 is the maximum speed, i.e. v3, you can understand it) v4=8

The length of the inclined plane is 8*8*

1-a1t1 2=4 (1) The first motion formula 21-a2t2 2=6 (2) The second motion formula 2a1t1=a2t2 (3) The final velocity of the first segment is equal to the initial velocity of the second segment.

Three unknowns, three equations

In favor of the method of using graphs, it is more intuitive.

The key to the second problem is t, that is, the time of the first segment and the time of the second segment are set to t1 t2, and the maximum speed v

t1=v/a1 (1)t2=v/a2 (2)1 1

a1t1^2+ -a2t2^2=s (3)2 2

Three unknowns, three equationsSolve the same way. Question 3.

The first question is easy to solve by simply drawing a V-T diagram...Oral arithmetic is OK: a1 = 1 2m s 2, a2 = 1 3m s 2.

The second question can also be a simple diagram, according to the average speed and time of each segment, v = (2a1a2s) (a1+a2).

The third question, I think the easiest way is to follow the displacement law of the uniform acceleration motion with zero initial velocity, that is, 1:3:5:7:9:11:13:15, which is very simple, and it takes eight seconds to see that the distance is 32m.

1,a1=1/2;a2=1/3;

2,v= [2a1a2s (a1+a2)] (open root number).

3. The object movement time is 8 seconds and the length is 32 meters.

After stopping pedaling, it is known by the kinetic energy theorem.

From the equation, it can be seen that the image is a parabola that passes through the opening of the origin upward, and because v is greater than 0, it is the right half, and it is finished.

The resistance does negative work, so that the velocity changes from v to 0

The kinetic energy theorem gives -w=0-mv 2 2

i.e. w=mv22

w and v are quadratic functions of the opening upwards.

So choose C and don't understand, you can ask

In fact, the bicycle is also subject to sliding friction during the movement, but the distance is very short, the greater the initial velocity of the deceleration, the longer the distance of the sliding friction work of the bicycle, and the sliding friction is much greater than that of rolling f, so with the increase of speed, the friction work is also getting bigger and bigger, the greater the slope of the graph line, choose C, I hope it will be useful to you.

It should be that the effect of air resistance is taken into account, and if the drag is constant, then it should be Figure B fv=p.

Now the air resistance decreases with a decrease in velocity, such as proportional to f=bv, where b is a constant. Then p=bv 2, which is Figure C.

The actual situation is of course more complicated than f=bv, but that's probably the case.

Pick B. You are tired of taking a look at this web page, and if you don't understand, I'll give you an answer.

Two small balls A and B, which can be regarded as particles, are connected by a rigid thin rod with negligible mass and placed in a smooth hemispherical surface, as shown in the figure The ratio of the mass of the ball A and B is known to be sqrt, and the length of the thin rod is a sqrt multiple of the radius of the spherical surface The two balls are in.

High School Physics. Jingyou.com.

The object is affected by its own gravity and air resistance, the gravitational force g direction is always downward, and the drag force f direction is opposite to the direction of motion. At the beginning, the object moves upwards, and the gravitational force and drag force are both downward, and the net force is f+g. When falling, the resistance is upward, and the resultant force is g-f.

It can be seen that the acceleration is relatively large when ascending, b pair. From h=at 2 2, it can be seen that when h is constant, a is large and t is small, so the rise time is less than the fall time, and a is wrong. Since the object does not do work from the beginning of the throw to the return to the original place, the gravitational force does not do work, and the resistance does negative work, so the kinetic energy decreases, and the velocity v2 of falling to the original place is less than the velocity v0 when it is thrown.

If the time in the ascending process is t1 and the time in the falling process is t2, the average velocity in the ascending stage is v0 t1

The average velocity of the fall phase is v2 t2

And because v0>v2

T1 is clear that the average velocity at the rise is greater than the average velocity at the descent, and C is right and D is wrong.

In summary, the answer is BC

1) Just through point C, there is Newton's second law: f = MV2 r, f = mg, so vc = 1m s; From the law of conservation of mechanical energy, we get: 1 2MVB2=1 2MVC2+mGH, H=2R, so: VB= 5M s

2) In the process of the racing car moving from point A to point B, the work done by the engine is equal to the work done by friction plus the kinetic energy of the car to point B, and by the law of conservation of energy, we get: pt-fl=1 2MVB2, and p=. Wish!

From the kinetic energy theorem, it can be seen that the work done by the combined external force is equal to the amount of kinetic energy change.

pt-fs=1 2mv 2 v is vb

In the process of A to B, the resistance does the work:

wf=f*l=2j

Kinetic energy theorem: pt-wf=mv*v 2-0

Bring in the data, get it. p=

From the kinetic energy theorem, pt-fl=1 2m(vb) 2-0 is substituted into the data to solve p=

Let the angle between the track and PQ be a, the 37-degree radian is written as b, L1=, and the distance from P to the inclined plane is L2=L1*cos(b). The length of the orbit is l3=l2 cos(b-a), and the acceleration is a=g*cos(a)time t=sqrt(2*l3 a)=sqrt(2*l1*cos(b) (g*cos(b-a)*cos(a)))sqrt(2*l1*cos(b) (g*so when b-2a=0, time is minimal:

t=sqrt(2*b

b The falling angle is an unknown quantity, and the time formula is found and the trigonometric function is simplified.

Using the momentum theorem, assuming that there is an average constant force that interacts, and then using the displacement formula of uniform acceleration to find the displacement in proportion to each other, see the figure for details.

x total = r

It can be obtained from a = f m, a car level, : a ball level = m car part 1: m ball part 1, x = (1 2) at 2, x car x ball = (1 2) (1 m car): 1 2) (1 m ball).

x total = x rook + x ball.

X car and X ball can be solved separately.