Ask a senior three physics question, a senior three physics question

There's no speed... Momentum is conserved.

Solution: The initial state of the whole device is stationary. From the conservation of momentum, the terminal state is known, and the whole device is at rest.

The gravitational potential energy is all converted into the internal energy of the system to obtain mgr=umgs

So s=r u

When the small object and the car are relatively stationary, the velocity of the car and the small object relative to the ground is 0, this is because the total momentum of the small object and the car is conserved in the horizontal direction is 0, you think that the gravitational potential energy of the small object is converted into kinetic energy, so the small object and the car will have velocity after being relatively stationary. Because the momentum of the whole system is not conserved in the vertical direction, but it is conserved in the horizontal direction.

Upstairs is right, the whole system is not affected by external forces, momentum is conserved, the initial and final momentum should be 0, and there is no velocity at the end.

Boy! The momentum in the horizontal direction is conserved in the system, so the momentum in the horizontal direction is zero at the beginning of the system, so the system is ultimately stationary!

The law of conservation of energy. The work done by gravity is equal to the work done by frictional resistance. i.e.: mgr=umgs;

The work done by frictional resistance is converted into kinetic energy, i.e.: umgs=1 2mv 2=1 2mv2

That is: mgr=umgs=1 2mv 2=1 2mv 21 2mv 2 is the kinetic energy obtained by the flatbed;

1 2mv 2 is the kinetic energy gained by a small object.

Your thinking is wrong in not understanding the conditions for the conservation of momentum. For the whole system, the initial and final states are not strong in terms of level!

1. The small object slides down from the top to o', the horizontal momentum of the system is conserved, and the mechanical energy of the system is conserved, that is, the gravitational potential energy of the small object is reduced by equal to the kinetic energy of the two objects, and the respective velocities can be calculated.

2. The small object and the trolley in the horizontal plane are decelerated, and after the small object is reduced, it accelerates under the action of friction until it is equal to the speed of the trolley. The momentum of the system is conserved, according to the kinetic energy theorem, the work done by friction is equal to the change in the kinetic energy of the system, and the two equations find r.

Mechanical energy is conserved, and only gravity does the work in the whole process, and the work done by gravity is converted into heat consumption, that is, UMGS. And the change in kinetic energy is only an intermediate quantity. Now the question asks:

The final distance from o when the small object is at rest relative to the car. So maybe, maybe, maybe you're getting it wrong.

First of all, the initial state of the whole device is stationary. It is obtained by the conservation of momentum. In the final state, the entire device is also stationary. Then there is the conservation of energy to get mgr=umgs. s is the distance of a small object on a flatbed.

The cars are stationary at first, and when they are relatively stationary again, both have a velocity of 0, do you consider the conservation of momentum?

Because the momentum of the system is conserved in the horizontal direction, and the initial momentum of the system is zero.

When small objects and cars are relatively stationary, they are relatively stationary with the ground.

Momentum is conserved, and the transformation of momentum and energy is grasped.

C, D are both correct.

The flat throwing point of the ball is based on the flat throwing point, and there is a "definite relationship" between the displacement vector r and the velocity vector v

tanv=2tanr where v is the sharp angle between the velocity vector v and the horizontal direction and r is the sharp angle between the displacement vector r and the horizontal.

At a certain time, according to the title, "all fall on the inclined plane", so 1 = 2 notice that d is talking about the size relationship of 1, 2, not its own size, increase, 1, 2, will change, but the size relationship of 1, 2 will not change, always equal.

The reason for choosing c, well, is to set the angle between the direction of velocity and the horizontal direction, because both times fall to the inclined plane, according to the reasoning formula tan = 2tan, a does not change.

So 1 = 2, and because = -a, the person in front of c is wrong!

Also: the derivation process of tan = 2tan.

tan = vy vx=gt v0, tan =sy sx=1 2gt2 v0t=1 2gt v0 (again because tan = gt v0) so tan = 1 2tan

It is clear that tan = 2tan, i.e. the tangent of the velocity angle is twice the tangent of the displacement angle.

C must be chosen for this question, everything else is wrong, and D is also wrong.

The method of exclusion excludes d d in the first place

Suppose that the infinitesimal to 0 is close to the horizontal, and the horizontal throw velocity v1 is infinitely small to 0, which is equivalent to free fall.

This is 1, which is close to the right angle, and then it slowly gets bigger, and we find that 1 is close to 90° minus, and if ab can also be excluded at the same time, because a is selected in the multiple choice question, then b must be right, so there is only c, and this is the elimination method in the answering technique.

Why choose C, the upstairs has already given you a clear answer, I will not repeat it, so we don't necessarily have to calculate the exact answer to some questions in the exam, and it is very effective to use the elimination method.

d.The magnitude relationship between 1 and 2 has nothing to do with the inclination angle of the inclined plane.

This one is correct.

The magnitude of the included angle is only related to the velocity, because the velocity cannot be determined, so only d is correct.

c d can make the angle falling on the inclined plane a According to (1 2gt 2) v0t =tana and tana=gt v0, the relationship between a and a can be obtained, and the rest is not explained.

I choose B....First of all, since they are both thrown horizontally and fall on the inclined plane, the ratio of the horizontal displacement to the vertical displacement of the two balls is the same, and they are both tangent values of the inclined plane. i.e. h s = tan .

h=1/2*gt^2,s=vt...v is the initial velocity, and the obvious comparison between the two results is (1 2gt) v....gt=vy...

So vy v=2tan

Choosing d only considers the moment of landing, because the air resistance is not calculated, the horizontal velocity of the first time is v1, the second throw is v2, the vertical velocity is the same as x, because the height is the same, so the final velocity in the vertical direction is also the same.

d.The magnitude relationship between 1 and 2 has nothing to do with the inclination angle of the inclined plane.

Pick D. The angle between velocity and horizontal is high, and the angle between displacement and horizontal is . tanβ=2tanα。The angle of the inclined plane is the angle of displacement, so choose d

I think it's what d did in high school before.

As the thin rod turns 90 degrees counterclockwise around its center, the electric potential of q decreases and the potential energy decreases. The electric potential of q becomes larger, the electric potential energy becomes smaller, the electric field force does positive work, and the electric potential energy decreases, so w2>0, since you know w1 and w3, I won't say more, presumably the answer is clear to you. Process analysis is very important for electric field problems.

Concepts are also important.

Only the W2 vertical line is a zero potential surface, according to the direction of the inductance line, the potential on the A side is less than 0, and the B side is greater than 0. If the rotating negative charge enters from 0 potential to greater than 0, the electric field force does positive work. If the positive current enters less than 0 from 0 potential, it is also positive work.