Senior 1 physics, I don t know if my question can be answered, thank you.

The weight is not in free fall, the weight gives the trolley traction, then the trolley is the upward pull force on the weight. Therefore, the weight force not only has gravity, but also an upward pulling force, so it is not a free-fall movement.

Because the mass of the trolley is not equal to the mass of the weight disc and weights, the acceleration of the trolley is not equal to the acceleration due to gravity.

Weights and weight discs cannot be considered free-fall motion. Because they also have to pull the trolley to do motion together, their common acceleration is the gravity of the weight and the weight disc divided by (weight + weight disc plus the mass of the trolley).

In the absence of thin wires, weights and weight discs can be considered to be in free fall.

However, the weight and the weight disc are tied together with the trolley, and only the gravity of the weight and the weight disc acts as the power of the whole system, and the gravity of the trolley does not play a role, because the gravitational force of the trolley is offset by the supporting force (and less friction).

If the table and the plank are removed at the same time, that is, the trolley loses its support, and at this time, the trolley will do a free fall motion with the weight and the weight disc with a = gravitational acceleration.

The analysis of this device is as follows:

**When the relationship between acceleration and force and mass, the friction should be balanced, and after the friction is balanced, it can be considered that the tensile force of the trolley is equal to the gravity of the hook code; During the experiment, the hook code moves together with the trolley, and the pulling force of the trolley is less than the gravity of the hook code, the larger the hook code mass, the greater the experimental error, and the smaller the hook code mass relative to the trolley mass, the smaller the error; When using the image method to process experimental data, it is most convenient when the image is a straight line

Weights and weight discs are subjected to an upward pull and are not in free fall.

Only the weights are in free fall, and the trolley just slides diagonally, got it?

Here it can be seen as a two-stage movement process, so the analysis is simple point 1The electron is decelerated to vt=0 under the action of the electric field, and the time it experiences is t 2, and the displacement in this process is s1

For electron analysis, the electric field force f=ee=ma

Kinematics: Acceleration a=(vt-v0) t=v0 (t 2)=2v0 t

Displacement formula 2as1=vt 2-v0 2=v0 22, the electron accelerates from rest to its original position under the action of an electric field, which is equivalent to the reverse process of 1 process. You don't have to analyze it anymore.

The whole process distance l=2s1

The specific solution formula:

ee=ma.

a=(vt-v0)/△t=2v0/t.

2as1=v0^2

The above formula can be combined to find the electric field strength e= distance l= 2

Hope it helps, thanks for adopting!

Let the electric field strength e, the acceleration of electron motion a=ee m, the electron first decelerates to a speed of 0, and then returns to do a uniform acceleration linear motion. The time of the deceleration motion process t 2 = v0 a

Combining the two solutions yields e=2mv0 et .

The displacement magnitude of the deceleration process is s=v0*v0 2a =v0t 4, so the distance traveled by the charged particles in the electric field is s=2s=v0t 2

D analysis: according to the definition formula of acceleration a=( vt-vo) t, the direction of velocity before collision is selected as the positive direction, v0=6 m s, vt=-3 m s, and the formula can be substituted to obtain a=-30 m s2, and the negative sign indicates that it is opposite to the selected positive direction, so d is correct. Note that the positive and negative signs in the vector indicate the direction, the positive sign indicates the same as the selected positive direction, and the negative sign indicates the opposite direction of the selected positive direction.

Momentum difference i=f*t = p = m*v.

a=f/m=v/t=30m/s^2

**The direction of the force is negative, so the acceleration is negative.

So choose D

Look at the picture and answer! The ab line is the tangent of the curve, that is, the acceleration a is the slope of ab, a=15-5 4-0=

Force analysis: f=mgcos, f=kv

cd is the asymptote of the curve, i.e., when v = 4, it moves at a uniform speed.

There is mgsin = kv2+ mgcos

Acceleration phase: a=mgsin -kv1- mgcos when v1=2, v2=4

Substituting Lianli can find k,

2. The acceleration of curvilinear motion is not necessarily perpendicular to velocity, such as flat throwing!

Compulsory 2 question.

The acceleration at other points is downward, because the title is talking about a linear motion with a uniform variable speed, and the acceleration is constant, and the magnitude and direction are the same.

Gravitational acceleration is related to gravity and goes all the way down.

If A can touch B for the first time when it swings to a vertical position (after T=T 4), it means that the two balls will touch in the shortest time.

Single pendulum period t = 2 pi root number (l g).

t=t 4=(丌 2) root number (l g).

On collision, the downward displacement of b is s=l

l=[(4t 2) (丌 2)]g=

When b slides, s=l=(1 2)at 2

a=(4/5)g

and ma=mg-f

Therefore f=mg-ma=mg-(4 5)mg=(1 5)mg The ratio of frictional resistance to its gravitational force is.

f/mg=1:5

A The mechanical energy is conserved when the ball is hemmed.

1 2) mv 2 = mgh = mgl (1-cosq) v = root number [2gl (1-cosq)].

Let the velocity of the b-ball be v1

v1 2=2al=2[(4 5)g]l=8gl 5v1=root(8gl 5).

The ratio is obtained.

v: v1 = root number [2gl(1-cosq)] root number (8gl 5) = root number [5(1-cosq) 4].

How much does the object in question 3 weigh?

Because it is a smooth plane, there is no friction, the small object slides on the plank, and is affected by the friction force given by the plank, because the force is mutual, it is said that the small object also gives the plank a force, and the friction force is reversed, so the plank also moves forward, and finally the small object and the plank move together at the same speed, and there is no relative motion between the two to move together on the plane.

First, find out the velocity of both moving together and conserve according to the momentum.

m1v1=（m1+m2）v

v=8m/s

This is the final velocity of the small block.

Then it is conserved according to the energy.

The decrease in the kinetic energy of the small block is equal to the work done by the combined external force, and the external force here is only friction, so the decrease in kinetic energy is equal to the work done by frictional force.

fs=1 2m1v1 2-1 2(m1+m2)v 2s is the distance at which a small object slides on the board.

When encountering this kind of problem in the future, the first thing that comes to mind is the momentum theorem and the conservation of energy.

The method is relatively simple, it is a solution idea, and you will do a few more questions.

Is the final speed upstairs 8 meters per second?

Isn't conservation with momentum 2 meters per second?

In short, the upstairs idea is right, but there may be something wrong with the results.

The acceleration of small blocks is -4m s2

The acceleration of the plank is 1m s2

So after 2 seconds it has the same velocity of 2m s

The glide distance is the displacement of the small block minus the displacement of the long plank s=12-2=2m

Action and reaction forces: 60n Acceleration: 60 50 = ; 60/75=;

Speed: v1 = , v2 =