The sophomore physics momentum is explained in detail

First of all, I think there should be friction between the slider and the board, and the bottom of the board should be smooth. If so, I want to talk about my thoughts, please forgive me for saying the right thing. The first step, of course, is that momentum is conserved, because the two are not affected by external forces as a whole, which gives them a common velocity.

The friction between the slider and the plank is resistance to the slider, but it is the power to the plank, that is to say, before reaching the common speed, the friction slows down the slider and accelerates the plank until the common velocity, at this time there is no relative sliding, and there is no friction. The key is to find the relative friction factor. Let a u, use kinematics to find the displacement of the slider and the plank relative to the ground, and subtract the two to l.

The second part is a little complicated, after finding U, or use kinematics to find the speed of the slider and the plank at 3 4L, and then the momentum of the slider and the other 1 4L is conserved, and the method is as before.

Solution: 1) When the board length is l: let the common ground velocity of the slider stop at the plank is v, and the friction force of the slider on the board is f

Conservation of momentum: mv0 = 3mv, therefore, v = v0 3

Function-energy conversion: 1 2 mv0 2 = 1 2 (m+2m)v 2 + fl = 1 6 mv0 2 + fl

Find f = mv0 2 (3l); fl = 1/3 mv0^2

2) When the board is divided into two sections, 3 4L and 1 4L, the mass of the board is divided into two parts: 3m 2 and m 2.

When the slider moves to 3 4L, let the slider speed be V1 and the plank speed V'

Conservation of momentum: mv0 = 2mv'+ mv1

Conversion of work and energy:

1/2 mv0^2 = mv'^2 + 1/2mv1^2 + 3/4fl = mv'^2 + 1/2 mv1^2 + 1/4 mv0^2

The two-formula solution gives v1 = 2 3v0, v'=1/6v0

3) When the slider slides on the rear 1 4 sections, it is assumed that the slider slides the distance on the board x and the overall speed when it finally stops on the board is v2.

Conservation of momentum: MV1 + 1 2 MV'= 3 2mv2, utilizing the result of 2), v2 = v0 2

Power and energy conversion: 1 2 mv1 2 + 1 4 mv'^2 = 1/2 (3/2m)v2^2 + fx

Find x = 1 8l

That is, the slider stops at the midpoint of the last 1 4 segments.

Just write down the formula.

mv0=(m+2m)v'

v'=v0/3

1 2mv0 2 = 1 2 (m + 2m) (v0 3) + fl (f is the friction force).

The solution yields f=mv0 2 3l

The next 4 equations are combined, and I don't want to ask for the result.

mv=mv1+2mv2 (v1 is the velocity of the small wooden block when it reaches 3 4, v2 is the velocity of the wooden block).

mv1+1 2mv2=(1 2m+m)v3 (v3 is the velocity of the second half of the block and the strip).

1/2mv0^2=1/2mv1^2+1/2(2m)v2^2+f*3/4l

1/2mv0^2=1/2*3/2mv2^2+1/2(1/2m+m)v3^2+f*s

s is the distance sought.

mv0=(m+m)v.

1 2mv0 2=1 2(m+m)v total 2+umglmv0=(3 4m+m)v.

1 2mv0 2=1 2(3 4m+m)v, a total of 2+umgl'

l'Do what you want, do the math yourself, and you should be able to understand the equation.

p=mv, its meaning is the scale of the intensity of the motion of the object, which is a physical quantity that is more basic than velocity and mass, and is often found to be more concise in the microscopic world to describe momentum because it uses momentum as the unit of motion.

Your question is incomplete......

In the process of the ball sliding on the surface, the car moves to the right, and when the ball slides down, the car will continue to move forward, so it will not return to the original position, A is wrong. From the ball to the highest point, knowing that the two have the same velocity, for the system composed of the car and the ball, the column formula of the law of conservation of momentum is mv=2mv', with a common velocity v'=v/2。The change in the momentum of the trolley is mv 2, and obviously, this increased momentum is the result of the pressure of the bulb, so b pairs.

For c, it is possible that the mechanical energy of the system does not increase because the law of conservation of momentum is satisfied, which occurs when the two surfaces are smooth. Since the original kinetic energy of the ball is mv2 2, and the kinetic energy of the system is 2m (v 2)2 2=mv2 4, so the momentum of the system decreases mv2 4, if the surface is smooth, the reduced kinetic energy is equal to the gravitational potential energy increased by the ball, i.e., mv2 4=mgh, and h=v2 4g. Obviously, this is the maximum, and if the surface is rough, the height is even smaller.

It's asking how fast the ball is when it first leaves the car, right? It is necessary to add that two surfaces are also smooth.

Using the conservation of momentum and the conservation of mechanical energy, let the velocity of the ball be v1 when it is just separated, and the velocity of the car is v2, then the direction of the velocity v at the beginning of the ball is the positive direction, yes.

mv=mv1+mv2 and mv2 2 mv1 2 2+mv2 2 2

As you can see from the force on the ball when it slides down the surface, v1 is not zero!

From the above two forms of synthesis, v1 v2 0

Because the two cars are subjected to the spring force fa=fb and the time ta=tb, so fa*ta = fb*tb, the momentum change is equal, so the ratio of the momentum change of car A to the momentum change of car B is 1

1 The so-called original direction does not refer to flying at 60 degrees, but the direction of flight of the grenade (hereinafter referred to as SLD) in front of **. According to the title, when reaching the highest point, SLD only has a forward horizontal velocity component, and no vertical motion component in the up and down directions. Therefore, the so-called "original direction" of the title refers to this forward horizontal direction.

2 According to my observation, this question is not clear, because the "original direction" can also be considered to be diagonally upward at an angle of 60 degrees to the horizontal direction, so the title should be written as "the most massive piece of it flies at a speed of 2v0 along the direction of motion before **". If you encounter this situation during the official exam, I recommend that you calculate both cases and explain both cases so that the examiner will definitely give you a full score.

3 Back to the topic. The initial velocity v0 of the sld is thrown diagonally upwards at an angle of 60 degrees to the horizontal, so the component of the velocity in the horizontal direction is v0cos60°=, and after that you have already assigned v in the problem'= Calculate, I won't write it. Direction is a matter because one of the pieces is horizontally forward, and the other is necessarily horizontally backward.

4 Let the chemical energy be e, then according to the conservation of energy: 3m*(, e=12*m*v0 2

5 The crux of this question is the so-called "original direction," and if I don't make it clear to you, you can ask additional questions.

First force analysis, the horizontal direction is not forced, and the vertical direction is subject to gravity.

At this time, the velocity should be divided into vertical direction and horizontal direction, the horizontal direction remains the same, the vertical speed has been decreasing, and the horizontal direction has been conserved momentum because it has not been subjected to force.

When the highest point is reached. That is to say, at this time, the velocity in the vertical direction is already 0, the reason is due to gravity, classmates, to carry out force analysis!。

When the highest point is reached, the vertical velocity is zero, only the horizontal velocity.

Both momentum and kinetic energy are related to the mass of the object and its state of motion.

The difference is that kinetic energy is reflected from the perspective of energy, is a scalar quantity, and has nothing to do with the direction of motion;

Momentum is a vector quantity that is related not only to the magnitude of v but also to the direction of motion. This is irreplaceable by kinetic energy.

Work is what causes the energy to be transformed, and the total work of the combined external force corresponds to the change in kinetic energy.

The impulse is what makes the momentum change, and the impulse of the resultant external force corresponds to the change in momentum. As long as the resultant external force is not zero, the momentum of the object must change, either in magnitude or direction.

Work is the cumulative effect of force on spatial displacement, and the change of kinetic energy involves f, displacement.

Impulse is the cumulative effect of force over time, and the change in momentum involves f, time.

These two concepts can describe the same object, the same process, the same motion in different ways, but from different angles, sometimes both methods can be used, and sometimes only one or the other can be used. Such as collision problems (momentum theorem, conservation of momentum).

Variable speed curve motion (kinetic energy theorem, conservation of mechanical energy).

After 6s, the ascent height h=420

m) shell velocity v = 40

m s) momentum p=2*mv=80m

direction upwards clear defeat).

Since the last piece of the light finally fell to the launch site, there was no horizontal component.

The first piece of ** to land gets v=8

m s) (direction upward).

Impulse i = 32m (direction down).

The momentum of the other excitation block is p=40m+32m=72m), and the velocity is v=pm=72

m s) (direction upward).

t=(36+2*849^

s) The answer is a bit weird, please check it yourself.

There is a question with the answer to this question!

When B's velocity is zero, if the ball is in A's hand, it is known by the conservation of momentum that A's velocity must be zero.

When B's velocity is zero, if the ball happens to be in the air, A's velocity must not be zero, otherwise momentum is not conserved.

When B's velocity is zero, there is no problem if the ball is in B's hand. Imagine that the velocity becomes zero after A throws once, and the velocity of B catches the ball with zero speed.

The answer to the question is incorrect, and the final position of the ball is uncertain.

The object of study is set as two people and a ball, and the total momentum is conserved, which can be solved in one question.

When the two complete a whole number of two-pass cycles, the ball can be in the first hand and the two are stationary.

Or when the last time the ball is passed from hand A, A only needs to control the power to be stationary after throwing the ball, the momentum in the two directions in the system is only the ball and B, and it is always equal, and B must be stationary after receiving the ball. The problem is solved.

If the ball is in B's hand, then where does the momentum of the ball from A to B come from? Then it is impossible for A to change its velocity to 0

I think it's okay, if A stops after the first pass, then B should also stop after receiving the ball.

"Passing the ball to each other while traveling, when B's velocity happens to be zero," is understood to mean that B's velocity is 0 first, and when the ball is in the air, A's velocity is not 0, and A's velocity is 0 after receiving the ball.

Consider B and "A and the ball (as C)" as a system because the resultant external force is zero, i.e., the momentum of the system is conserved. Because the initial state C and B have equal mass, and the initial velocity is equal in magnitude and opposite directions, the initial momentum is zero, so when B velocity is zero, C velocity is zero.

I think this should be possible, let the mass of A be m1, B is m2, and the ball is m0, assuming that A is at rest when A throws the ball first(m1+m0)v=m0v, then v=(m1+m0)v m0 B, when B receives the ball, m0v-m2v=0, and the momentum of B and the ball are both 0

The momentum of the whole bench is conserved: in the end all the balls are combined.

mv0=(m+2m+……nm)v

v=[1/(1+2+ …The Rough Hall....n)] *v0 = 0 mechanical energy lost: e = (1 2) mv0 2

n If there is no comma base limit, then it is infinite, the mass of all the balls on the right is infinite, the momentum after collision is 0, then the velocity is 0, and all the mechanical energy is lost.

is the initial mechanical energy: (1 2)mv0 2

dSince momentum is conserved, everyone is at rest at the beginning, so the sum of the final momentum is 0

Let the mass of the person add the car be m and the mass of the ball be m, and the following relationship is obtained:

mv2+mv=mv1, so v1 > v2, and because the ball has been unable to catch up with B.

So v<=v2