The question of momentum and energy, the relationship between momentum and energy

Let the velocity of the block after the bullet pass through the block be v

Due to the conservation of momentum mv=mv+mv 2

This is then solved with v=

Due to the conservation of energy.

Note, e is the lost mechanical energy (complement: yes, the internal energy converted into bullets and wooden blocks, since both are warmed up).

Then v is brought into the above equation, and e= is obtained

This is the mechanical energy lost in the whole process, and I don't know if you mean this by "mechanical energy lost by bullets". If it refers to the poor kinetic energy of the bullet, it is relatively simple'=

This, momentum conservation thia!

Because when the bullet and the wooden block act, there is no external force in the horizontal direction of the system [The friction between the bullet and the wooden block is the internal force of the system. ], so momentum is conserved. mv=mv+m(v 2), v=mv 2m can be solved.

Energy is conserved, and δe is the internal energy converted into work done by frictional force. δe=

The gravitational potential energy of the bullet is not considered, only the amount of change in kinetic energy is sought.

e=△ek=mv2^2/2-mv1^2/2

The conservation of momentum is derived from Newton's second law, but it is a universal law that applies to objects at any speed.

There is no necessary relationship between energy and momentum, to talk about the relationship, you can deduce a formula yourself, which is about momentum and kinetic energy: p=mv, e=1 2 (mv square), and Simultaneous can get p=under the root sign (2me).

Calculate "momentum" from "kinetic energy" :p= (2*m*ek) calculate "kinetic energy" from "momentum": ek=p 2 (2*m) (momentum p=m*v, kinetic energy ek=(1 2)*m*v 2).

Understand the formulas of kinetic energy and momentum.

Kinetic energy = 1 2mV2 , is a scalar quantity.

Momentum = mv, which is a vector and v is directional.

If there is kinetic energy, then v is not zero, and momentum must not be zero.

The imitation body of the object grip beam with momentum must have kinetic energy, but the kinetic energy of the momentum change is not the same, for example, the momentum is changing in a uniform circular motion, and the momentum is changing all the time but the kinetic energy is unchanged.

I don't know where your words come from, I hope the wrong Duan Xian words can be corrected.

Hello. I'll give my opinion based on what you have to describe.

The coefficient of sliding friction between a and b is = friction force fab= mg= pair b kinetic energy theorem ek car = fs=1 2mv 2=6j s=s0=1m so.

v car = 12m s that is, the kinetic energy of the car is 6j when the wall car collides with the wall, the small ball gets the heat of kinetic energy added q + ek ball = 1 2mv0 2-6j = 18j When the car hits the wall after the ball hits the wall, the ball energy ek ball = 18J-FS (relative displacement) S (relative displacement) = l fs (relative displacement) = heat generated in the process =

ek ball = so v1 = root number 3 ek ball = < so the balls don't slide out of the cart they have to work together.

For the ball and the trolley, the momentum is conserved, mv1 = (m+m)v, total v=3 times, root number, 3 4 m s

The answer is a bit strange.

Question 1; A bullet [m] with a mass of 10 g is shot into a wooden block [m] with a mass of 24 g stationary on a horizontal table at a speed of 300 m s [v1] and remains in the block. After the bullet is left in the block, what is the speed of the block [v2]? If the bullet penetrates the block, the speed of the bullet is 100m s [v3], what is the speed of the block [v4]?

Let me tell you this: in the process of the bullet entering the wooden block, there is friction between it and the wooden block, and heat is generated.

So it is not satisfied with the conservation of mechanical energy, but it satisfies the conservation of momentum.

If the question is the loss of mechanical energy, or the internal energy is the square that is taken—it can be obtained.

Question 2; A sandbag with a mass of m1 is hung from the thin line, forming a single pendulum with a length L. A bullet with mass m is fired horizontally into the sandbag and remains in the sandbag, swinging with the sandbag. In order to know that the maximum declination angle of the cycloid when the sandbag swings is o, find the velocity of the bullet before it enters the sandbag.

Let me tell you this: the same, there is a loss of mechanical energy in the process of "injecting" the bullet, which is converted into internal energy; "The system of the sandbag and the bullet during the ascent of the sandbag" is conserved in the process of swinging with the sandbag.

The process of injection "satisfies the conservation of momentum mv0=(m1+m)v."

In the process of swinging with the sandbag, the square of the conservation of mechanical energy = (m1+m)gl(1-coso) is obtained from the above solution v0=(m1+m)*(square root of 2gl(1-coso)) m

You can see that this kind of problem is subdivided into the process, and the conditions that the process meets are distinguished.

Premise: If you can figure out or the question gives the friction and even the change of internal energy in the collision process, then: there is no difference between impulse and momentum, you can answer it, and the answer is the same!

So why did we learn what impulse momentum theorem, kinetic energy theorem or something? Since it's the same, why study two? This is because for the collision problem, the instantaneous high occurrence, the internal force or something is not so easy to know, and even if you know, it is basically a variable force, just ask:

Is it easy to change force to do work? It's not good for the ball, so the impulse is used to solve it sideways. That's why it's also important to learn impulse.

The teacher should have talked about it, so you have to be clear about what you are learning for!

For example: on a smooth plane, a sufficiently long plank m rests on it, and the block m is placed on it, the coefficient of friction is u, and the initial velocity of the block is v, try to find: 1, what is the common velocity of the two when they are relatively stationary?

2, How long does it take to reach the common speed?

3. Is kinetic energy conserved in this process? Why?

The question should be simple and informative, but there is one requirement: you must use the theory you have learned to solve it from several different angles! Only then will you be able to appreciate the similarities and differences between theories.

Solution 1: Use Newton's kinematics knowledge to answer! That is, we use the knowledge of: f=ma, kinematics: v=v1+at, and displacement formula s=v1t+!

Solution 2: Use momentum energy to solve the problem when finding the common velocity (you can't just use impulse or kinematics to solve it), and solution 3: use the impulse momentum theorem to solve it (again, the first two knowledge points can't be used).

If you know all three solutions, then please experience the similarities and differences between the starting point and methods of the three solutions, and summarize the advantages and disadvantages of the three solutions for this problem.

On the basis of the above, try to comprehensively string together various knowledge points and solve them comprehensively, that is, to find a better solution to this problem. This is the comprehensive application of knowledge points.

If you have done all the above steps and asked you to do the same, then what are the differences between your previous question and the one I gave? And these differences determine what is right for you.

Of course, if you don't know how to do this, that's a different story.

The total momentum of the car and trailer remains the same, the total mechanical energy of the car and trailer increases, and the mechanical energy of the car increases. , the momentum of the car increases.

Analysis: If the external forces of traction and resistance remain unchanged, the resultant external force on the system is still zero.

], so momentum is conserved.

After decoupling: the force analysis shows that the car accelerates, the trailer decelerates, and the speed of the car continues to increase, so the increase in the momentum of the car is related to the inference of whether the mechanical energy increases:

Let the traction force be f, the car resistance is f1, and the trailer is f2 before taking off; f=f1+f2, after taking off (f-f1)*s1-f2*s2=kinetic energy increment e, substitute (f1+f2-f1)s1-f2s2=e

f2（s1-s2)=e.

Since the car accelerates and the trailer slows down, the distance traveled by the car is greater than that of the trailer, i.e., S1>S2, so the mechanical energy of the system increases instead of decreasing.

The total kinetic energy, momentum are unchanged,

The total kinetic energy and momentum of the car increases, and the trailer gradually decreases.

The momentum of the system consisting of the trailer and the car is conserved, so the total momentum is constant.

The traction force is only on the car, but both cars have resistance, the original traction force is equal to the resistance of the two cars f=f1+f2, after decoupling, the resultant force of the car is f-f1, the car will accelerate the movement, the net force of the trailer is f2, the trailer slows down, the impulse of the two cars in t time is equal and the direction is opposite, so the total momentum remains unchanged.

The moment the trailer stops moving, the speed is approximately 0. The kinetic energy when the two are separated is 1 2 (m1 m2)v, according to the momentum is constant, then the velocity v2 of the car at the moment of stopping motion is [(m1 m2) m1] v, and the kinetic energy when bringing in and stopping is 1 2 m1 v2, and the total kinetic energy can be increased.

Choose A. The object of study is the small ball, the energy of the small ball is converted into elastic potential energy and gravitational potential energy, the kinetic energy of the small ball decreases, and the kinetic energy is not conserved. (Because the study is a process, not the beginning and end of the state point).

Momentum is also not conserved because momentum p=mv, velocity decreases, momentum decreases.

I don't know how to ask questions.

Mechanical energy is of course conserved because only the elastic force of the spring and the gravitational force of the ball do the work in motion. Both spring elasticity and gravity just change the mechanical energy from one to the other, so the mechanical energy is conserved.

Momentum is certainly not conserved. At the beginning of motion, the momentum of the ball is in the direction of downward direction (direction of velocity is downward), and the momentum is upward after bounce back (direction of velocity is upward). The direction of momentum is changed, so momentum changes, and momentum is not conserved. Pick B.

The requirement for conservation of momentum is that it is not acted upon by external forces and is acted upon by gravity during the fall. Kinetic energy does not seem to talk about the question of whether it is conserved or not. The gravitational potential energy and kinetic energy of the ball are the same at the highest point, so the mechanical energy is conserved.

Because the velocity of the ball changes all the time during this whole process, the kinetic energy is not conserved, and the momentum is not conserved.

In the process of contact with the spring, part of the mechanical energy of the ball is transmitted to the spring, and the mechanical energy is not conserved

All I know is that the mechanical energy doesn't change, the momentum doesn't know, and we don't have an elective.

The gun carriage is affected by the recoil force of the shell diagonally to the lower left, and this force is broken down into a vertical downward component and a horizontal to the left. For the gun carriage, it is subjected to a downward component force in the vertical direction, and at the same time is supported by the ground in the vertical direction, and these two forces are a pair of balanced forces, so the external force of the gun carriage in the vertical direction is equal to zero, and there is no acceleration in the vertical direction. So there was no weightlessness in the vertical direction of the gun carriage.

The system composed of the gun carriage and the shell is equal to zero in the horizontal direction, so the momentum of the system is conserved in the horizontal direction. But in the vertical direction, the cannonball is affected by gravity, and the momentum of the cannonball is constantly changing, and its change is equal to the product of gravity and time, so the momentum of the cannonball is not conserved in the vertical direction. It can be seen that the total momentum of the whole system is not conserved.

Let's look at the question of kinetic energy. This kind of ** type of problem can be classified as a non-fully elastic collision problem, and the kinetic energy of the system is partially lost when the system undergoes a non-completely elastic collision.

1: The force is constant, but the momentum in the vertical direction is conserved, because the ground is an ideal benchmark with infinite mass, and any large momentum change acting on the ground cannot be reflected, so the ground momentum is considered to be unchanged. The gun carriage could not create a vertical downward velocity.

2: This is a process of conversion of several energies into each other, and it is obviously changed in terms of kinetic energy, so it is not conserved.

3: To add to that, Comrade Qin Qin Ding Ding said that momentum is not conserved, which is not quite appropriate. Because momentum is conserved in any situation.

1) Although the gun carriage has a downward acceleration, the ground does not accelerate downward with the gun carriage, so there is no weightlessness in the vertical direction.

2) The kinetic energy of the system composed of the gun carriage and the shell is not conserved, because the gunpowder in the gun chamber is burned when the gun is fired, and chemical energy is converted into mechanical energy.

The momentum of the system composed of the artillery carriage and the shell is also not conserved, and the net force it receives in the vertical direction is not 0, and the ground support force is the gravity of the system.