The total momentum of physics is not conserved, and the exercise is conserved in a certain direction

The small ball and the circular groove are selected as the system, and the horizontal direction is not subject to external force, so the momentum in the horizontal direction is conserved, but the momentum in the vertical direction is not conserved, and gradually decreases, so the overall kinetic energy is decreasing (because the momentum in the vertical direction decreases).

The method is: mvo = (m+m)v (conservation of momentum in the horizontal direction) mvo 2 2=(m+m)v 2 2+mgh (conservation of mechanical energy) can be calculated h=

Fig! Moreover, there is no necessary connection between the conservation of momentum and the conservation of kinetic energy.

If you calculate the example of a 10m s ball colliding with the same stationary ball and sticking together after touching, here is the conservation of momentum, and the kinetic energy is not conserved, because the kinetic energy is converted into other energy.

In fact, I have always believed that conservation of kinetic energy is not easy to use at all, he has too many constraints, and the most commonly used is conservation of energy, which is the immutable truth of the universe.

The premise of momentum conservation is that the sum force in this direction is zero, and now the groove is smooth and not subject to friction, so momentum is conserved.

The premise is very important, and the formula that does not look at the premise is laborious and does not get the correct answer.

The sum force in the vertical direction is varied, very complex, and what can be determined is that it must not be zero, so it is not conserved in the vertical direction.

Hehe, because. Think of the slot and the ball as a whole (system). The system is not subjected to force in the horizontal direction.

The reason is that the ground is smooth! Then the momentum of the system is conserved in the horizontal direction, and notice that it is only in the horizontal direction. In addition, the groove surface is smooth, and the total mechanical energy is conserved.

The solution of two equations is!

Non-collision conditions: v B v A.

And the so-called is exactly a critical point, which means that it is satisfied on the left side, and it is not satisfied on the right, so this is the critical point.

For this question, v B "v A" will definitely not hit when v, v B so v B = v A, it is such a critical point. So just not colliding is talking about this kind of tipping point situation.

This is a relatively simple way to do this by means of a change of momentum.

When A pushes out the box, the velocity of the box determines the amount of change in the momentum of A and B after that, that is, the velocity of the box is the smallest, and the change in the momentum of A and B after that is also the smallest.

If they do not collide, then v B v A, and the change in velocity is caused by momentum, the change in momentum of B is 30*(2+v B), the change in momentum of A is 30*(2-v A), and the momentum change of the box is 15*(2-v B).

Add these together, i.e., the total change in momentum is: 150 + 15V B - 30V A.

From this equation, the smaller v B and the larger v A, the smaller the value of the equation, but due to the non-collision condition v B v A, the change in the momentum of the system is minimal only when v B = v A, that is, the velocity of the box is the smallest at this time, just the condition of not colliding.

Therefore, the condition for A and B not to collide is v A = v B.

Because the velocity is the same after the collision, there is a loss of mechanical energy during the collision, which is analyzed in two stages:

One. Blocks fall to collision (conserved with momentum).

Two. After the collision to the bounce (conservation of mechanical energy).

When the mass at a is m, the shed argues for containment.

Just about to collide with the speed chain laughter vo = 6gxo

Conservation by momentum.

The common velocity of the two is v1=vo2=

At this point, the spring has elastic potential energy and the block has kinetic energy. The gravitational potential energy of the block increases after reaching point o, and the kinetic energy is 0 (exactly reached).

Conservation of mechanical energy.

E bomb + 1 2 * (2m) * v1 2 = 2mgxo can represent e bomb.

When the mass changes to 2 m.

After the collision, the two have a common velocity of v1'=2vo/3=

VO is above.,Because it's hard to type, I can't write it.。

E bomb + ek = ep gravitational potential + ek end.

E bomb +1 2*3m*v1'Square = 3mgxO + EK end.

The end of the EK is the kinetic energy shared by both when they reach the O point.

The speed is the same at this point.

After the O point, due to the spring stretching, the lower block slows down, and the upper one does a straight throwing motion.

End of ek = 1 2*3m*v end 2

h=v-end 2 2g=

I calculated that the result was like this. Give points quickly.

The v0 direction is the positive direction. If the solution is positive, it is the positive direction of the year, and the negative number is the opposite of the square finches.

Conservation of momentum: m1*v0=m1*v1+m2*v2

Conservation of kinetic energy: 1 2*m1*v0 2=1 2*m1*v1 2+1 2*m2*v2 2

Bring the first formula to v1=(m1*v0-m2*v2) m1 into the second formula.

v2 of the solution = 2m1v0 (m1+m2).

Bring v2 to the first equation to solve v1, v1=(m1-m2)v0 (m1+m2).

If the velocity of m2 is v

Replace the above formula M1 with M2, M2 with M1, take the town v0 with V, V1 with V2, V1 with V2, and V1 with V2.

Hee-hee. There is a formula for the elastic forward collision of two objects:

v1'=(m1-m2)v1+2*m2*v2)/(m1+m2)

v2'=(m2-m1)v2+2*m1*v1)/(m1+m2)

The problem of physical elements (force, motion) is not all vector in the same straight line (force and fortune motion are vectors) can be decomposed by Cartesian coordinates, the principle of selecting the coordinate direction, because there is no friction in the horizontal direction, that is, there is no external force, so the horizontal momentum is conserved, and the vertical momentum is not conserved, so the horizontal and vertical, coordinate decomposition is selected, so that the source solution in the horizontal direction can be conserved by momentum. If the direction of the initial velocity of the bullet is positive, then m bullet x v bullet = m stone x v'Stone - M bullet x v'Bullet level. In the case that the angle of ** back is not known, 300ms is not helpful for solving the above equation.

And v'If the level of the stone and v' bullet is unknown, the above equation cannot be solved, and an additional condition is required, such as the conservation of energy, that is, the conservation of kinetic energy, that is, 1 2 x m bullet x v bullet square = 1 2 x m stone x v'Stone square + 1 2 x m bullet x v'Bullet squared, v'The bullet is known to be 300m s, so v'The stone can be solved, substituting the first spine equation v'The bullet level is solvable, so v'The angle between the bullet and the horizontal can be solved using the inverse cosine function.

This question may seem complicated, but it is actually a fairly basic understanding and application. If you can't think of this decomposition method and the application of momentum conservation, it means that you don't understand enough of the vector decomposition and coordinate decomposition of physics. The mechanical factors in physics can be decomposed in Cartesian coordinates, and the physical factors decomposed in any direction still follow various physical laws in that direction, just to see whether the conditions are met.

The decomposition of this problem satisfies the conservation of momentum in the horizontal direction and the conservation of momentum in the vertical direction, but still satisfies the momentum theorem, from which the impulse of the table against the stone can be calculated.

The first floor is very well analyzed, top!

1. The applicable conditions of the law of conservation of momentum.

The internal force does not change the total momentum of the system, but the external force can change the total momentum of the system, and this law can be applied in the following three cases:1The system is not subject to external forces or the vector sum of the external forces is zero; 2.

The external force of the system is much smaller than the internal force, such as the collision or instant, the external force can be ignored; 3.If the vector sum of the external force or the external force in a direction of the system is zero, or the external force is less than the internal force, the momentum in that direction is conserved (partial momentum conservation).

2. Different expressions and meanings of the law of conservation of momentum.

The total momentum p before the interaction of the system is equal to the total momentum p') after the interaction;

2.p=0 (the increment of the total momentum of the system is equal to zero); 3.p1 = - p2 (in a system of two objects, each in opposite increments).

Among them, the form of 1 is the most commonly used, and there are three common forms in practical application: m1v1 + m2v2 = m1v1' + m2v2' (suitable for a system composed of two objects moving before and after action).0=m1v1+m2v2 (applicable to the system composed of two objects at rest, such as **, recoil, etc., the rate and displacement of the two are inversely proportional to their respective masses).

m1v1+m2v2=(m1+m2)v (applicable to the case where two objects are combined or have a common velocity after action).

3. Apply the basic steps of the law of conservation of momentum to solve the problem.

1.When analyzing whether the total momentum of the interacting objects is conserved, the objects under study are usually referred to as systems. It is important to clarify which objects the system under study consists of.

2.It is necessary to analyze the forces on the objects in the system and figure out which are the forces that interact with each other within the system, that is, the internal forces; Which are the forces exerted by the objects outside the system on the objects inside the system, i.e., the external forces. On the basis of the force, according to the condition of conservation of momentum, judge whether to apply the law of conservation of momentum.

3.Clarify the interaction process under study, and determine the beginning and end states of the process, that is, the magnitude or expression of the initial and final momentum of each object in the system. Note that after selecting the positive direction of a known quantity, all known quantities in the same direction as the selected positive direction will have a positive value, and the reverse value will be negative.

4.Establish the equation of conservation of momentum, substitute the known quantity, solve the quantity to be solved, if the calculation result is positive, it means that the direction of the quantity is the same as the positive direction, if it is negative, it is opposite to the selected positive direction.

Personal cheats: Replace the momentum formula and the mechanical energy conservation formula (the amount of increase = the amount of loss) and it's OK.

One of the conditions for the conservation of momentum The system is not subject to or is subject to zero resultant external forces, and the second condition for the conservation of momentum is the second condition for the conservation of momentum In ** collisions or other scenarios, the external forces on the object are negligible.

Conservation of Momentum Condition 3 Conservation in a certain direction In the current knowledge of physics, there are only two directions of motion, and these two directions are perpendicular to each other (in these two directions, the force of motion, etc.), whether it is horizontal or vertical, as long as the two directions are perpendicular, they can be considered separately, but the horizontal and vertical direction is generally chosen, because the gravitational force of an object is always vertically downward.

So later you may encounter conservation of momentum in the horizontal direction, and the vertical direction can be ignored at this time.

Generally, problems like collisions are to be solved by conservation of momentum The condition is that in the case of not being subject to external forces, the so-called absence from external forces means that the whole system is not subject to external forces, not the forces within the system, and the forces within the system can be ignored Generally, it will be solved in combination with the conservation of energy.

Specifically, the condition for the conservation of momentum is that the system is subjected to zero or no resultant force. In the case of ** and collision, because the internal force of the system is much greater than the external force, so in these cases, we generally think that the momentum is conserved, and the conservation of momentum is not necessarily only in the horizontal direction, as long as the net force of the system in a certain direction is zero or not affected by the external force, then it can be considered that the momentum in this direction is conserved.