Physical momentum problem, help! Sophomore physical momentum small problem Hurry online and wait!

It's a single pendulum, swinging from the lowest point to 30 degrees, swinging back, but in the opposite direction. Note that at this time, the second bullet has not yet hit the sandbag (critical), and the second formula in the analysis is a negative sign before v, and the positive direction is v1 and v2. From the conservation of mechanical energy, kinetic energy is converted into potential energy, and the position is the same, so the initial velocity is the same.

+ lb = h under column

hectares = 1 2 gt? 2 lbs = VOT-1 2gt

The two equations have vot = h t = (h g) > momentum m = ft for the positive direction, which establishes.

The momentum of the moment before the collision of a ball horse = -mgt va = -

B ball MB = 3MVO 3mgt = 0 VB = 0 > (VA, VB).

Momentum after collision is conserved on the velocity of the ball before hitting:

Momentum, MA+MB=-M GH=MVA'+3 mvb“①

Energy saving: 1 2 mV2 +1 2 mV2 = 1 2 mV2

1/2×m×va'^ 2 +1 / 2×3mvb'^ 2②

Solution, remove the redundant root VA'= 1/2√gh

So in the last ball rises the height = 1 2 (va'^ 2）/ g = 1/8×h

va'^ 2 = va'×va'empathy).

The maximum elastic potential energy is, of course, when the spring compression is the shortest, and of course m and m have the same velocity. If the two velocities are different at this time, then the spring must be in the stage of continuing to compress or elongate, which is not the moment when the elastic potential energy is the largest.

Since the problem is that the elastic potential energy of the system is the largest when the velocity of m is the smallest, then you only need to determine whether the spring is the maximum compression when the velocity of m is the smallest, that is, whether m and m have the same velocity (except in the last relative resting state).

Obviously, when m speed is the smallest, it is not the moment when the spring compresses the longest!

Analysis: The title has clearly stated that it is possible to reach the left end of the car relative to the stationary position of the car (this has ruled out many possible scenarios). Therefore, we analyze the movement of the trolley and the wooden block.

At the beginning, the wooden block began to move at the initial velocity v, so there was friction, the trolley accelerated to the right, and the wooden block decelerated to the right, until the wooden block touched the spring, as soon as it touched the spring, so because the thrust of the spring on the trolley was to the right, the acceleration of the trolley was greater than before, and in the same way, the acceleration of the deceleration motion of the wooden block was also greater (don't worry that the wooden block has been relatively stationary with the trolley before it hits the spring, because the topic has made the final state clear!). So peace of mind analysis is OK). This compression process continues until the velocity of the two is equal, when the spring compression is the largest and the elastic potential energy is the largest, but is the right velocity of the block the smallest?

No. The analysis is as follows: assuming that the spring elastic force at this time is f, according to the title, the spring will inevitably elongate after that, so at this time for the wooden block, the elastic force given by the spring to him is definitely greater than the friction force f of the car to her, f to the left, f to the right, so the net force of the wooden block at this time is to the left, but at this time, according to the conservation of momentum, it is easy to know that the speed of the wooden block is to the right, so the wooden block will continue to slow down after this! At this point, we can know :

The motion velocity of the wooden block m is the smallest, and the elastic potential energy of the system is the largest.

After that, the wooden block decelerates, the trolley continues to accelerate, and continues until the spring elastic force f is equal to the frictional force, the trolley will stop accelerating, and the wooden block will stop decelerating (this moment is the time when the speed of the wooden block is the smallest, because after this moment, the wooden block will accelerate again, and the speed will gradually increase, and at this time, it is obvious that the spring has released part of the elastic potential energy, so the elastic potential energy of the spring is not the largest at this time), and then the friction force is greater than the spring elastic force, so after this time the trolley will slow down, and the wooden block will accelerate, Continue until the speed of the block catches up with the speed of the trolley, at which point the block moves to the leftmost end of the trolley relative to the trolley.

When the elastic potential energy is the highest, it is when the spring compression is the most powerful, and the velocity of m and m is equal. The next moment, due to the elastic force, the m-velocity continues to increase and the m-velocity continues to decrease.

At the moment when the DAB ball lands, the momentum in the horizontal direction is conserved, the lateral velocity is zero and the mechanical energy is conserved, the kinetic energy of ball A is zero, and the gravitational potential energy of ball B is all converted into kinetic energy of ball B bar + WG = ek, and w bar = 0

The landing time of ball B is greater than the free fall time of ball B at the same height.

i-rod + ig=p (vector sum), i-bar ≠0

The answer is A, yes.

The isochronous circle is to make a circle with AC as the diameter, and the midpoint of AC is the center of the circle, and then as shown in the figure, I point A slides statically, and the time spent to any point on the circumference is equal, and it is easy to prove that AC takes the shortest time, so when A= is on, point B is just on the circumference, which is in line with the topic.

The center of the isochronous circle is just below the release point.

r can be adjusted, but the smallest r corresponds to the shortest time, when the isochronous circle is tangent to the inclined plane, r is the shortest, and any shorter will not fall on the inclined plane.

Of course, it is pulled in the same direction, because the object squeezes the upper plate upward, and the upper plate will have an indication. At the same time, shouldn't the upper plate give a downward reaction force to the object that is pressing upwards against him? That's going down.

Simple analysis, the total momentum to the right, only the last two ships and people are running at the same speed to the right when the person jumps out horizontally for the ground with the smallest velocity, let the velocity at this time v, let the right be positive, then according to the conservation of momentum, 1 3m*v0=7 3m*vThe process of jumping in the middle is also conserved in momentum, after the person jumps out, the speed of boat A is v to the right, boat B is v0 to the left, and the person is set to v1 to the right, then the above formula can also be written as 1 3m*v0=7 3m*v=mv1+mv-mv0. Solving the system of equations has v1=25 21*v0

Hello! When the person jumps out horizontally, the velocity for the ground is v1, the common speed of the two ships is v2, and the speed of the first ship is in the positive direction.

For A and man, momentum is conserved.

There is (m+m3)v0=mv2+mv1 3

For B and man, momentum is conserved.

There is -mv0+mv1 3=(m+m 3)v2 solution v2=v0 7 v1=v0(25 7) landlord, the person you asked about the equation is right, but the solution is wrong, hope to adopt.

The original momentum 4mv0 3-mvo=mv0 3 to find the critical problem of velocity can be seen as two ships moving in the agreed direction at exactly the same speed, then they do not collide, and the subsequent momentum is 7mv 3

Because the system (A, B, ship and man) can be regarded as only subject to gravity, then the momentum is conserved, there is mv0 3=7mv 3, and v=v0 7 is pushed, and the speed of the person at the moment of jumping out is v1, then there is 4mv0 3=mv+mv1 3

So v1 = 25v0 7

First of all, it is a critical problem, and the final result must be covelocity, you can first establish the conservation of momentum (m+m 3)v0=m 3v1+mv with people and A as the system, and then use B and man-made systems m 3v1-mv0=-(m+m 3)v to solve the problem. I hope to adopt!

Let the velocity of the jumping ship be v, after jumping the ship, the velocity of ship A is v1, and the direction is unchanged, then it is conserved by momentum:

m+m 3)v0=mv1+mv 3 solution: v1=(4v0-v) 3

After the person jumps on the B boat, he moves in the opposite direction with the B boat, and the velocity is v2, which is conserved by the momentum:

mv 3-mv0=(m+m 3)v2 solution: v2=(v-v0) 4

In order to make the two ships not collide, then v1 v2 satisfies v1 v2, that is: (4v0-v) 3 (v-v0) 4 solution: v 19v0 7, so the minimum speed that you don't want to install is: 19v0 7

The horizontal bounce rate v of the person is at least 7v0

Conservation of energy: In the whole process, only friction generates heat, which is regarded as the reduction of kinetic energy. There is also an energy loss if there is a collision, but the kinetic energy loss due to the collision is obviously not taken into account here.

The whole process does not take into account the conservation of momentum. Conservation of momentum and conservation of kinetic energy are different. The fundamental difference is:

Conservation of momentum does not have to take into account the loss of energy, or even if there is a loss of energy, it does not affect the application of the law of conservation of momentum. Conservation of kinetic energy, on the other hand, requires a clear understanding of what energy is lost. The frictional energy loss is better said, but the loss cannot be considered.

Conservation of kinetic energy cannot be used.

The second place is the same, the kinetic energy theorem "The work done by a force on an object in a process is equal to the change in kinetic energy in that process." "In the whole process, only friction is done, so you only need to look at the amount of work done by friction and the change in kinetic energy.

Frictional heat generation is where the system loses mechanical energy in addition to the mechanical energy lost by collision.

The heat generated by friction is the frictional force multiplied by the displacement of an object relative to the object in contact.

The displacement of a relative to c is the length of c. That is: Q = mg(L) After the collision, A decelerates and BC accelerates, and the kinetic energy decreased by A is greater than the kinetic energy increased by BC.

The frictional force of A to BC multiplied by the relative displacement of BC is the positive work done by friction on the BC system, which is the kinetic energy increased by the BC system.

I've actually struggled, you think, according to what you learned about the conservation of mechanical energy, why not conserve it, because friction generates heat, and when I see this kind of problem in the future, the reduction of kinetic energy is equal to friction heat.

1.With the relative velocity of the big reed, the air in front of the sail changes to v1-v2 with respect to the speed of the ship, and the time is t, and the momentum changes ps(v1-v2) t(v1-v2).

That is, PST(v1-v2) = ft So f=ps(v1-v2) 2, f=kv

This is a difficult question.

It needs to be understood with the same image, because the description is more troublesome than the uplift, just simply describe the drawing of 2 figures, one is the f-t diagram, and the other is the v-t diagram, because the ball is thrown and thrown back at the same point, the area on the x-axis of the v-t diagram is the same as the area below, and because the f=kv two graphs are similar, the area of the upper and lower parts of the f-t diagram is also equal.

Therefore, the total impulse of f (friction) is 0

So mgt=m(v1-v0).

t=(v1-v0)/g

Forehead... Can I get more points?

(m+m)v=mv1+mv2

Large momentum: (m+m)v-mv2

Small momentum: (m+m)v-mv1

Momentum is conserved and is not subject to external forces.

It's all momentum. The momentum that is large from (m+m)v=mv1+mv2: (m+m)v-mv2 The small momentum: (m+m)v-mv1

Because I don't know whether the external force does the work, this is the first judgment; When the work is done by external forces, it is a different matter...

Momentum is conserved.

m+m)v=mv1+mv2

Large momentum: (m+m)v-mv2

Small momentum: (m+m)v-mv1

Momentum is conserved and is not subject to external forces.