High 2 physical momentum problem, urgent master help

v1>v2

Then v1 rises higher than v2, and obviously we can analyze from the process:

After V2 rises, it decelerates to the highest point, and then falls to the throwing point, at this time, the speed is V3, and the time is t2

After V1 rises, it first decelerates to V2 and then reaches the highest point and falls, and then accelerates to V3 (this process is the process when V2 is thrown, and the time is t2), and then accelerates to fall to the throwing point, and the whole process takes time t1

Obviously: t1>t2

I don't know what the two upstairs think?

For shuangwhywhy:

False: The same s of the two throws is correct! But shuangwhywhy is wrong to separate the ascent from the descent!

Below I will follow the idea of shuangwhywhy to send an analysis:

1.When selling with v1, it rises to the highest point h1, and it takes h1 = (1 2)(a1)(t1) 2

When falling to the throw point, it takes h1=(1 2)(a2)(t2) 2

Total time t1 = t1 + t2

2.When selling with v2, it rises to the highest point h2, and the time taken is h2=(1 2)(a3)(t3) 2

When falling to the throw point, it takes h2=(1 2)(a4)(t4) 2

Total time t2 = t3 + t4

Knowing h1>h2, a1>a3, how do you compare the size of t1 with t2?

In the same way, how do you compare the size of t3 to t4?

It is worth noting: s = (1 2)a * t 2 is only suitable for movements with uniform speed change or uniform change in acceleration, do not use it indiscriminately!

In order to better understand the interpretation of the ** I draw (below is a picture of the speed of the two processes).

img] Note: v3 is the speed at which v1 is thrown to the landing point.

v4 is the speed at which v2 throws to the landing point.

Obviously. t1=t3

t2=t2-t1

t1>t2

This process is easy to analyze, when ascending, due to gravity downward, the air resistance is downward, and the acceleration is large.

When descending, the gravitational force is downward, the air resistance is upward, and the acceleration is small.

Obviously, it takes less time to move the same distance and the acceleration is large.

So t1

v1 > v2 don't need to be considered.

According to the equation s = (1 2)a * t 2, it has nothing to do with the velocities v1 and v2.

S is the same, A is larger when ascending, and A is smaller when descending, then T1 I recommend Deadcrane1 to throw a feather up hard to try it!

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.

The idea you understand is not the same as the way you answer the question.

The first question is the law of conservation of momentum.

The second question uses the law of conservation of energy. And not the kinetic energy theorem that you understand!!

For the second question, you can also use the kinetic energy theorem that you understand, but you need to isolate the bullet. The displacement you get in that way is what you call more than l. In the process, you have to find the bullet movement displacement s, you have to use some kinematics formulas, and you have to quote acceleration.

Of course, it is not as convenient as conservation of energy. After that, the sub-resistance size is directly obtained. (The final kinetic energy minus the initial kinetic energy is the amount of change in kinetic energy, and the kinetic energy theorem is that the work done by the combined external force is equal to the amount of change in kinetic energy.)

You solidify the basics! ）

(1) Let the right direction be positive, and the velocity of the wooden block is v, and mvo=3mv+2 5mvo, v=v0 5 is obtained by the law of conservation of momentum

2) If the displacement of the plank is S, then the displacement of the bullet is S+L

Use the kinetic energy theorem for the plank: fs = 3m(v0 5) Use the kinetic energy theorem for the bullet: -f(s+l) = m(2v0 5) -mv0 f and s can be obtained from the above two equations

The formula in your answer is an intermediate process derived from the subtraction of the above two formulas. and not the primitive formula derived from the kinetic energy theorem. You don't have to memorize it. It doesn't have to be understood.

This equation is based on a system of bullets and wooden blocks. In this process, heat is generated, q=fl. According to the conservation of energy, the heat produced is converted by the reduced mechanical energy of the system.

Therefore, the kinetic energy of the original bullet minus the kinetic energy of the bullet and the kinetic energy of the block is the heat generated by this process, which is equal to FL

I don't understand that the distance traveled by the bullet should be more than L relative to the ground, so what do you say about this L.

The l here does not refer to the distance at which the bullet moves relative to the ground, but the distance at which the resistance passes! There is also the calculation of the bullet kinetic energy variable, the kinetic energy of the beginning and the end is subtracted, how can the kinetic energy of a wooden block be subtracted in the middle?

Because the initial energy of the bullet = the kinetic energy converted into the wooden block + the work done to overcome the resistance is converted into internal energy + the remaining kinetic energy of the bullet - it can be explained by the law of energy conversion and conservation! 】

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

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.

d a does not slip away from bThen the last two move together, there is mv + m (-v) search bucket = (m + m) v after.

After v = 2 m s

The observer standing on the ground sees that A is doing an acceleration for a period of time, and at this time the small plank begins to move to the left, and the final velocity is 2m s, so its speed cannot be greater than 2m chant s

The speed of jumping on the ground is v, and the momentum generated when jumping is MV, so the momentum generated when jumping on the boat is not divided between the ship and the people.

mv1+mv2=0 where v1 is negative.

mv2+(-mv1)=mv, that is, v2 < v generated by jumping on the ship, so you can't jump ashore.

I don't know if it's right, I hope it helps you hehe.

The bow of the boat is l from the river bank, and if the boat is fixed during his jump, then he can jump ashore. Now the point is that the boat is not fixed, then according to the theorem of conservation of momentum, the athlete and the ship are not subjected to force in the horizontal direction (not counting the water resistance), so the athlete and the ship are conserved in momentum. That is to say, you said 0=mv1+mv2, v1 and v2 are in opposite directions, so the boat moves backwards, and the distance from the river bank is greater than l, so the athlete can't jump.

S1=R refers to: when M1 rolls down, M3 moves horizontally to the left, M2 does not move (the horizontal plane is smooth, there is no force in the horizontal direction of M2), when M1 collides with M2, the horizontal displacement of M1 is 1 4 The horizontal radius of the arc ) S1+S3 is meaningless, learn the layered diagram, and the kinematic problem is very easy to analyze.

According to your description, I also think it's s1+s3=r, isn't that obvious.

Let there be no relative slip between M2 and M3, then.

Conservation of momentum: m1v0=(m1+m2+m3)v---1

No kinetic energy loss: (1 2) m1v0 2 = (1 2) (m1 + m2 + m3) v 2 + (1 2) kx 2---2

The condition that there is no relative slip is that the above equations can be solved by kx coupling.

v0 means that the velocity of the block m1 moving to the right should be at least m2g [(m1+m2+m3) k(m11m2+m1m3)] 1 2), and there is a relative slide between the small block m2 and the lower block m3.