Two questions about momentum, question about momentum

1. The car will not continue to move to the right, but it will swing from side to side in the balance position. Because when a person pulls the hammer to the left, the hammer has a momentum to the left, and according to the momentum, the car will move to the right; When a person hits the car with a hammer to the right, the hammer has momentum to the right, and in the same way, the car will move to the left and return to its original position. Then the person keeps repeating the above process, and the car keeps rocking from side to side.

Second, it's like a person wearing skates and pushing a strong wall. The person exerts a force f on the wall, and at the same time the right reaction force of the wall – f is applied to the person. A force analysis of the person shows that the person is only subjected to the reaction force f of the wall, so the person moves in the opposite direction.

And this system does not satisfy the condition of conservation of momentum, because whether it is a system of people alone, or when people and walls are a system, the resultant external force on them is not zero but f (when people and walls are a system: the only external force on which the system is subjected to the ground and faces the wall - f). Conservation of momentum requires that the net external force on the system be zero, so the second problem is not the conservation of momentum.

I guess I can figure it out.

1. No. When struck with a sledgehammer, the action time is short (i=ft), and the impulse gained by the object is small, so that the momentum of the object changes less.

So the flatbed will only sway from side to side, and it won't move to the right...

2. Conservation. The law of conservation of momentum applies if the system is not subject to an external force (or the system is not subject to an external force in a certain direction).

When people leave each other, the system of these two people is not affected by external forces, so it is conserved...

Categories: Education, Science, >> Learning Aid.

Problem description: The length of a small car placed on a smooth non-horizontal surface is l, and the mass is equal to mStanding a person at one end of the car, the mass of the person is equal to m, and both the person and the car remain stationary at the beginning. When a person walks from one end of the car to the other end of the car, the distance the car retreats is .

Analysis: The velocity of the person is v1, and the speed of the car is v2, which is conserved by momentum.

mv2=mv1

v1=mv2/m

Since the velocity calculated by the law of conservation of momentum is the velocity of the person relative to the car, the velocity of the person relative to the ground v=v1+v2=mv2 m+v2

Therefore, when a person walks to the other end of the car, it takes a total time t=l v=ml (mv2+mv2), so the distance of the car backwards = v2*t=ml (m+m) is selected a

WF is the work of friction, which is the heat generated. And energy is conserved, so the initial kinetic energy minus the last kinetic energy is the work done by friction. And the final kinetic energy is made up of two parts, the first part is the kinetic energy of the plank, which is 1 2mv1, and the other part is the kinetic energy of the slider, which is 1 2m(1 3v0), so both parts have to be subtracted.

So it's not plus.

Solution: Take the position of the trolley.

bai is the gravitational potential energy.

Du zero potential energy position, then the speed at which the luggage bag just falls on the DAO trolley is set to v, and the root is capacitated according to the conservation of energy

1 2mv0 +mgh = 1 2mv substituting the known amount, you can get:

v=4m/s

1) Let the speed of the car be V1 when the luggage bag and the car remain relatively stationary, according to the momentum theorem

mv=（m+m）v1

Substituting a known quantity, we can find:

v1=4/3m/s

2) The frictional force experienced by the luggage bag during the sliding process on the trolley is:

f=μmg=

The acceleration of the sliding process of the luggage bag on the trolley is:

a=f/m=80/20=4m/s

If the time it takes for the car to stand still relative to the car from the luggage bag to the small lane is t, then there is:

at=v-v1

Substituting a known quantity, we can find:

t=2/3s

Conservation of energy 1 2mv2+mgh 1 2v2 to obtain v=4m s2 conservation of momentum mv=(m+m)v to obtain v=4 3m s a mg m=4m s2

t=(v-v)/a=2/3s

Momentum is conserved. Since the momentum of the two after the collision is equal, the total amount is mv, so the velocity of the object m after the collision is, that is, the momentum of the object m after the collision is, so v is the velocity of the object m after the collision.

v=mv 2m can be obtained

And because there may be energy loss due to collisions, there is according to the conservation of energy.

Energy lost e=

We know that if the two are stuck together after the collision, that is to say, when the two have the same velocity, this collision is called a completely inelastic collision, and the energy of the loss of the by-permeability chain is the largest, then there is conservation according to the momentum.

mv= can be obtained m=

The calculation yields e=

Therefore, there is Yunsun i.e. m m 2

When the bullet is shot into M1, the action time is extremely short, M and M1 have the same velocity, the momentum is conserved, the spring is compressed, M2 accelerates, when M1 and M2 have the same velocity, the spring is compressed to the shortest, the spring begins to elongate, M2 still accelerates, when the spring elongates to the original length, the M2 velocity increases to the maximum.

1] m is instantaneously shot into m1 and reaches the same velocity v1 with the momentum conserved.

mvo=(m+m1)v1 ①

m, m1, m2 reach the same velocity is v2, the maximum length of the spring is compressed is x, and the momentum of this process is conserved: (m+m1)v1=(m+m1+m2)v2

Conservation of system energy: (m+m1)v1 = (m+m1+m2)v2 + kx

The formula V1 and V2 are represented by VO, and then substituted into the formula to solve X

x=mvo√[m2/k(m+m1)(m+m1+m2)]

2] When the spring is compressed, when the original length is restored again, the speed of m and m1 is v3, and the speed of m2 is maximum v4

Conservation of momentum: (m+m1)v1=(m+m1)v3+m2v4

Conservation of energy: (m+m1)v1 = (m+m1)v3 + m2v4

Use v1 and v3 to denote v1 and v3: v1=mvo (m+m1); v3=(mvo-m2v4)/(m+m1)

Re-substitution: Finally get.

v4=2mvo (m+m1+m2) This is the maximum speed, and the minimum speed is of course v2

The bullet injection process is extremely short, and the momentum of the system composed of bullet and wooden block is conserved, mv0=(m1+m0)v'

The system composed of m1m0 and m2 moves on the smooth horizontal plane, and the momentum of the system composed of the three is balanced, and the compression length of the spring is maximum (m1+m0)v when the speed of the three is the same'=(m1+m0+m2)v'', the kinetic energy lost in this process is converted into elastic potential energy, (1 2)kx 2=(1 2)(m1+m0)v'^2-（1/2）(m1+m0+m2)v''2, x is the maximum compression length of the spring. The minimum velocity of the block m2 relative to the ground is 0, and the momentum is conserved at the maximum velocity (m1+m0)v'=(m1+m0)v1+m2v2, （1/2）(m1+m0)v'2=(1 2)(m1+m0)v1 2+(1 2)m2v2 2, simultaneous solution v2 is the maximum speed.

The first question: bai can be seen from the meaning of the question, after the collision, the wooden block 1 and the bullet have the same velocity dao to push the spring, and the spring pushes the version of the wooden block 2The mass of the spring is not taken into account. While.

When the wood block 1 and the wood block 2 have the same velocity, the spring compresses to the shortest.

Conservation of momentum MVO=(M+M1+M2)V1

The conservation of energy 1 2MVO = 1 2 (m+m1+m2)v1 +1 2kx x is its maximum compression.

Question 2: There is conservation of momentum.

First calculate the velocity v1 when the hammer just fell 5 meters (before touching), which is obtained by v1 2=2gh.

v1 root number (2gh) root number (2*10*5) 10 ms fall time t1 is obtained by v1 g*t1.

t1 v1 g 10 10 1 sec.

Then calculate the upward velocity v2 when just touching, which is obtained by v2 2 2gh.

v2 root number (2gh) root number (2*10* m s upward movement time t2 is obtained by v2 g*t2.

t2 v2 g 2 10 seconds.

So the collision time of the hammer with the helmet is t3=t, total t1 t2 seconds.

In the process of collision, the momentum theorem is used, and the vertical upward direction is taken as the positive direction.

f－mg)*t3＝mv2－（－mv1）

The average impulse of the silver is f [(mv1 mv2) t3 ] mg

i.e. f [ 4*10+4*2) ] 4*10 520 N.

The answer I calculated doesn't seem to be the same as yours.

1 The initial state of the slider and the trolley is defeated and returned to stationary speed: 0 (common speed) That's fine.

2 At the end of the state slider is stationary relative to the car: finally it returns to B relative to the car stationary (common velocity), the key is why the speed is 0

Because at the beginning, the system composed of the trolley, spring and ball is stationary relative to the horizontal plane, and the horizontal plane is used as a reference, and the system momentum p1 = 0

Let the final system velocity be v, and the system will not be subjected to external forces in the horizontal direction, so the momentum will be conserved in the horizontal direction.

p1=p2=（m+m)v=0

The velocity of the final state is v=0

Therefore, the two are at the same speed and the speed is 0

Momentum is conserved because the first two balls are at rest and the total momentum is 0, so the total momentum of the band digging is 0 at any time, so the momentum of the big ball is opposite to the direction of the small fool.

Let the velocity of the ball be v'The large ball scatters hail as v

mv'=-2mv

And the displacement s=vt hence.

mv't=-2mvt

ms1=-2ms2

Hence s1 s2=-2

And the small ball moves with respect to the big ball so s1-s2=r and therefore s2=r 3

The landlord gives a little hard work.。。。