Physics problem of kinetic momentum mechanics, a physics problem about momentum

Explain why the kinetic energy theorem is wrong, not that the kinetic energy theorem is wrong, but that you have chosen the wrong research object, where the angular velocity of bc is the same, you can establish the proportional relationship between the linear velocities between them, and you can calculate the ratio of kinetic energy at the lowest point.

In the second step, the reduced gravitational potential energy of BC is converted into kinetic energy, and the kinetic energy and velocity of C can be calculated by distributing the reduced gravitational potential energy according to the proportion of kinetic energy that we just calculated.

It is true that the last position cannot be found, because the line is oblique.

The team will answer for you

When the trolley is not fixed.

1 Assumes that the spring bounce is instantaneous, then by the momentum theorem, mv1=-mv2, it is clear that the car and the ball move for the same time when the ball falls, therefore, s=s2=ms1 m

That is, the landing point is s1+s2=(m+m)s m

2. Assuming that the spring bounce is a process, during which the elastic force changes, and thus the acceleration of the ball and the car changes, and the magnitude of the two elastic forces is equal and the direction is opposite, so that the acceleration of the car and the ball is also in line with MA1=MA2, and the acceleration end also has MV1'=mv2', the ratio of the horizontal displacement of the ball and the car in both processes is m m, therefore, the landing point is also at the (m+m)s m position.

Situation 1: The car is fixed.

Conservation of energy: EP= MV1 EP spring potential energy, V1 ball flat tossing initial velocity.

s=v0t case 2: the car is not fixed.

Conservation of energy: EP= MV2 + MV V2 The initial velocity of the MV V2 ball, V is the velocity of the car.

Conservation of momentum: mv2=mv (v=mv2 m) gets: mv0 = mv2 +m(mv2 m) v2=v0 m (m+m).

The car moves in the opposite direction of the ball, and the velocity of the ball relative to the car v phase = v2 + v = (m + m) v2 m = v0 (m + m) m

s'=v phase, t=v0t, (m+m), m=s, (m+m), m (flat throwing height, same falling time).

1 Assumes that the spring bounce is instantaneous, then by the momentum theorem, mv1=-mv2, it is clear that the car and the ball move for the same time when the ball falls, therefore, s=s2=ms1 m

That is, the landing point is s1+s2=(m+m)s m

2. Assuming that the spring bounce is a process, during which the elastic force changes, and thus the acceleration of the ball and the car changes, and the magnitude of the two elastic forces is equal and the direction is opposite, so that the acceleration of the car and the ball is also in line with MA1=MA2, and the acceleration end also has MV1'=mv2', the ratio of the horizontal displacement of the ball and the car in both processes is m m, therefore, the landing point is also at the (m+m)s m position.

I've done this question, but I've also done it wrong. Forget now. The answer is 9 2ek

Because in the air, it can be regarded as the conservation of momentum, then mv (velocity before explosion) = -1 2mv (after explosion) + 1 2mv (after explosion), and by the title ek = 1 2mv (velocity before explosion), ek 2 = 1 2 1 2mv (after explosion), this piece of kinetic energy in front of ek = 1 2 1 2mv (after explosion), substitute into various formulas, ek = 9 2ek. LZ bring it yourself, it may be more annoying, but you can bring it out and try it.

Apply the kinetic energy theorem and the law of conservation of momentum.

Let the velocity of the shell be v, ** at the beginning, the forward part of the velocity is v1, and the backward velocity is v2, so it is known: **into two pieces of the same mass) solve: how much is it.

Law of Conservation of Momentum:

m*v=.

In the process, v, v1, and v2 can be represented by ek.

Don't consider the gravity of the rain when it is still the first time, because the rain will flow away after falling on the roof.

The rain of t collides from 5m s to zero, so there is.

f-δ mg) δ t δ mv, f δ mv δ t δ mgm s25 t, δ t is very short, δ mg is zero, f s25 m s25 t meaning: 24 hours of rainwater thickness divided by 24 hours to obtain the thickness per unit time, and then calculate the mass after δ t.

There should be a problem with this question. I believe that it is impossible for anyone to calculate the pressure on the roof of the first rain based on the conditions of the question.

1. Solution: Let the force of the bullet on B during the passage of A be F, the velocity of the bullet after passing A is V1, and the final velocity of A is V2 (also for B to be bulleted.

When hit. ), and the final velocity of b is v3, then.

For the process of the bullet passing through A, use it for A.

Momentum theorem. (f-f)t=2×v2-0

For the process of the bullet crossing A, the momentum theorem is used for the bullet: -ft=

It is used for the whole process.

Momentum is conserved. The process of the bullet from A to the time it stays in B is conserved with momentum:

The above equation 4 is obtained by the four unknowns, f, v1, v2, v3, and the four equations.

v1=500m

s,v2=6m/s,v3=,f=1800n

2. Solution: This problem can be simplified to the fact that the traction force accelerates the particles to the same speed as the spacecraft within s=10 3m, that is, f(s v)=mv-0, and brings in the known, f (1000 10000)=10 4 2 10 (-7) 10000

The solution is f=200n.

3. Solution: both.

Tie together. , because the acceleration is a, then the sum of the collected external forces is (m+m)a

After the thin wire is broken, although there is no internal force between them, all the external forces are combined.

Still. unchanged, so the momentum theorem can be applied to the whole process of a system composed of two objects:

m+m)a×

2t（mv+m×0）-（m+m）×0

The solution yields v=2(m+m)at m

4. Solution: (1) If the wooden block does not fall, the momentum conservation can be directly applied to the two after the collision, and to the left.

Positive direction. i.e. mv-mv=-(m+m)v1, so v1=(m-m)v (m+m).

2) With. Relative motion.

method. The initial velocity of the wooden block relative to the trolley is 2v, and the final velocity is 0, relative acceleration.

For -g, 0-2v=- gt, so time=2v g

All the questions are done, please identify.

1 The momentum of the system composed of wood and object is conserved, so m wood * 0 + m matter * 6 = (m + m) * v

Solving this equation yields v=1m s

2. Analyze the motion process of the object separately: the acceleration of the object a=(6-1) t, so a=

According to f friction = m object * a

f = dynamic friction factor u * (mg).

So u=take 10 and don't go.

(1) Momentum equilibrium: solution v=1m s (2) kinetic energy theorem: f=umg ft=mv-0 solution u=