Find a physics master, fast momentum problem

Hello! Solution: Analyze the state of the spring when A starts moving.

Since the ground is smooth, the compression spring does work on the spring

The elastic potential energy acquired by the spring is w

After the force f is released, the spring does work on b, a is affected by the elastic force of the wall and the elastic force of the spring, the two are balanced, and a does not move.

When the spring returns to its original length (the elastic potential energy is 0), the kinetic energy obtained by b is w and the spring force on a is 0

Since B continues to move to the right, and the spring is stretched again, at this moment, A only receives the spring force to the right, and the wall has 0 elastic force on A, that is, A starts to move to the right.

1) From the above analysis, it can be obtained that when a starts to move, the spring potential energy is 0, and the kinetic energy of b is w

The kinetic energy from B to the right is W, and A, B, and Springs are considered as a whole.

The impulse i of the wall against a is equal to the momentum gained by the system (i.e., the momentum gained by b).

For B: W=1 2V2(3M).

So the momentum is v*(3m) = root number(6mw).

So the impulse i = the root number (6mw).

2) From the above analysis, it is clear that a starts to accelerate at 0 speed.

So a minimum velocity is 0

Then, analyze the movement of A to the right with B under the tension of the spring.

Since the ground is smooth, a,b forms an overall momentum that is conserved and eventually a,b reaches a common velocity v'

There is m(b)v+m(a) 0 = [m(b)+m(a)] v'

Since w=1 2v 2(3m).

So v'= root number (3w 8m).

The self-calculation may be wrong.

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1) Impulse = change of momentum, a has no change, for b, all the elastic potential energy is converted into the kinetic energy of b, w = find v, then impulse = mbv.

2) The minimum speed of A is 0, and the minimum speed of B is 1 4V

6.Solution: (1) For a system composed of m and m, the momentum and mechanical energy are conserved before and after the collision:

mv=mv'+mu <1> '^2+ <2>==>v'=[(m-m)/(m+m)]*v；u=[2m/(m+m)]*v

2) After the first collision of the two, the velocity u of m is greater than the velocity v of m'After m hits the wall, **, there is no loss of mechanical energy according to the problem, so the distance of their respective sliding is m:s+(2 3)s=5s 3;m:s 3, the time taken is the same, so the ratio of the velocities is.

u:v'=5:1 <3> substituting the above result: 2m=5(m-m)==>m:m=3:5

The conditions of this question are not very rigorous, strictly speaking, it is impossible to find out, unless the conditions are supplemented.

Solution: (1) Let the masses of the neutron and the carbon nucleus be m and m respectively, the velocity of the neutron before the collision is v0, and the velocity of the neutron and the carbon nucleus after the collision are v and v respectively According to the law of conservation of momentum, it can be obtained: mv0=mv+mv---

According to the conservation of kinetic energy, we get: (2).

Set e1, e2 ,..., en denotes the first and second ...... of neutrons, respectivelyKinetic energy after the nth collision.

(1) Since there is no energy loss in the collision and time is not counted, then a returns at the original velocity, which can be obtained by the conservation of momentum

m2 v0-m1 v0=(m1+m2)v: v=1m The direction of the s is the same as b, to the left.

2) When the velocity of a relative to the ground is 0, it is farthest away from c.

In this process, A is only subjected to the friction of B to its left in the horizontal direction.

f=μ×m2×g=15n

Then a does a uniform deceleration motion of acceleration to the left, so that there is.

a=f/m1=15m/s^2

s=（v/2）×（v/a）=2/15m

3) Assuming that it does not slip out, the energy lost by friction is:

f×l=(1/2)m1×v0^2+(1/2)m2×v0^2-(1/2)(m1+m2)v^2

v is the common velocity).

Get: l=<

So it won't slip out.