A sophomore physics momentum question, and a high school physics momentum question

The answer should be m (5gr).15mg

1.At a, the momentum theorem is determined by i=mv

From b to d, the mechanical energy is conserved: 1 2 mv 2=mg2r + 1 2 mv 2 at point d, Newton's second law: mg=mv 2 r

Solution: v= (5gr), i=m (5gr)2At a, by the momentum theorem: 2i=mv'

V'=2 (5gr), from b to d, mechanical energy is conserved: 1 2 mV' 2=mg2r+1 2 mV' 2

At point d, Newton's second law: f+mg=mv'2 r is solved: f=15mg

The first question is, when you think you can reach the highest point d, the velocity is zero. Note that the velocity is zero, and the object cannot reach point d.

1.Let the initial velocity v0 and the highest point velocity v, just to reach the highest point, then v= gr (gravity provides centripetal force).

Conservation of mechanical energy: (mv0 2) 2 = mg2r + mv 2) 2 gives v0= 5gr

Therefore i=m 5gr

2.At an impulse of 2i, v0 = 2 5gr

Initial kinetic energy = 10mgr

Final kinetic energy = (mv 2) 2

Conservation of mechanical energy: 10mgr = mg2r + mv 2) 2 to get mv 2 = 16mgr, centripetal force is 16mg pressure is 15mg I don't know what your calculation is, both times are conservation of mechanical energy, note that the gravitational potential energy corresponds to the height of 2r

Let the initial velocity of c be vo=10m s, and when it just slides to b, the velocity of c is v1, the common velocity of ab is v2, and the common velocity of c and b is v3=

When C just slides to B: m1VO=M1V1 + M2+M3) V2C just slides to B, B continues to accelerate, A final velocity is V2, and finally CB moves together:

m1v1+m3v2=(m1+m3)v3

The joint solution yields: v1= v2=

final velocity of a: v2=

The velocity of the iron block when it just slides on b: v1=

There is an idea, as follows:

Calculate the height that the block can reach after the first rebound and the momentum a of the first collision.

Compare this height to the total height and set it to k, then the total impulse of the four times should be a*(1+k+k*k+k*k*k*k).

I haven't done it for a long time, try to use the conservation of energy to see it as a whole, and frictional heat generation to see it as internal energy.

There is a misconception that you know:

The mass is different, but the initial kinetic energy is the same, indicating that the velocity is different.

The mass is large and the speed is small.

Second, the action time is the same but the acceleration is different, and the distance traveled is different, so the work of the force is also different.

The big ball runs slowly, so the action distance is short and the work is less.

The ball runs fast, the distance is long, and the work is big, and you just got the problem reversed!

In this way, p1> p2, e1

f is the same and acts for the same time, this variable of momentum should be the same, right.

But their initial momentum is the same, but their initial momentum is not the same.

The square of momentum is equal to the square of the mass multiplied by the square of the velocity, and the deformation is 2 times the mass multiplied by the kinetic energy. The initial kinetic energy of m1 and m2 is the same, that is, the initial momentum m1 is greater than m2, and the change is the same, then p1 > p2, due to v1< v2 (initial) and a1

Let's explain the momentum first From the momentum theorem ft = p we can know that the increment of the two momentum is the same, so it is enough to judge the initial momentum, and from the equality of kinetic energy, we can know that the initial momentum must be m1 large, so p1 > p2

Let's explain kinetic energy again At the beginning, the kinetic energy is equal m1>m2 v2 can be seen v2>v1 The same horizontal force acts for the same time, m1 acceleration can be known.

a1 is greater than m2 acceleration a2 , and the velocity after force f action must v11< v22 is based on the kinetic energy theorem, the initial kinetic energy is the same, so it is enough to judge the increase in kinetic energy, fs1=ft(v1+v11)*(1 2) (here you use an average velocity to find the displacement, you can ask your teacher, the computer is not convenient to type. In the same way, fs2 can be compared. It's written more concisely, I hope you can understand.

The amount of change in momentum is the same, but at the beginning m1 and m2 are the same kinetic energy. So 1 2m1v1 2=1 2m2v2 2, because m1 > m2, so v1m2v2, so the initial momentum of m1 is greater than the initial momentum of m2. Add the same amount of change, then p1 > p2

Since the initial velocity of m1 is less than m2, and because f is the same, m1 > m2, then a1< a2, so in the same time, the distance s1 traveled by m1 must be less than the distance s2 of m2, according to w=fs, then the kinetic energy change of m1 is smaller than that of m2, and their initial kinetic energy is the same, so e1

For the same initial kinetic energy, momentum is equal to 2mek (this should be known, it can be deduced), that is, the momentum with large mass is large, and because the same force acts for the same time, the momentum change is the same, so in line with the result, the kinetic energy is equal to the square of the momentum divided by twice the mass, and it can be launched.

Sorry, mistaken.

The system composed of small balls and large balls is not subject to external forces in the horizontal direction at any time, so the momentum is conserved at any time, and the direction of movement of the ball is positive, and the average velocity of the large ball is v and the average velocity of the small ball is v', and the duration of this process is t

There are 2m*v-m*v'=0, the solution is 2v=v', so 2vt=v't, that is, the horizontal displacement of the small ball is twice that of the large ball.

Before the movement, the ball is on the far left end of the big ball; After the movement, the ball moves to the left more than the ball in the bottom of the ball, i.e. the ball moves to the left more than the ball1 2

2r=rs large + s small |=|S big + 2s big |=3s large=r answer. The big ball moved r

I don't know the second question, probably the answer is wrong.