High 2 physics Mass m 0 5kg and an object with an unknown mass M collide and merge into one

The first 4 seconds m go to the collision m velocity is 16 4 = 4 m s after combining (4-12 seconds) the velocity is (24-16) (12-4) = 1m s listed by the conservation of momentum.

mv1=(m+m)v2 v1=4 v2=1 m=then we get m=

The magnitude of the impulse is the momentum of the change in m.

The momentum of the change in m*v2=m is m(v2-v1)=

As can be seen from the figure, the objects are moving at a uniform speed before and after the collision.

The velocity of m before the collision can be calculated as 16m 4s = 4 m s, then its momentum can be calculated as *4 = 2 kg*m s, and the collision occurs at 4 seconds, then the velocity after the collision can be calculated: (24-16) (12-4) = 1 m s

The law of conservation of momentum, which can calculate what is the total mass after a collision?

You can find out what m is, m * 1 m s is the magnitude of the impulse obtained.

The change in momentum of m = 1 - 4 =

m1=,v1=4,v2=1, according to the conservation of energy, so m is equal to, the impulse is mv=, and the momentum change is m(v2-v1)=

After the collision, the two objects move in the same direction, mv=m(v1)+m(v2), m(v1)=m(v2).

Derives (v1) = v 2

v2)=mv/(2m)

Energy Relationship Satisfaction:

m（v）2≥m（v1）2+m（v2）2

Substituting v1 and v2 into to obtain m m 3

mv=mv+mv1

mv=mv1

v1=1/2v

1/2mv^2>1/2mv1^2+1/2mv^23/4mv^2>mv^2

3/4mv^2>m^2/mv1^2

3/4mv^2>m^2/4m v^2

m/m<3

I don't know if you can understand it.

It should also be greater than 1.

Solution: The momentum is conserved when the collision in the branch is conserved, so mv mv1 mv2 and the momentum of the two are equal.

mv1＝mv2 ①

Then, when colliding, the mechanical energy can not be increased in the middle of the cherry blossoms.

Organized by simplification3

So the answer is ab

I can't copy the characters over, so please see **.

In addition to the upstairs method, there is actually an easier way to solve the problem.

Let the original momentum of m be p, and the respective momentum of m and m after the collision is p

Conservation according to momentum: p=2p

Because the title says that "the momentum of the two is exactly equal", since momentum is a vector quantity, the direction of their velocity can only be the same, and m cannot move in the opposite direction. In addition, due to the bridge tremor during the collision, the latter cannot move beyond the former, that is, the velocity of m cannot be greater than the velocity of m. There are only three scenarios.

The following are three types of situations:

1.In a fully elastic collision, the mechanical energy of the system is conserved:

p^2/2m=p^2/2m+p^2/2m ②

Simultaneous solution m m = 3:1

2.Completely inelastic collision (with the largest amount of energy loss, and the two blocks are moving together).

Then since they are stuck together in motion, their momentum is equal and their velocity is also equal (common velocity), then m m is naturally 1:1

3.Inelastic collision, at this time there is a partial loss of energy, because the amount of energy loss is uncertain, then m m has a value range.

The range is 1 and 1, 2, and 3 points, and in order for m and m to meet the requirements of the problem, their value range is: 1 m m 3

Conservation of momentum: mv = mv'+mv'';

mv' =mv''

Energy guards the bridge of Heng Xuan:

1/2mv^2 = 1/2mv'^2 +1/2mv''2 so the lack of 3 4mv 2 = mv''2 vs. (mv''Divided.

m = 3m i.e.: m:m = 1:3

The momentum is conserved in the collision and we get m*v m*v1 m*v2 2m*v1 2m*v2

If it is elastic positive collision, there is mv 2 2 = mv1 2 2 + mv2 2 2 , and m m 3 is obtained

If it is an inelastic positive collision, MV 2 2 MV1 2 2 + MV2 2 2 , m m 3

Summing up the above, we get m m 3

Conservation of momentum: mv = mv'+mv'';

mv' =mv''

Conservation of energy: 1 2mV2 = 1 2mV'^2 +1/2mv''2So3 4mv 2 = mv''2 vs. (mv''Divided.

m =3m

Assuming that the velocities of the two objects after the collision of m and m are v1 and v2 respectively, it can be known according to the known conditions, the principles of conservation of momentum and energy

mv=mv1+mv2,1 2 mv =1 2 mv1 +1 2 mv2 ,mv1=mv2,it can be deduced:

m/m =3

A simple math would be fine.

Let me tell you a rule: the relative velocity of two objects before and after the collision does not increase (for a fully elastic collision, the relative velocity does not change; General inelastic collision with reduced relative velocity).

According to the conservation of momentum, the total momentum before and after the collision is constant, and the momentum of the two is equal after the collision, so the velocity of the object of m becomes v 2, and for the object of m, according to mv 2=mu, the velocity u = mv 2m is obtained

You and v are obviously in the same direction, so according to the law of the tomb mentioned at the beginning, you can make a loss of 0 u-v 2 v

Solving this inequality gives us the range of values of m m.

1≤m/m≤3

Fully elastic collision: energy is conserved, momentum is conserved. (If the collided ball is stationary, the ball has a velocity of the same size as the collision ball, and the collision ball stops).

If two objects with mass m collide with initial velocity as v1, let the velocity after the collision be 0 and the velocity of the collided object v2

m1v1+0

0+m2v2

m1v1^2

1/2m2v2^2

v1=v2

The first 4 seconds. m to hit m

The velocity is 16 4 = 4m s

After combining to disturb Hu Qi (4-12 seconds).

The velocity is (24-16) (12-4)=1m s and is listed by the momentum guard Liang Heng.

mv1=(m+m)v2

v1=4v2=1

m=then we get m=

The magnitude of the impulse is the slow momentum of the change of m.

m*v2=The momentum of the change in m is.

m(v2-v1)=

Let me tell you a rule: the relative velocity of two objects before and after a collision does not increase (for a fully elastic collision, the relative velocity does not change; General inelastic collision with reduced relative velocity).

According to the conservation of momentum, the total momentum before and after the collision is constant, and the momentum of the two is equal after the collision, so the velocity of the object of m becomes v 2, and for the object of m, according to mv 2=mu, the velocity u = mv 2m is obtained

U and v are obviously in the same direction, so according to the law mentioned at the beginning, we can get the inequality of 0 u-v 2 v to get the range of m m.

1≤m/m≤3

An object at rest indicates that it is subjected to a balancing force.

Therefore, A is subjected to the frictional force of B, and the direction is to the left, just equilibrium the mold rent.

The action is reciprocal (Newton's third law), so b is subject to the frictional force of a, the direction is to the right, and the magnitude is f.

AB analyzes the whole and finds that the two forces F in the horizontal direction are balanced, so there is no friction between B and the horizontal plane.