3 5 Physics Thank you, masters. Write about the answering process

Before the line is broken, the three objects move together at V1 speed, according to the conservation of momentum, there is MV0=3MV1, so there is V1 (1 3)V0, when the line is broken, the spring has impulse at both ends, so it can be regarded as the total momentum is still unchanged, let C leave the spring, the speed of AB is V2, the direction is to the right, then there are 2MV2 + MV0 MV0, V2 0, that is, when C leaves the spring, A and B are stationary. Before the collision, the kinetic energy of the system e0=(1 2)m(v0) 2;The kinetic energy of the system before the line is broken is e1 (1 2)*3m*(v1) 2=(1 6)m(v0) 2;When c leaves the spring, the kinetic energy of the system is e2 (1 2)*2m*(v2) 2+(1 2)*m*(v0) 2 (1 2)*m*(v0) 2. We can see that before the line is broken after the collision, the total kinetic energy of the system is reduced to e1 = (1 6) m (v0) 2, and after the broken wire c leaves the spring, the kinetic energy e e2 = (1 2) m (v0) 2, and the system kinetic energy returns to the initial kinetic energy, indicating that after the spring ** is broken, the kinetic energy is added to the system, and the potential energy of the spring is e, then there is e e2 e1 (1 2) m(v0) 2 (1 6) m(v0) 2=(1 3) m(v0) 2.

That is, the answer is (1 3)m(v0) 2.

Momentum is conserved, energy is conserved, and it's gone.

1.According to the momentum theorem, let the bullet velocity be v1, and the velocity of the bullet and sandbag be v2

m*v1=(m+m1)v2

v2=v1*m/(m1+m)

When the sandbag swings Hu Zhao, the energy is conserved, and the potential energy is 0 when the horizontal position is set. 1/2(m1+m)(v2)^2

g=(m1+m)*g*l(1-coso).

The graphics are fine, but it is easy to see that when the angle is O, the projection in the vertical direction is L*Coso

So. 1/2*(v2)^2=l(1-coso)=2glsin(o/2)^2

v2=2sin(o 2)*(gl)square root) into (1)v1*m (m1+m).

2sin(o 2)*(gl)square root)).

v1 = 2sin(o 2)*(gl) square root)(m1+m) m

The cross-sectional area of the solution pipe is s, p1=76-10=66cmhg, v1=30s, v2=(40*2+30)s=110s

p1v1=p2v2，p2=66•30s/110s=18cmhg76•70s=（18-1）•（l-1）•10sl=532/17+1=

Answer: Slightly.

1 All 1 is wrong, beta particles are neutrons in the nucleus that decay into a proton and an electron, so the atomic number after decay is increased by one.

2. Wrong, generally choose elements with a short half-life.

3 False, gamma rays are photon streams, not charged, no ionization, should be strong penetrating ability. 4 pairs.

First, after the nucleus emits the particle, the atomic number increases by one, because the particle is emitted by neutrons, and the neutron becomes a proton, an electron, and many neutrinos, therefore, after the nucleus decays, the number of protons is more than the original, that is, 1 is wrong.

Second, if the radiotracer atom has a long half-life, it will cause radioactive contamination for a long time, so it is not suitable, and it is necessary to choose a moderate half-life according to the actual needs, and 2 is not suitable.

Thirdly, rays are electromagnetic waves, i.e., photons, photons are not charged and do not have ionizing effect, 3 false fourth, true.

Let the bullet mass m, B mass 4m, and A mass 3m

1) Momentum is conserved, mv0 = (m+3m)v1

v1=v0/4

2) When the spring compression is maximum, 3 objects are at the same velocity.

mv0=(m+3m+4m)v2

v2=v0/8

3) Internal energy is generated during the process of the bullet embedding A.

Conservation of energy. e=

4) The maximum elastic potential energy of the spring.

It is the state of 2 hours, the three objects are isovelocity, and the energy is conserved.

e1=mv0^2/8-mv0^2/16=mv0^2/16

The AC collision process should satisfy not only the conservation of momentum but also the conservation of energy. In this problem, the two small steel balls should collide elastically with no loss of mechanical energy. In this way, the sum of the kinetic energy of the two balls at the moment after the collision is equal to the incident ball kinetic energy at the moment before the collision.

The equation for the conservation of momentum and the equation for the conservation of energy are listed to give the ac. Also keep in mind the conclusion: an elastic collision of objects of equal mass is a velocity exchange.

Also remember the formula for elastic collision in Elective 3-5: equations (1) and (2) at the bottom of page 18.

AC, because the elasticity of the steel ball is very good, the action time of the two balls is very short when they collide, so the mechanical energy is not lost after the collision, so the speed is exchanged after the collision, and the law of conservation of momentum is followed.

As to why the exchange velocity occurs: the exchange velocity generally occurs in a fully elastic collision, regardless of the energy loss. An object with mass m and velocity v0 collides with another object with a mass m and is at rest in a completely elastic collision (positive collision), and the velocity of the two is exchanged.

That is, an object with a velocity of v0 at the beginning is stationary, and an object at rest at the beginning gains a velocity v0.

AC is selected when the resistance in the air is ignored, D is selected otherwise

I don't know who AB is in front and who is behind, so option A is boring and can't be determined, when A is behind, option A is right!

The impulse is a vector quantity, and in the process of exploding, the ** force on a b is an internal force, which is equal in magnitude, opposite in direction, and equal in time. Change option b to: In the process of exploding, the impulse of the ** force received by a b must be the same!

AB does not know before and after, and cannot judge.

d must be correct.

Ask the landlord, do you know the front and back? Have you forgotten the conditions?

A. The speed of side-rush B is not combustible and must be the same as the original, and it can be reversed.

b. The impulse is related to the force and time exerted, but the force exerted on skin A and B is definitely different.

c and d have the same landing time depending on its horizontal velocity.

So just choose D

Obviously, D is chosen, and in the vertical direction A and B are only used by gravity for concealment, so the vertical and square directions are all free falling, so the time is the same!!

C, why? The fixed slippery wooden block does not slide relative to the ground, and the displacement of the bullet in the wooden block is longer than that of the ground, and the displacement of the non-fixed content is longer, we know that the initial velocity is the same, the displacement is long, the time is long, and the resistance to do more work is more, so it is not fixed for a longer time, and the internal energy of transformation is more, so choose C

Internal energy value = drag multiplied by the relative early kinetic distance of the slide (in this problem, the sliding distance of the accommodating limb is the same, which is the length of the block itself), so e1 = e2. Taking the bullet as the object of study, s=v0*t+, because s1 is not equal to s2, t1 is not equal to t2. It can be seen that choose A