First, the motion of the particle (1)- linear motion

1) moving in a straight line at a uniform speed

1. average speed Vping = s/t (definition) 2. Useful inference VT2-VO2 = 2as.

3. Intermediate speed vt/2 = Vping = (vt+VO)/2 4. Final speed vt = VO+AT.

5. Intermediate position speed vs/2 = [(VO2+VT2)/2] 1/26. Displacement S = V level T = VOT+AT2/2 = vt/2t.

7. Acceleration A =(vt-Vo)/t {With Vo as the positive direction, A and Vo are in the same direction (accelerating) a>0; On the other hand, a < 0}

8. It is inferred experimentally that δs = at2 {δs is the displacement difference of continuous adjacent equal time (t)}

9. Main physical quantity and unit: initial velocity (VO): m/s; Acceleration (a): m/s2; Terminal speed (Vt): m/s; Time (t) seconds (s); Displacement (s): m; Distance: meters; Speed unit conversion:1m/s = 3.6km/h.

Note: (1) average speed is vector; (2) When the speed of the object is high, the acceleration is not necessarily high; (3)a=(Vt-Vo)/t is only a measure, not a judgment;

(4) Other related contents: particle, displacement and distance, reference system, time and time; Speed and speed. Instantaneous velocity.

2) Free falling body movement

1. Initial velocity Vo = 0.2. Final speed VT = GT 3. Falling height H = GT2/2 (calculated downward from VO position) 4. Inference Vt2=2gh.

Note: (1) Free falling body is a uniformly accelerated linear motion with zero initial velocity, which follows the law of uniformly variable linear motion;

(2) A = G = 9.8m/S2 ≈ 10m/S2 (the gravity acceleration near the equator is small, and the mountain is smaller than the flat, and the direction is vertical downward).

(3) Vertical throwing.

1. Displacement S = VOT-GT2/22. The final speed vt = VO-gt (g = 9.8m /S2 ≈10m/S2).

3. Useful inference VT2-VO2 =-2G4. Maximum lifting height hm = VO2/2g (from the throwing point)

5. Round trip time t = 2vo/g (time from throwing back to original position)

Note: (1) Full-course treatment: it is a linear motion with uniform deceleration, with positive upward direction and negative acceleration;

(2) Segmented processing: the upward motion is a linear motion with uniform deceleration, and the downward motion is a free-falling motion, which is symmetrical;

(3) The process of ascending and descending is symmetrical, for example, at the same point, the speed is equal and the direction is opposite.

Second, the motion of particles (2)-curve motion, gravity

1) flat throwing motion

1. Horizontal speed: VX = VO 2. Vertical speed: vy = GT.

3. horizontal displacement: x = vot4. Vertical displacement: y = gt2/2.

5. Exercise time t = (2 y/g) 1/2 (usually expressed as (2h/g) 1/2)

6. Closing speed vt = (vx2+vy2)1/2 = [VO2+(gt) 2]1/2.

The angle β between the closing speed direction and the horizontal plane: tgβ = vy/VX = gt/v0.

7. Joint displacement: s = (x2+y2) 1/2,

Angle α between displacement direction and horizontal plane: tgα = y/x = gt/2vo.

8. Horizontal acceleration: ax = 0；; Vertical acceleration: ay = g

Note: (1) Flat throwing motion is a curve motion with uniform change, with acceleration of g, which can usually be regarded as the synthesis of uniform linear motion in horizontal direction and free falling motion in vertical direction;

(2) The movement time is determined by the falling height h(y) and has nothing to do with the horizontal throwing speed;

(3) The relationship between θ and β is TGβ= 2tgα；; ;

(4) The time t of flat throwing is the key to solving the problem; (5) An object moving along a curve must have acceleration. When the direction of velocity and the direction of resultant force (acceleration) are not in a straight line, the object moves in a curve.

2) Uniform circular motion

1. linear velocity v = s/t = 2π r/t 2. Angular velocity ω = φ/t = 2π/t = 2π f.

3. centripetal acceleration a = v2/r = ω 2r = (2π/t) 2R4. Centripetal force f center = mv2/r = mω 2r = mr (2π/t) 2 = mω v = f.

5. Period and frequency: t = 1/f 6. Relationship between angular velocity and linear velocity: v = ω r.

7. The relationship between angular velocity and rotational speed ω = 2 π n (frequency and rotational speed have the same meaning here).

8. Main physical quantities and units: arc length (s): (m); Angle (φ): radian (rad); Frequency (f); Hz； Period (t): seconds (s); Rotation speed (n); r/s； Radius (r): meter (m); Linear speed (v): m/s; Angular velocity (ω): radians per second; Centripetal acceleration: m/s2.

Note: (1) The centripetal force can be provided by a specific force, resultant force or component force, and the direction is always perpendicular to the speed direction and points to the center of the circle;

(2) The centripetal force of an object moving in a uniform circular motion is equal to the resultant force. The centripetal force only changes the direction of the speed without changing the size of the speed, so the kinetic energy of the object remains unchanged, and the centripetal force does not do work, but the momentum is constantly changing.

3) Gravity

1. Kepler's third law: t2/r3 = k (= 4π 2/gm) {r: orbital radius, t: period, k: constant (independent of the mass of the planet, depending on the mass of the central celestial body)}.

2. Law of gravitation: f = GM1m2/R2 (g = 6.67×10-1n? M2/kg2, the direction is on their connection)

3. Gravity and gravity acceleration on celestial bodies: GMM/R2 = mg; G = GM/R2 {R: celestial radius (m), m: celestial mass (kg)}

4. Orbital velocity, angular velocity and period of the satellite: v = (GM/R)1/2; ω=(GM/R3) 1/2； T = 2π (R3/GM) 1/2 {m: mass of central celestial body}

5. The first (second and third) cosmic velocity V 1 = (G and r)1/2 = (GM/r)1/2 = 7.9 km/s; V2 = 1 1.2km/s； V3 =16.7km/s

6. Geosynchronous satellite GMM/(R+H) 2 = M4 π 2 (R+H)/T2 {H ≈ 36,000 km, H: height from the earth's surface, R: radius of the earth}

Note: (1) The centripetal force required for celestial movement is provided by gravity, and the F direction = F million;

(2) The mass density of celestial bodies can be estimated by applying the law of universal gravitation.

(3) Geosynchronous satellites can only run over the equator, and the running period is the same as the earth's rotation period;

(4) When the orbit radius of the satellite decreases, the potential energy decreases, the kinetic energy increases, the speed increases and the period decreases.

(5) The maximum circling speed and minimum launching speed of the Earth satellite are 7.9 km/s. ..

III. Force (common force, composition and decomposition of force)

(1) ordinary force

1. Gravity G = mg (vertical downward direction, G = 9.8m/S2 ≈ 10m/S2, the point of action is at the center of gravity, which is applicable to the vicinity of the earth's surface).

Hooke's law f = kx {direction is along the direction of recovery deformation, k: stiffness coefficient (N/m), x: deformation variable (m)}

3. Sliding friction force f =μFN {opposite to the relative motion direction of the object, μ: friction coefficient, FN: positive pressure (n)}

4. Static friction force 0≤f Static ≤fm (contrary to the relative motion trend of objects, fm is the maximum static friction force)

5. Gravity F = GM1m2/R2 (g = 6.67×10-11n? M2/kg2, the direction is on their connection)

6. Electrostatic force F = kq1Q2/R2 (k = 9.0×109N? M2/C2, the direction is on their connecting line)

7. electric field force f = eq (e: field strength N/C, q: electric quantity c, the electric field force applied to the positive charge is in the same direction as the field strength).

8. Ampere force f = bilsin θ (θ is the angle between b and l, when L⊥B: f = Bil, when B//L: f = 0).

9. Lorentz force f = qvbin θ (θ is the included angle between B and V, when V⊥B: f = qvb, when V//B: f = 0).

Note: (1) The stiffness coefficient k is determined by the spring itself;

(2) The friction coefficient μ has nothing to do with pressure and contact area, but is determined by the material characteristics and surface conditions of the contact surface.

(3)fm is slightly larger than μFN, which is generally considered as FM ≈ μ fn;

(4) Other related contents: static friction (magnitude and direction);

(5) Symbol and unit of physical quantity B: magnetic induction intensity (T), L: effective length (M), I: current intensity (A), V: charged particle velocity (m/s), Q: charged particle (charged body) electric quantity (C);

(6) The directions of Ampere force and Lorentz force are determined by the left-hand rule.

2) Composition and decomposition of force

1. The resultant force on the same straight line has the same direction: f = f 1+F2, and the opposite direction: f = f 1-F2 (f 1 > F2).

2. Composition of mutually angled forces:

When f = (f12+f22+2f1f2cos α)1/2 (cosine theorem) f1⊥ F2: f = (f12+f22)/kloc.

3. resultant force range: | f1-F2 |≤ f≤| f1+F2 |

4. Orthogonal decomposition of force: FX = FCOS β, FY = FSIN β (β is the included angle between the resultant force and the X axis TG β = FY/FX).

Note: The synthesis and decomposition of (1) force (vector) follows the parallelogram rule;

(2) The relationship between resultant force and components is equivalent substitution, and resultant force can be used to replace the * * * interaction of components, and vice versa;

(3) In addition to the formula method, it can also be solved by drawing method. At this time, we must choose the scale and draw strictly;

(4) When the values of F 1 and F2 are constant, the greater the included angle (α angle) of F 1 and F2, the smaller the resultant force;

(5) The combination of forces on the same straight line can be taken along the positive direction of the straight line, and the direction of forces is represented by symbols, which is simplified as algebraic operation.

Four. Dynamics (motion and force)

1. Newton's First Law of Motion (Law of Inertia): An object has inertia and always maintains a uniform linear motion state or a static state until an external force forces it to change this state.

2. Newton's second law of motion: f = ma or a = f/ma (determined by external force and consistent with the direction of external force)

3. Newton's third law of motion: f =-F' (the negative sign indicates that the directions are opposite, and f and f' interact, and the balance force is different from the reaction force. Practical application: recoil movement).

4. The balance f of * * * point force is equal to 0, which summarizes the {orthogonal decomposition method and the intersection principle of three forces}.

5. Overweight: FN>g, weightlessness: fn

6. Applicable conditions of Newton's law of motion: it is suitable for solving low-speed motion problems, for macroscopic objects, for dealing with high-speed problems, and for microscopic particles.

Note: the equilibrium state means that the object is at rest or moving in a straight line at a uniform speed, or rotating at a uniform speed.