The first chapter describes the movement
I. Particles, reference systems and coordinate systems
1.CD2.B3.C4 cloud land shore 5. Bc6.d7.a8.2km-3km 0
East 59. C 10。 The mapping of (1) 2025152 (2) 45 east by north is slightly 1 1.
Second, time and displacement.
1.ac2.ad3.a4.bc5.bc6.c7.acabod8.60m schematic diagram
9.6mx axis is 4m, MX axis is 20m 10. C 1 1。 The distance is 900 meters and the displacement is 500 meters, 500 meters and 500 meters.
12. The distance of the center point is equal to the displacement. The distance of a point on the side is greater than the displacement 13. (1) Distance (2) Reflection on displacement.
Third, the description of the speed of movement-speed
1.CD2 . B3 . C4 . 3m/s53m/s25m/s 5.06 . ac7 . CD8 . d
9.CD 10。 ABC 1 1。 Distance 100m, average speed 0 12. Different 1463km is distance, not displacement. Measure the length of the two places from the map, and then calculate the straight-line distance from the scale of 1080 km.
13. Measure the length of the line segment between the moving distance and the vehicle length from the figure, and calculate the actual displacement as 13 5m, v ≈13504m/s = 338m/s121km/h > 80km/.
V. Description of speed change-acceleration
/kloc-0 1.C2.BD3.B4.D5 Aircraft train ball 6.9 8m/s2 vertically downward 7.d..
8.Ab9. 1 50- 1 5 10。 C11.509m/S2-6m/S2 is opposite to the initial velocity.
12.5 2m/s2 13。
Chapter one review questions
1.A2.D3.CD4.ACD5.BD6.D7.ABC 8。 D9.A 10.200m 1。 T20 ~ t1and T2 ~ t 312. Turn left 0 30 80.
(3) Down 5m(4)-35m 125m 14. Distance 80m, displacement 10m, left direction15.12m/s ≤ VB ≤ 200m/s.
Research on uniform linear motion. The relationship between speed and time of uniform linear motion
1. A BD2. D3。 ACD4。 BCD5。 C6。 B7。 Uniform acceleration straight line uniform deceleration straight line Dongdong Dongdong.
8.53-39.200m/s 2 10.72s 1 1。 (1) as shown in the figure.
(2)2m/s2(3)2m/s2, and (4) do uniform deceleration linear motion.
Thirdly, the relationship between displacement and time of uniform linear motion.
1.C2
8 . 29 . 1 10.79 s253m/s 1 1。 (1) 8m (2) 72m (3) Yes, find the "area".
12.( 1)69 4(2)2 9 km (3)429 8
Fourth, the relationship between displacement and speed of uniform linear motion
1.AB2 b 3 . C4 . c 5.0 1286. 187.58. 16
9. Speed (km/h) reaction distance (m) braking distance (m) total stopping distance (m) 405 6813616 77288 710. ( 1) 25× 65438
Application of the Law of Uniform Linear Motion in Special Topics
1.D2
8. 1250m9.4250 10。 (1) t1=10st2 =15s (truncation) (2) v =1m/s (3) x.2a indicates that the acceleration direction is wrong, and b indicates that the train has stopped at the 2s terminal. ③ According to the kinematic formula v=v0+at, it can be known that the time from braking to stopping is t=v-v0a=0-30-8s=3.75s, so the displacement within 2s is x = v0t+12at2 = 30× 2m+12× (-8 )×.
Five, free fall motion
1.D2 . cd3 . C4 . bd5 . v = gtx = 12g T2 v2 = 2gx 6。 AB7.D
8.33109.1.510. Uniformly accelerated straight line 0.1930 3850 5750 7689 581.1.4
12. (1) 5m10m/s (2)15m (3) 80m/S320m (4) 70m/s250m Tip: Considering that it takes a certain time for sound propagation, the time for the stone to land is less than 8s, so the landing speed and the height of the wave crest are both. According to the altitude calculated above, the sound propagation time can be 0.9s, so the estimated landing speed and peak altitude are about V ′ = gt ′ 70m/s, and H ′ =12gt ′ 2250m13. (1) 28.3m/s (2) omitted.
Six, Galileo's study of free fall.
1. The speed at which an object falls has nothing to do with its weight. It takes force to maintain the motion of an object. 2.C3.CD4.C Tip: Although there was no instrument to accurately measure time in Galileo era, it was the most difficult to verify the relationship between v∝t (in fact, the meaning of instantaneous velocity was not clear at that time). 5.C6. Accelerate the displacement of the motorcycle in adjacent equal time intervals.
7. What is the nature of a falling body? V∝t Mathematical Reasoning x∝t2 Experiment verifies that the reasonable extrapolation of xt2 in inclined plane experiment increases with the increase of inclination angle. When the inclination angle is equal to 90, it becomes a free fall, and it is concluded that the free fall is a motion with a uniform increase in speed.
8.6s0 82m9。 (1)④(2) Omit 10. (1) Production and observation capacity (2)1592 ~1610 years (3) After several days of exploration,
Chapter II Review Questions
1.CD2 . bc3 . C4 . b5 . D6 . D7 . c8 . 4829 . 69 . 2
10. Uniform deceleration straight line 30-14001.02s100012 6022 6030 000 3250 625.
13.( 1)8m/S20m(2)4 s4m/s 14。 (1) 150km/h overspeed (2) You can't just measure the average speed.
15.20m 16。 (1) 48m/S2 (2) 36m
Chapter III Interaction
First of all, the basic interaction of gravity
1.bd2.abc3.ad4.bd5.b6.bcd7. spring balance with direction of action point n.
8. Book desktop book force object 9.10×10-310.98601. Omit 12. Omit 13.22L 14.
Second, flexibility.
/kloc-0 1.B2.CD3.B4.BC5.CD6.D7 The mobile desktop is deformed and cannot be seen by the microscope. 8.09. The spring is stretched (or shortened) 1m, resulting in an elastic force of 500N, which almost straightens the spring (exceeding the elastic limit) by 46,636 cm13.58666666 (1) for short (2) the original length of the spring (3). Experiments show that
Third, the friction (a)
1.d2.ad3.d4.bd5.bd6.a7.b8.a9.d10.0/.4 Wood block quality.
Friction (2)
1.b2.d3.CD4.d5.b6.bcd7. (1) Static1nLeft (2) Maximum static 21n.
Left (3) slip 2N0 4(4) slip 2N(5) slip 2N8 20 left 9.μ g10.04
1 1.03 168n 12。 (1) Omitting (2) has better braking effect. Because the groove on the tread tire can discharge the moisture between the tire and the ground, so that the tire can keep good contact with the ground and generate enough friction (3) Friction is independent of the contact area, so the braking ability of tires without tread but with different widths is the same.
Special stress analysis
1 . a2 . D3 . C4 . C5 . a6 . a7 . D8 . BCD 9 . d 10。
1 1. 12. Gravity, static friction, magnetic force and supporting force; Draft 13. Sketch, 4 sketches (F inclines right and left).
Iv. resultant force (1)
1.AC2 . bc3 . B4 . C5 . B6 . a 7.6n 14n 8。 B9.6N4N
10. (1)10n (2) 0 draft 1 1. ( 1) 0 (2) 180 (3) 90.
The composition of the troops (2)
1.300 omitted 2. D3.D4.C5.8666.CD7.ABD8 Vertical upward 20
The direction of 1039. Omitted 10. 102N is in the third quadrant, making an angle of 45 with the negative direction of the X axis.
Five, the decomposition of force
1.abc2.b3.ad4.d5.gsinθ GCOS θ 6.507.b8.slightly 9.320N
10.( 1) Decomposition diagram of force (2) Omit 1 1. (1) In order to show the function of force more clearly, (2) it is explained that the gravity of the cylinder has two functions, one is to press the inclined plane and the other is to press the baffle. (3) When the included angle increases, the deformation (pressure) of the soft foam on the inclined plane increases, and so does the deformation (pressure) of the soft foam on the baffle.
Chapter III Review Questions
1.B2 . bc3 . a4 . b5 . a6 . D7 . b8 . d9 . BD 10。 advertisement
1 1.A 12。 Static 30N sliding 40n13.833N43N14.20n15. (1) omitted.
(2)2003n 16.300n/m 0375 17。 (1) Omit (2)
Chapter IV Newton's Law of Motion
First of all, Newton's first law
1 . C2 . bc3 . D4 . D5 . a6 . 23 142 1347.8 . CD
9. This idea is not feasible. Because everything on the earth (including the atmosphere around the earth) rotates with the earth, after the balloon is lifted off, it still maintains its original rotation speed due to inertia. When the influence of other relative movements with the earth (such as wind) is ignored, the hot air balloon launched is in a relatively static state with the ground below, so it is impossible to make it orbit the earth.
Third, Newton's second law.
1.2 or above. AB3.0 05m/s2 accelerated evenly by 4.2 125. ① ④ ⑤ 6.Ad7.c。
8.BCD9.( 1) Stationary (2) Make uniform acceleration to the left at the acceleration of 1m/s2; (3) Make acceleration to the left first, and the acceleration will gradually increase from 0 to1m/s2; After deceleration, the acceleration gradually decreases from 1m/s2 to 0; Finally, the train moves in a straight line at a constant speed of 10. The total train mass m= 1 0× 105kg, the maximum running speed v=75m/s, the continuous traction force F= 1 57× 105N, and the train resistance FF = 0 1 mg = according to Newton's second law. Train acceleration A = F-FFM = 1.57×65433 According to the kinematic formula, t = v-v0a = 75-00.59s =127s11.745n.
12.150n ≤ f ≤180n. Hint: when F= 150N, it can move upward at a uniform speed; When F= 180N, the tension between A and B reaches the maximum.
Fourthly, mechanical unit system.
1.A2.C3 meters (m), kilograms (kg), seconds (s), amperes (a), kelvin (k), moles (mol), Candeira (cd).
4.ABC5.C6.D7.BC8.D9 is correct, because the unit of A is m/s, which is the same as the speed unit.
10.3× 103kg 1 1 . kg/m3
Fifth, Newton's third law.
/kloc-0 1.D2.D3.BC4.A5.D6 The supporting force between the two places and people on the ground, and the friction force between the ground and people on the ground are 7.c8.. Ab9。 D 10。 (1) 250n (2) The reason why children pull adults with the same force as adults pull children.
Six, using Newton's law to solve the problem (1)
1.D2.C3.B4.D5.A6.2 53 0 On the contrary.
7. As shown in the figure, the branch f =19.6 n.
F = 2.2n。
M = 2.0kg kg
a= 1. 1m/s2
μ=0.2 1
8. as shown in the figure, g = 784 n.
F branch =679N
F resistance = 20.8 newtons
A = 5.26m m/s 2
S = 38m
9. According to Newton's second law, mg tan 15 = ma, so a = gtan15 = 27m/S210. ( 1)75m/s 2 1.5× 104n。
(2) 2) The key of 27m is to control the speed.
Seven, using Newton's law to solve the problem (2)
(Lesson 65438 +0: Equilibrium Conditions of Point Force)
1.AB2. Reference binding force (external force) 3. B4。 B5。 D6。 C7。 A8。 B9。 The tension of OA rope is 17N, and that of OB rope is 8 5N 10. (1) When the force is pushed horizontally, (2) When it is pushed obliquely downward, it is pulled obliquely upward to save labor. (4) Although the horizontal component is small, the friction force is also reduced because of the upward component of the pulling force. (5) The analysis problem should be comprehensive. First, the stress analysis can help us to fully understand the problem, which direction has the minimum tension, what is the direction of the minimum tension, and what methods can be used to study this problem.
(Lesson 2: Overweight and Weightlessness)
1.d2.b3.d4.d5.c6.1250n7.0 ~ 5s, 48kg (or 480N)8. A
9.( 1) Accelerations a 1 and a2 during acceleration and deceleration during ascent and a3 and A4 during descent (2) The experimental principle is Newton's second law: f-g = ma experimental equipment is a spring scale and a heavy object; The measured data to be recorded are: the gravity g of the heavy object when the elevator is stationary, and the indexes F 1, F2, F3 and F4 of the corresponding spring scale when the elevator accelerates and decelerates up and down. According to Newton's second law, the corresponding elevator acceleration a 1, a2, a3, a4 10. 1 74 is multiplied by 1 1. (1) is in a state of complete weightlessness at this time, and the head can't feel the force of the helmet. (2) helmet mass m = gg = 3.
Chapter four review questions
1.kgmsnf = ma2.0816g unchanged 3.6438+005.b.51030001
6.ad 7 . c8 . C9 . CD 10 . b 1 1。 According to Newton's second law, F-mg=ma, so a=Fm-g, and the substitution value is a = 32m/s2 12. ( 1).
13.( 1) f = 4n, a = 1m/S2 (2) t = 1s (3) The shortest baggage transit time is 2s.
Comprehensive exercise (1)
1.ABC 2 . D3 . a4 . b5 . B6 . D7 . c8 . ad 9 . BD 10。 Alternating current
1 1.5 horizontal and right F0 12. -225 13.8025 14.0 75 120 15.-4m/S204。
16.50 2 17.( 1)43 3N25N(2)200N 18。 ( 1)a=5m/s2，t= 1 26s(2)v=6 3m/s
19. (1)17.32m/s (2)150m/S2 (3) 8000n (4) is not 20. ( 1) 130 ~ 65438.0~3s:a 1t=v0，a 1= 1 6m/s2，FN-mg=ma 1， FN=58N. weighing scale reading is 58n (2) s = s 1+S2+S3 = v0t2+v0 (t1+t2)/2 = 696m,1The actual height from floor to floor 25 is 24, and h = s24 =.
Comprehensive exercise (2)
1.ABC 2 . C3 . B4 . b5 . C6 . a7 . D8 . a9 . b 10。 C
1 1.D 12。 A
Stress at point 13. A as shown in figure 14. (1) Squat (2) BCD 15.438+0.
16.( 1) As shown in the right figure.
(2) Friction of a moving car (3)C 17. V = 50km/h > 40km/h, so the vehicle overspeed is 18. ( 1)70m/S2(2)02S(3) 1 28×6558。
(2) start the standby reducer; A≥ 1m/s2 (it indicates that it takes 2 seconds for electromagnetic wave to propagate, 3 seconds for scientists to analyze, judge and input commands, during which the detector has advanced 10m).