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【AP物理C】【真题】2001年力学解答题答案

AP Physics C: Mechanics

2001 Scoring Guidelines

employment policies are guided by that principle.

The College Board is a national nonprofit membership association dedicated to preparing, inspiring, and connecting students to college and opportunity. Founded in 1900, the association is composed of more than 3,900 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 22,000 high schools, and 3,500 colleges, through major programs and services in college admission, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT?, the PSAT/NMSQT?, the Advanced Placement Program? (AP?), and Pacesetter?. The College Board is committed to the principles of equity and excellence, and that commitment is embodied in all of its

programs, services, activities, and concerns.

trademarks of the College Entrance Examination Board.

General Notes about 2001 AP Physics Solutions

1. The solutions contain the most common method(s) of solving the free-response questions, and the

allocation of points for these solutions. Other methods of solution also receive appropriate credit for correct work.

2. Generally, double penalty for errors is avoided. For example, if an incorrect answer to part (a) is

correctly substituted into an otherwise correct solution to part (b), full credit will usually be awarded.

3. An exception to this may be cases when the numerical answer to a later part should be easily

recognized as wrong, e.g., a speed faster than the speed of light in vacuum.

4. Implicit statements of concepts normally receive credit. For example, if use of the equation

expressing a particular concept is worth one point, and a solution contains the application of the equation to the problem but does not separately list the basic equation, the point is still awarded.

Question 1

15 points total

1. (a) 6 points Distribution of Points

The average acceleration is the change in velocity divided by the time interval For correct subtraction to find the time interval 1 point 0.370.330.04s ?=?=?=f i t t t From graph: 0.22 m s υ=i

For getting i υ in the range 0.20.25 m υ<≤i 1 point From graph:0.18 m s υ=?f

For getting υf in the range 0.150.20 m s υ?≥>?f

1 point For getting υ? consistent with the student’s values of υi and υf , including subtracting in

the correct direction

1 point 0.18 m s 0.2

2 m s 0.40 m υυυ?=?=??=?f i

0.4m s

0.04s

υ??==

?a t

For correct substitution of values in the above equation 1 point

210 m =?a

For showing deceleration (e.g., with a minus sign) 1 point

Note: There were three alternate methods for solving parts (b) and (c) that could receive full credit.

Method 1.

1. (b) 3 points

For any indication of the concept of finding the area under the curve in the second graph 1 point

?=ò

p F dt or ?=p the area under the F vs. t curve

0.6 N s ??=p or kg m

0.6

??=p For correct numerical value of 0.6 1 point For correct units 1 point

1. (c) 2 points

For any statement of the correct equation for the change in momentum 1 point υ?=?p m

For correct substitution of values consistent with those obtained above

1 point 0.6 N s 1.5 kg 0.4 m s

υ??===?p m

Question 1 (cont.)

Method 2.

1. (b) 3 points Distribution of Points

Expressing the change in momentum in terms of the average force:

?=?p F t

For some method of using the graph to find the average force for the four non-zero

intervals such as indicating that the area is equivalent to 6 boxes each with a height

of 10 N, so that 60415 N ==F

1 point

()()15 N 0.04 s =0.6 N s ?

?=p or kg m

0.6 s

??=p

For correct numerical value of 0.6 1 point For correct units 1 point

1. (c) 2 points

Expressing the average force in terms of the average acceleration: =F ma

For correct equation (=F ma also accepted)

1 point For correct substitution of values consistent with those obtained above

1 point 2

15 N 1.5 kg 10 m s =

==F m a

Method 3. Student solved part (c) first and went back to part (b)

1. (c) 3 points

For some method of using the graph to find the average force for the four non-zero

intervals such as indicating that the area is equivalent to 6 boxes each with a height of 10 N, so that 60415 N ==F

1 point

For a correct expression for Newton’s second law

1 point =F ma

For correct substitution of values consistent with those obtained above

1 point 215 N 1.5 kg 10 m s

==F m a

Question 1 (cont.)

Method 3. (continued)

1. (b) 2 points Distribution of Points

()()1.5 kg 0.4 m s =0.6 N s υ??=?=p m or kg m

0.6

??=p For correct substitution of values consistent with those obtained above 1 point For correct units 1 point

1. (d) 4 points

For a correct statement of energy change 1 point ?=?f i E E E

For a kinetic energy equation

1 point 21

υ=

E m For correct substitution of values consistent with those obtained above including the

squared velocities

1 point ()()()()22111.5 kg 0.2

2 m 1.5 kg 0.18 m s 22?=

??E 0.012 J ?=E For correct units 1 point

Question 2

15 points total

2. (a) i. 3 points Distribution of Points

There were two methods generally used to solve this part.

Method 1.

=c F ma

J GM m

F R For a statement of at least one of Newton’s laws above

1 point Equating the two equations above and substituting expression for centripetal force:

J c GM m

ma R

For substituting the expression for centripetal force

1 point 22υ=J GM m

m R R

For a solution for υ that follows algebraically from previous work

1 point υ=

Method 2. 2=

J GM a R or 2

GM g R

For statement of either of the above, which are derived from Newton’s laws 1 point For a correct statement of centripetal acceleration

1 point 2

υ=c a R

Equating the two expressions above for c a and solving for υ:

υ= For a solution for υ that follows algebraically from previous work. 1 point

Two points were awarded for an approach that started with 2122υ=

=J GM m K m R

and solved for υ as long as there was no sign error in the equation and there were no incorrect statements regarding energy prior to the equation.

Question 2 (cont.)

2. (a) ii. 3 points Distribution

of Points

There were three methods generally used to solve this part.

Method 1.

2πυ==d R t T

For an expression for υ in terms of the period T

1 point For substitution of 2πR for the length d of the orbital path

1 point Solving for T gives 2πυ

For correct substitution for υ from (a) i.

1 point =T

Method 2.

22π

πω

For the equation for T in terms of ω

1 point For substitution of for υωR in the equation 1 point For correct substitution for υ from (a) i.

1 point =T

Method 3.

2222υπω??

====?÷è?J GmM m F m R m R R T R 2 points

222

4π=J

GmM mR T R

For 232

4π=J

R T GM

1 point Note: Direct use of 23

4π=J

R T GM was awarded 1 point only, if it was defined as

Kepler’s law or Law of Orbits.

Question 2 (cont.)

2. (b) 3 points Distribution of Points

T from (a) ii. or derivation of this relationship

1 point =T

For correct solution for R or 3R , numerical or symbolic, from above equation

1 point 2

24π=J GM T R or 1/3

4π??=?÷è?

J GM T R or ()()()1/3

112227426.6710 N m kg 1.910 kg 3.5510 s 4π??éù

=×××êú??

For a correct answer

1 point

81.5910 m =×R

Note: If J R was subtracted from R the answer point was only awarded if the difference

was clearly indicated to be the height of the orbit above the surface.

2. (c) i. 3 points

For stating that the orbit is an ellipse

1 point For diagram with orbit drawn completely outside the circle with point of contact only at

point P and major axis along PJ .

Partial credit of 1 point awarded for any path or orbit completely outside the circle. No points were awarded in any part of path or orbit was inside the circle. 2 points

Question 2 (cont.)

2. (c) ii. 3 points

Distribution

of Points For stating that the orbit is an ellipse 1 point For diagram with orbit drawn completely inside the circle with point of contact only at

2 points

point P and major axis along PJ.

Partial credit of 1 point awarded for any path or orbit completely inside the circle.

No points were awarded if any part of path or orbit was outside the circle.

Note: Three points may also be awarded in this part for a path in which the satellite “crashes” into Jupiter only if there is specific reference to the scale of the orbit from

part (b) and the given radius of Jupiter.

Question 3

15 points total

3. (a) 3 points Distribution

of Points For a correct formula for the rotational inertia 1 point 2=?

I mr

For a sum containing a term of the form 2mL (may include extra incorrect terms, but

the point was not awarded if the expression does not contain an 2mL term) 1 point 22=+I mL mL

For the correct answer 1 point 22=I mL

3. (b) 6 points

For a correct expression of Newton’s 2nd law 1 point

=F ma

For correct substitutions into Newton’s law 1 point 44?=mg T ma

For a correct formula for torque 1 point τα=I or Tr α=I Tr

α=I T r

From Newton’s 2nd law equation above: 44=?T mg ma

Substituting into the torque equation: 44α

=?I mg ma r

For substituting the expression for I from part (a) into Newton’s law 1 point 2244α

=?mL mg ma r

For the expression α=a r

1 point

Substituting this expression into the previous equation:

244=?mL a

mg ma r

For the correct answer

1 point 2

22=+gr a L r

Note: For the solution 44?=

mg T

a m

, obtained by solving 44?=mg T ma for a directly, a maximum of 3 points was awarded for part (b) as follows; 1 point for Newton’s law, 1 point for substitutions, and 1 point for answer.

Question 3 (cont.)

3. (c) 3 points Distribution

of Points For correctly checking the space in front of “Equal to 4mgD” 1 point For correct justification, such as “The kinetic energy gained by the two smaller blocks

2 points

comes from the decrease in the potential energy of the 4m block.” OR “Total energy

is conserved.”

Note: No points awarded for part (c) if wrong box was checked.

3. (d) 3 points

For correctly checking the space in front of “Less” 1 point For correct justification, such as “The small blocks rise and gain potential energy. The

2 points

total energy available is still 4mgD. Therefore the kinetic energy must be less than

in part (c).”

Note: No points awarded for part (d) if wrong box was checked.

大学物理力学题库及答案

一、选择题：（每题3分） 1、某质点作直线运动的运动学方程为x ＝3t -5t 3 + 6 (SI)，则该质点作 (A) 匀加速直线运动，加速度沿x 轴正方向． (B) 匀加速直线运动，加速度沿x 轴负方向． (C) 变加速直线运动，加速度沿x 轴正方向． (D) 变加速直线运动，加速度沿x 轴负方向．［ D ］ 2、一质点沿x 轴作直线运动，其v -t 曲线如图所示，如t =0时，质点位于坐标原点，则t =4.5 s 时，质点在x 轴上的位置为 (A) 5m ． (B) 2m ． (C) 0． (D) -2 m ． (E) -5 m. ［ B ］ 3、图中p 是一圆的竖直直径pc 的上端点，一质点从p 开始分别沿不同的弦无摩擦下滑时，到达各弦的下端所用的时间相比较是 (A) 到a 用的时间最短． (B) 到b 用的时间最短． (C) 到c 用的时间最短． (D) 所用时间都一样．［ D ］ 4、一质点作直线运动，某时刻的瞬时速度=v 2 m/s ，瞬时加速度2/2s m a -=，则一秒钟后质点的速度 (A) 等于零． (B) 等于-2 m/s ． (C) 等于2 m/s ． (D) 不能确定．［ D ］ 5、一质点在平面上运动，已知质点位置矢量的表示式为 j bt i at r 22+=（其中 a 、 b 为常量）, 则该质点作 (A) 匀速直线运动． (B) 变速直线运动． (C) 抛物线运动． (D)一般曲线运动．［ B ］ 6、一运动质点在某瞬时位于矢径()y x r , 的端点处, 其速度大小为 (A) t r d d (B) t r d d (C) t r d d (D) 22d d d d ?? ? ??+??? ??t y t x [ D ] 1 4.5432.52-112 t (s) v (m/s) O c b a p

【AP物理C】【真题】解答题C2004

AP? Physics C 2004 Free Response Questions The College Board is a nonprofit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 4,500 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 23,000 high schools, and 3,500 colleges, through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT?, the PSAT/NMSQT?, and the Advanced Placement Program? (AP?). The College Board is committed to the principles of equity and excellence, and that commitment is embodied in all of its programs, services, activities, and concerns. For further information, visit https://www.doczj.com/doc/3011146868.html, College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board.

大学物理力学一、二章作业答案

第一章质点运动学一、选择题 1、一质点在xoy 平面内运动，其运动方程为2 ,ct b y at x +==，式中a 、b 、c 均为常数。当运动质点的运动方向与x 轴成450角时，它的速率为[ B ]。 A ．a ； B ．a 2； C ．2c ； D ．224c a +。 2、设木块沿光滑斜面从下端开始往上滑动，然后下滑，则表示木块速度与时间关系的曲线是图1-1中的[ D ]。 3、一质点的运动方程是j t R i t R r ωωsin cos +=，R 、ω为正常数。从t ＝ωπ/到t =ωπ/2时间内该质点的路程是[ B ]。 A ．2R ； B ．R π； C ． 0； D ．ωπR 。 4、质量为0.25kg 的质点，受i t F =(N)的力作用，t =0时该质点以v =2j m/s 的速度通过坐标原点，该质点任意时刻的位置矢量是[ B ]。 A ．22 t i +2j m ； B ． j t i t 23 23+m ； C ．j t i t 343243+； D ．条件不足，无法确定。二、填空题 1、一质点沿x 轴运动，其运动方程为2 25t t x -+=（x 以米为单位，t 以秒为单位）。质点的初速度为 2m/s ，第4秒末的速度为 -6m/s ，第4秒末的加速度为 -2m/s 2 。 2、一质点以π（m/s ）的匀速率作半径为5m 的圆周运动。该质点在5s 内的平均速度的大小为 2m/s ，平均加速度的大小为 22 m /5 s π 。 3、一质点沿半径为0.1m 的圆周运动，其运动方程为2 2t +=θ（式中的θ以弧度计， t 以秒计），质点在第一秒末的速度为 0.2m/s ，切向加速度为 0.2m/s 2 。 4、一质点沿半径1m 的圆周运动，运动方程为θ=2+3t 3，其中θ以弧度计，t 以秒计。

AP物理C力学模拟卷Free Response Question

Free Response Time:45 minutes. You may refer to the Constants sheet and Equations sheet in the Appendixes. You may also use a calculator on this portion of the exam. CM 1 Two 5-kg masses are connected by a light string over two massless, frictionless pulleys. Each block sits on a frictionless inclined plane, as shown above. The blocks are released from rest. a.Determine the magnitude of the acceleration of the blocks. b.Determine the tension in the string. Now assume that the 30° incline is rough, so that the coefficient of friction between the block and the plane is 0.10. The 60° incline is s till frictionless. 1.Determine the magnitude of the acceleration of the blocks. 2.Determine the tension in the string. CM 2

大学物理考试题目及答案2

1.1下面几个质点运动学方程，哪个是匀变速直线运动？（1）x=4t-3；（2）x=-4t 3+3t 2+6；（3）x=-2t 2+8t+4；（4）x=2/t 2-4/t 。给出这个匀变速直线运动在t=3s 时的速度和加速度，并说明该时刻运动是加速的还是减速的。（x 单位为m ，t 单位为s ）解：匀变速直线运动即加速度为不等于零的常数时的运动。加速度又是位移对时间的两阶导数。于是可得（3）为匀变速直线运动。其速度和加速度表达式分别为 22484dx v t dt d x a dt ==+== t=3s 时的速度和加速度分别为v =20m/s ，a =4m/s 2。因加速度为正所以是加速的 1.3 一质点沿x 轴作直线运动，t 时刻的坐标为x = 4.5 t 2 – 2 t 3 (SI) ．试求： (1) 第2秒内的平均速度；(2)第2秒末的瞬时速度； (3) 第2秒内的路程．解：(1) 5.0/-==??t x v m/s (2) v = d x /d t = 9t - 6t 2 v (2) =-6 m/s (3) 由v =9t - 6t 2 可得：当t<1.5s 时，v>0; 当t>1.5s 时，v<0. 所以 S = |x (1.5)-x (1)| + |x (2)-x (1.5)| = 2.25 m

1.8 已知一质点作直线运动，其加速度为 a ＝4+3t 2s m -?，开始运动时，x ＝5 m ，v =0，求该质点在t ＝10s 时的速度和位置．解：∵ t t v a 34d d +== 分离变量，得 t t v d )34(d += 积分，得 12234c t t v ++= 由题知，0=t ,00=v ,∴01=c 故 22 3 4t t v += 又因为 22 34d d t t t x v +== 分离变量， t t t x d )2 34(d 2+= 积分得 232212c t t x ++= 由题知 0=t ,50=x ,∴52=c 故 52 1232++=t t x 所以s 10=t 时 m 70551021102s m 190102310432101 210=+?+?=?=?+?=-x v 2.8 一颗子弹由枪口射出时速率为10s m -?v ，当子弹在枪筒内被加速时，它所受的合力为 F =(bt a -)N(b a ,为常数)，其中t 以秒为单位： (1)假设子弹运行到枪口处合力刚好为零，试计算子弹走完枪筒全长所需时间；(2)求子弹所受的冲量；(3)求子弹的质量．解: (1)由题意，子弹到枪口时，有 0)(=-=bt a F ,得b a t = (2)子弹所受的冲量

大学物理学第二章课后答案

习题2 选择题 (1) 一质点作匀速率圆周运动时， (A)它的动量不变，对圆心的角动量也不变。 (B)它的动量不变，对圆心的角动量不断改变。 (C)它的动量不断改变，对圆心的角动量不变。 (D)它的动量不断改变，对圆心的角动量也不断改变。 [答案：C] (2) 质点系的内力可以改变 (A)系统的总质量。 (B)系统的总动量。 (C)系统的总动能。 (D)系统的总角动量。 [答案：C] (3) 对功的概念有以下几种说法： ①保守力作正功时，系统内相应的势能增加。 ②质点运动经一闭合路径，保守力对质点作的功为零。 ③作用力与反作用力大小相等、方向相反，所以两者所作功的代数和必为零。在上述说法中： (A)①、②是正确的。 (B)②、③是正确的。 (C)只有②是正确的。 (D)只有③是正确的。 [答案：C] 填空题 (1) 某质点在力i x F )54( （SI ）的作用下沿x 轴作直线运动。在从x=0移动到x=10m 的过程中，力F 所做功为。 [答案：290J ] (2) 质量为m 的物体在水平面上作直线运动，当速度为v 时仅在摩擦力作用下开始作匀减速运动，经过距离s 后速度减为零。则物体加速度的大小为，物体与水平面间的摩擦系数为。 [答案：2 2 ;22v v s gs ] (3) 在光滑的水平面内有两个物体A 和B ，已知m A =2m B 。（a ）物体A 以一定的动能E k 与静止的物体B 发生完全弹性碰撞，则碰撞后两物体的总动能为；（b ）物体A 以一定的动能E k 与静止的物体B 发生完全非弹性碰撞，则碰撞后两物体的总动能为。

[答案：2; 3 k k E E ] 在下列情况下，说明质点所受合力的特点：（1）质点作匀速直线运动；（2）质点作匀减速直线运动；（3）质点作匀速圆周运动；（4）质点作匀加速圆周运动。解：（1）所受合力为零；（2）所受合力为大小、方向均保持不变的力，其方向与运动方向相反；（3）所受合力为大小保持不变、方向不断改变总是指向圆心的力；（4）所受合力为大小和方向均不断变化的力，其切向力的方向与运动方向相同，大小恒定；法向力方向指向圆心。举例说明以下两种说法是不正确的：（1）物体受到的摩擦力的方向总是与物体的运动方向相反；（2）摩擦力总是阻碍物体运动的。解：（1）人走路时，所受地面的摩擦力与人的运动方向相同；（2）车作加速运动时，放在车上的物体受到车子对它的摩擦力，该摩擦力是引起物体相对地面运动的原因。质点系动量守恒的条件是什么？在什么情况下，即使外力不为零，也可用动量守恒定律近似求解？解：质点系动量守恒的条件是质点系所受合外力为零。当系统只受有限大小的外力作用，且作用时间很短时，有限大小外力的冲量可忽略，故也可用动量守恒定律近似求解。在经典力学中，下列哪些物理量与参考系的选取有关：质量、动量、冲量、动能、势能、功？解：在经典力学中，动量、动能、势能、功与参考系的选取有关。一细绳跨过一定滑轮，绳的一边悬有一质量为1m 的物体，另一边穿在质量为2m 的圆柱体的竖直细孔中，圆柱可沿绳子滑动．今看到绳子从圆柱细孔中加速上升，柱体相对于绳子以匀加速度a 下滑，求1m ，2m 相对于地面的加速度、绳的张力及柱体与绳子间的摩擦力(绳轻且不可伸长，滑轮的质量及轮与轴间的摩擦不计)．解：因绳不可伸长，故滑轮两边绳子的加速度均为1a ，其对于2m 则为牵连加速度，又知2m 对绳子的相对加速度为a ，故2m 对地加速度，题图由图(b)可知，为 a a a 12 ① 又因绳的质量不计，所以圆柱体受到的摩擦力f 在数值上等于绳的张力T ，由牛顿定律，

【AP物理C】【真题】解答题C2005

AP? Physics C 2005 Free response Questions These materials were produced by Educational Testing Service? (ETS?), which develops and administers the examinations of the Advanced Placement Program for the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opportunity, and their programs, services, and employment policies are guided by that principle. The College Board is a national nonprofit membership association dedicated to preparing, inspiring, and connecting students to college and opportunity. Founded in 1900, the association is composed of more than 4,200 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 22,000 high schools, and 3,500 colleges, through major programs and services in college admission, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT?, the PSAT/NMSQT?, and the Advanced Placement Program? (AP?). The College Board is committed to the principles of equity and excellence, and that commitment is embodied in all of its programs, services, activities, and concerns. Copyright ? 2005 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, SAT, and the acorn logo are registered trademarks of the College Entrance Examination Board. APIEL is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark jointly owned by the College Entrance Examination Board and the National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service.

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AP Physics C: Mechanics 2001 Free-Response Questions These materials were produced by Educational Testing Service (ETS), which develops and administers the examinations of the Advanced Placement Program for the College Board. The College Board and Educational Testing Service (ETS) are dedicated to the principle of equal opportunity, and their programs, services, and employment policies are guided by that principle. The College Board is a national nonprofit membership association dedicated to preparing, inspiring, and connecting students to college and opportunity. Founded in 1900, the association is composed of more than 3,900 schools, colleges, universities, and other educational organizations. Each year, the College Board serves over three million students and their parents, 22,000 high schools, and 3,500 colleges, through major programs and services in college admission, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT?, the PSAT/NMSQT?, the Advanced Placement Program? (AP?), and Pacesetter?. The College Board is committed to the principles of equity and excellence, and that commitment is embodied in all of its programs, services, activities, and concerns. Copyright ? 2001 by College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, and the acorn logo are registered trademarks of the College Entrance Examination Board.

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大学物理学第二习题解答

大学物理学第二习题解答 TPMK standardization office【 TPMK5AB- TPMK08- TPMK2C- TPMK18】

(8) “物体做曲线运动时，速度方向一定在运动轨道的切线方向，法向分速度恒为零，因此其法向加速度也一定为零.”这种说法正确吗？ (9) 任意平面曲线运动的加速度的方向总指向曲线凹进那一侧，为什么？ (10) 质点沿圆周运动，且速率随时间均匀增大，n a 、t a 、a 三者的大小是否随时间改变？ (11) 一个人在以恒定速度运动的火车上竖直向上抛出一石子，此石子能否落回他的手中？如果石子抛出后，火车以恒定加速度前进，结果又如何？ 1.2 一质点沿x 轴运动，坐标与时间的变化关系为224t t x -=，式中t x ,分别以 m 、s 为单位，试计算：(1)在最初s 2内的位移、平均速度和s 2末的瞬时速度； (2)s 1末到s 3末的平均加速度；(3)s 3末的瞬时加速度。解： (1) 最初s 2内的位移为为： (2)(0)000(/)x x x m s ?=-=-= 最初s 2内的平均速度为： 0 0(/)2 ave x v m s t ?= ==? t 时刻的瞬时速度为：()44dx v t t dt = =- s 2末的瞬时速度为：(2)4424/v m s =-?=- (2) s 1末到s 3末的平均加速度为：2(3)(1)804/22 ave v v v a m s t ?---= ===-? (3) s 3末的瞬时加速度为：2(44) 4(/)dv d t a m s dt dt -===-。 1.3 质点作直线运动，初速度为零，初始加速度为0a ，质点出发后，每经过τ时间，加速度均匀增加b 。求经过t 时间后，质点的速度和位移。解: 由题意知，加速度和时间的关系为 0b a a t τ =+

大学物理第二章质点动力学习题答案

习题二 2-1 质量为m 的子弹以速率0v 水平射入沙土中，设子弹所受阻力与速度反向，大小与速度成正比，比例系数为k ，忽略子弹的重力，求：(1)子弹射入沙土后，速度大小随时间的变化关系； (2)子弹射入沙土的最大深度。 [解] 设任意时刻子弹的速度为v ，子弹进入沙土的最大深度为s ，由题意知，子弹所受的阻力 f = - kv (1) 由牛顿第二定律 t v m ma f d d == 即 t v m kv d d ==- 所以 t m k v v d d -= 对等式两边积分 ??-=t v v t m k v v 0 d d 0 得 t m k v v -=0ln 因此 t m k e v v -=0 (2) 由牛顿第二定律 x v mv t x x v m t v m ma f d d d d d d d d ==== 即 x v mv kv d d =- 所以 v x m k d d =- 对上式两边积分 ??=-00 0d d v s v x m k 得到 0v s m k -=- 即 k mv s 0 = 2-2 质量为m 的小球，在水中受到的浮力为F ，当它从静止开始沉降时，受到水的粘滞阻力为f ＝kv (k 为常数)。若从沉降开始计时，试证明小球在水中竖直沉降的速率v 与时间的关系为 ??? ? ??--= -m kt e k F mg v 1 [证明] 任意时刻t 小球的受力如图所示，取向下为y 轴的正方向，开始沉降处为坐标原点。由牛顿第二定律得 t v m ma f F mg d d ==--

即 t v m ma kv F mg d d ==-- 整理得 m t kv F mg v d d =-- 对上式两边积分 ??=--t v m t kv F mg v 00 d d 得 m kt F mg kv F mg -=---ln 即 ??? ? ??--= -m kt e k F mg v 1 2-3 跳伞运动员与装备的质量共为m ，从伞塔上跳出后立即张伞，受空气的阻力与速率的平方成正比，即2kv F =。求跳伞员的运动速率v 随时间t 变化的规律和极限速率T v 。 [解] 设运动员在任一时刻的速率为v ，极限速率为T v ，当运动员受的空气阻力等于运动员及装备的重力时，速率达到极限。此时 2 T kv mg = 即 k mg v = T 有牛顿第二定律 t v m kv mg d d 2=- 整理得 m t kv mg v d d 2= - 对上式两边积分 mgk m t kv mg v t v 21d d 00 2?? =- 得 m t v k mg v k mg = +-ln 整理得 T 22221 111v e e k mg e e v kg m t kg m t kg m t kg m t +-=+-=