劉宗儒物理段考試題（一）
【範圍】
：第一章～第五章
測驗時間：80 分鐘
第一部份：單選題（每題 7 分，共 70 分）
1. An object starts from rest at the origin and moves along the x axis with a constant
acceleration of 4 m/s2. Its average velocity as it goes from x=2 m to x=8 m is：
(A) 1 m/s
(B) 2 m/s (C) 3 m/s (D) 5 m/s (E) 6 m/s
2. A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an
angle of 45°. With what speed was it thrown? Use g=10 m/s2.
20 m
45
(A) 14 m/s
(B) 20 m/s
(C) 28 m/s
(D) 32 m/s (E) 40 m/s
3. The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of
6.0 m/s2. At an instant when the magnitude of the total acceleration is 10.0 m/s2, what is
the speed of the particle?
(A) 3.0 m/s
(B) 4.0 m/s
(C) 5.0 m/s
(D) 6.0 m/s
(E) 8.0 m/s
4. Two blocks are accelerated across a horizontal frictionless surface as shown below.
Frictional forces keep the two blocks from sliding relative to each other, and the two move
with the same acceleration. If F=1.2 N and M=1.0 kg, what is the horizontal component
(fricitonal force) of the force of the small block on the large block?
(A) 0.48 N to the right
(B) 0.48 N to the left
2M
(C) 0.72 N to the right
F
3M
(D) 0.72 N to the left
(E) 0.65 N to the right
5. A certain pendulum consists of a 1.5 kg mass swinging at the end of a string (length =2.0
m). At the lowest point in the swing the tension in the string is equal to 20 N. To what
maximum height above this lowest point will the mass rise during its oscillation? Take
g=10 m/s2.
(A) 77 cm (B) 50 cm (C) 63 cm (D) 33 cm (E) 95 cm
－1－
6. Block A, with a mass of 10 kg, rests on a 30° incline. The coefficient of kinetic friction is
0.20. The attached string is parallel to the incline and passes over a massless, frictionless
pulley at the top. Block B, with a mass of 8.0 kg, is attached to the dangling end of the
string (take 3  1.7 and g=10 m/s2). The acceleration of B is：
(A) 0.72 m/s2, up
(B) 0.72 m/s2, down
(C) 2.6 m/s2, up
(D) 2.6 m/s2, down (E) 0 m/s2
7. Which of the following values is closest to the minimum power in hp required for a 1000kg car that is climbing a 30° hill at a steady 108 km/h? Assume that the average drag force
on the car at that speed is 500 N throughout. (g=10 m/s2, 1 hp =750 W)
(A) 5
(B) 20 (C) 40 (D) 100
(E) 220
8. Suppose the velocity of a falling sphere of mass m due to the gravity and the air resistance
is given by v(t )  v (1  e t ) with the condition that the air resistance vanishes when v=0.
The formula of the air drag force is then given by
(B) m v (C) m v 2 (D) 2m v2 (E) m v2 / 2
(A) 2m v
9. Continued to the above question, v 
(A) g / 
(B) 2 g /  (C) g / 2 (D)  / g (E) 2 / g
10. A ball of mass m, at one end of a string of length L, rotates in a vertical circle just fast
enough to prevent the string from going slack at the top of the circle. Assuming mechanical
energy is conserved, the speed of the ball at the bottom of the circle is:
(A) 2gL (B) 3gL (C) 4gL (D) 5gL (E) 7gL
11. The acceleration of an object that oscillates on the x axis is a  kx , where k is a constant.
If its velocity V=15 m/s at x=0 and V=0 at x=3 m, what is the speed of the object at x=2
m?
(A) 2 5 m/s
(B) 3 5 m/s
(C) 4 5 m/s
(D) 5 5 m/s
(E) 6 5 m/s
12. The total energy E=K+V of a particle traveling under the conservative force field is a
constant of motion, where K is the kinetic energy and V is the potential energy. In a onedimensional case, say the x-axis, the conservative force is related to potential energy by
which equation?
d 2V
d 2V
(A) F   2 (B) F  2
dx
dx
(C) F  
dV
dx
－2－
(D) F 
dV
dx
(E) none of the above.
13. Which of the following best describes the direction of the acceleration of a pendulum bob
at points “a” through “e”?
(A)
(B)
e
(C)
e
a
a
d
d
b
(D)
a
d
c
b
c
e
b
c
(E)
e
e
a
d
a
b
d
b
c
c
14. An object is placed on a rough inclined plane. According to
Coulomb's law of friction, which f-θ diagram is reasonable

for the relationship between the magnitude of the friction
force (f) on an object and the angle (θ) of the inclined plane?
(A) f
(B)
0
(C)
90

0
(D)
f
0
90
0
90
f
90
f

0
90
(E) f

15. A toy car is running on a banked circular track of radius 10 m,
as shown. If the car weighs 5 kg and on wet ice, find the
maximum velocity for the car to keep on the track without
skid.
(Gravitational acceleration g=10 m/s2, cos30°=0.87, sin30°=0.5)
(A) 4.5 m/s
(B) 5.4 m/s
(C) 7.6 m/s

(D) 9.4 m/s
－3－
(E) 12.6 m/s

16. Aluminum Rod #1 has a length L and a diameter d. Aluminum Rod #2 has a length 2L and
a diameter 2d. If Rod #1 is under tension T and Rod #2 is under tension 2T, how do the
changes in length of the two rods compare according to elasticity theory?
(A) They are the same.
(B) Rod #1 has double the change in length that Rod #2 has.
(C) Rod #2 has double the change in length that Rod #1 has.
(D) Rod #1 has quadruple the change in length that Rod #2 has.
(E) Rod #2 has quadruple the change in length that Rod #1 has.
17. The total mechanical energy of a 2.00 kg particle moving along the x axis is 8.00 J. The
potential energy is given as U ( x)  0.50 x 4  4.00 x 2 (J) , with x in meters. What is the
maximum velocity of the particle?
(A) 2.0 m/s
(B) 3.0 m/s
(C) 4.0 m/s
(D) 5.0 m/s
(E) 6.0 m/s
18. A small block of mass m rests on the sloping side of
a triangular block of mass M which itself rests on a
m
horizontal table as shown in figure. Assuming all
M

surfaces are frictionless, determine the magnitude of
F
the force F that must be applied to M so that m
remains in a fixed position relative to M.
(A) mgsinθ (B) mgtanθ (C) (m+M)gtanθ (D) (m+M)gsinθ (E) None of these.
19. A vertical force applied tangentially to a uniform cylinder of
weight 100 N as shown in the figure. The coefficient of static
F
friction between the cylinder and all surfaces is 0.4. Find the
magnitude of the maximum force F that can be applied without
causing the cylinder to rotate:
(A) 12.3 N (B) 24.6 N (C) 32.6 N (D) 48.9 N (E) 57.7 N
20. A boy wants to row across a river in the shortest
possible distance. He can row at 2 m/s in still water and
the river is flowing at 1 m/s. At what angle θ should he
point the bow (front) of his boat?
(A) 30° (B) 45° (C) 53° (D) 60° (E) 90°
－4－

1m/s
劉宗儒物理段考試題（一）答案卷
得
考生姓名：_________________
單選題（每題 5 分）
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【11】
【2】
【12】
【3】
【13】
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【14】
【5】
【15】
【6】
【16】
【7】
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【8】
【18】
【9】
【19】
【10】
【20】
－5－
分
劉宗儒物理段考試題（一）解答
【範圍】
：第一章～第五章
1. Ans : (E)。
1
Sol : 由 x  x0  V0t  at 2 得：
2
1
2  0  0  t   4  t12  t1  1
2
1
8  0  0  t   4  t22  t2  2
2
 t  t2  t1  1
x 8  2
 Vav 

 6 (m/s)
t
1
2. Ans : (B)。
Sol : ∵ ax  0
∴ Vx  V0
∵ ay  g  10 m/s2
∴由 Vy2  V02y  2ay 得： Vy2  02  2 10  20  Vy  20 (m/s)
∴ Vx  Vy  V0  20 (m/s)
∵落地成 45°角
3. Ans : (B)。
Sol : 依題意知： at  6.0 (m/s2)
故得： an 
由 an 
a  at2  10.02  6.02  8.0(m/s 2)
2
v2
v2
得： 8.0 
 v  4.0 (m/s)
r
2.0
4. Ans : (B)。
Sol : 考慮整體：
由 F  ma 得： 1.2  (2.0  3.0)  a  a  0.24 (m/s2)
考慮 large block：
由 F  ma 得： f  3.0  0.24  0.72 (N)()
5. Ans : (D)。
v2
20
 v2 
2.0
3
2
1
v
20 / 3

 0.33 (m) = 33 (cm)
由「力學能守恒」得： mv 2  mgh  h 
2
2 g 2  10
Sol : 令 F  man 得： 20  1.5 g  1.5 
6. Ans : (B)。
Sol : 設 B 物之加速度為 a()
－6－
由 F  ma 得：
(B)： 8.0 g  T  8.0  a ……………………
(A)： T  5.0 g  0.20  5.0 3g  10  a ……
  a  0.72 (m/s2)( )
7. Ans : (E)。
Sol : 設引擎產生之驅動力為 F
由「沿斜面合力＝0」得： F  1000g  sin 30  500 5500(N)
車速：108 km/hr = 30 m/s
由 P=FV 得： P  5500  30 / 750  220 (hp)
8. Ans : (B)。
Sol : 令： f d  km vn
dv d
 [v (1  e t )]   v e t   (v  v)
dt dt
由 F  ma 得：
f d  mg  ma  f d  ma  mg  m (v  v)  mg  m v  mg  m v
a
 km vn  m v  mg  m v
m v  mg  0  v  g / 

 k  1
n  1

故得： f d  m v
9. Ans : (A)。
Sol : 詳如上題解中。
10. Ans : (D)。
Sol : 由「力學能守恒」得：
1 2
1
mvT  mg  2 L  mvB2  vB  vT2  4 gL
2
2
∵剛好能維持鉛直圓周運動時，最高點速率為： vT  gL
∴此時最低點速率為： vB  5 gL
11. Ans : (D)。
Sol : (1)先求 k：
由 VdV  adx 得：

0
VdV  k
15

3
xdx  k 
0
225
9
(2)再求 x=2m 處之速率 v：
由 VdV  adx 得：

V
VdV  
0
225
9
 speed v  5 5 (m/s)
－7－

2
3
xdx  V  5 5 (m/s)
12. Ans : (C)。
Sol : 一維保守力與位能之關係為： F  
dV
dx
13. Ans : (D)。
Sol : a 點、e 點為端點，速度為 0, 故無法線加速度；又重力有切線分量，故有切
線加速度。
b 點，速度最大，故法線加速度最大；又重力無切線分量存在，故無切線加
速度，
14. Ans : (A)。
Sol : 當θ小於某一臨界角時，物體靜止不動，此時： f  mg sin 
當θ大於某一臨界角時，物體靜止下滑，此時： f  k N  k mg cos
N
15. Ans : (C)。
Sol : Fn  man  N sin 30  m
v2
r
r  10m
Fy  0  N cos30  mg

sin 30
v2
0.5
v2



 v  7.6 (m/s)
cos30
gr
0.87 10  10
mg
16. Ans : (A)。
FL
得：
AY
T L
4TL
，
L1 

2
 2

d
Y
d Y
4
 L1  L2
Sol : 由 L 
L2 

4
2T  2 L
(2d ) 2  Y

4TL
 d 2Y
17. Ans : (C)。
Sol :
dU
d 2U
 2.00 x3  8.00 x
 6 . 0 x02  8 . 0 0
dx
dx 2
dU
 0  2.00 x3  8.00 x  0  x  0,  2
令：
dx
d 2U
∵
 0 ∴在 x  2 處，位能有極小值。
dx 2 x 2
故得： U min  0.50  (2)4  4.00  (2)2  8.00(J)
由「力學能守恒」得： K max  E  U min
1
2
  2.00  vmax
 (8.00)  8.00
2
 vmax  4.00 (m/s)
－8－
18. Ans : (C)。
Sol : 參考下圖：
a
a
F
y
mg

R
 N
mg
x
Mg
考慮 block：
mg
cos
由 Fx  max 得： N sin  ma  a  g tan
由 Fy  0 得： N cos  mg  N 
考慮(block+wedge)：
由 Fx  max 得： F  ( M  m) a  ( M  m) g tan
19. Ans : (C)。
Sol : 參考如右之自由體圖：
由  A  0 得：
100 N
N1  r  0.4 N1  r  0.4N2  2r  100 r  0
 1.4 N1  0.8 N 2  100 …………
由 Fx  0 得： N 2  0.4 N1 ……
聯立，得：
2500
1000
N、 N 2 
N
N1 
43
43
由 Fy  0 得：
F  N1  0.4 N 2  100  F 
0.4N 2
F
A
N2
C
y
0.4N1
x
N1
1400
 32.56  32.6 (N)
43
20. Ans : (D)。
Sol : 最短距離渡河，則船之絕對速度方向應朝正對岸，如圖示。
V水  1
V船 / 水  2
V船

故得：   60
－9－