國立中山大學 102 學年度普通物理(一)期中考
102.11.20
答題須知：
一、請由答題紙第六列開始填寫，並標明題號。考試時間為 19：00~21：30，20：00 以後方可交卷。
二、請詳述相關物理觀念及計算過程，記得寫上單位，建議以M.K.S制表示。遇力學題目時，請畫出力學圖解圖，未畫者將
扣分。
三、重力加速度值請統一使用g = 9.81 (m/s2)。
1. (10%) A ball is to be shot from level ground upward at an angle  (Fig. 1a). Fig. 1b shows the y
component vy of the ball’s velocity as a function of the distance x. The scaling is set by vys = 5.0
m/s and xs = 20 m. What is the launch angle ?
xs
Fig. 1
From Fig. 1(b), we have initial speed of the ball along y-direction, v0 y  5.0 m/s.
In x-direction, the ball travels in a constant speed, x  v0 x t . --- (2-1) [2%]
In y-direction, the ball is in constant acceleration motion, v y  v0 y  gt . --- (2-2) [2%]
Let (2-1) in (2-2), v y  v0 y  g
x
. [2%]
v0 x
From Fig. 1(b), we have the slope of 
g
1
, which equals to 
. [1%]
2
v0 x
 v0 x  2 g  19.6 m/s [1%]
Thus the launching angle is tan  
v0 y
v0 x
   tan 1 (
v0 y
v0 x
)  14 o. [2%]
2. (12%) Two objects, A and B, are connected by hinges to a rigid rod that
has a length L. The objects slide along perpendicular guide rails as
shown in Fig. 2. Assume object A slides to the left with a constant speed
v and 0 < θ < /2. (a) Find the velocity vB of object B as a function of the
angle θ. (b) Describe vB relative to v when object A is approaching the
origin. (Is vB always smaller than v, larger than v, or the same as v, or
does it have some other relationship?)
Fig. 2
Solution:
(a)
x  L cos  , y  L sin  [2%]
vA 
dx dx d
d
d
v

  L sin 
 v 

--- (1) [2%]
dt d dt
dt
dt L sin 
vB 
dy dy d
d

 L cos 
--- (2) [2%]
dt d dt
dt
Put (1) in (2):
v B  L cos 
v
 v cot  [2%]
L sin 
(b)
When A approaches to the origin, θ goes up to 90o ( cot  tends to zero). [2%] The result shows
v B decreases from a finite value to zero. [2%]
3. (12%) A spring (k＝200 N/m) is fixed at the top of a frictionless plane
inclined at angle θ＝40° (Fig. 3). A 1.0 kg block is projected up the
plane, from an initial position that is distance d＝0.60 m from the end
of the relaxed spring, with an initial kinetic energy of 16 J. (a) What is
the kinetic energy of the block at the instant it has compressed the
spring 0.20 m? (b) With what kinetic energy must the block be
projected up the plane if it is to stop momentarily when it has
compressed the spring by 0.40 m?
Fig. 3
令 block 在 A 點時位能為 0(即 UA = 0 J )
(a)
Considering the law of the energy conservation,
[2分]
[2分]
[2分]
(b)
Based on the law of the energy conservation,
[2 分]
4. (12%) A block of mass m = 0.5 kg is held against a massless spring which is compressed by A =
15 cm from its equilibrium position O to point P as shown in Fig. 4. Take the friction coefficient
to be μk = 0.4 and the spring constant to be k = 120 N/m. When the system is released from point
P, find:
(a) The work done by the spring on the block
until the block reaches to the point O.
(b) The work done by friction on the block
until the block reaches to the point O.
m
(c) The speed at which the block reaches to
P
O
the point O.
Fig. 4
(d) How far the block travels from the point
O before coming to a stop.
令向右為正
[2 分]
(a)
(b)
[1 分]
[2 分]
(c) Based on the law of the energy conservation,
Wsp＋ Wf =
1
mv 2 [1 分]
2
1
1.35  0.294   0.5  v 2
2
[2 分]
(d)
[1 分]
[2 分]
5. (12%) A 5.20 g bullet moving at 700 m/s strikes a 700 g wooden block at rest on a frictionless
surface. The bullet emerges, traveling in the same direction with its speed reduced to 450 m/s. (a)
What is the resulting speed of the block? (b) What is the speed of the bullet-block center of
mass?
【解】(a) 在子彈射入木塊旋即穿出之碰撞過程中,
系統之力圖如右圖所示.
系統所受外力合力之 x 分量為
零, 即
 Fx  0
dPx
0
dt
故系統總動量之 x 分量為定值
即
Px  定值
亦即系統總動量之 x 分量守恆.
Px = mv + MV = MV + mv
(A)
3
3
式中, m  5.20 10 kg 及 M  700 10 kg 分別為子彈與木塊之質量
v  700 m/s 與 V  0 分別為碰撞前子彈與木塊之速度
由於
F
x

v '  450 m/s 與未知之 V ' 分別為碰撞後子彈與木塊之速度
以上數據代入(A)式得
V '
m(v  v ')
( 5.20 103 kg)(700 m/s  450 m/s)

 1.857 m/s
M
700 103 kg
vcm 
(b) 質心之速度
m v
m
i i
i
碰撞前, 後系統質心速度之 x 分量為
vcm, x
mv
mv ' MV '


mM
mM
3
(5.20 10 kg)(700 m/s)

 5.162 m/s
(5.20  700) 103 kg
6. (14%) The three balls in the overhead view of the figure 5 below are
identical. Ball 2 and 3 touch each other and are aligned perpendicular
to the path of ball 1. The velocity of ball 1 has magnitude vo  10 m/s
and is directed at the contact point of ball 1 and 2. Find the velocities
of all three balls after the collision.
Fig. 5
三個球一模一樣. 球 1 沿著圖上的路徑,
與球碰撞之瞬間,
三個球之球心連線構成正三角形.
故    2  3  60o ,
此二角正分別是碰撞後球 2 及
球 3 之運動方向.
由於此系統除了於碰撞時
受到彼此之衝力外, 不受外力作用,
F
故
ext
F
由
ext
已知碰撞前
假設碰撞後
 P  定值
x
P
y

0
dPtotal
 0 得知, Ptotal  定值 , 即系統之總動量 Ptotal 守恆.
dt
ˆ o , v2  v3  0 ; vo  10 m/s
v1  iv
ˆ '2 cos2  ˆjv '2 sin 2 ； v '3  iv
ˆ '3 cos3  ˆjv '3 sin 3
ˆ '1 ； v '2  iv
v '1  iv
vo  v '1  v '2 cos 2  v '3 cos 3  v '1 
 定值
由於碰撞為完全彈性
0  v '2 sin  2  v '3 sin 3 
3
(v '2  v '3 )
2
1
(v '2  v '3 )
2
(3)
v '2  v '3
由(2)式得
(4)代入(1)式
(2)
1
1
1
1
m(vo ) 2  m(v '1 ) 2  m(v '2 ) 2  m(v '3 ) 2
2
2
2
2
(vo ) 2  (v '1 ) 2  (v '2 ) 2  (v '3 ) 2
(4)代入(3)式
(1)
(4)
(vo ) 2  (v '1 ) 2  2(v '2 ) 2
vo  v '1 
(5)
3
(v '2  v '3 )  v '1  3v '2
2
(6)
將(6)式移項得 v'1  v0  3v'2 ，代入(5)式
v02  (v'1 ) 2  2(v'2 ) 2  v02  3(v'2 ) 2  2 3v0 v'2 2(v'2 ) 2  v02  2 3v0 v'2 5(v'2 ) 2
5(v'2 ) 2  2 3v0 v'2  0 ， v'2 (5v'2 2 3v0 )  0 ， v'2  0 or
v'2  v'3  4 3 m/s ； v1 = 10－12= 2 m/s
2
3v0
5
7. (12%)A sphere, a cylinder, and a shell, all with the same radius R and same mass M, roll down
along an inclined plane at angle θstarting at a height H from rest. Find in each case their
acceleration and the velocity when they arrive at the base of the plane.
【Note: Ic (Shell) =
1
2
2
MR2, Ic (Disk) = MR2, Ic (Sphere) = MR2】
3
2
5
選取斜面為 X 軸，其正向為 Y 軸
方法一:
由  Fix  Ma x &  i  I
i
i
Mg sin   f s  Macom
可得:
------------------
------------------ [1%]
Rf s  I com
又 
acom
R
聯立後可得 acom 
及 v 2  2ad  2a
v 
[1%]
g sin 
I
1  com2
MR
-----------------
[2%]
h
sin 
2 gh
I
1  com2
MR
------------------[2%]
方法二:
由能量守恆:
可得 v 
Mgh  12 Mvcom  12 I com 2 ，& vcom  R
2
2 gh
I
1  com2
MR
及由 v 2  2ad  2a
--------- [2%]
-------------- [2%]
h
g sin 
可得 acom 
I
sin 
1  com2
MR
--------------[2%]
--------------------------------------------------------------------------------------------------------------------將 I cylinder 
acylinder 
1
2
2
MR 2 , I sphere  MR 2 , I shell  MR 2 分別代入可得
2
5
3
2
5
3
g sin  , asphere  g sin  , ashell  g sin 
3
7
5
及 vcylinder 
4 gh
10 gh
, vsphere 
, vshell 
3
7
6 gh
5
--------[3%]
--------[3%]
8. (16%) In Fig. 6, two blocks m1 and m2 (m2  m1 ) , are connected by a
massless cord that is wrapped around a uniform disk of mass M and
radius R. The disk can rotate without friction about a fixed horizontal axis
through its center; the cord cannot slip on the disk. The system is released
from rest. Find (a) the magnitude of the acceleration of the blocks, (b)
tension T1, (c) tension T2, and (d) the magnitude of the pulley's angular
acceleration?
依第二運動定律: 可得
T1  m1 g  m1a1
m2 g  T2  m2 a2
T2 R  T1 R  I
} --------(3% for each) 9%
其中 a1  a2  R =a
(a) disk 的轉動慣量為: I  21 MR 2 , 故可解得
1
m2 g  m1 g  (m1  m2  M )a
2
(m2  m1 ) g
--------- [2%]
a
(m1  m2  12 M )
(b) a 值代入上述可得:
T1 
m1 (2m2  M2 )
g
m1  m2  M2
--------- [2%]
及
(c) T2 
(d)  
m2 (2m1  M2 )
g
m1  m2  M2
(m2  m1 ) g
R(m1  m2  12 M )
-------------- [2%]
-------------- [1%]
Fig. 6