國立中山大學 100 學年度普通物理(一)期中考
100.11.16
答題須知：
一、 作答時，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、 請詳列相關公式及計算過程，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。
1. (10%) In Fig. 1, a car is driven at constant
speed over a circular hill (radius = R) and
then into a circular valley (radius = 2R).
At the top of the hill, the normal force on
the driver from the car seat is zero. The
Fig .1
driver's mass is 100 kg. What is the
magnitude of the normal force on the driver from the seat when the car passes
through the bottom of the valley? Assume gravitational constant g = 10 m/sec2.
【1 分】
【1 分】
【2 分】
代
【2 分】
【2 分】
【2 分】
1
2. (10%) In Fig. 2, a ball is shot directly upward from
the ground with an initial speed of v0 = 7.00 m/s.
Simultaneously, a construction elevator cab begins
to move upward from the ground with a constant
speed of vc = 3.00 m/s. What maximum height does
Fig .2
the ball reach relative to (a) the ground and (b) the
cab floor? At what rate does the speed of the ball change relative to (c) the ground
and (d) the cab floor? Assume gravitational constant g = 10 m/sec2.
【1 分】
【2 分】
【2 分】
【2 分】
【1 分】
【1 分】
【1 分】
2
3. (10%) A block of mass m lies on a horizontal
frictionless face and is attached to one end of a
horizontal spring whose other end is fixed.
The block is initially at rest at the position
where the spring is unstretched (x = 0) when a
constant horizontal force in the positive
direction of the x axis is applied to it. A plot of
the resulting kinetic energy of the block versus
its position x is shown in Fig. 3. The scale of
the figure's vertical axis is set by KS. (a) What
is the magnitude of horizontal force? (b) What
Fig .3
is the value of the spring constant?
Assume that the magnitude of the horizontal force is F and the value of the spring
constant is k.
Consider the work-kinetic energy theorem
K = Wa (applied work)＋Ws (work done by spring)
K = Fx －
1 2
kx
2
【2 分】
For x = 1, K = Ks
Ks = F －
1
k12
2
 2Ks = 2F－k
(1)
【2 分】
 0 = 4F－4k
(2)
【2 分】
For x = 2, K = 0
0 = 2F －
1
k22
2
(a) (1)×4－(2)
4Ks = 2F  F = 2Ks
(b) From (2)
k = F = 2Ks
【2 分】
【2 分】
3
4. (15%) A crate of mass m hangs from the end of a rope of length
L = 12.0 m. You push horizontally on the crate with a varying
force to move it distance d = 6.00 m to the side (Fig. 4). (a) What
is the magnitude of the pushing force when the crate is in this
final position? During the crate's displacement, what are (b) the
work done by the gravitational force on the crate, and (c) the
work done by the pull on the crate from the rope? (d) Knowing
that the crate is motionless before and after its displacement,
what's the work for the pushing force does on the crate?
Fig .4
(a) In the final position, the forces acting on the object are shown as follow
T cos = mg
T sin = F
T
【2 分】
d
1
 F = mg tan = mg
=
mg
2
2
3
L d

F
【1 分】
mg
(b) The vertical displacement h can be expressed as
h = L–Lcos = L(1–cos) 【2 分】
The work done by the gravitational force is
Wg = Fgd = - mgh = - mgL(1–cos) =  mg  12(1 
3
)  6mg (2  3 )
2
【2 分】
(c) The tension force is always perpendicular to the direction of motion of the object.
Thus, Wt = 0.
【3 分】
(d) By the work-kinetic energy theorem
K = Wa＋Wg＋Wt 【2 分】
The object are both motionless initially and finally, K = 0
0  Wa  6mg(2  3)  0
Wa  6mg(2  3)
【3 分】
4
5. (10%) In Fig. 5, a block of mass m = 3.0 kg slides from rest
a distance d down a frictionless incline at angle θ = 30.0°
where it runs into a spring that can be compressed 10.0 cm
by a force of 29.4 N. When the block momentarily stops, it
has compressed the spring by 20.0 cm. What are (a)
distance d and (b) the distance between the point of the first
block–spring contact and the point where the block's speed
is greatest?
(a) Spring constant k 
F
= 294 N/m
x
d
Fig .5
【1 分】
When the block stops, the gravitational potential energy completely transfers
to elastic potential energy
mg (d  x) sin  
1 2
kx
2
【2 分】
3  9.8(d  0.2) 
1 1
  294  (0.2) 2
2 2
d＋0.2 = 0.4
 d = 0.2 (m) 【2 分】
x
d
-----------------------------------------------------------------------------------------Or, considering the conservation of energies,
Ki＋Ug,i＋Us,i = Kf＋Ug,f＋Us,f
Set x = 0 as the reference of potential energies
1
0  mgd sin   0  mgx sin   kx2
2
3  9.8  d 
【2 分】
1
1 1
 3  9.8  0.2    294  (0.2) 2 
2
2 2
d = 0.2 (m)
【2 分】
(b) The velocity is greatest while the block is at its equilibrium position
 mg sin = kx
【3 分】
29.4 
1
 294  x  x = 0.05 (m)
2
【2 分】
-----------------------------------------------------------------------------------------Or, considering 0  mgd sin  
v 2  2 gd sin   2 gx sin  
1 2
1
mv  mgx sin   kx2
2
2
k 2
x
m
【3 分】
The velocity reach maximum when
d (v 2 )
k
 2 g sin   2 x  0
dx
m
mg
x
sin  = 0.05 (m)
【2 分】
k
5
6. (10%) In the overhead view of Fig.6, a 300g ball with a
speed v of 10.0m/s strikes a wall at an angleθof 30o
and then rebounds with the same speed and angle. It is
in contact with the wall for 10 ms. In unit-vector
notation, what are (a) the impulse on the ball from the
wall and (b) the average force on the wall from the
ball?
m  300 g  0.3kg
v  10.0m/s
  30o
Fig .6
t  10ms
(a) 衝量
t2
I   F  dt  mv2  mv1
t1
ˆ cos   ˆjv sin  )  m(iv
ˆ cos   ˆjv sin  )
 m(iv
 2 ˆjmv sin   ˆj 2(0.3kg)(10.0m/s)sin30o  ˆj3kg  m/s 【5 分】
(b) 球所受的平均力 Fave 為
t2
I   F  dt  Fave t  mv2  mv1
t1
Fave 
mv2  mv1 ˆj 3kg  m/s ˆ

 j300N
t
10 103 s
球受的平均衝力 【4 分】
'
牆所受的平均力 Fave
'
Fave
  Fave   ˆj 300N
【1 分】
6
7. (10%) A rifle bullet of mass 0.01 kg
strikes and embeds itself in a block of
mass 0.99 kg, which rests on a
horizontal frictionless surface and is
attached to a coil spring, as shown in
Fig. 7
Fig. 7. The impact compresses the
spring 10 cm. The force constant of the spring is 100 N/m. (a) Find the maximum
potential energy of the spring. (b) Find the velocity of the block just after the
impact. (c) What is the initial velocity of the bullet?
m  0.01 kg
(a) 彈簧最大位能
子彈質量
EP 2 
木塊質量
M  0.99 kg
彈簧力常數
k  100 N/m
1 2 1
kx2  (100N/m)(0.1m) 2  0.5J 【3 分】 (a)
2
2
(b) 子彈與木塊碰撞之瞬間, 子彈嵌進木塊,
此時之力圖如圖(b)所示. 因此時彈簧未及壓縮,
(b)
故彈力 Fe  0 .
假設碰撞後瞬間, 系統之瞬時速度為 v '
(c)
因此碰撞時外力為零., 系統之總動量保持常數
(即總動量守恆)
mv  (m  M )v '
碰撞後系統壓縮彈簧後之過程, 應用功能定律
1
1
1
 1

Wnc   (m  M )v22  kx22    (m  M )v12  kx12 
2
2
2
 2

v1  v ' 
mv
mM
x1  0
v2  0
(1)
(2)
x2  10cm  0.1m
Wnc   (m  M ) g  ds   N  ds  0
代入(2)式
mv 2
 1
 1

0  0  kx22    (m  M )(
)  0
2
mM

 2

mv  k (m  M ) x2
v' 
k ( m  M ) x2
mv
k
100 N / m


x2 
(0.1m)  1m / s
mM
mM
mM
(0.01  0.99)kg
【5 分】
(c) 子彈的初速
v
mM
(0.01  0.99)kg
v' 
(1m/s)  100m/s
m
0.01kg
【2 分】
7
8. (10%) Two identical blocks, each of mass M, are connected by a light string over
a pulley of radius R and rotational inertia I, as shown in Fig.
8. The string does not slip on the pulley; it is not known
whether there is friction between the table and the sliding
block; the pulley's axis is frictionless. When this system is
released from rest, the pulley turns through an angle  in
time t, and the acceleration of the blocks is constant. What
Fig. 8
are (a) the magnitude of the pulley's angular acceleration,
(b) the magnitude of either block's acceleration, (c) string tension T1 and T2? All
answers are to be expressed in terms of M, I, R, , g and t.
畫力之圖解 【2 分】
M
1
(a)   0 t  t 2
2
∵0 = 0
∴ 
T2
2
【2 分】
t2
T2
, I
T1
T1
(b) M 之加速度
a
2
2 R
a  R  R( 2 )  2 【2 分】
t
t
Mg
(c) 由右圖：T1 – Mg = –Ma
∴T1 = M( g – a) = M( g –
2 R
) 【2 分】
t2
∵(T1 –T2)R = I
∴ T2  T1 
I
2 R
I 2
2
I
 M ( g  2 )  ( 2 )  Mg  2 ( MR  ) 【2 分】
R
R t
R
t
t
8
9. (15%) A hoop, a disk, and a sphere, have the same mass M and
radius R, roll without slipping from rest down a ramp whose
length is L and the ramp angle is , as shown in Fig. 9. (a) What
are the speeds of these objects at the bottom of the ramp? (b)
What are the accelerations of center of mass of these objects? (只
寫答案不計分)
【Note: Ic (Hoop) = MR2, Ic (Disk) =
Fig. 9
1
2
MR2, Ic (Sphere) = MR2】
2
5
滾動物體所受的外力計有重力 Mg ,
正向力 N 及靜摩擦力 f s 如圖 9 所示.
由於非保守力所作的功 Wnc  W f  WN  0
s
(∵ f s 無位移, N 與位移垂直)
故滾動物體之機械能守恆.
令斜面頂端與底端之位置分別為 1 與 2. 則
1
1
1
1
( Mv22  I C22 )  Mgy2  ( Mv12  I C12 )  Mgy1
2
2
2
2
v2
1
1
2
Mvcm
 I C ( cm2 )  0  0  MgL sin 
2
2
R

MgL sin  
∴ MgL sin  
2

∴ vcm
1
1
2
I C  2  Mvcm
2
2
v
I v2
1
1
1
2
2
I C ( cm ) 2  Mvcm
 ( C cm
)  Mvcm
【4 分】
2
2
R
2
2
2R
2MgL sin  2 gL sin 

IC
I
M
1 C 2
2
R
MR
∴ vcm 
2 gL sin 
【4 分】
IC
1
MR 2
(a) Hoop：Ic = MR2 ∴ vcm  gL sin  【1 分】
Disk：Ic =
1
2 gL sin 
4
MR2 ∴ vcm 

gL sin  【1 分】
1
2
3
1
2
Sphere：Ic =
2
2 gL sin 
10
MR2 ∴ vcm 

gL sin  【1 分】
2
5
7
1
5
9
(b) 應用牛頓第二定律
F
 Mg sin   f s  Macm
(1)
F
 N  Mg cos  0
(2)

 f s R  I cm
(3)
x
y
cm
acm  R
(4)
(4)式代入(3)
fs 
I cm
acm
R2
(5)
(5)式代入(1)式
Mg sin  
I cm
acm  Macm
R2
得物體的質心加速度為
Hoop： acm 
Disk： acm
acm 
Mg sin 
I
M  cm2
R
(6)
【1 分】
gL sin  1
 g sin  【1 分】
2L
2
4
gL sin 
2
3

 g sin  【1 分】
2L
3
Sphere： acm
10
gL sin 
5
 7
 g sin  【1 分】
2L
7
---------------------------------------------------------------------------------------------------------------(a)小題也可由以下方法求得：
由運動學求物體滾至底端之速度, 由(6)式得知, acm 為定值, 故物體作等加速運動
從等加速運動公式
v22  vo2  2acm ( x2  x1 )  0  2acm x2
物體在斜面底端之質心速度 vcm 為
vcm  2acm L
Hoop:
vcm  gL sin 
Disk:
vcm 
4
gL sin 
3
Sphere:
vcm 
10
gL sin 
7
10