【CH09】Center of Mass and Momentum Homework of ... - 物理學系

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【CH09】Center of Mass and Momentum
Homework of Chapter 9:
3, 5, 11, 13, 15, 16, 21, 27, 31, 33, 39, 43, 45, 53, 55, 63, 67, 73, 79
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【Problem 9-3】
What are (a) the x coordinate and (b) the y coordinate of the center of mass for the uniform plate
shown in Fig. 9-38 if L= 5cm?
圖 9-38 為均勻平板,假設 L= 5cm,問(a)x
座標(b)y 座標的質量中心?
(圖 9-38)
:因為是 uniform plate(均勻平版),所以質量與面積成正比。
第一像限:質量: 2
4L 。質量中心: (2 L,2.5 L )
第二像限:質量: 2
6L 。質量中心: ( − L,1.5 L)
第三像限:質量: 2
8L 。質量中心: ( −L, − 2 L)
第四像限:質量: 2
4L 。質量中心: ( L, − 3 L)
2 2 2 2
(4 L )(2 L) + (6 L )( − L) + (8 L )( − L) + (4 L )( L)
(a) xCM
=
2 2 2 2
4L + 6L + 8L + 4L
−2
= L=− 0.45cm
22
2 2 2 2
(4 L )(2.5 L) + (6 L )(1.5 L) + (8 L )( − 2 L) + (4 L )( −3 L)
−9
(b) yCM = = L=−2cm 2 2 2 2
4L + 6L + 8L + 4L 22
【Problem 9-5】
In the ammonia ( NH 3 ) molecule of Fig. 9-40, three hydrogen (H) atoms form an equilateral triangle,
−11
with the center of the triangle at distance d = 9.4× 10 m from each hydrogen atom. The nitrogen
(N) atom is at the apex of a pyramid, with the three hydrogen atoms forming the base. The
nitrogen-to-hydrogen atomic mass ratio is 13.9, and the nitrogen-to-hydrogen distance is
−11
L= 10.14× 10 m.
What are the (a) x and (b) y coordinates of the molecule’s center of mass?
NH 3 分子,如圖 9-40,三個 H 原子形成等邊三角形,從每一個氫原子到三角形中心的距離
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−11
為 d = 9.4× 10 m,N
原子在以 H 原子為底的三角錐的頂點,N 對 H 原子質量比為 13.9,N
−11
對 H 距離為 L= 10.14× 10 m,(a)x
座標(b)y 座標的問分子質量中心?
(圖 9-40)
:(a) 因為是對稱的,所以 x cm = 0
(b) 3 mHycm = mN( yN− ycm)
其中, y N 指的 N 原子到三個 H 原子平面的距離。
由圖 9-40 可知道
−11 2 −11 2 −11
yN= (10.14× 10 ) − (9.4× 10 ) = 3.803× 10 m
−11
mNyN (14.0067)(3.803× 10 )
−11
ycm = = = 3.13× 10 m
m + 3m 14.0067+ 3(1.00797)
N H
【Problem 9-11】
A stone is dropped at t = 0 . A second stone, with twice the mass of the first, is dropped from the
same point at t = 100ms.
(a) How far below the release point is the center of mass of the two stones
at t = 300ms?
(Neither stone has yet reached the ground.) (b) How fast is the center of mass of the
two-stone system moving at that time?
一顆石頭,在 t = 0 時開始掉落,第二顆石頭質量是第一顆石頭的 2 倍,在 t = 100ms時同一
地點掉落,(a)在 t = 300ms,兩顆石頭的質量中心為何?(b)此時質量中心的速度為何?
:已知 m2 = 2m1
(a) 在時間 t = 300ms時
第一顆石頭( m 1 )位置在
1 2 1 −3
2
(9.8)(300 10 ) 0.441
y1= gt
2
=
2
× = m
第二顆石頭( m 2 )位置在
1 2 1 −3 −3
2
y2= gt = (9.8)(300× 10 − 100× 10 ) = 0.196m
2 2
my 1 1+ my 2 2 m1(0.441) + 2 m1(0.196)
ycm
= = = 0.28
m1+ m2 m1+ 2m1
−3
(b) v1 = gt = (9.8)(300× 10 )
−3 −3 −3
v2 = g( t−
100× 10 ) = (9.8)(300× 10 − 100× 10 )

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−3
質量中心在 t = 300× 10 s時的速度為
−3 −3 −3
mv 1 1+ mv 2 2 m1(9.8)(300× 10 ) + 2 m1(9.8)(300×
10 − 100× 10 )
vcm
=
=
m1+ m2
m1+ 2m1
= 2.3 m/ s
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【Problem 9-13】
Figure 9-44 shows an arrangement with an air track, in which a cart is connected by a cord to a
hanging block. The cart has mass m1= 0.6kg,
and its center is initially at xy coordinates (-0.5m, 0
m); the block has mass m2= 0.4kg,
and its center is initially at xy coordinates (0, -0.1m). The mass
of the cord and pulley are negligible. The cart is released from rest, and both cart and block move
until the cart hits the pulley. The friction between the cart and the air track and between the pulley
and its axle is negligible. (a) In unit-vector notation, what is the acceleration of the center of mass of
the cart–block system? (b) What is the velocity of the com as a function of time t? (c) Sketch the
path taken by the com. (d) If the path is curved, determine whether it bulges upward to the right or
downward to the left, and if it is straight, find the angle between it and the x axis.
圖 9-44,滑車以一條繩子和積木連接,滑車質量 m1= 0.6kg,一開始滑車中心位置為(-0.5m,
0 m),積木質量 m2= 0.4kg,一開始積木中心位置為(0,
-0.1m)。假設繩子和滑輪無質量。滑
車從靜止釋放,滑車和積木滑動,直到滑車碰到滑輪,滑車和氣軌無摩擦,(a)滑車、積木
系統的質量中心,(b)以 時間 t 為函數寫出質量中心的速度,(c)畫出質量中心的運動軌跡,
(d)如果軌跡是一條曲線,那曲線是向上突起或向下凹?如果軌跡是一條直線,那它和 x
軸的夾角為何?
(圖 9-44)
mg ( m m) a
mg 2 (0.4)(9.8)
2
a = = = 3.92 m/ s
m1+ m2
0.6 + 0.4
ma
1 1+ ma 2 2 (0.6)(3.92 i)
+ (0.4)( −3.92
j)
a 2.35 1.57
cm = = = i− j
m1+ m2
0.6 + 0.4
(b) v (2.35 1.57
cm = acmt = i− j) t
(c) The last sentence of our answer for part (b) implies that the path of the center-of-mass is
a straight line.
:(a) 2 = 1+ 2
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(d)
−1 −1.57
0
tan =− 33.4
2.35
【Problem 9-15】
A shell is shot with an initial velocity 0 v of 20 m/s, at an angle of
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0
θ 0 = 60 with the horizontal. At
the top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 9-46). One
fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the
gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?
0
一顆子彈,以 v0= 20 m/ s,水平夾角
θ 0 = 60 的初速射出,在射程頂端,子彈分成 2 塊相同
質量的碎片,如圖 9-46,爆破後,其中一塊碎片速度立刻變 0,直接往下垂直掉落。問另一
片碎片離發射的槍多遠?假設地形是水平,空氣阻力忽略不計。
v = v − gt = v sinθ
− gt
: y 0y 0 0
v0sinθ
0
v y = 0 時, t =
g
(圖 9-46)
v
20
x = v t = v t = = = m
2 2
0x( 0cos θ0) 0 sinθ0cosθ0 g
0 0
sin 60 cos 60 2
9.8
17.7
0 y
1 2
2 2
v0
sin θ0
2 2 0
(20 )(sin 60 )
15.3
y = v t− gt
2
=
2 g
=
(2)(9.8)
= m
m
mv cosθ
= V
2
0
V0 = 2v0cosθ0 = 2(20) cos 60 = 20 m/ s
在最高點時, x = 17.7m
, y = 15.3m。
1 2
y = y0− gt
2
y = 0時,
t =
2y0
g
x = x0 + V0t = x0 + V0 2y0 = 17.7 + (20)
g
2(15.3)
= 53m
9.8
後半段分裂成兩段,因為動量守恆,所以 0 0 0
【Problem 9-16】
Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg
canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3 m apart and
symmetrically located with respect to the canoe’s center. If the canoe moves 40 cm horizontally

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relative to a pier post, what is Carmelita’s mass?
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里卡多(質量 80 公斤)和卡梅莉塔(比較輕),都喜歡在黃昏默塞德湖在 30 公斤獨木舟。
當獨木舟是靜止在平靜的水面上,他們交換座位,此時相隔 3 米,位於對稱方面的獨木舟
的中心。如果獨木舟水平移動 40cm 相對於一個碼頭後,卡梅莉塔的質量為何?
:Ricardo 質量: M R
Carmelita 質量: M C
L L
M R( − x) = mx+ MC( + x)
2 2
40
x = = 20cm = 0.2m
2
L
3
MR( −x) −mx 80( −0.2) −(30)(0.2)
M 2 2
C = = = 58kg
L 3
+ x
+ 0.2
2 2
【Problem 9-21】
0
A 0.3 kg softball has a velocity of 15 m/s at an angle of 35 below the horizontal just before
making contact with the bat. What is the magnitude of the change in momentum of the ball while in
contact with the bat if the ball leaves with a velocity of (a) 20m/s, vertical downward, and (b) 20m/s,
horizontally back toward the pitcher?
一 0.3 公斤的壘球以速度 15m/s 水平向下 0
35 接觸球棒。問壘球碰觸到球棒時,壘球動量變
化為何?如果球離開的速度(a)20m/s,垂直向下,和(b)20 米/秒,水平朝向投手?
0 0 0
:一開始動量 p
0 = (0.3 kg)(15 m / s)( ∠ 215 ) = 4.5cos 215 i + 4.5sin 215 j
(a)20m/s垂直向下
0 0 0
動量: p = (0.3 kg)(20 m / s)( ∠− 90) = 6cos( − 90) i + 6sin( −90)
j
f
−1 −3.42
0
p
f − p0= 3.69 i−3.42 j(
動量變化: 5 kg ⋅ m / s
(b) 20m/s 水平朝向投手
2 2
(3.69) + (3.42) = 5 , tan
3.69
= − 43 )
動量: p
0
= (0.3 kg)(20 m / s)( ∠ 0 ) = 6 i + 0j
f
−1 −2.58
0
p
f − p0= 9.68 i−2.58 j(
動量變化:10 kg ⋅ m / s
2 2
(9.68) + (2.58) = 10 , tan
9.68
= − 15 )
【Problem 9-27】
A 1.2 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an initial
speed of 10 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact
with the floor for 0.02 s, what is the magnitude of the average force on the floor from the ball?
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一顆球 1.2 公斤,垂直掉落到樓層板上,撞擊時速度為 25m/s,反彈後最初速度為 10m/s(a)
球接觸時,衝量為何?(b)如果球在接觸地面時間為 0.02 秒,球作用在樓層版上的平均力
大小為何?
:初速: v 25 /
i =− m sj
末速: v 10 /
f = m sj
(a) J = mv f − mvi = (1.2)(10) −(1.2)( − 2.5) = 42 kg ⋅m/
s
J 42
3
(b) Favg = = = 2.1× 10 N
Δt
0.02
【Problem 9-31】
Figure 9-52 shows a 0.3kg baseball just before and just after it collides with a bat. Just before, the
ball has velocity v1 0
of magnitude 12 m/s and angleθ
1 = 35 . Just after, it is traveling directly
upward with velocity 2 v of magnitude 10.0 m/s. The duration of the collision is 2ms. What are the
(a) magnitude and (b) direction (relative to the positive direction of the x axis) of the impulse on the
ball from the bat? What are the (c) magnitude and (d) direction of the average force on the ball from
the bat?
圖 9-52 顯示了一個 0.3 公斤的棒球被球棒打到的前後,一開始球速為 v1 ,大小為 12m/s,角
0
度 θ 1 = 35 。被球棒打到後,筆直以 v2 向上飛出, 大小為 10m/s。球棒接觸到球的時間為 2ms。
問衝量的(a)大小(b)方向,還有平均施力的(c)大小(d)方向?
(圖 9-52)
0 0 0
: p (0.3)(12)( 215 ) 3.6cos 215 3.6sin 215
i = ∠ = i+ j
0 0 0
p (0.3)(10)( 90 ) 3cos90 3sin 90
f = ∠ = i+ j
J =Δ p= p 2.95 5.06
f − pi = 2 2
−15.06
0
i+ j(
(2.95) + (5.06) = 5.85 , tan = 59.8 )
2.95
(a) J = Δ p= 5.86 kg⋅ m/ s = 5.86N⋅
s
(b) +x 軸夾 59.8 度
5.86N⋅s
3
(c) J = FavgΔ t = 5.86N⋅
s ⇒ Favg = ≈ 2.93× 10 N
−3
2.00× 10 s
(d) +x 軸夾 59.8 度

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【Problem 9-33】
Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator
cab free-falls from a height of 36 m. During the collision at the bottom of the elevator shaft, a 90 kg
passenger is stopped in 5.0 ms. (Assume that neither the passenger nor the cab rebounds.) What are
the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the
passenger were to jump upward with a speed of 7.0 m/s relative to the cab floor just before the cab
hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force
(assuming the same stopping time)?
在電梯撞到前跳起來。在電纜線和安全系統故障後,電梯從 36m 高的地方自由落體落下,
在撞到電梯井底部之前,一位 90 公斤的乘客停在 5ms。問乘客碰撞的(a)衝量(b)平均
施力大小為何?如果在電梯碰到電梯井底部時,乘客以向上速度 7m/s 跳起來碰到電梯天花
板,問(c)衝量(d)平均力大小為何?
1 2 2
:(a) mv = mgh ⇒ v = 2gh = 2(9.8 m/s )(36 m) = 26.6 m/s.
2
3
J = | Δ p| = m| Δ v| = mv=
(90kg)(26.6m/s) ≈ 2.39× 10 N⋅ s.
−3
(b) With duration of Δ t = 5.0× 10 s for the collision, the average force is
3
J 2.39× 10 N ⋅s
avg −3
5
F = = ≈ 4.78× 10 N.
Δ t 5.0× 10 s
(c) If the passenger were to jump upward with a speed of v′ = 7.0 m/s , then the resulting
downward velocity would be
v′′ = v− v′
= 26.6 m/s − 7.0 m/s = 19.6 m/s,
and the magnitude of the impulse becomes
3
J′′ = | Δ p′′ | = m| Δ v′′ | = mv′′
= (90kg)(19.6m/s) ≈ 1.76× 10 N⋅ s.
(d)
3
J′′
1.76× 10 N ⋅s
F′′
avg = = ≈ ×
−3
Δ t 5.0× 10 s
5
3.52 10 N.
【Problem 9-39】
A 91 kg man lying on a surface of negligible friction shoves a 68 g stone away from himself, giving
it a speed of 4 m/s. What speed does the man acquire as a result?
一 91 公斤男子倒臥在一個無摩擦的表面,猛推一個 68 公克的石頭遠離自己,給石頭的速
度為 4m/s,結果這個人獲得怎樣的速度?
: mv m m= mv s s
ms s s (0.068)(4)
−3
vm= = = 3× 10 m/ s
m 91
m
【Problem 9-43】
Figure 9-58 shows a two-ended “rocket” that is initially stationary on a frictionless floor, with its
center at the origin of an x axis. The rocket consists of a central block C (of mass M = 6kg
) and
blocks L and R (each of mass m= 2kg)
on the left and right sides. Small explosions can shoot
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either of the side blocks away from block C and along the x axis. Here is the sequence: (1) At time
t = 0 , block L is shot to the left with a speed of 3 m/s relative to the velocity that the explosion
gives the rest of the rocket. (2) Next, at time t = 0.8s,
block R is shot to the right with a speed of 3
m/s relative to the velocity that block C then has. At t = 2.8s,
what are (a) the velocity of block C
and (b) the position of its center?
圖 9-58 顯示了一個兩端的火箭,一開始被放在無摩擦平面上,中心在 x 軸原點。火箭包含
有中間的積木 C(質量 M = 6kg
),和左右兩個積木 L 和 R(質量都是 m= 2kg)。一個小爆
炸可以把兩個小積木沿著 x 軸推離積木 C。接著(1)在 t = 0 ,積木 L 以相對速度 3m/s 被
推到左邊,爆炸使得火箭靜止(2)接下來, t = 0.8s時,積木
R 以相對速度 3m/s 被推到積
木 C 的左邊,在 t = 2.8s,(a)問積木
C 的速度,和(b)中心位置?
(圖 9-58)
:(a) With SI units understood, the velocity of block L (in the frame of reference indicated in
the figure that goes with the problem) is (v1 – 3)i ^ . Thus, momentum conservation
(for the explosion at t = 0) gives
mL (v1 – 3) + (mC + mR)v1 = 0
3 mL 3(2 kg)
which leads to v1 =
mL + mC +
=
mR 10 kg
= 0.60 m/s.
Next, at t = 0.80 s, momentum conservation (for the second explosion) gives
mC v2 + mR (v2 + 3) = (mC + mR)v1 = (8 kg)(0.60 m/s) = 4.8 kg·m/s.
This yields v2 = – 0.15. Thus, the velocity of block C after the second explosion
is v2 = –(0.15 m/s)i ^ .
(b) Between t = 0 and t = 0.80 s, the block moves v1Δt = (0.60 m/s)(0.80 s) = 0.48 m.
Between t = 0.80 s and t = 2.80 s, it moves an additional
v2Δt = (– 0.15 m/s)(2.00 s) = – 0.30 m.
Its net displacement since t = 0 is therefore 0.48 m – 0.30 m = 0.18 m.
【Problem 9-45】
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the
explosion, one piece, of mass m, moves with velocity ( −30
m/ s) i and a second piece, also of mass
m, moves with velocity ( − 30 m/ s) j.
The third piece has mass 3m. Just after the explosion, what
are the (a) magnitude and (b) direction of the velocity of the third piece?
一個導管靜止在 xy 座標的原點,爆炸分成三塊。爆炸後,其中一塊質量 m,移動速度
( −30 m/ s) i。第二塊質量一樣是
m,速度 ( − 30 m/ s) j。第三塊質量
3m,問第三塊速度(a)
大小(b)方向?
:已知 m1m
v m s
i
= , 1 = ( −30
/ )

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m m = ,
v
2 = ( −30
m/ s) j
2
(a) mv0 = m1v1+ m2v2 + m3v3 0 = m( − 30 i) m( 30
+ − j) + 3mv3
v
3 = 10 −110
0
i+ 10 j(
v3= 14 m/ s,
tan = 45 )
10
(b)
0
45
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【Problem 9-53】
In Fig. 9-61a, a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The
bullet passes through block 1 (mass 1.2 kg) and embeds itself in block 2 (mass 1.8 kg). The blocks
end up with speeds v1= 0.63 m/ s and v2= 1.4 m/ s (Fig. 9-61b). Neglecting the material removed
from block 1 by the bullet, find the speed of the bullet as it (a) leaves and (b) enters block 1.
圖 9-61a,一個 3.5 公克的子彈,水平撞擊兩個靜止的積木,子彈穿過第一塊積木(質量 1.2
公斤),停留在第二塊積木(質量 1.8 公斤)裡,之後積木速度分別為 v1= 0.63 m/ s和
v2= 1.4 m/ s,如圖
9-61b,被子彈移除的第一塊積木的質量忽略不計,當子彈(a)離開(b)
進入…第一塊積木時,問此時子彈速度?
(圖 9-61)
:(a) (0.0035 kg) v = (1.8035 kg)(1.4 m / s)
⇒ v= 721 m/ s
(b) (0.0035 kg) v0 = (1.2 kg)(0.63 m/ s) + (0.00350 kg)(721 m/ s)
⇒ v= 937 m/ s
【Problem 9-55】
In Fig. 9-63, a ball of m mass m= 60g
is shot with speed vi= 22 m/ s into the barrel of a spring
gun of mass M = 240g
initially at rest on a frictionless surface. The ball sticks in the barrel at the
point of maximum compression of the spring. Assume that the increase in thermal energy due to
friction between the ball and the barrel is negligible. (a) What is the speed of the spring gun after
the ball stops in the barrel? (b) What fraction of the initial kinetic energy of the ball is stored in the
spring?
圖 9-63,球 m 質量 m= 60g,以速率
v = 22 m/ s,被射進去一個質量
M = 240g
的彈簧槍裡,
i
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彈簧槍一開始靜止在無摩擦平面,球將彈簧壓縮到最大,假設球與彈簧槍之間,因摩擦產
生的熱能忽略不計,(a)球在彈簧槍裡靜止後,彈簧槍的速率為何?(b)球儲存在彈簧槍
裡的動能比例為何?
(圖 9-63)
:(a) mvi = ( M + m) v
mvi
(60)(22)
v= = = 4.4 m/ s.
m+ M 60+240
1 2
(b) Ki = mvi
2
2 2
1 2 1 mvi
K f = ( M + m) v =
2 2 ( M + m)
2 2
1 2 1 mvi
1 2 m 1 2 M
Us = Ki − K f = mvi
− = mvi
(1 − ) = mvi ( )
2 2 M + m 2 M + m 2 M + m
U s M 240
= = = 0.8
K m+ M 60+240
i
【Problem 9-63】
A body of mass 2kg makes an elastic collision with another body at rest and continues to move in
the original direction but with one-fourth of its original speed. (a) What is the mass of the other
body? (b) What is the speed of the two-body center of mass if the initial speed of the 2kg body was
4m/s?
一 2 公斤物體和另一靜止物體做彈性碰撞,並繼續朝原來的方向前進,速度減為原來的四
分之一。(a)問另一物體質量?(b)原來 2 公斤物體的初速為 4m/s,問質量中心速度?
:(a) 1f1i 1
v = v
4
m1−m2 v1f= v1i(請看最後補充資料:一維完全彈性碰撞)
m1+ m2
1
v 1i 1i
1i − v v − v
1f
4 3
m2 = m1 = (2) = (2) = 1.2kg
v 1 1i + v1f
v
5
1i + v1i
4
mvi + mv i = ( m + m ) vcm
mv 1 1i + mv 2 2i
(2)(4) + (1.2)(0)
vcm = = = 2.5 m/ s
m + m 2+ 1.2
(b) 1 1 2 2 1 2
1 2

Fundamentals of Physics
Halliday & Resnic
東海大學物理系
【Problem 9-67】
Block 1 of mass m 1 slides along a frictionless floor and into a one-dimensional elastic collision
with stationary block 2 of mass m2 = 3m1.
Prior to the collision, the center of mass of the two-block
system had a speed of 3m/s. Afterward, what are the speeds of (a) the center of mass and (b) block
2?
積木一質量 m 1 ,沿著無摩擦地板與積木 2 質量 m2 3m1
= 做一維彈性碰撞。碰撞之前,質量
中心速度 3m/s,後來(a)質量中心速度為何和(b)積木二速度為何?
:(a) The center of mass velocity does not change in the absence of external forces. In this
collision, only forces of one block on the other (both being part of the same system) are
exerted, so the center of mass velocity is 3.00 m/s before and after the collision.
(b) ( m1+ m2) vcm = mv 1 1i+ m2(0)
( m1+ m2) vcm m1+ 3m1
v1i= = (3) = 12 m/ s
m1 m1
2m1 2m1
v2f = v1i = (12) = 6 m/ s
m + m m + 3m
1 2 1 1
【Problem 9-73】
In Fig. 9-23, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The
0
alpha particle is scattered at angle θ 1 = 64 and the oxygen nucleus recoils with speed
5
0
1.2× 10 m/ s and at angleθ
2 = 51 . In atomic mass units, the mass of the alpha particle is 4u and
the mass of the oxygen nucleus is 16u. What are the (a) final and (b) initial speeds of the alpha
particle?
圖 9-23 中,被拋射的物體 1 為阿發粒子,被打中的粒子 2 是氧分子,阿發粒子散射角度為
0
5
0
θ 1 = 64 ,氧分子速率 1.2× 10 m/ s,角度
θ 2 = 51 ,阿發粒子質量為 4u,氧分子質量為 16u,
問(a)一開始(b)最後,阿發粒子的速率?
(圖 9-23)
:已知 v2f 5
0
0
= 1.2× 10 m/ s、
θ 1 = 64 、 θ 2 = 51
mv = mv cosθ + mv cosθ
1 1i 1 1f 1 2 2 f 2
Fundamentals of Physics
Halliday & Resnic
東海大學物理系
0= mv 1 1f sinθ1− mv 2 2 f sinθ2
5 0
mv 2 2 f sin θ2
(16)(1.2 × 10 )(sin 51 )
5
(a) v1f= = = 4.15× 10 m/ s
0
m1sin
θ1
(4)(sin 64 )
5 0 5 0
mv 1 1f cosθ1+ mv 2 2 f cosθ
2 (4)(4.15× 10 )(cos64 ) + (16)(1.2 × 10 )(cos51 )
(b) v1i
=
=
m
(4)
1
5
4.84 10 m/ s
= ×
【Problem 9-79】
In Fig. 9-72, two long barges are moving in the same direction in still water, one with a speed of 10
km/h and the other with a speed of 20 km/h. While they are passing each other, coal is shoveled
from the slower to the faster one at a rate of 1000 kg/min. How much additional force must be
provided by the driving engines of (a) the faster barge and (b) the slower barge if neither is to
change speed? Assume that the shoveling is always perfectly sideways and that the frictional forces
between the barges and the water do not depend on the mass of the barges.
(圖 9-72)
如圖 9-72,兩艄船在靜止水面上同方向行駛,其中一艄速率為 10km/h,另一艄為 20km/h,
當他們相遇時,較慢的船將煤炭以 1000kg/min 速度剷到較快的船上,如果兩艄船都不改變
速率,問引擎要加給多大的力給(a)較快的船(b)較慢的船?假設剷的方向總是側向船,
船與水的摩擦與船的質量無關。
:(a) We consider what must happen to the coal that lands on the faster barge during one
minute (Δt = 60s). In that time, a total of m = 1000 kg of coal must experience a change
of velocity
Δv = 20 km h − 10 km h = 10 km h = 2. 8 m s,
where rightwards is considered the positive direction. The rate of change in momentum
for the coal is therefore
Δp mΔv ( 1000 kg)( 2.8 m/s)
= = = 46 N
Δt Δt
60 s
which, by Eq. 9-23, must equal the force exerted by the (faster) barge on the coal. The
processes (the shoveling, the barge motions) are constant, so there is no ambiguity in
equating Δp
dp
with
Δt
dt .

Fundamentals of Physics
Halliday & Resnic
東海大學物理系
(b) The problem states that the frictional forces acting on the barges does not depend on
mass, so the loss of mass from the slower barge does not affect its motion (so no extra
force is required as a result of the shoveling).
【補充資料】關於一維完全彈性碰撞
【University Physics, Harris Benson】
【Problem 9-7】
A particle of mass 1 m moving at velocity 1 1 u = u i makes a one-dimensional elastic collision with
a particle of mass 2 m moving at velocity 2 2 u = u i. The final velocities are 1 1 v = v i and 2 2 v = v i.
( m1− m2) u1+ 2m2u2
Show that v1
=
m1+ m2
2 mu 1 1+ ( m2−m1) u2
v2
=
m + m
1 2
一個粒子(質量 m 1 ,速度 1 = u1
碰撞後兩者分別為 1 = v1
=
u i),和另一個粒子(質量 m 2 , u2 = u2i)做完全彈性碰撞,
v i和 v2 v2i.,請推導出
( m1− m2) u1+ 2m2u2
2 mu 1 1+ ( m2−m1) u2
v1
=
, v2
=
。
m1+ m2
m1+ m2
1 2 1 2 1 2 1 2
:能量守恆: mu 1 1+ mu 2 2= mv 1 1+ mv 2 2……………....○1
2 2 2 2
⇒
m ( u − v ) = m ( v − u )
2 2 2 2
1 1 1 2 2 2
m ( u − v )( u + v ) = m ( v − u )( v + u ) …...○3
⇒ 1 1 1 1 1 2 2 2 2 2
動量守恆: mu 1 1+ mu 2 2= mv 1 1+ mv 2 2…………..………….....○2
m ( u − v ) = m ( v −u ) …..............................○4
⇒ 1 1 1 2 2 2
由○3 、○4 式: u1+ v1 = u2 + v2
代回○2 式 mu 1 1+ mu 2 2= mv 1 1+ m2( u1+ v1− u2)
( m1− m2) u1+ 2 m2u2 = ( m1+ m2) v1
( m1− m2) u1+ 2m2u2
v1
=
m + m
1 2
Fundamentals of Physics
Halliday & Resnic
2 mu + ( m − m) u
1 1 2 1 2
同理 v2
=
m1+ m2
【補充資料】關於二維完全彈性碰撞
( m − m ) u + 2m
u
v1
=
m1+ m2
2 mu + ( m − m) u
v2
=
m + m
1 2 1 2 2
1 1 2 1 2
1 2
東海大學物理系