1
Theoretical Mechanics
2
3
In the preceding chapters the velocities of all points of a
body under the action of forces always change continuously and
gradually, In the present chapter we will study another mechanical
phenomenon——impact,When bodies collide the velocities of all
point of a body suffer a finite change in a very small time
interval,The main subject here is to study collision
phenomena,the fundamental theorems of the theory of
impact,the loss in the kinetic energy of bodies during impact,
the center of impact,etc,
4
在前面讨论的问题中，物体在力的作用下，运动速度都
是连续地、逐渐地改变的。本章研究另一种力学现象 ——碰
撞,物体发生碰撞时，会在非常短促的时间内，运动速度突
然发生有限的改变。 本章研究的主要内容有 碰撞现象的特征，
用于碰撞过程的基本定理，碰撞过程中的动能损失，撞击中
心。
5
§ 19–1 Impact phenomenon and its fundamental
characteristics,force of impact,
§ 19-2 Fundamental theorems of the theory of
impact
§ 19–3 Impact of a particle on a fixed plate,
Coefficient of restitution,
§ 19–4 Direct Central impact of two bodies.Loss
in kinetic energy,
§ 19–5 The action of an impact force on a rigid
body rotating about a fixed axis.The center
of impact
Small Summary
Chapter 19, Impact
6
§ 19–1 碰撞现象及其基本特征 ? 碰撞力
§ 19-2 用于碰撞过程的基本定理
§ 19–3 质点对固定面的碰撞 ? 恢复系数
§ 19–4 两物体的对心正碰撞 ? 动能损失
§ 19–5 碰撞冲量对绕定轴转动刚体的作用 ?
撞击中心
小结
第十九章 碰撞
7
§ 19-1 Impact phenomenon and its fundamental
characteristics, force of impact
Fundamental characteristics of an impact phenomenon：
The velocities or moment a of the points of a body suffer finite
changes in a very small time interval during impact,The impact time
is very short,measured to be of the order of one thousandth or even
of one ten-thousandth of a second,Therefore,the values of the
acceleration and of the acting forces are very large,
Impact,The phenomenon that the velocities of the points of a
body suffer sharp changes in a very small time interval if the moving
body is suddenly subjected to a collision (including switched on or
switched off forces of constraints)is called impact,
Force of impact, The forces occurring during an impact
are called forces of impact,They are also called instantaneous
forces because the action time is very short,
8
§ 19-1 碰撞现象及其基本特征 ? 碰撞力
碰撞现象的基本特征,物体的运动速度或动量在极短的
时间内发生有限的改变。碰撞时间之短往往以千分之一秒甚
至万分之一秒来度量。因此加速度非常大，作用力的数值也
非常大。
碰撞,运动着的物体在突然受到冲击（包括突然受到约
束或解除约束）时，其运动速度发生急剧的变化，这种现象
称为碰撞。
碰撞力（瞬时力）,在碰撞过程中出现的数值很大的力
称为碰撞力；由于其作用时间非常短促，所以也称为瞬时力。
9
Let a hammer of weigh 10N hit a block of iron with v1=6m/s,the
impact time is ? =1/1000s and the hammer is rebounced with v2=1.5m/s
after impact,Determine the average value of the force of impact,
The general time-behavior of the change of the force of impact
is shown in diagram,
The average value of the force of impact is,
being 765 times larger than the weight of the hammer,
N7 6 5 0/* ?? ?SF
Illustration of the characteristics of the forces of
impact by the example of a hammer hitting iron,
Svmvm ?? 12,
s N 65,7 ? S ) 6 5, 1 ( 10 ? + S
g
Take the hammer as the object to be studied,
and
Hence,
,
10
设榔头重 10N，以 v1=6m/s的速度撞击铁块，碰撞时间 ?
=1/1000s,碰撞后榔头以 v2=1.5m/s的速度回跳。求榔头打击铁块
的力的平均值。
Svmvm ?? 12 的投影形式得
sN 65.7 ; )65.1(10 ????+ SSg
碰撞力的变化大致情况如图所示。
平均打击力,是榔头重的 765倍。 N7 6 5 0/* ?? ?SF
以榔头打铁为例说明碰撞力的特征,
以榔头为研究对象，根据动量定理
11
We can see that in the case of a large velocity of motion the
magnitude of the instantaneous force can is very large up to a surprising
value even if the mass of the body is very small,
When a flying bird of weight 17.8N collides with a flying plane,the
velocity of which is 800km/h (for a modern plane this is only of medium
velocity) the force of impact can reach 3.56?105N.I.e.,the force of impact
is 20 thousand times larger than the weight of the bird.This is the reason
of so called bird disaster in aviation,
As to impacts,Harmful aspects are,bird disasters”,damages of
mechanisms and apparatuses caused by impact etc,Advantageous
aspects are to use to use impacts to perform useful work as,for
example,forging,driving piles etc,
The study of impact,phenomena has the aim to undress and laws
in order to make positive use of its advantageous aspects and in order
to avoid its dangerous and harmful aspects,
12
可见，即使是很小的物体，当运动速度很高时，瞬时力
可以达到惊人的程度。有关资料介绍，一只重 17.8N的飞鸟
与飞机相撞，如果飞机速度是 800km/h，（对现代飞机来说，
这只是中等速度），碰撞力可高达 3.56?105N，即为鸟重的 2
万倍！这是航空上所谓“鸟祸”的原因之一。
害的一面,,鸟祸”、机械、仪器及其它物品由于碰撞损坏等。
利的一面,利用碰撞进行工作，如锻打金属，用锤打桩等。
研究碰撞现象，就是为了掌握其规律，以利用其有利的一
面，而避免其危害。
13
§ 19-2 Fundamental theorems of the
theory of impact
(1) Compared with forces of impact ordinary forces as
gravity,elastic forces etc.are is very small during the impact, Such
forces can be neglected,But it is must be noted that the action of
conventional forces on the change of the state of motion of bodies
before and after an impact cannot be neglected,
(2) Because the impact time is very small but the changes of
velocities are finite quantities,the displacement of a body during an
impact is a very small quantity which can be neglected in practice,I,
e.,the position of the body at the beginning of an impact is the same
as that at the end of the impact,
Two basic premises,
14
§ 19-2 用于碰撞过程的基本定理
(1)在碰撞过程中，重力、弹性力等普通力与碰撞力相比
小得多，其冲量可以忽略不计。但必须注意，在碰撞前和碰撞
后，普通力对物体运动状态的改变作用不可忽略。
(2)由于碰撞时间极短，而速度又是有限量，所以物体在
碰撞过程的位移很小，可以忽略不计，即认为物体在碰撞开始
时和碰撞结束时的位置相同。
两个基本假设,
15
Two basic theorems,
In theoretical mechanics we are mainly concerned with the
change of the velocity of a body under the action of a force of
impact,So,the momentum and the angular momentum theorem
become the main tools for investigating impact phenomena,
1,The momentum theorem applied to an impact,Theorem of
impulse,
Let the mass of a particle be m,its velocity before the impact
is,and at the end of the impact,the impact impulse is,
Neglecting the moment a of the convention al
forces,momentum the theorem in integral form becomes,
v u S
1)-( 1 9 Svmum ??
16
两个基本定理,
在理论力学中，我们关心的主要是由于碰撞冲量的作用而
使物体运动速度发生的变化。因此，动量定理和动量矩定理就
成了研究碰撞问题的主要工具。
1、用于碰撞过程的动量定理 ——冲量定理。
设质点的质量为 m，碰撞开始时的速度,结束瞬时的速
度,碰撞冲量,不计普通力的冲量，则质点动量定理
的积分形式为,
v
u S
1)-( 1 9 Svmum ??
17
For a system of n particles,dividing the impulses of impact
acting on the i-the particle into an external impulse of impact
and an internal impulses of impact,we have )(e
iS )(iiS
),,2,1(, )()( niSSvmum iieiiiii ??+??
Let the total mass of the system of particles be M,and
are the velocities at the end and the beginning of the
impact,Applying the theorem of the motion of the center of mass
to the,to the equation (19-2)we get
C u C v
??? )( eiCC SvMuM （ 19-3）
Adding these n equations and using (internal impulses of
impact always appear pair wise)we get 0)( ?? iiS
?????
??
)(
11
e
i
n
i ii
n
i ii
Svmum
（ 19-2）
This is the th orem of impulse,
18
对于有 n个质点组成的质点系，将作用于第 i 个质点上的
碰撞冲量分为外碰撞冲量 和内碰撞冲量,则有,
)(eiS )(iiS
),,2,1( )()( niSSvmum iieiiiii ??+??
将这 n个方程相加,且 (内碰撞冲量总是成对出现的 ),故 0)( ?? i
iS
?????
??
)(
11
e
i
n
i ii
n
i ii
Svmum 冲量定理 （ 19-2）
设质点系总质量 M,分别为碰撞结束和碰撞开始
时质心的速度，则利用质心运动定理，上式可写成,
CC vu 和
??? )( eiCC SvMuM （ 19-3）
19
The change in the momentum of a system of particles during
an impact is equal to the vector sum of all the external impulses
acting on the system,The projection forms of the expressions (19-1),(19-2) and (19-
3) are the same as in the case of the ordinary momentum theorem,
The difference is that the impulses of the conventional forces are
neglected here,
2,The angular momentum theorem applied to an impact,Theorem of the moment of the impulse,
We know from assumption (2) that the position of a particle
does not change,From expression (19-1) we then have,
.Srvmrumr ?????
Because,represent the angular moment
a of the particle with respect to point O at the beginning and at the
end of the impact, is the moment of the impact
impulse with respect to the point O
21,OO lumrlvmr ????
)( SmSr O??
s
)(12 Smll OOO ?? （ 19-4）
20
碰撞时质点系动量的改变等于作用在质点系上所有外碰
撞冲量的矢量和。
由假设 (2)知，碰撞过程中，质点的矢径 保持不变，
则由 (19-1)式，有,
r
式（ 19-1),(19-2)和 (19-3)都写成投影形式，形式上与普
通的动量定理相同，所不同的是在这里都不计普通力的冲量。
2、用于碰撞过程的动量矩定理 ——冲量矩定理
Srvmrumr ?????
而 ； 为碰撞始末时质点对
O点的动量矩。 是碰撞冲量 对 O点的矩，所以,21
,OO lumrlvmr ???? 21 OO ll 和
)( SmSr O?? s
)(12 Smll OOO ?? （ 19-4）
21
The change of the angular momentum of a particle during an
impact with respect to any center is equal to the moment of the
impact impulse acting on that particle taken with respect to the
same center,
For a system of particles the sum of the moments of the internal impact with respect,impulses to any center is zero,Thus we have,
)( )(12 eOOO SmLL ??? This expression is the of the
moment of the impulse,theorem
（ 19-5）
The expressions (19-4) and (19-5) can be expressed in projection
forms,The impulses of the conventional forces are neglected,
The change of the total angular momentum during impact of a
system with respect to any center is equal to the vector sum of the
moments of all external impulses acting on the system taken with
respect to the same center,
22
碰撞时，质点对任一固定点动量矩的改变，等于作用于
该质点的碰撞冲量对同一点之矩。
对于质点系,由于内碰撞冲量对任一点的矩之和等于零,于是有
)( )(12 eOOO SmLL ??? 冲量矩定理 （ 19-5）
式 (19-4),(19-5)也可写成投影形式，且式中均不计普通
力的冲量矩。
在碰撞过程中，质点系对任一固定点的动量矩的改变，等
于作用于质点系的外碰撞冲量，对同一点之矩的矢量和。
23
§ 19-3 Impact of a particle on a fixed
plate, Coefficient of restitution,
Let a small sphere (taken as a particle ) fall along the vertical
direction onto a fixed horizontal plate as is shown in the diagram,
Ple
ase
lo
ok
at
th
e an
im
atio
n
24
§ 19-3 质点对固定面的碰撞 ? 恢复系数
设一小球（可视为质点）沿铅直方向落到水平的固定平面
上，如图所示。
请
看
动
画
25
The first stage refers to the
time during which the small
sphere contacts with the plate to
reach its largest deformation,
During this stage the kinetic
energy of the small sphere
decreases,while the deformation
quantity of the small sphere
is,Projecting the theorem of
impulse on the axis y we have,
1)(0 Smv ???
1S
The course of the impact can be divided into two stages,
The second stage rulers to the time during which the internal elastic
deformation of the sphere forces it to escape contact,During this stage,
the elastic deformation restores gradually the kinetic energy, Let the
impact impulse be,then,Therefore,
2S 20 Smu ??
1
2
S
S
v
u ??
26
第一阶段,开始接触
至变形达到最大。该阶段
中，小球动能减小,变形增
大。设碰撞冲量为,则
应用冲量定理在 y 轴投影
式 1)(0 Smv ???
1S
碰撞过程分为两个阶段,
第二阶段,由弹性变形开始恢复到脱离接触。该阶段中，
小球动能增大，变形（弹性）逐渐恢复。设碰撞冲量为,
则,
2S
20 Smu ??
1
2
S
S
v
u ??
27
For a given material the ratio (of the absolute values of u to v is
constant and is called the coefficient of restitution,
vuk ?
The coefficient of restitution for
different bodies has to be
determined experimentally,
In general 0<k<1,We refer to the list in this book about the
coefficients of restitution for various materials,
If k=1,a limiting (ideal) condition,the impact is called perfectly
elastic impact,
If k=0,the other limiting condition,the impact is called perfectly
inelastic or plastic impact,
28
对于给定材料,|u|与 |v|的比值是不变的，该比值称为 恢复系数 。
vuk ? ——由实验测定
一般 0<k<1，各种材料的恢复系数，可查阅书中表。
k=1 理想情况 ——完全弹性碰撞。
k=0 极限情况 ——非弹性碰撞或塑性碰撞。
29
The impact of two colliding bodies is called to be a central
impact if the common normal to the surfaces of the bodies
through their point of contact pass through their centers of mass,
§ 19-4 Direct central impact of two
bodies, Loss in kinetic energy,
A central impact during which the velocities of the
approaching centers of mass are directed along a straight line
passing through the centers of mass is called a direct central
impact (direct impact),otherwise it is called an oblique central
impact (oblique impact),
30
对心碰撞,碰撞时两物体质心的连线与接触点公法线重合。
§ 19-4 两物体的对心正碰撞 ? 动能损失
对心正碰撞与对心斜碰撞,碰撞时两质心的速度也都沿两质
心连线方向，则称为对心正碰撞（正碰撞），否则称为对心
斜碰撞（斜碰撞）。
31
Please look at the animation
32
请看动画
33
1,The velocities of the two centers of mass at the end of
the direct impact
Example,The velocities of the approach of
two colliding bodies are and,
The velocities of their separation are denoted
by and (along the ling through the two
centers of mass),
Determine the velocities of separation of the
two centers of mass,
2 v 1 v ) ( 2 1 v v ?
2 u 1 u
34
1、正碰撞结束时两质心的速度
例如,两物体碰撞
碰撞前,
碰撞结束,（沿质心连线）
分析碰撞结束时两质心的速度。
)(,2121 vvvv ?
21,uu
35
Analysis,
( 2 )
21
12
vv
uuk
?
??
Choose the system of the two bodies as the
object to restudied,
From the theorem of impulse we obtain,
( 1 ), 0)()( 22112211 ?+?+ vmvmumum
A complementary equation is
(the, Applying the momentum theorem,we obtain,
for the first stage
2 2 2 2 1 1 ) (,) ( S u u m S u u m ? ? ? ? ?
1 2 2 1 1 1 ) (,) ( S v u m S v u m ? ? ? ? ?
2 1
1 2
1
1
2
2
1
2
v v
u u
v u
u u
v u
u u
S
S k
?
? ?
?
? ?
?
? ? ?
for the second stage,
there fore
,
,
,
36
研究对象：两物体组成的质点系。
由冲量定理，得,
( 1 ) 0)()( 22112211 ?+?+ vmvmumum
( 2 )
21
12
vv
uuk
?
??
分析,
列出补充方程,
（分别以两物体为研究对象，应用动量定理可得出。具体地
对于第一阶段,
对于第二阶段,
222211
122111
)(,)(
)(,)(
SuumSuum
SvumSvum
?????
?????
21
12
1
1
2
2
1
2
vv
uu
vu
uu
vu
uu
S
Sk
?
??
?
??
?
????
37
The coefficient of restitution k for a direct impact of two bodies
is equal to the ratio of the relative velocities of the bodies),
),( ) 1 ( 2 1
2 1
1
2 2 v v m m
m k v u ?
+ + + ?
2 1 m m
m
+ ),( ) 1 ( 2 1
2
1 1 v v k v u ? + ? ?
Combining the expressions (1) and (2) and solving them we can obtain,
).(2,)(2 21
21
1
2221
21
2
11 vvmm
mvuvv
mm
mvu ?
++??+??
For a perfectly elastic impact ( k=1 )we get
1,2 2 1 and v u v u ? ? 2 1 m m ? If then Thus,in the case of a perfectly
elastic impact two bodies of equal mass exchange their velocities,
38
对于两物体正碰撞的情况，恢复系数等于两物体在碰撞
结束与碰撞开始时，质心的相对速度大小的比值。 )
)()1(
)()1(
21
21
1
22
21
21
2
11
vv
mm
m
kvu
vv
mm
m
kvu
?
+
++?
?
+
+??
联立 (1),(2)式，解得,
对于完全弹性碰撞（ k=1）,
)(2 ; )(2 21
21
1
2221
21
2
11 vvmm
mvuvv
mm
mvu ?
++??+??
122121,,vuvumm ??? 则若 (碰撞后两物体交换速度 )
39
For plastic impact impact（ k =0） we get
21
221121
mm
vmvmuuu
+
+???
In the several case (0<k <1）,we get
2211,vuvu ??
2,Loss in kinetic energy during a direct impact
At the beginning of an impact the kinetic energy is
At the end of an impact the kinetic energy is 2 2,2 2 1 1 2 2 1 2 1 u m u m T + ?
2
2,2
2
1 1 1 2 1 2 1 v m v m T + ?
The loss of kinetic energy is,
))((
2
1))((
2
1
)(
2
1)(
2
1
2222211111
2
2
2
22
2
1
2
1121
uvuvmuvuvm
uvmuvmTTT
+?++??
?+?????
40
对于塑性碰撞（ k =0）,
21
221121
mm
vmvmuuu
+
+???
对于一般情况（ 0<k <1）,2211,vuvu ??
2、正碰撞过程中的动能损失
碰撞开始,
碰撞结束,2
22
2
112
2
22
2
111
2
1
2
1
2
1
2
1
umumT
vmvmT
+?
+?
则动能损失,
))((
2
1))((
2
1
)(
2
1)(
2
1
2222211111
2
2
2
22
2
1
2
1121
uvuvmuvuvm
uvmuvmTTT
+?++??
?+?????
41
We know from the formula of the velocities of the two centers of
mass at the end of impact that
).()1( nd )()1( 21
21
1
2221
21
2
11 vvmm
mkuvavv
mm
mkuv ?
++????++??
Substituting these results into the expression for,we obtain,
)].())[(()1(21 221221
21
21 uvuvvv
mm
mmkT +?+?
++??
2
2 1
2
2 1
2 1
2 1 ),)( 1 ( ) ( 2 v v k m m
m m T T T ? ?
+ ? ? ? ?
2 1 2 1 ).Therefore,( v v k u u ? ? ? ?
T?
Again
42
由正碰撞结束时两质心的速度公式知,
)()1( ; )()1( 21
21
1
2221
21
2
11 vvmm
mkuvvv
mm
mkuv ?
++????++??
代入上式中，得,
)]())[(()1(21 221221
21
21 uvuvvvmm mmkT +?+?++??
2
21
2
21
21
21
2121
))(1(
)(2
)(
vvk
mm
mm
TTT
vvkuu
??
+
?????
?????又
43
The loss in kinetic energy of a system of bodies during a plastic
impact equals to the kinetic energy the system would have if the its
bodies would move with the lost velocities,If v2=0 then
.
1
1
2
1
)(2
1
2
121
22
11
2
1
21
21 T
m
mmm
mvmv
mm
mmT
+
?
+
?
+
??
The kinetic energy of the system and from the conservation law of the
mechanical energy the velocities of separation can be determined,
.021 ???? TTT
(1) For perfectly elastic impact（ k =1）,
(3) For a general impact (0<k<1)
.021 ???? TTT If is always positive
.)(
2
1
)(
2
1
,)(
)(2
2
212
2
111
2
21
21
21
21
uvmuvmT
vv
mm
mm
TTT
?+???
?
+
????(2) For a plastic impact（ k =0),
or
44
系统动能没有损失,可以利用机械能守恒定律求碰撞后的速度。
021 ???? TTT
2
212
2
111
2
21
21
21
21
)(
2
1)(
2
1
)(
)(2
uvmuvmT
vv
mm
mm
TTT
?+???
?
+
????
或
塑性碰撞时损失的动能等于速度损耗的动能。
1
2
121
22
11
2
1
21
21
1
1
2
1
)(2 T
m
mmm
mvmv
mm
mmT
+
?+?+??
)))(1()(2 ( 2212
21
21
21 vvkmm
mmTTT ??
+????
(1) 对于完全弹性碰撞（ k =1）,
(2) 对于塑性碰撞 （ k =0）,
若 v2=0，则
(3) 对于弹性碰撞 （ 0<k <1 ）,
021 ???? TTT (恒为正值）
45
[Example 1] There is a driving pile mechanism, The hammer has mass m
1 and the height that the hammer falls is h, The pile of mass m
2 sinks by ?,The two bodies perform a plastic impact,Determine the velocity of the pile at the end of impact and
the average resistance of the soil against the pile,
According to the theorem of kinetic energy calculate the average
resistance R of the soil against the pile when it sinks,
Solution, At the beginning of the impact
the velocity of pile is and
the velocity of pile
After the plastic impact
ghv 21?.02?v
.2 21 121 ghmmmuuu +? ??
ghv 21 ?
.02 ?v
.2
21
1
21
gh
mm
m
uuu
+
?
??
46
[例 1] 打桩机。锤,m1，下落高度 h； 桩,m2，下沉 ? 。两
者塑性碰撞。求碰撞后桩的速度和泥土对桩的平均阻力。
解,碰撞开始时,
锤速,
桩速
塑性碰撞后,
ghv 21 ?
02 ?v
ghmm m
uuu
2
21
1
21
+?
??
根据动能定理,计算下沉 ? 过程中,泥土对桩的平均阻力 R。
47
.
)(
,)()2(
2
0
21
2
1
21
21
2
21
121
?
?
mm
ghm
gmgmR
h e u c eRgmgmgh
mm
mmm
+
++?
?+?
+
+
?
Beaux the first two terms of the
right side of the expression are smaller
than the third term they can be neglected
and the above expression can be
approximated as
.)(
21
2
1
?mm
ghmR
+?
48
?
?
)(
)()2(
2
0
21
2
1
21
21
2
21
121
mm
ghm
gmgmR
Rgmgmgh
mm
mmm
+
++?
?+?
+
+
?
由于右端前两项远
比第三项小,往往可以
略去,于是上式可写为,
?)( 21
2
1
mm
ghmR
+?
49
[Example 2] A steam hammer is forging metal,The mass of the
steam hammer is m1=1000kg,the total mass of the forged body and
its anvil block is m2=15000kg,the coefficient of restitution k =0.6,
Determine the efficiency of the steam hammer,
If the forged body is heated its coeficient of restitution tends to
decrease,When the temperature reaches a certain value the hammer
does not rebounce any longer,In this case we can approximately get
k =0.Thus,the efficiency of the hammer becomes
%.9494.0
21
2 ??
+? mm
m?
1T
T???
%,60 6, 0 ) 6, 0 1 ( 15000 1000 15000 ) 1 ( 2 2
2 1
2 ? ? ?
+ ? ? + ? k m m
m ?
Solution, The definition of the efficiency of the steam
hammer is
,2 1 and 0 2 1 1 1 2 v m T v ? ?,)1(
1
2
21
2 Tk
mm
mT ?
+??
Beaux we have
There fore
50
[例 2] 汽锤锻压金属。汽锤 m
1=1000kg，锤件与砧块总质量
m2=15000kg，恢复系数 k =0.6，求汽锤的效率。
1T
T???
故因,21,0 21112 vmTv ??
1221 2 )1( Tkmm
mT ?
+??
%606.0)6.01(150001000 15000)1( 22
21
2 ????+??+? kmm m?于是
若将锻件加热，可使 k 减小。当达到一定温度时，可使
锤不回跳，此时可近似认为 k =0，于是汽锤效率
%9494.0
21
2 ??+? mm m?
解,汽锤效率定义为
51
§ 19-5 The action of an impact force on a rigid body
rotating about a fixed axis,The center of impact
Assume that a rigid body rotating about a fixed axis,the
moment of inertia of which is IZ,is subjected to the action of an
external impact impulse,
At the beginning of the impact we have,
at the end of the impact we have,
Projecting the theorem of the moment of the impulse on the axis z
we get
)(eiS
w 2 2 z z I L ?
1 1 w z z I L ?
),,2,1( ni ??
z
n
i
e
i z
n
i
e
i z z z
I
S m
S m I I
?
?
? ?
? ?
1
) (
1 2
1
) (
2 1
) (
),(
w w
w w ?
?
Therefore,,
52
§ 19-5 碰撞冲量对绕定轴转动刚体的作用 ? 撞击中心
设刚体绕固定轴 z 转动，转动惯量为 IZ，受到外碰撞冲量
的作用。
碰撞开始时
碰撞结束时
由冲量矩定理在 z 轴上的投影式，有,
)(eiS ),,2,1( ni ??
22
11
w
w
zz
zz
IL
IL
?
?
z
n
i
e
iz
n
i
e
izzz
I
Sm
SmII
?
?
?
?
???
??
1
)(
12
1
)(
21
)(
)(
ww
ww
53
Below we will study the calculation and the condition of
elimination of reaction forces on bearings during an impact
The change in the angular velocity of a rigid body during
an impact is equal to the ratio of the sum of the moments of all
external impact impulses to the moment of inertia of the
body,both moments taken with respect to the axis of rotation,
Let a rigid body have a plane of symmetry,It rotates about a
fixed axis z perpendicular to this plane,The mass of the rigid body
is M,the center of mass is C and OC= a,The impact impulse is
applied to the
point K in the plane of symmetry and
OK=l, Then we have
.c o s12
zI
Sl ?ww ??
54
下面研究碰撞时轴承反力的碰撞冲量 的计算及消除条件,
OS
设刚体有对称面，绕垂直此平
面的固定轴 Oz转动，质量 M，质心
C点且 OC= a,作用在对称平面
内 K点,OK=l，则有
s
zI
Sl ?ww co s12 ??
碰撞时刚体角速度的改变，等于作用于刚体的外碰撞冲
量对转轴之矩的代数和 除以刚体对该轴的转动惯量。
55
Applying the theorem of impulse we get
.0s i n
),()(c o s 12
?+
????+
?
ww?
SS
MavuMSS
Oy
CxCxOx
.s i n
,c o s)( 12
?
?ww
SS
SMaS
Oy
Ox
??
???
Thus,
If we want that the rigid body is not subjected to the action of
impact impulses at the bearings,i,e.,
0,0 ?? oyox SS
56
应用冲量定理，有
0s i n
)()(c o s 12
?+
????+
?
ww?
SS
MavuMSS
Oy
CxCxOx
?
?ww
s i n
c o s)( 12
SS
SMaS
Oy
Ox
??
???
于是
若使刚体在轴承处不受碰撞冲量作用,即使 0,0 ??
oyox SS
57
Thus,in order to prevent the generation of reaction
forces a the in bearings of a rotating body (possessing a plane
of symmetry perpendicular to the plane through the axis of
rotation z and the center of mass C of the body some-thing is
missing here,lying in the plane of symmetry and applied to
the center of impact,
then it is necessary that
In practice,the concept of the center of impact has many applications,
sin ?0 ? ? ? S S Oy 0 cos ) ( 1 2 ? ? ? ? ? w w S Ma S Ox
,s inc o s ?? SSlIMa
z
? ?? 0?
Ma
Il z? The point K that satisfies the formula is called the
center of impact,
mast be satisfied
simultaneously,
and
Ma
Il z?
58
同时成立与 0s i n 0c o s)( 12 ??????? ??ww SSSMaS OyOx
Ma
IlSSl
I
Ma z
z
??,s inc o s ??
0??
Ma
Il z? 满足 的点 k 称为撞击中心。
欲使转动刚体（具有与 转轴垂直的对称面）的轴承处不
产生碰撞冲量，必须使碰撞冲量（作用在刚体对称面内）垂
直于转轴 O与质心 C的连线，并作用于撞击中心。
则须
撞击中心的概念在实践中有许多应用。
59
[Example 3] A homogeneous rod of mass M and length 2a can rotate about a fixed axis through the point O which is perpendicular to
the plane of the paper as shown in the diagram, The rod falls without
initial velocity from a horizonal position,colliding with a fixed
material block of mass m, The coeficient of restitution is k,
Determine the angular velocity of the rod at the end of the impact,the
impulsive reaction force a the bearing during impact and the position
of the center of impact,
Solution, From the theorem of kinetic
energy at the beginning of impact we get
.)2(3121021 21221 ww ????? aMIM g a O.23
1 ag?w
At the end of the impact we have
.2312 agkk ?? ww
Solving for we get 1w
.)2(3121021 21221 ww ????? aMIM g a O
.231 ag?w
1w
a
gkk
2
3
12 ?? ww
60
[例 3] 均质杆质量 M，长 2a，可绕通过 O点且垂直于图面
的轴转动，如图所示，杆由水平无初速落下，撞到一质量为 m
的固定物块。设恢复系数为 k，求碰撞后杆的角速度，碰撞时
轴承的碰撞冲量及撞击中心的位置。
解,碰撞开始时，由动能定理,
21221 )2(3121021 ww ????? aMIM g a O
a
g
2
3
1 ?w
碰撞结束时,
a
gkk
2
3
12 ?? ww
求得,
61
From I Sl
O
) ( 1 2 ? ? ? w w
.233 )1(4)(
2
21 a
g
l
kMa
l
IS O +?+? ww
we obtain
According to the impulse theorem,we have
.0
,)( 12
?
????
Oy
Ox
S
SSaaM ww
0,)3 4 )( ( 2 2 1 ? ? + ? Oy Ox S a l a a M S w w Hence,
The position of the center of impact is
0),? Ox S 3 4 ? al (obtained tom
62
得由 )( 12
OI
Sl??? ww
a
g
l
kMa
l
IS O
2
3
3
)1(4)( 2
21
+?+? ww
根据冲量定理，得,
0
)( 12
?
????
Oy
Ox
S
SSaaM ww
0,)34)(( 221 ??+? OyOx SalaaMS ww则
撞击中心的位置,
),0 ( 34 得出令 ?? OxSal
63
2.Two basic premises concerning impact problems,
(1)The conventional forces are much smaller than the forces of
impact,they can be neglected during the impact,
(2) The displacement of the points of a body during the time of
impact can be neglected,
Summary
1.The main properties of impact phenomena,
The impact time is extreme by small,while the impact forces are
very large, The changes of all quantities occur in a very small time
interval during impact,
3.Two fundamental theorems for impact problems are the
theorem of impulse and the theorem of the moment of the impulse,
4.The co efficient of restitution k during an impact of two bodies
equals to the ratio of the absolute values of the relative velocities
of the two bodies along the common nomal at the end of impact to
that at the beginning of the impact,
64
2、研究碰撞问题的两个基本假设,
(1) 在碰撞过程中，普通力远小于碰撞力，可以忽略不计；
(2) 物体在碰撞过程中不发生位移。
小 结
1、碰撞现象的主要特征,
碰撞过程时间极短，碰撞力非常大，它使物体的速度在极
短的时间内发生有限的变化。
3、研究碰撞问题的两个基本定理是冲量定理和冲量矩定理。
4、两物体碰撞的恢复系数 k 等于碰撞结束和开始时，两物体
接触点沿公法线方向相对速度大小的比值。
65
Au impact is called to be a perfectly elastic one if k=1.In the of
k=0 if is called a perfectly inelastic or a plastic one,
5,There is a loss of kinetic energy during an impact, the
magnitude of which depends on the value of the coefficient of
restitution,
6.An external impact impulse acting on a rigid body rotating
about a fixed axis will causes a sharp change of the angular
velocity of the body and reaction forces at the bearings,
The reaction forces at the bearings equal to zero if the
external impact impulse lies in the plane of symmetry
perpendicular to the axis of rotation and is applied to the center of
impact and directed perpendicular to the plane through the axis of
rotation and the center of mass of the body,
66
当外碰撞冲量作用在刚体的垂直于转轴的对称面内的撞
击中心，且垂直于质心与轴心的连线时，可使轴承的反碰撞
冲量等于零。
0<k <1 为弹性碰撞,k =1为完全弹性碰撞,k =0为非弹性
碰撞或塑性碰撞。
5、在碰撞过程中有动能损失,动能损失的多少取决于恢复系
数 k 值的大小。
6、作用于绕定轴转动刚体上的外碰撞冲量将引起刚体角速度
的突变，并引起轴承的反碰撞冲量。
67
68
Theoretical Mechanics
2
3
In the preceding chapters the velocities of all points of a
body under the action of forces always change continuously and
gradually, In the present chapter we will study another mechanical
phenomenon——impact,When bodies collide the velocities of all
point of a body suffer a finite change in a very small time
interval,The main subject here is to study collision
phenomena,the fundamental theorems of the theory of
impact,the loss in the kinetic energy of bodies during impact,
the center of impact,etc,
4
在前面讨论的问题中，物体在力的作用下，运动速度都
是连续地、逐渐地改变的。本章研究另一种力学现象 ——碰
撞,物体发生碰撞时，会在非常短促的时间内，运动速度突
然发生有限的改变。 本章研究的主要内容有 碰撞现象的特征，
用于碰撞过程的基本定理，碰撞过程中的动能损失，撞击中
心。
5
§ 19–1 Impact phenomenon and its fundamental
characteristics,force of impact,
§ 19-2 Fundamental theorems of the theory of
impact
§ 19–3 Impact of a particle on a fixed plate,
Coefficient of restitution,
§ 19–4 Direct Central impact of two bodies.Loss
in kinetic energy,
§ 19–5 The action of an impact force on a rigid
body rotating about a fixed axis.The center
of impact
Small Summary
Chapter 19, Impact
6
§ 19–1 碰撞现象及其基本特征 ? 碰撞力
§ 19-2 用于碰撞过程的基本定理
§ 19–3 质点对固定面的碰撞 ? 恢复系数
§ 19–4 两物体的对心正碰撞 ? 动能损失
§ 19–5 碰撞冲量对绕定轴转动刚体的作用 ?
撞击中心
小结
第十九章 碰撞
7
§ 19-1 Impact phenomenon and its fundamental
characteristics, force of impact
Fundamental characteristics of an impact phenomenon：
The velocities or moment a of the points of a body suffer finite
changes in a very small time interval during impact,The impact time
is very short,measured to be of the order of one thousandth or even
of one ten-thousandth of a second,Therefore,the values of the
acceleration and of the acting forces are very large,
Impact,The phenomenon that the velocities of the points of a
body suffer sharp changes in a very small time interval if the moving
body is suddenly subjected to a collision (including switched on or
switched off forces of constraints)is called impact,
Force of impact, The forces occurring during an impact
are called forces of impact,They are also called instantaneous
forces because the action time is very short,
8
§ 19-1 碰撞现象及其基本特征 ? 碰撞力
碰撞现象的基本特征,物体的运动速度或动量在极短的
时间内发生有限的改变。碰撞时间之短往往以千分之一秒甚
至万分之一秒来度量。因此加速度非常大，作用力的数值也
非常大。
碰撞,运动着的物体在突然受到冲击（包括突然受到约
束或解除约束）时，其运动速度发生急剧的变化，这种现象
称为碰撞。
碰撞力（瞬时力）,在碰撞过程中出现的数值很大的力
称为碰撞力；由于其作用时间非常短促，所以也称为瞬时力。
9
Let a hammer of weigh 10N hit a block of iron with v1=6m/s,the
impact time is ? =1/1000s and the hammer is rebounced with v2=1.5m/s
after impact,Determine the average value of the force of impact,
The general time-behavior of the change of the force of impact
is shown in diagram,
The average value of the force of impact is,
being 765 times larger than the weight of the hammer,
N7 6 5 0/* ?? ?SF
Illustration of the characteristics of the forces of
impact by the example of a hammer hitting iron,
Svmvm ?? 12,
s N 65,7 ? S ) 6 5, 1 ( 10 ? + S
g
Take the hammer as the object to be studied,
and
Hence,
,
10
设榔头重 10N，以 v1=6m/s的速度撞击铁块，碰撞时间 ?
=1/1000s,碰撞后榔头以 v2=1.5m/s的速度回跳。求榔头打击铁块
的力的平均值。
Svmvm ?? 12 的投影形式得
sN 65.7 ; )65.1(10 ????+ SSg
碰撞力的变化大致情况如图所示。
平均打击力,是榔头重的 765倍。 N7 6 5 0/* ?? ?SF
以榔头打铁为例说明碰撞力的特征,
以榔头为研究对象，根据动量定理
11
We can see that in the case of a large velocity of motion the
magnitude of the instantaneous force can is very large up to a surprising
value even if the mass of the body is very small,
When a flying bird of weight 17.8N collides with a flying plane,the
velocity of which is 800km/h (for a modern plane this is only of medium
velocity) the force of impact can reach 3.56?105N.I.e.,the force of impact
is 20 thousand times larger than the weight of the bird.This is the reason
of so called bird disaster in aviation,
As to impacts,Harmful aspects are,bird disasters”,damages of
mechanisms and apparatuses caused by impact etc,Advantageous
aspects are to use to use impacts to perform useful work as,for
example,forging,driving piles etc,
The study of impact,phenomena has the aim to undress and laws
in order to make positive use of its advantageous aspects and in order
to avoid its dangerous and harmful aspects,
12
可见，即使是很小的物体，当运动速度很高时，瞬时力
可以达到惊人的程度。有关资料介绍，一只重 17.8N的飞鸟
与飞机相撞，如果飞机速度是 800km/h，（对现代飞机来说，
这只是中等速度），碰撞力可高达 3.56?105N，即为鸟重的 2
万倍！这是航空上所谓“鸟祸”的原因之一。
害的一面,,鸟祸”、机械、仪器及其它物品由于碰撞损坏等。
利的一面,利用碰撞进行工作，如锻打金属，用锤打桩等。
研究碰撞现象，就是为了掌握其规律，以利用其有利的一
面，而避免其危害。
13
§ 19-2 Fundamental theorems of the
theory of impact
(1) Compared with forces of impact ordinary forces as
gravity,elastic forces etc.are is very small during the impact, Such
forces can be neglected,But it is must be noted that the action of
conventional forces on the change of the state of motion of bodies
before and after an impact cannot be neglected,
(2) Because the impact time is very small but the changes of
velocities are finite quantities,the displacement of a body during an
impact is a very small quantity which can be neglected in practice,I,
e.,the position of the body at the beginning of an impact is the same
as that at the end of the impact,
Two basic premises,
14
§ 19-2 用于碰撞过程的基本定理
(1)在碰撞过程中，重力、弹性力等普通力与碰撞力相比
小得多，其冲量可以忽略不计。但必须注意，在碰撞前和碰撞
后，普通力对物体运动状态的改变作用不可忽略。
(2)由于碰撞时间极短，而速度又是有限量，所以物体在
碰撞过程的位移很小，可以忽略不计，即认为物体在碰撞开始
时和碰撞结束时的位置相同。
两个基本假设,
15
Two basic theorems,
In theoretical mechanics we are mainly concerned with the
change of the velocity of a body under the action of a force of
impact,So,the momentum and the angular momentum theorem
become the main tools for investigating impact phenomena,
1,The momentum theorem applied to an impact,Theorem of
impulse,
Let the mass of a particle be m,its velocity before the impact
is,and at the end of the impact,the impact impulse is,
Neglecting the moment a of the convention al
forces,momentum the theorem in integral form becomes,
v u S
1)-( 1 9 Svmum ??
16
两个基本定理,
在理论力学中，我们关心的主要是由于碰撞冲量的作用而
使物体运动速度发生的变化。因此，动量定理和动量矩定理就
成了研究碰撞问题的主要工具。
1、用于碰撞过程的动量定理 ——冲量定理。
设质点的质量为 m，碰撞开始时的速度,结束瞬时的速
度,碰撞冲量,不计普通力的冲量，则质点动量定理
的积分形式为,
v
u S
1)-( 1 9 Svmum ??
17
For a system of n particles,dividing the impulses of impact
acting on the i-the particle into an external impulse of impact
and an internal impulses of impact,we have )(e
iS )(iiS
),,2,1(, )()( niSSvmum iieiiiii ??+??
Let the total mass of the system of particles be M,and
are the velocities at the end and the beginning of the
impact,Applying the theorem of the motion of the center of mass
to the,to the equation (19-2)we get
C u C v
??? )( eiCC SvMuM （ 19-3）
Adding these n equations and using (internal impulses of
impact always appear pair wise)we get 0)( ?? iiS
?????
??
)(
11
e
i
n
i ii
n
i ii
Svmum
（ 19-2）
This is the th orem of impulse,
18
对于有 n个质点组成的质点系，将作用于第 i 个质点上的
碰撞冲量分为外碰撞冲量 和内碰撞冲量,则有,
)(eiS )(iiS
),,2,1( )()( niSSvmum iieiiiii ??+??
将这 n个方程相加,且 (内碰撞冲量总是成对出现的 ),故 0)( ?? i
iS
?????
??
)(
11
e
i
n
i ii
n
i ii
Svmum 冲量定理 （ 19-2）
设质点系总质量 M,分别为碰撞结束和碰撞开始
时质心的速度，则利用质心运动定理，上式可写成,
CC vu 和
??? )( eiCC SvMuM （ 19-3）
19
The change in the momentum of a system of particles during
an impact is equal to the vector sum of all the external impulses
acting on the system,The projection forms of the expressions (19-1),(19-2) and (19-
3) are the same as in the case of the ordinary momentum theorem,
The difference is that the impulses of the conventional forces are
neglected here,
2,The angular momentum theorem applied to an impact,Theorem of the moment of the impulse,
We know from assumption (2) that the position of a particle
does not change,From expression (19-1) we then have,
.Srvmrumr ?????
Because,represent the angular moment
a of the particle with respect to point O at the beginning and at the
end of the impact, is the moment of the impact
impulse with respect to the point O
21,OO lumrlvmr ????
)( SmSr O??
s
)(12 Smll OOO ?? （ 19-4）
20
碰撞时质点系动量的改变等于作用在质点系上所有外碰
撞冲量的矢量和。
由假设 (2)知，碰撞过程中，质点的矢径 保持不变，
则由 (19-1)式，有,
r
式（ 19-1),(19-2)和 (19-3)都写成投影形式，形式上与普
通的动量定理相同，所不同的是在这里都不计普通力的冲量。
2、用于碰撞过程的动量矩定理 ——冲量矩定理
Srvmrumr ?????
而 ； 为碰撞始末时质点对
O点的动量矩。 是碰撞冲量 对 O点的矩，所以,21
,OO lumrlvmr ???? 21 OO ll 和
)( SmSr O?? s
)(12 Smll OOO ?? （ 19-4）
21
The change of the angular momentum of a particle during an
impact with respect to any center is equal to the moment of the
impact impulse acting on that particle taken with respect to the
same center,
For a system of particles the sum of the moments of the internal impact with respect,impulses to any center is zero,Thus we have,
)( )(12 eOOO SmLL ??? This expression is the of the
moment of the impulse,theorem
（ 19-5）
The expressions (19-4) and (19-5) can be expressed in projection
forms,The impulses of the conventional forces are neglected,
The change of the total angular momentum during impact of a
system with respect to any center is equal to the vector sum of the
moments of all external impulses acting on the system taken with
respect to the same center,
22
碰撞时，质点对任一固定点动量矩的改变，等于作用于
该质点的碰撞冲量对同一点之矩。
对于质点系,由于内碰撞冲量对任一点的矩之和等于零,于是有
)( )(12 eOOO SmLL ??? 冲量矩定理 （ 19-5）
式 (19-4),(19-5)也可写成投影形式，且式中均不计普通
力的冲量矩。
在碰撞过程中，质点系对任一固定点的动量矩的改变，等
于作用于质点系的外碰撞冲量，对同一点之矩的矢量和。
23
§ 19-3 Impact of a particle on a fixed
plate, Coefficient of restitution,
Let a small sphere (taken as a particle ) fall along the vertical
direction onto a fixed horizontal plate as is shown in the diagram,
Ple
ase
lo
ok
at
th
e an
im
atio
n
24
§ 19-3 质点对固定面的碰撞 ? 恢复系数
设一小球（可视为质点）沿铅直方向落到水平的固定平面
上，如图所示。
请
看
动
画
25
The first stage refers to the
time during which the small
sphere contacts with the plate to
reach its largest deformation,
During this stage the kinetic
energy of the small sphere
decreases,while the deformation
quantity of the small sphere
is,Projecting the theorem of
impulse on the axis y we have,
1)(0 Smv ???
1S
The course of the impact can be divided into two stages,
The second stage rulers to the time during which the internal elastic
deformation of the sphere forces it to escape contact,During this stage,
the elastic deformation restores gradually the kinetic energy, Let the
impact impulse be,then,Therefore,
2S 20 Smu ??
1
2
S
S
v
u ??
26
第一阶段,开始接触
至变形达到最大。该阶段
中，小球动能减小,变形增
大。设碰撞冲量为,则
应用冲量定理在 y 轴投影
式 1)(0 Smv ???
1S
碰撞过程分为两个阶段,
第二阶段,由弹性变形开始恢复到脱离接触。该阶段中，
小球动能增大，变形（弹性）逐渐恢复。设碰撞冲量为,
则,
2S
20 Smu ??
1
2
S
S
v
u ??
27
For a given material the ratio (of the absolute values of u to v is
constant and is called the coefficient of restitution,
vuk ?
The coefficient of restitution for
different bodies has to be
determined experimentally,
In general 0<k<1,We refer to the list in this book about the
coefficients of restitution for various materials,
If k=1,a limiting (ideal) condition,the impact is called perfectly
elastic impact,
If k=0,the other limiting condition,the impact is called perfectly
inelastic or plastic impact,
28
对于给定材料,|u|与 |v|的比值是不变的，该比值称为 恢复系数 。
vuk ? ——由实验测定
一般 0<k<1，各种材料的恢复系数，可查阅书中表。
k=1 理想情况 ——完全弹性碰撞。
k=0 极限情况 ——非弹性碰撞或塑性碰撞。
29
The impact of two colliding bodies is called to be a central
impact if the common normal to the surfaces of the bodies
through their point of contact pass through their centers of mass,
§ 19-4 Direct central impact of two
bodies, Loss in kinetic energy,
A central impact during which the velocities of the
approaching centers of mass are directed along a straight line
passing through the centers of mass is called a direct central
impact (direct impact),otherwise it is called an oblique central
impact (oblique impact),
30
对心碰撞,碰撞时两物体质心的连线与接触点公法线重合。
§ 19-4 两物体的对心正碰撞 ? 动能损失
对心正碰撞与对心斜碰撞,碰撞时两质心的速度也都沿两质
心连线方向，则称为对心正碰撞（正碰撞），否则称为对心
斜碰撞（斜碰撞）。
31
Please look at the animation
32
请看动画
33
1,The velocities of the two centers of mass at the end of
the direct impact
Example,The velocities of the approach of
two colliding bodies are and,
The velocities of their separation are denoted
by and (along the ling through the two
centers of mass),
Determine the velocities of separation of the
two centers of mass,
2 v 1 v ) ( 2 1 v v ?
2 u 1 u
34
1、正碰撞结束时两质心的速度
例如,两物体碰撞
碰撞前,
碰撞结束,（沿质心连线）
分析碰撞结束时两质心的速度。
)(,2121 vvvv ?
21,uu
35
Analysis,
( 2 )
21
12
vv
uuk
?
??
Choose the system of the two bodies as the
object to restudied,
From the theorem of impulse we obtain,
( 1 ), 0)()( 22112211 ?+?+ vmvmumum
A complementary equation is
(the, Applying the momentum theorem,we obtain,
for the first stage
2 2 2 2 1 1 ) (,) ( S u u m S u u m ? ? ? ? ?
1 2 2 1 1 1 ) (,) ( S v u m S v u m ? ? ? ? ?
2 1
1 2
1
1
2
2
1
2
v v
u u
v u
u u
v u
u u
S
S k
?
? ?
?
? ?
?
? ? ?
for the second stage,
there fore
,
,
,
36
研究对象：两物体组成的质点系。
由冲量定理，得,
( 1 ) 0)()( 22112211 ?+?+ vmvmumum
( 2 )
21
12
vv
uuk
?
??
分析,
列出补充方程,
（分别以两物体为研究对象，应用动量定理可得出。具体地
对于第一阶段,
对于第二阶段,
222211
122111
)(,)(
)(,)(
SuumSuum
SvumSvum
?????
?????
21
12
1
1
2
2
1
2
vv
uu
vu
uu
vu
uu
S
Sk
?
??
?
??
?
????
37
The coefficient of restitution k for a direct impact of two bodies
is equal to the ratio of the relative velocities of the bodies),
),( ) 1 ( 2 1
2 1
1
2 2 v v m m
m k v u ?
+ + + ?
2 1 m m
m
+ ),( ) 1 ( 2 1
2
1 1 v v k v u ? + ? ?
Combining the expressions (1) and (2) and solving them we can obtain,
).(2,)(2 21
21
1
2221
21
2
11 vvmm
mvuvv
mm
mvu ?
++??+??
For a perfectly elastic impact ( k=1 )we get
1,2 2 1 and v u v u ? ? 2 1 m m ? If then Thus,in the case of a perfectly
elastic impact two bodies of equal mass exchange their velocities,
38
对于两物体正碰撞的情况，恢复系数等于两物体在碰撞
结束与碰撞开始时，质心的相对速度大小的比值。 )
)()1(
)()1(
21
21
1
22
21
21
2
11
vv
mm
m
kvu
vv
mm
m
kvu
?
+
++?
?
+
+??
联立 (1),(2)式，解得,
对于完全弹性碰撞（ k=1）,
)(2 ; )(2 21
21
1
2221
21
2
11 vvmm
mvuvv
mm
mvu ?
++??+??
122121,,vuvumm ??? 则若 (碰撞后两物体交换速度 )
39
For plastic impact impact（ k =0） we get
21
221121
mm
vmvmuuu
+
+???
In the several case (0<k <1）,we get
2211,vuvu ??
2,Loss in kinetic energy during a direct impact
At the beginning of an impact the kinetic energy is
At the end of an impact the kinetic energy is 2 2,2 2 1 1 2 2 1 2 1 u m u m T + ?
2
2,2
2
1 1 1 2 1 2 1 v m v m T + ?
The loss of kinetic energy is,
))((
2
1))((
2
1
)(
2
1)(
2
1
2222211111
2
2
2
22
2
1
2
1121
uvuvmuvuvm
uvmuvmTTT
+?++??
?+?????
40
对于塑性碰撞（ k =0）,
21
221121
mm
vmvmuuu
+
+???
对于一般情况（ 0<k <1）,2211,vuvu ??
2、正碰撞过程中的动能损失
碰撞开始,
碰撞结束,2
22
2
112
2
22
2
111
2
1
2
1
2
1
2
1
umumT
vmvmT
+?
+?
则动能损失,
))((
2
1))((
2
1
)(
2
1)(
2
1
2222211111
2
2
2
22
2
1
2
1121
uvuvmuvuvm
uvmuvmTTT
+?++??
?+?????
41
We know from the formula of the velocities of the two centers of
mass at the end of impact that
).()1( nd )()1( 21
21
1
2221
21
2
11 vvmm
mkuvavv
mm
mkuv ?
++????++??
Substituting these results into the expression for,we obtain,
)].())[(()1(21 221221
21
21 uvuvvv
mm
mmkT +?+?
++??
2
2 1
2
2 1
2 1
2 1 ),)( 1 ( ) ( 2 v v k m m
m m T T T ? ?
+ ? ? ? ?
2 1 2 1 ).Therefore,( v v k u u ? ? ? ?
T?
Again
42
由正碰撞结束时两质心的速度公式知,
)()1( ; )()1( 21
21
1
2221
21
2
11 vvmm
mkuvvv
mm
mkuv ?
++????++??
代入上式中，得,
)]())[(()1(21 221221
21
21 uvuvvvmm mmkT +?+?++??
2
21
2
21
21
21
2121
))(1(
)(2
)(
vvk
mm
mm
TTT
vvkuu
??
+
?????
?????又
43
The loss in kinetic energy of a system of bodies during a plastic
impact equals to the kinetic energy the system would have if the its
bodies would move with the lost velocities,If v2=0 then
.
1
1
2
1
)(2
1
2
121
22
11
2
1
21
21 T
m
mmm
mvmv
mm
mmT
+
?
+
?
+
??
The kinetic energy of the system and from the conservation law of the
mechanical energy the velocities of separation can be determined,
.021 ???? TTT
(1) For perfectly elastic impact（ k =1）,
(3) For a general impact (0<k<1)
.021 ???? TTT If is always positive
.)(
2
1
)(
2
1
,)(
)(2
2
212
2
111
2
21
21
21
21
uvmuvmT
vv
mm
mm
TTT
?+???
?
+
????(2) For a plastic impact（ k =0),
or
44
系统动能没有损失,可以利用机械能守恒定律求碰撞后的速度。
021 ???? TTT
2
212
2
111
2
21
21
21
21
)(
2
1)(
2
1
)(
)(2
uvmuvmT
vv
mm
mm
TTT
?+???
?
+
????
或
塑性碰撞时损失的动能等于速度损耗的动能。
1
2
121
22
11
2
1
21
21
1
1
2
1
)(2 T
m
mmm
mvmv
mm
mmT
+
?+?+??
)))(1()(2 ( 2212
21
21
21 vvkmm
mmTTT ??
+????
(1) 对于完全弹性碰撞（ k =1）,
(2) 对于塑性碰撞 （ k =0）,
若 v2=0，则
(3) 对于弹性碰撞 （ 0<k <1 ）,
021 ???? TTT (恒为正值）
45
[Example 1] There is a driving pile mechanism, The hammer has mass m
1 and the height that the hammer falls is h, The pile of mass m
2 sinks by ?,The two bodies perform a plastic impact,Determine the velocity of the pile at the end of impact and
the average resistance of the soil against the pile,
According to the theorem of kinetic energy calculate the average
resistance R of the soil against the pile when it sinks,
Solution, At the beginning of the impact
the velocity of pile is and
the velocity of pile
After the plastic impact
ghv 21?.02?v
.2 21 121 ghmmmuuu +? ??
ghv 21 ?
.02 ?v
.2
21
1
21
gh
mm
m
uuu
+
?
??
46
[例 1] 打桩机。锤,m1，下落高度 h； 桩,m2，下沉 ? 。两
者塑性碰撞。求碰撞后桩的速度和泥土对桩的平均阻力。
解,碰撞开始时,
锤速,
桩速
塑性碰撞后,
ghv 21 ?
02 ?v
ghmm m
uuu
2
21
1
21
+?
??
根据动能定理,计算下沉 ? 过程中,泥土对桩的平均阻力 R。
47
.
)(
,)()2(
2
0
21
2
1
21
21
2
21
121
?
?
mm
ghm
gmgmR
h e u c eRgmgmgh
mm
mmm
+
++?
?+?
+
+
?
Beaux the first two terms of the
right side of the expression are smaller
than the third term they can be neglected
and the above expression can be
approximated as
.)(
21
2
1
?mm
ghmR
+?
48
?
?
)(
)()2(
2
0
21
2
1
21
21
2
21
121
mm
ghm
gmgmR
Rgmgmgh
mm
mmm
+
++?
?+?
+
+
?
由于右端前两项远
比第三项小,往往可以
略去,于是上式可写为,
?)( 21
2
1
mm
ghmR
+?
49
[Example 2] A steam hammer is forging metal,The mass of the
steam hammer is m1=1000kg,the total mass of the forged body and
its anvil block is m2=15000kg,the coefficient of restitution k =0.6,
Determine the efficiency of the steam hammer,
If the forged body is heated its coeficient of restitution tends to
decrease,When the temperature reaches a certain value the hammer
does not rebounce any longer,In this case we can approximately get
k =0.Thus,the efficiency of the hammer becomes
%.9494.0
21
2 ??
+? mm
m?
1T
T???
%,60 6, 0 ) 6, 0 1 ( 15000 1000 15000 ) 1 ( 2 2
2 1
2 ? ? ?
+ ? ? + ? k m m
m ?
Solution, The definition of the efficiency of the steam
hammer is
,2 1 and 0 2 1 1 1 2 v m T v ? ?,)1(
1
2
21
2 Tk
mm
mT ?
+??
Beaux we have
There fore
50
[例 2] 汽锤锻压金属。汽锤 m
1=1000kg，锤件与砧块总质量
m2=15000kg，恢复系数 k =0.6，求汽锤的效率。
1T
T???
故因,21,0 21112 vmTv ??
1221 2 )1( Tkmm
mT ?
+??
%606.0)6.01(150001000 15000)1( 22
21
2 ????+??+? kmm m?于是
若将锻件加热，可使 k 减小。当达到一定温度时，可使
锤不回跳，此时可近似认为 k =0，于是汽锤效率
%9494.0
21
2 ??+? mm m?
解,汽锤效率定义为
51
§ 19-5 The action of an impact force on a rigid body
rotating about a fixed axis,The center of impact
Assume that a rigid body rotating about a fixed axis,the
moment of inertia of which is IZ,is subjected to the action of an
external impact impulse,
At the beginning of the impact we have,
at the end of the impact we have,
Projecting the theorem of the moment of the impulse on the axis z
we get
)(eiS
w 2 2 z z I L ?
1 1 w z z I L ?
),,2,1( ni ??
z
n
i
e
i z
n
i
e
i z z z
I
S m
S m I I
?
?
? ?
? ?
1
) (
1 2
1
) (
2 1
) (
),(
w w
w w ?
?
Therefore,,
52
§ 19-5 碰撞冲量对绕定轴转动刚体的作用 ? 撞击中心
设刚体绕固定轴 z 转动，转动惯量为 IZ，受到外碰撞冲量
的作用。
碰撞开始时
碰撞结束时
由冲量矩定理在 z 轴上的投影式，有,
)(eiS ),,2,1( ni ??
22
11
w
w
zz
zz
IL
IL
?
?
z
n
i
e
iz
n
i
e
izzz
I
Sm
SmII
?
?
?
?
???
??
1
)(
12
1
)(
21
)(
)(
ww
ww
53
Below we will study the calculation and the condition of
elimination of reaction forces on bearings during an impact
The change in the angular velocity of a rigid body during
an impact is equal to the ratio of the sum of the moments of all
external impact impulses to the moment of inertia of the
body,both moments taken with respect to the axis of rotation,
Let a rigid body have a plane of symmetry,It rotates about a
fixed axis z perpendicular to this plane,The mass of the rigid body
is M,the center of mass is C and OC= a,The impact impulse is
applied to the
point K in the plane of symmetry and
OK=l, Then we have
.c o s12
zI
Sl ?ww ??
54
下面研究碰撞时轴承反力的碰撞冲量 的计算及消除条件,
OS
设刚体有对称面，绕垂直此平
面的固定轴 Oz转动，质量 M，质心
C点且 OC= a,作用在对称平面
内 K点,OK=l，则有
s
zI
Sl ?ww co s12 ??
碰撞时刚体角速度的改变，等于作用于刚体的外碰撞冲
量对转轴之矩的代数和 除以刚体对该轴的转动惯量。
55
Applying the theorem of impulse we get
.0s i n
),()(c o s 12
?+
????+
?
ww?
SS
MavuMSS
Oy
CxCxOx
.s i n
,c o s)( 12
?
?ww
SS
SMaS
Oy
Ox
??
???
Thus,
If we want that the rigid body is not subjected to the action of
impact impulses at the bearings,i,e.,
0,0 ?? oyox SS
56
应用冲量定理，有
0s i n
)()(c o s 12
?+
????+
?
ww?
SS
MavuMSS
Oy
CxCxOx
?
?ww
s i n
c o s)( 12
SS
SMaS
Oy
Ox
??
???
于是
若使刚体在轴承处不受碰撞冲量作用,即使 0,0 ??
oyox SS
57
Thus,in order to prevent the generation of reaction
forces a the in bearings of a rotating body (possessing a plane
of symmetry perpendicular to the plane through the axis of
rotation z and the center of mass C of the body some-thing is
missing here,lying in the plane of symmetry and applied to
the center of impact,
then it is necessary that
In practice,the concept of the center of impact has many applications,
sin ?0 ? ? ? S S Oy 0 cos ) ( 1 2 ? ? ? ? ? w w S Ma S Ox
,s inc o s ?? SSlIMa
z
? ?? 0?
Ma
Il z? The point K that satisfies the formula is called the
center of impact,
mast be satisfied
simultaneously,
and
Ma
Il z?
58
同时成立与 0s i n 0c o s)( 12 ??????? ??ww SSSMaS OyOx
Ma
IlSSl
I
Ma z
z
??,s inc o s ??
0??
Ma
Il z? 满足 的点 k 称为撞击中心。
欲使转动刚体（具有与 转轴垂直的对称面）的轴承处不
产生碰撞冲量，必须使碰撞冲量（作用在刚体对称面内）垂
直于转轴 O与质心 C的连线，并作用于撞击中心。
则须
撞击中心的概念在实践中有许多应用。
59
[Example 3] A homogeneous rod of mass M and length 2a can rotate about a fixed axis through the point O which is perpendicular to
the plane of the paper as shown in the diagram, The rod falls without
initial velocity from a horizonal position,colliding with a fixed
material block of mass m, The coeficient of restitution is k,
Determine the angular velocity of the rod at the end of the impact,the
impulsive reaction force a the bearing during impact and the position
of the center of impact,
Solution, From the theorem of kinetic
energy at the beginning of impact we get
.)2(3121021 21221 ww ????? aMIM g a O.23
1 ag?w
At the end of the impact we have
.2312 agkk ?? ww
Solving for we get 1w
.)2(3121021 21221 ww ????? aMIM g a O
.231 ag?w
1w
a
gkk
2
3
12 ?? ww
60
[例 3] 均质杆质量 M，长 2a，可绕通过 O点且垂直于图面
的轴转动，如图所示，杆由水平无初速落下，撞到一质量为 m
的固定物块。设恢复系数为 k，求碰撞后杆的角速度，碰撞时
轴承的碰撞冲量及撞击中心的位置。
解,碰撞开始时，由动能定理,
21221 )2(3121021 ww ????? aMIM g a O
a
g
2
3
1 ?w
碰撞结束时,
a
gkk
2
3
12 ?? ww
求得,
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From I Sl
O
) ( 1 2 ? ? ? w w
.233 )1(4)(
2
21 a
g
l
kMa
l
IS O +?+? ww
we obtain
According to the impulse theorem,we have
.0
,)( 12
?
????
Oy
Ox
S
SSaaM ww
0,)3 4 )( ( 2 2 1 ? ? + ? Oy Ox S a l a a M S w w Hence,
The position of the center of impact is
0),? Ox S 3 4 ? al (obtained tom
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得由 )( 12
OI
Sl??? ww
a
g
l
kMa
l
IS O
2
3
3
)1(4)( 2
21
+?+? ww
根据冲量定理，得,
0
)( 12
?
????
Oy
Ox
S
SSaaM ww
0,)34)(( 221 ??+? OyOx SalaaMS ww则
撞击中心的位置,
),0 ( 34 得出令 ?? OxSal
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2.Two basic premises concerning impact problems,
(1)The conventional forces are much smaller than the forces of
impact,they can be neglected during the impact,
(2) The displacement of the points of a body during the time of
impact can be neglected,
Summary
1.The main properties of impact phenomena,
The impact time is extreme by small,while the impact forces are
very large, The changes of all quantities occur in a very small time
interval during impact,
3.Two fundamental theorems for impact problems are the
theorem of impulse and the theorem of the moment of the impulse,
4.The co efficient of restitution k during an impact of two bodies
equals to the ratio of the absolute values of the relative velocities
of the two bodies along the common nomal at the end of impact to
that at the beginning of the impact,
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2、研究碰撞问题的两个基本假设,
(1) 在碰撞过程中，普通力远小于碰撞力，可以忽略不计；
(2) 物体在碰撞过程中不发生位移。
小 结
1、碰撞现象的主要特征,
碰撞过程时间极短，碰撞力非常大，它使物体的速度在极
短的时间内发生有限的变化。
3、研究碰撞问题的两个基本定理是冲量定理和冲量矩定理。
4、两物体碰撞的恢复系数 k 等于碰撞结束和开始时，两物体
接触点沿公法线方向相对速度大小的比值。
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Au impact is called to be a perfectly elastic one if k=1.In the of
k=0 if is called a perfectly inelastic or a plastic one,
5,There is a loss of kinetic energy during an impact, the
magnitude of which depends on the value of the coefficient of
restitution,
6.An external impact impulse acting on a rigid body rotating
about a fixed axis will causes a sharp change of the angular
velocity of the body and reaction forces at the bearings,
The reaction forces at the bearings equal to zero if the
external impact impulse lies in the plane of symmetry
perpendicular to the axis of rotation and is applied to the center of
impact and directed perpendicular to the plane through the axis of
rotation and the center of mass of the body,
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当外碰撞冲量作用在刚体的垂直于转轴的对称面内的撞
击中心，且垂直于质心与轴心的连线时，可使轴承的反碰撞
冲量等于零。
0<k <1 为弹性碰撞,k =1为完全弹性碰撞,k =0为非弹性
碰撞或塑性碰撞。
5、在碰撞过程中有动能损失,动能损失的多少取决于恢复系
数 k 值的大小。
6、作用于绕定轴转动刚体上的外碰撞冲量将引起刚体角速度
的突变，并引起轴承的反碰撞冲量。
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