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§9.1 Momentum and Newton’s second law of
motion
Collisions:
All around us (cars, buses)
Grand astronomical scale (stars, galexies)
Nuclear, atomic and molecular physics
The impulse-momentum theorem relates the
change in the momentum of a system to the
total force on it and the time interval during
which this total forces acts.
1
§9.1 Momentum and Newton’s second law of
motion
1. momentum
(1)a single particle of massr m
r
r
dr
Define: p = m v = m
dt
(2)a group of particles of mass m1, m2, …, mi, …
r
r
r
p = ∑ pi = ∑ m i v i
Position vector
i
of center of mass
i
2. the center of mass
Total mass
r
r
d
r
The total momentum of p = M CM ?
a system of particle：
dt r
r
r
r
dr
d ⎛ mr ⎞
p = ∑ pi = ∑ mi i = M ⎜⎜ ∑ i i ⎟⎟
dt
dt ⎝ i M ⎠
i
i
§9.1 Momentum and Newton’s second law of
motion
(1)The position vector of mass
center (weighted average)
Define: r
r
mr
rCM = ∑ i i
M
i
m r m r
m r
= 1 r1 + 2 r2 + L + N rN
M
M
M
That is
z
m1
r
r1
O
x
m2
C
r
r 2 rr
c
r mN
rN
y
r
r
r
r
m 1 r1 + m 2 r2 + L + m N rN
rCM =
m1 + m 2 + L + m N
Total mass of the particle system:
M = mtotal = m1 + m 2 + L + m N
2
§9.1 Momentum and Newton’s second law of
motion
The components in the Cartesian coordinate
system:
For the system composed of single particles
N
∑
r
r CM =
i=1
N
r
m i ri
x CM =
M
∑
m i xi
i =1
M
N
z
m2
m1
r
r1
r
r2
r
rc
m i yi
i =1
M
N
r
rN m N
O
x
y CM =
C
∑
y
z CM =
∑
i=1
m izi
M
§9.1 Momentum and Newton’s second law of
motion
For the common objects as essentially
continuous distribution of matter:
z
dm( x, y, z )
r
r
x o
x CM =
M
y
r
r CM =
∫
r
rdm
M
y CM =
z CM =
∫ xdm
∫
M
yd m
∫
M
zd m
M
3
§9.1 Momentum and Newton’s second law of
motion
Example 1: P420 56(b)
y
3m
l
Solution:
m
l
l
2m x
m × 0 + 2ml + 3ml / 2 7 l
=
6m
12
m × 0 + 2m × 0 + 3ml sin 60o 3 3l
=
=
6m
12
xCM =
yCM
§9.1 Momentum and Newton’s second law of
motion
Example 2: A thin strip of material is bent
into the shape of a semicircle of radius R.
Find its center of mass.
y
Solution:
dφ dm
since the symmetry,
R
φ
we obtained
x CM = 0
1
1 π
M
y CM =
y
d
m
=
(
R
sin
)
dφ
φ
M ∫
M ∫0
π
R π
2R
= ∫ sin φ d φ =
= 0 .637 R
π
0
x
π
4
§9.1 Momentum and Newton’s second law of
motion
(2) Momentum of system of particles
r
r
r
r
r
dri
drCM
d ⎛ mi ri ⎞
⎟⎟ = M
= M ⎜⎜ ∑
p = ∑ pi = ∑ mi
d
d
dt
t
t⎝ i M ⎠
i
i
r
r
r
drCM
ptotal = M
= MvCM
dt
(3) Kinetic energy of system of particles
r r
r
rri = rrCM + rir′
v i = vCM + v i′
z
O
r
ri
mi r
ri′
r
rCM
x
C
y
§9.1 Momentum and Newton’s second law of
motion
The total Kinetic energy of the whole system
1
1 r r
KE total = ∑ m i v i2 = ∑ m i v i ⋅ v i
i 2
i 2
r
r
r
r
1
= ∑ m i (vCM + v i′ ) ⋅ (vCM + v i′ )
i 2
r
r
r
∑i m i v i′ = 0
1
2
2
= ∑ m i (vCM + v ′i + 2vCM ⋅ v i′ )
i 2
r
r
1
1
2
= ∑ m i vCM
+ ∑ m i v i′ 2 + vCM ⋅ ∑ m i v i′
2 i
2 i
i
KE total =
1
1
2
MvCM
+ ∑ m i v ′i 2
2
i 2
5
§9.1 Momentum and Newton’s second law of
motion
3. Momentum and Newton’s second law
(1)dynamics of a single particle
r
r
r
r
dp d(m v )
=
= ma
F total =
dt
dt
(2)dynamics of a system of particles
(a)The velocity of the center of mass:r
r
mi vi
r
r
r
v
dm
r
mi dri ∑
drCM d mi ri
∫
i
或
vCM =
= ∑
=∑
=
M dt
M
M
dt
dt i M
(b)The acceleration of the center of mass:
r
r
m i ai
r
∑
2r
dm
a
r
d v CM d rCM
∫
i
a CM =
=
=
or
M
M
dt
dt 2
§9.1 Momentum and Newton’s second law of
motion
(c)Momentum of a system of particles
r
r
r
drCM
ptotal = M
= MvCM
dt
(d) Newton’s second law for the system of particles
r
m
v
d
(
r
r
∑
i
i)
r
r
dptotal
dvCM
i
Ftotal =
=
=M
= MaCM
dt
dt
dt
The longer the total force acts on a system,
the longer the system is subjected to an
acceleration and the greater the change in its
velocity.
6
§9.1 Momentum and Newton’s second law of
motion
(e) Internal and external forces
The forces from other particles within the
system of particles is called internal force.
The forces from sources outside the system
of particles is called external force.
The total force of the system:
r
r r
r
r
Ftotal = F1 + F2 + L + Fi + L + FN
r
r
r
r
= F1in + F1ex + F2 in + F2ex + L
r
r
r
= F1ex + + F2ex + L + Fi ex + L
§9.1 Momentum and Newton’s second law of
motion
According to
Newton’s third law
r
r
Fin = ∑ Fi in ≡ 0
i
because
r
F1ex
m1
r
F2ex
r
r F
F12 21
r
F3ex
r
F13 r
F31
m2
r
F23
r
F32
m3
r
r
r
r
d p1
F 1 = F 1 ex + F 1 in =
d rt
r
r
r
dp2
F 2 = F 2 ex + F 2 in =
dt
LLL
r
r
r
r
dpN
F N = F N ex + F N in =
dt
7
§9.1 Momentum and Newton’s second law of
motion
r
r
Consider Fin = ∑ Fi in ≡ 0
i
The sum of the equations all above is
r
r
r
r
r
d r
F1 ex + F2 ex + L + F N ex =
( p1 + p 2 + L + p N )
dt
r
r
r
r
dptotal
dvCM
Ftotal =
=M
= MaCM
dt
dt
§9.2 Impulse-momentum theorem
1. Impulse-momentum theorem of a single particle
r r
r
r
p
d
because
= Ftotal
Ftotal dt = dp
dt
r r
let dI = Ftotal dt --differential impulse
The impulse of the total force on the particle
r
r
r
tf r
pf r
r
r
r
I = ∫ dI = ∫ Ftotal (t)dt = ∫r dp = p f − pi = ∆p
ti
pi
8
§9.2 Impulse-momentum theorem
2. The Cartesian components of the impulse
I
I
x
=
y
=
∫
∫
∫
Iz =
t
f
ti
t
f
ti
t
f
ti
total
dt = ∆px
y total
dt = ∆p y
Fx
F
F z total d t = ∆ p z
Fx total
Fx ave
t
ti
O
tf
3. Impulse and average impulse force
r
r
t2 r
I = ∫ Ftotal dt = Fave ∆t
t1
t2
Ix = ∫ Fx aved t = Fx ave∆t I y = Fy ave∆t Iz = Fz ave∆t
t1
§9.2 Impulse-momentum theorem
4. Impulse-momentum theorem of a system of
particles
r
r
dptotal
For a particle system
= Ftotal
dr t
r
I total =
∫
tf
ti
r
F total d t =
∫
pf
r
pi
r
r
d p total = ∆ p total
Components in the Cartesian coordinate system:
tf
I x total = ∫ Fx totaldt =∆p x total
ti
tf
I y total = ∫ F y totaldt = ∆p y total
ti
tf
I z total = ∫ Fz totaldt = ∆pz total
ti
notice：
r
Fin total =
N
∑
i =1
r
F i in ≡ 0
r
tf r
I in total = ∫ Fin totaldt ≡ 0
ti
9
§9.2 Impulse-momentum theorem
Example 1: throw a ball of mass 0.40kg against
a brick wall. Find (a)the impulse of the force on
the ball;(b)the average horizontal force exerted
on the ball, if the ball is in contact with the wall
for 0.01s.
Solution: (a) I x = pxf − pxi
v1 = −30m/s
p xi = mv 1 = 0.40 × ( −30)
v 2 = 20m/s
= −12kg ⋅ m/s
p xf = mv 2 = 0.40 × 20
x
= 8.0kg ⋅ m/s
I x = p xf − p xi = 8.0 − (12) = 20 kg ⋅ m/s
§9.2 Impulse-momentum theorem
t
(b) I x = ∫ f Fx dt = Fx ave ∆t
ti
Fx ave =
Ix
20
=
= 2000N
∆t 0.01
35o
Example 2: a base ball of mass 0.14kg is moving
horizontally at speed of 42m/s when it is truck
by the bat. It leaves the bat in a direction at an
angle φ=35º above its incident path and with a
speed of 50m/s. (a)Find the impulse of the force
exerted on the ball. (b) Assuming the collision
last for 1.5 ms, what is the average force?
(c)Find the change in the momentum of the bat.
10
§9.2 Impulse-momentum theorem
r r
r
r
Solution : (a) I = p f − pi = ∆p
I x = p fx − pix = ∆px = mv fx − (−mvix )
= mv f cosφ + mvi = 11.6kg ⋅ m/s
r
pi
r
pf
yr
θ
I
φ
x
I y = p fy − piy = ∆p y = mv fy − mv iy
= mv f sin φ − 0 = 4.0kg ⋅ m/s
Magnitude I =
I x2 + I y2 = 12.3kg ⋅ m/s
θ = tan
direction
−1
Iy
Ix
= 19
o
§9.2 Impulse-momentum theorem
r
r
I
12.3
(b) Fav =
=
= 8200N
∆t 0.0015
θ = tan
−1
Iy
Ix
= 19
o
r
pi
r α
∆pbat
r
pf
yr
θ
I
φ
x
(c) According to the Newton’s third law, the
change in momentum of the bat is equal and
opposite to that of the ball.
Magnitude
∆p = I = I x2 + I y2 = 12.3kg ⋅ m/s
direction
α = 180o − tan −1
Iy
Ix
= 171o
11
§9.2 Impulse-momentum theorem
Example 3: Coal drops from a stationary
hopper of height 2.0 m at rate of 40 kg/s on
to conveyer belt moving with a speed of 2.0
m/s, find the average force exerted on the
belt by the coal in the process of the
transportation.
h
r
v
A
§9.2 Impulse-momentum theorem
Solution:
Choose the mass element
of the coal as the particle
y
△m
r
p1
∆px = p2 x − p1 x = p2 = ∆mv
∆py = p2 y − p1 y = p1 = ∆mv′
= ∆m 2 gh
v = 2.0m/s
r
p2
r r
r
p2 − p1 = ∆p
α
β
x
According to the impulse-momentum theorem
r
r
tf r
r
I = ∫ Fdt = Fav ∆t = ∆p
ti
12
§9.2 Impulse-momentum theorem
∆p x p2 ∆mv
=
=
= qv = 80 ( N )
∆t
∆t
∆t
∆p y p1 ∆m
=
=
=
= q 2 gh = 125.2 ( N )
∆t ∆t ∆t
r
Fx av =
F y av
y
Fav = Fx av + F y av = 149 ( N )
△m
r
p1
The angle with respect
to the x axis
α = tan -1
F y av
Fx av
= tan -1
p2
r
r r
p2 − p1 = ∆p
α
β
x
125.2
= 57.4o
80
§9.2 Impulse-momentum theorem
the average force exerted on the belt by the coal
F ′ = 149 ( N )
β = 180o − 57.4o = 122.6o
y
△m
r
p1
r
p2
r r
r
p2 − p1 = ∆p
α
β
x
13
§9.3 the rocket: a system with variable mass
A rocket is propelled forward by
rearward ejection of burned fuel
(gases)that initially was in rocket.
The forward force in the rocket is
the reaction to the backward
force on the ejected material. The
total mass is constant , but the
mass of the rocket itself decreases
as material is ejected.
Assuming: initial mass of the system is m0;
initial speed of the system is v0;
final mass of the system is m´;
speed of ejected gases with respect
to the rocket is ve.
§9.3 the rocket: a system with variable mass
The system: rocket and exhausted gases
At instant t:
The total mass of the rocket: m
m
The total momentum of the rocket:
r
r
pi = m v
At instant t+dt:
mass of the rocket m + dm (dm < 0)
Mass of the exhausted gases − dm
r
r
m + dm
Velocity of the rocket v + dv
Velocity of the exhausted gases
r r
ve + v
− dm
r
v
r
r
v + dv
r
ve
14
§9.3 the rocket: a system with variable mass
The total momentum of the system
r
r r
r r
p f = (m + dm)(v + dv ) + (−dm)(ve + v )
r
r
r
r r
r
= mv + mdv + dmv + dmdv − vedm − dmv
r
r r
= mv + mdv − vedm
The change of the momentum in m + d m
time interval dt
r
r r
− dm
dp = m dv − v e dm
r
r
v + dv
r
ve
Ignore the resistant force of air, according to
the impulse –momentum theorem
r r
r
dI = Ftotaldt = dp
§9.3 the rocket: a system with variable mass
r
r
r
dp
dv r dm
Ftotal =
=m
− ve
dt
dt
dt
r
r
r dm
dv
=m
Ftotal + v e
dt
dt
r
r dm
Fthrust = v e
--thrust of the rocket
dt
1For ra vertical flying
F total = − mg ˆj
r r
r
r
d
I
= Ftotaldt = dp
ˆ
ve = −ve j
r
d v = d vˆj
15
§9.3 the rocket: a system with variable mass
then
− mg dt = m dv + v e dm
Integrate the both sides of the equation
(assuming that all fuel is exhausted at
instant t ´)
t′
− ∫ gd t =
0
∫
vm
v0
m′
dm
m0 m
dv + v e ∫
The maximum speed of the rocket
m
v m − v 0 = v e ln 0 − g t ′
m′
m
v m = v 0 + v e ln 0 − g t ′
m′
§9.3 the rocket: a system with variable mass
2For a horizontal flying
r
r
r
F x total = 0 , v e = − v e iˆ , d v = d v iˆ
r
r
r dm
dv
dm dv
Q Ftotal + v e
=m
=
∴ −
dt
dt
m
ve
v m = v 0 + v e ln
m0
m′
If v0=0, then
v m = v e ln
m0
m′
16
§9.3 the rocket: a system with variable mass
3Step rocket:
vm = v0 + ve1lnN1 + ve 2lnN2 + L+ venlnNn
if
v0 = 0
v e 1 = v e 2 = v e 3 = 2500 m ⋅ s -1
N1 = N 2 = N 3 = 6
vm = 2500 ⋅ ln6 3 = 13440 m ⋅ s -1
then
It is enough for the launch of a satellite.
Question:
What will be the effect of the
air resistant force?
§9.3 the rocket: a system with variable mass
三
17
§9.3 the rocket: a system with variable mass
长征2号C火箭
光荣的长征火箭家族
中国已经自行研制了四大系列12
种型号的运载火箭：
长征1号系列：发射近地轨道小
卫星.
长征2号系列：发射近地轨道中
、大型卫星，和其它航天器.
长征3号系列：发射地球同步高
轨道卫星和航天器.
长征4号系列：发射太阳同步轨
道卫星.
§9.3 the rocket: a system with variable mass
1970年4月……2003年5月：发射70次，将54颗国产卫
星，27颗外国卫星，4艘神舟号无人飞船送入太空。
成功率91％（美国德尔塔火箭：94％，欧空局阿丽亚
娜火箭：93％，俄罗斯质子号火箭：90％）。
2003年10月15日：长征2号F运载火箭成功发射神舟5
号载人飞船。
宇航员杨力伟
长征3号A火箭发射的东
方红三号通信卫星
18
§9.3 the rocket: a system with variable mass
神州1号
神州4号
神州2号
神
州
5
号
升
空
神州3号
§9.3 the rocket: a system with variable mass
10月15日，我国在
酒泉卫星发射中心进行
首次载人航天飞行。９
时整，“神舟”五号载
人飞船发射升空。
10月１６日６时２３
分，“神舟”五号载人飞船
在内蒙古主着陆场成功着
陆，实际着陆点与理论着陆
点相差４.8公里。返回舱完
好无损。航天英雄杨利伟自
主出舱。我国首次载人航天
飞行取得圆满成功。
19
神州5号拍摄的中国版图
沙漠化的中国!!!
从图片上我
们可以清晰看
到我国大部分
土地没有被绿
色植被所覆
盖，而是以赤
裸裸的黄色直
接面向宇宙，
多年的干旱和
毫无节制的滥
砍滥伐使我们
的绿色极度匮
乏！
§9.3 the rocket: a system with variable mass
2005年10月12日：长征2号F型运载火箭成功发射神舟6号载人飞船。
报道：“我们在神舟五号的基础上继续攻克多项载人航天的基本技术，
第一次进行了真正有人参与的空间科学实验。”
神舟6号矗立在发射台上
宇航员费俊龙、聂海胜
20
§9.3 the rocket: a system with variable mass
神舟载人飞
船在组装调
试。最上部
为轨道舱、
中部灰黑色
圆柱体为返
回舱、下部
白色段为推
进舱。
§9.3 the rocket: a system with variable mass
21
§9.3 the rocket: a system with variable mass
Example 1: A spaceship with a total mass of
13600 kg is moving relative to a certain inertial
reference frame with a speed of 960 m/s in a
region of space of negligible gravity. It fires its
rocket engines to give an acceleration parallel
to the initial velocity. The rocket eject gas at a
constant rate of 146 kg/s with a constant speed
(relative to the spaceship) of 1520 m/s, and they
are fired until 9100 kg of fuel has been burned
and ejected. (a) what is the thrust produced by
the rockets? (b) What is the velocity of the
spaceship after the rockets have fired?
§9.3 the rocket: a system with variable mass
Solution: (a) The thrust is given by
dm
Fthrust = v e
= 1520 × 146 = 2.22 × 105 (N)
dt
(b)choose the positive x direction to be that of
the spaceship’s initial velocity, then we have
vf
m f dm
dv
dm
, ∫ dv = v e ∫
m
= ve
vi
mi
m
dt
dt
mf
v f − v i = v e ln
mi
4500
v f = 960 + ( −1520) ln
= 2640m/s
13600
22
§9.4 conservation of momentum
The law of conservation : r
r
dptotal
For a particle system
= Ftotal
r
t
d
r
tf r
pf
r
r
I total = ∫ F total d t = ∫ r d p total = ∆ p total
ti
rp i
r
dptotal
=0
If Ftotal = 0, then
dt
r
or I total =
r
ptotal = constant
∫
tf
ti
r
F total d t =
∫
r
pf
r
pi
r
r
d p total = ∆ p total = 0
r
r
p total, i = p total, f
Notice: 1The condition of conservation!!!
§9.4 conservation of momentum
2for an isolated particle system
r
r
ptotal = Mv c = constant
The center of mass will remain constant velocity.
3the component forms of conservation of
momentum
Fx total = 0
px total = ∑ mnvnx = constant
n
Fy total = 0
p y total = ∑ mnvny = constant
Fz total = 0
pz total = ∑ mnvnz = constant
n
n
4if the internal force is much larger than the
external forces
23
§9.4 conservation of momentum
Example 1: The scattering of alpha. The alpha
r
collide with the Oxygen atom.
v2
r
Find the speed ratio of
v
m 1
alpha before and after the
θ
scattering, θ = 72o , β = 41o .
M βr
v
Solution:
For the system of alpha and Oxygen atom,
there is no external forces, the total momentum
is conserved.
r
r
r
pa = m v1 pO = 0
Before collision:
r
r
r
r
After collision:
pa = m v 2 pO = Mv
§9.4 conservation of momentum
Conservation of momentum
r
r
r
m v1 = m v 2 + Mv
y
In Cartesian coordinate
system
mv 1 = mv 2 cosθ + Mv cos β
o
r
mv2
θ
β
r
mv1 x
r
Mv
0 = mv 2 sin θ − Mv sin β
the speed ratio of alpha before and after the
scattering
v2
sin β
sin 41o
=
=
= 0.71
v1 sin(θ + β ) sin 72o + 41o
(
)
24
§9.4 conservation of momentum
Example 2: The ballistic pendulum was used to
measure the speed of the bullets before
electronic timing devices were developed. The
version shown in figure consists of a large
block of wood of mass M=5.4 kg, A bullet of
mass m= 9.5 g is fired into the block, coming
quickly to rest. The
block and bullet then swing
upward, their center of mass
rising a vertical maximum
distance h=6.3 cm before the
pendulum comes to rest,
what is the speed of the bullet
just prior to the collision?
§9.4 conservation of momentum
Solution:
Two steps:
The bullet-block collision
The bullet-block rise
Step 1: the horizontal total force is zero, the
momentum of system is conserved.
m
mv = ( m + M )V ⇒ V =
v
m+M
Step 2: the mechanical energy of the bulletblock-earth system is conserved.
1
( m + M )V 2 = ( m + M ) gh
2
m+M
then v =
2 gh = 630m/s
m
25
§9.4 conservation of momentum
Example 3: on a frictionless horizontal table,
both ends of a spring connect block A and B
respectively. The blocks have same mass M. a
bullet with mass m and the initial speed v0
impacted into the block A and stopped in it,
find the maximum compression distance of
the spring.
r
v0
A
B
§9.4 conservation of momentum
Solution: 1The collision of the bullet and A
mv 0
mv 0 = ( m + M )V0 ⇒ V0 =
m+M
2 The collision of the bullet with A and B
mv 0
mv 0 = ( m + 2 M )V ⇒ V =
m + 2M
3the instant when A and B get the same speed
1 2
1
1
( m + M )V02 = ( m + 2 M )V 2 + kxmax
2
2
2
M
]1 2
xmax = mv 0 [
k ( m + M )( m + 2 M )
26
§9.4 conservation of momentum
问题讨论:
一绳跨过一定滑轮，两端分别系有质量m 及M的物体，
且M>m。最初M 静止在桌上，抬高m,使绳处于松弛状
态。当m自由下落距离h后，绳才被拉紧，求此时两物
体的速率v 和M 所能上升的最大高度（不计滑轮和绳的
质量、轴承摩擦及绳的伸长）。
分析运动过程
m
h
M
当m自由下落h距离，绳被拉紧
的瞬间,m和M获得相同的运动
速率v。此后m向下减速运动,M
向上减速运动。
M上升的最大高度为：
v2
H =
2a
分两个阶段求解
§9.4 conservation of momentum
第一阶段：绳拉紧，求共同速率 v
m
M
h
解1：Q M > m ∴ m 不能提起 M ,
共同速率 v = 0
解2：
绳拉紧时冲力很大，忽略重力，
m + M 系统动量守恒
m 2 gh = ( m + M )v
; v=
m 2 gh
m+M
解3： 动量是矢量，以向下为正，系统动量守恒：
m 2 gh = mv + M ( − v )
; v=
m 2 gh
m−M
以上三种解法均不对！
27
§9.4 conservation of momentum
正确解法：
Ny
+
绳拉紧时冲力很大，轮轴反作
r
用力 N 不能忽略 ，m + M 系
统动量不守恒，应分别对它们
用动量定理；
Nx
m
h
M
r
设冲力为 F ，取向上为正方向
+
F
F
Mg
mg
I1 = ∫ (F − mg )dt = −mv − ( −m 2 gh )
I 2 = ∫ (F − Mg)dt = Mv − 0 = Mv
§9.4 conservation of momentum
忽略重力，则有 I 1 = I 2
+
− mv − ( − m 2 gh ) = Mv
v=
m 2 gh
M+m
M
m
第二阶段：
M 与 m 有大小相等，方向相反的加速度
绳拉力为 T ，画出 m 与 M 的受力图
T
a
a ，设
T
a
Mg
mg
28
§9.4 conservation of momentum
T
T
a
a
mg
Mg
⎧ Mg − T = Ma
⎨
⎩T − mg = ma
由牛顿运动定律:
a=
解得:
( M − m )g
M+m
M 上升的最大高度为
H =
m ( 2 gh 2
v2
m 2h
2( M − m ) g
=(
) (
)=
M +m
M +m
2a
M 2 − m2
§9.4 conservation of momentum
类似问题：
C
B
A
29
§9.5 Collisions
If the impulse force is much larger than any
external forces (such as gravitational force,
friction, ……), as is the case in most of the
collision, we can neglect the external forces
entirely and treat the system as an isolated
system. Then the momentum is conserved in
the collision.
1. The conservation of momentum in collision
If a system has zero total force on it, then
r
r
r
r
r
ptotal = p1bfr + p2 bfr = p1aft + p2 aft
The individual momenta of the particles do
change, but the total momenta of the system
of colliding particles does not.
§9.5 Collisions
2. Elastic collisions
(1)If the total kinetic energy of the particles
is not changed, then the collision is called
the elastic collision.
∆KE total = 0 J
(2)Elastic collision occur when
the forces between the colliding
bodies are conservative.
Wtotal = ∆KE total = 0 J
During the brief collision, the
kinetic energy is stored in the
system as potential energy, then
dumped back into kinetic energy
after the instant of the collision.
30
§9.5 Collisions
For one dimensional motion,rshowrthat two
particles with relative speed v A1 − v B1will leave
each other with same relative speed after an
elastic collision, that means
r
r
r
r
v A1 − v B1 = −(v A 2 − v B 2 )
Solution:
The momentum and mechanical energy is
conserved in an elastic collision
1
1
1
1
mAv A2 1 + mBvB2 1 = mAv A2 2 + mBvB2 2
2 r
2r
r2
r 2
mAv A1 + mBvB1 = mAv A2 + mBvB2
§9.5 Collisions
Solving the equations, one can get
r
2m B r
m − mB r
v A2 = A
v A1 +
v B1
m A + mB
m A + mB
r
2m A r
m − mA r
vB2 = B
v B1 +
v A1
m A + mB
m A + mB
The relative speed after collision:
r
r
m − mA r
2m B
v A2 − v B 2 = (
− B
)v B1
m A + mB m A + mB
m − mB
2m A r
+( A
−
)v A1
m A + mB m A + mB
r
r
= − ( v A1 − v B 1 )
31
§9.5 Collisions
3.One-dimensional collisions in the center-ofmass reference frame
The velocity of the CM frame relative to the
lab frame v S ′S
r
p Ai
mA
r
pBi
mB
x
The total initial momentum of two bodies in the
cm frame
pi = m A (v Ai − v S ′S ) + m B (v Bi − v S ′S )
Define the CM frame to be the frame in which
the initial momentum of the two bodies system
is zero.
§9.5 Collisions
pi = 0 ⇒ v S ′S =
m Av Ai + m B v Bi
m A + mB
In the CM frame, before the collision,
r
p′Bi
r
p′Ai
mA
mB
x
The total momentum is conserved, then we must
haver
r
r
r
r
p′f = p′Af + p′Bf = 0,
r
p′Af
mA
mB
p′Af = − p′Bf
r
p′Bf
x
32
§9.5 Collisions
r
p′Af
mA
mB
r
p′Af m A
mB
mA
mB
r
p′Af
mA
mB
r
p′Bf
r
p′Bf
x
elastic
x
inelastic
Completely
inelastic
x
r
p′Bf
x explosive
§9.5 Collisions
For elastic collision(in CM frame):
v ′Af = −v ′Ai , v ′Ai = v Ai − v S ′S , v ′Af = v Af − v S ′S
After the collision:
v Af = −(v Ai − v S ′S ) + v S ′S
= − v Ai + 2v S ′S
m v + m B v Bi
= − v Ai + 2 A Ai
m A + mB
m − mB
2m B
= A
v Ai +
v Bi
m A + mB
m A + mB
33
§9.5 Collisions
Following the same way,
v Bf =
2m A
m − mA
v Ai + B
v Bi
m A + mB
m A + mB
Discussion:
1equal masses
m A = mB
v Af = v Bi v Bf = v Ai
2target particle at rest
v Bi = 0
m − mB
v Af = A
v Ai
m A + mB
v Bf =
2m A
v Ai
m A + mB
§9.5 Collisions
3massive target
m B >> m A
v Af ≈ − v Ai + 2v Bi
v Bf ≈ v Bi
4massive projectile
m A >> m B
v Af ≈ v Ai v Bf ≈ 2v Ai − v Bi
34
§9.5 Collisions
Example 1: a ping-pang ball and a bowling ball
r
r
v
v
before
A
after
r
r
vA = −v
A
vA =
B
before
x
A
r
vB
x
B
r
vA
after
m A − mB
v,
m A + mB
A
vB =
B
B
r
vB
x
x
2m A
v
m A + mB
Example 2: if mA=mB
v A = 0,
vB = v
Chapter 9 impulse, momentum, and collisions
Example 3:The mass of Saturn is 5.67×1026 kg,
its speed relative to the sun is 9.6 km/s; one
spacecraft of mass 150 kg, its speed relative to
the sun is 10.4 km/s. the spacecraft is moving
toward the Saturn, due to the gravitation of
Saturn, the spacecraft rounds the Saturn and
departs in the opposite direction—the slingshot
effect. Find the speed of the spacecraft relative
to the sun.
r
v A2
r
v B1
r
v A1
35
Chapter 9 impulse, momentum, and collisions
r
m − mB
Solution: v A 2 = A
m A + mB
r
m − mA
vB2 = B
m A + mB
Due to
r
v A1 +
2m B
m A + mB
r
2m A
v B1 +
m A + mB
r
v B1
r
v A1
m B = M >> m A = m
r
r
r
v A 2 ≈ − v A1 + 2 v B 1
r
r
v B 2 ≈ v B1
+
v A 2 ≈ − v A1 − 2v B 1 = −10.4 − 2 × 9.6
= −29.6(km/s)
v A 2 > v A1
§9.5 Collisions
Example 4: Show that two balls with same
mass will separate perpendicularly to each
other after an elastic collision which is not
head-on, if one ball is at rest before collision.
Solution:
Conservation of momentum and energy
r
r
r
r r r
m v 0 = m v 1 + m v 2 ⇒ v 0 = v1 + v 2
1
1
1
1
mv 02 = mv 12 + mv 22 ⇒ v 02 = v12 + v 22 2
2
r r 2r r 2 r r
v 0 ⋅ v 0 = ( v1 + v 2 ) ⋅ ( v 1 + v 2 ) 3
r r
4
v 02 = v12 + v 22 + 2v1 ⋅ v 2
36
§9.5 Collisions
Compare 2 and 4 ,
2
v 02 = v12 + v 22
r r
v 02 = v12 + v 22 + 2v1 ⋅ v 2 4
r r
2v 1 ⋅ v 2 = 0
we have
r r
That means v1 ⋅ v 2 = v1v 2 cosθ = 0
Then
θ=
π
2
§9.5 Collisions
4. Inelastic collisions
If the total energy of the particles involved
in a collision is not conserved, the collision is
called an inelastic collision. If the particles
stick together after collision, the collision is
called a completely inelastic collision.
Wtotal = ∆KE total
From
For a collision which there are no external
forces, we have
Wtotal = 0
∆KE total ≠ 0
Why?
37
§9.5 Collisions
Notice: (a)The total momentum of the system
is conserved in both elastic and
inelastic collisions.
(b)The total kinetic energy is
conserved only for elastic collisions.
(c)If the particles stick together after
the collision, it is called a completely
inelastic collision.
Example 1:
Example 2:
Example 3:
Example 4:
P392 9.9
P394 9.10
P395 9.11
disintegrations and explosions(§9.6)
38