C8-Linear Momentum

•
second thelawmotion
must be used. of Examples
a rocket, where the mass decreases due to burnt
include:
Fnet =
∆p fuel
• the motion of a rocket, where the mass decreases due to burnt fuel
Definition
ejected
INTRODUCTION of
away Momentum
from the rocket ∆t
ejected away from the rocket
•
•When sand
sand falling
Newton falling
onfirst on a conveyor
a conveyor
expressed belthis
so Second
the mass belt sohethe
increases
Law, mass
didn’t writeincreases
Fnet =
The average net force on
of the momentum of the
•
• a droplet
ma. aInstead,
dropletof
water condenses.
water
he of falling
expressed
through
water the falling mist
law
and increasing
through
in the mist
words,
in mass
Theand as more
increasing
alteration of in mass as more
Wewater
motion is ne
defi condenses.
… aproportional
new concept, to linear
the … momentum,
force impressed.…p, to beBythe
“motion,”
product of he
It is easy to see that if the mas
the usual ma:
We define a new concept, linear momentum, p, to be the product o
the mass of a body times its velocity:
meant the product of mass and velocity, a vector quantity known as
∆p p − p
the
linear
mass of a body times its velocity: Fnet = = final initial
p = mvmomentum which is denoted by p: ∆t ∆t
mvfinal − mvinitial
=
Momentum ∆t
p−1= mv is a vector and has the direction of the velocity. Its unit is
kg m s or the equivalent N s. v −v
= m final initial
∆t
So Newton’s original formulation of the Second Law read p F, or,
In Momentum
terms of momentum, is a vector
Newton’sand haslaw
second theis: direction of the velocity. m∆v Its unit is
equivalently, F p. But a large force that acts for a short period of =
−1 ∆t
time
kgFm cans produce
∆p or thethe same
equivalent change Nin linear
s.
∆𝒑 𝑑𝒑 momentum as a small
force acting
net =
∆tfor a greater 𝑭"#$period
= limof time.=Knowing this, if we take the
average force that acts over the ∆$→+ ∆𝑡 interval
time 𝑑𝑡 t, we can turn the Fnet = ma
proportion above into an equation:
 b The forces on the ball are its weight and the reaction from the floor, R.

Definition
F = R − mg
net of Momentum
This is also the force that produces the change in momentum:
–1
2.47 A 0.50
∆p
kg ball is dropped from rest above a hard floor. When it reaches the floor it has a velocity of 4.0 m s .
FnetThe
= ball then bounces vertically upwards. Figure 2.77 is the graph of velocity against time for the ball. The
∆t
positive direction for velocity is upwards. downwards
Substituting in this equation:
a Find the magnitude of the momentum change of the ball during the bounce.
3.0 ball stayed in contact with the floor for 0.15 s. What average force did the floor exert on the ball?
Fnetb=The = 20 N
0.15
v /m s–1
We need
4.0
to find R, so:
Exam tip
reaction force R
R = 20 + 5.0 = 25 N. This is a very tricky problem with lots of
possibilities for error. A lot of people forget
The average force exerted on the ball by the floor is 25 N.
0 to include the minus sign in the rebound
t velocity and also forget the weight, so they
weight mg
–2.0 answer incorrectly that R = 20 N.
Figure 2.77

a The momentum when the ball hits the注意公式里的是合力哦！

floor is: 0.50 × 4.0 = 2.0 N s
The momentum when the ball rebounds from the floor is: 0.50 × (−2.0) = −1.0 N s
ometimes this is referred to as the impulse–momentum theorem,
Impulse-Momentum
IMPULSE Theorem
ut it’s just another way of writing Newton’s Second Law. If F varies
ithThe
time over ofthe
product interval
force and theduring which
time during it acts,
which it actsthen the as
is known impulse
elivered
impulse;
mpulse andbyforce–time
thea force
it’s vector F = F(t)that’s
quantity
graphs fromdenoted
time tby=J:t1 to t = t2 is given by
emay
following
rearrange thedefinite
equation: integral:
J=
∆p
Fnet =
In∆t terms of impulse, Newton’s Second Law can be written in yet
get: another form:

np =the
Δ FnetΔequation
t sheet for the free-response
J= p section, this information
F/N 4

ill be represented
e quantity asimpulse
FnetΔt is called the follows:
of the force, and is usually 3
Sometimes this is referred to as the impulse–momentum theorem,
noted by J. It is the product of the average force times the time
but it’s just another way of writing Newton’s
which the force acts. The impulse is also equal to the change in
Second Law.
2 If F varies

withNotice
mentum. time that
over the isinterval
impulse a vector whoseduring which
direction it same
is the acts,as then 1the impulse
t of delivered
the force (or by
the the force
change F = F(t) from time t = t1 to t = t2 is given by
in momentum).
the following
jump from definite
a height of,integral:
0 t / ms
When you say, 1 m, you will land on the 0 1 2 3 4
a graph
und of force-versus-time
with a speed −1
of about 4.5 m s . Assuming isyour
given,
mass isthen the impulse of force F
60 kg, your
a
as
 −3
J / t = (12 kg · m / s) / (8 × 10 s) = 1500 N [≈ 340 lb]
Impulse-Momentum Theorem：例题
Example 3 An 80 kg stuntman jumps out of a window
that’s 45 m above the ground.
(a) How fast is he falling when he reaches ground
level?
(b) He lands on a large, air-filled target, coming to
rest in 1.5 s. What average force does he feel
while coming to rest?
(c) What if he had instead landed on the ground
(impact time = 10 ms)?
Impulse-Momentum Theorem：例题答案
gs in cars, for instance.
Impulse-Momentum Theorem：例题
Example 4 A small block of mass m = 0.07 kg,
initially at rest, is struck by an impulsive force F of
duration 10 ms whose strength varies with time
according to the following graph:

What is the resulting speed of the block?

Impulse-Momentum Theorem：例题答案
Conservation of Momentum：练习

2004真题
Conservation of Momentum：练习答案

2004真题

A
Conservation of Momentum：练习
2004真题

没有题干哦～

补充问题：
1. Impulse ?
2. Final speed ?
Conservation of Momentum：练习答案
2004真题

没有题干哦～
Conservation of Momentum：练习

2009真题
Conservation of Momentum：练习答案

2009真题

B
 3
⇒ Fmax =and
0.227 net =≈02.3
so if×F10 × 102 that
it follows N ∆p = 0. There is no change in momentum.
Conservation
This is expressed of
as the Momentum
law of conservation of momentum:

When the net force on a system is zero the momentum does not
Conservation of momentum
change, i.e. it stays the same. We say it is conserved. 6.0 m s–1 8.0 kg
Consider a system with momentum p. The net force on the system is:
Notice that ‘system’ may refer to a single body or a collection of many 4.0 kg
∆p different bodies.
Fnet =
∆t Let us consider the blue block of mass 4.0 kg moving at speed 6.0 m s−1
to the right shown in Figure 2.80. The blue block collides with the red
and so if Fnetblock
= 0 of
it mass
follows
8.0 kgthat p=
that∆is 0. There
initially is After
at rest. no change in momentum.
the collision the two
This is expressed
blocks as theofflaw
move of conservation of momentum:
together.
As the blocks collide, each will exert a force on the other. By Newton’s system

third law, the magnitude of the force on each block is the same. There
When the are net forcethat
no forces oncome
a system is zero
from outside the the momentum
system, does
i.e. no external not
forces.
change, i.e.
Youitmight
stayssaythethatsame. We say
the weights of theit blocks
is conserved.
are forces that come from 6.0 m s–1 8.0 kg
the outside. That is correct, but the weights are cancelled by the normal
Notice that reaction
‘system’forces
mayfromreferthetotable. So the
a single net external
body force on the
or a collection of system
many is 4.0 kg
zero. Hence we expect that the total momentum will stay the same. Figure 2.80 In a collision with no external
different bodies.
The total momentum before the collision is: −1
forces acting, the total momentum of the
Let us consider the blue block of mass 4.0 kg moving at speed 6.0 m s system stays the same.
4.0 × 6.0
to the right shown in+Figure
8.0 × 0 =2.80.
24 NThe
s blue block collides with the red
block of mass
The8.0 kgmomentum
total that is initially at collision
after the rest. After
is: the collision the two
 lastic or not), Conservation of Linear Momentum
before states
Kinetic energy and momentum
Conservation of Momentum-Collision
–1 –1
before 10 m s 10 m s8 kg 12 kg 10 m s–1
–1 –1
We have seen
10 m that,
s in a collision or explosion
10 m s where no external forces are 10 m s–1
before
present, the total momentum
Let us
12 kgexamine these collisions from
8 kg
of the system is conserved.You canaftereasily
4 mthe
s point
1 m
of
s
vie –1
–1

total yourself
8 kg
convince pbefore 12 kg
thatcollision =collisions
in the three total pafter
In all cases s the total kinetic energy
colision
4m
illustrated in Figure 2.82
1 ms
–1
6 ms before
–2 m s–1the collisio –1
–1
after –1 –1 1 –1 2
momentum is conserved.4The
ms
incoming body 6 ms
1 m s–1 has mass 8.0 kg and–2
the
m s–1
8 ms
after
other a mass of 12 kg. 1 EK = 12 × 8.0 × 102 2Figure= 400
2.82 JMomentum is conserved
3 in these three collisions.

mple 6 1
Two balls roll toward
Figure 2.82 each other. The
2Momentum is conserved in these three
red
3 collisions.
Let us after
examinethe thesecollision
collisions from the pointcase
of vie
Figure 2.82 Momentum is conserved in theseThe total kinetic energy in each
hasbefore
a mass10 mofs 0.5 kg and a speed
–1
three collisions.
10 m s –1 of 4 m/s justthe total
In all cases 10 m kinetic
s –1
energy before the collisi
Let us examine these collisions from the point of view of energy.
re Let
impact. 8 kg The green
12 kg In all ball
cases thehas
case
us examine these collisions from the point of view oftotal a
1:
kineticmass
E =
energy
K 2energy.
of
1 before
× 20E0.3
K×
the
= 1 2kg
24 = 160
collision
× 8.0 × 10 2
is: =J400 J
a speed
In all casesofthe2total
m/s.kineticAfter
energy the
E4Km=s 12 ×before
8.0 × 10
–1 head-on
the2 collision is:collision, the
= 400 J 6 m sThe total kinetic energy after the collision
–1
8 ms in each case
–1

ballEKafter
continues forward
J The total with a speed EK =of2 ×1.7 8.0 ×m/s.
–1 1 –2 m 2s –1 1 2
= 12 × 8.0 × 102 = 400
case 2:
1 ms
1 + 21 × 12 × 2
6 = 220 J
kinetic energy after the collision case in each
1: case
EK =is: 2 × 20 × 4 = 160 J
the speed 1 of the green ball 2 1 after the1 collision. 3 2 11 2 12 × 62 = 220 J
The total kinetic energy after the casecollision
1: EK =in case
× each
20 ×3: EK = 2 × 8.0 × 2 +
case
4 2
=is:
160 J case 2: E K 22 ×
= ×8.0 12× ×12 +812 ×= 400 J
theFigure
collision elastic?
2.82 Momentum
1
2
is conserved in these three collisions.
case 1: EK = 2 × 20 × 42 = 160 J 2: EK = 12 × 8.0 × 12 + 12 × 12 × 62 = 220
case caseJ 3: EK = 12 × 8.0 × 22 + 12 × 12 × 82 = 400 J
1 2 1
We 1
thus observe
2 1
that
2
whereas momentum is conserved
caseLet usEexamine
K = 2 × 8.0
2:completely + 2 case
× 123:
× 1inelastic
these collisions 62E=K 220
× from =the 8.0 × 2of+view
2 ×J point
inelastic 2 × 12of We
× energy.
8 = 400thus
J observe that whereas momentum is conserved
Perfectly elastic
kinetic energy is not. When
kinetic energy kinetic
is not. When energy
kinetic is conserved
energy is conserve
rst case
remember that
× 2 +momentum J is
is said to be elastic. Whenthe
a vector quantity, so
In all cases 1the total kinetic
2 We energy
1 thus observe 2 before the collision is:
3: EK = 2 × 8.0 = 400that
2 × 12 × 8 collision
whereas momentum is conserved inbe allelastic.
cases,
collision is said to it When
is notit is(cases 1 1an
not (cases an
 total pbefore collision = total pafter colision
Conservation of Momentum-Collision：例题
Example 6 Two balls roll toward each other. The red
ball has a mass of 0.5 kg and a speed of 4 m/s just
before impact. The green ball has a mass of 0.3 kg
and a speed of 2 m/s. After the head-on collision, the
red ball continues forward with a speed of 1.7 m/s.
Find the speed of the green ball after the collision.
Was the collision elastic?

on. First remember that momentum is a vector quantity, so the

on of the velocity is crucial. Since the balls roll toward each
one ball has a positive velocity while the other has a negative
y. Let’s call the red ball’s velocity before the collision positive;
例题答案
of the collision.
Conservation of Momentum-Collision：例题
Example 7 Two balls roll toward each other. The red
ball has a mass of 0.5 kg and a speed of 4 m/s just
before impact. The green ball has a mass of 0.3 kg
and a speed of 2 m/s. If the collision is completely
inelastic, determine the velocity of the composite
object after the collision.

on. If the collision is completely inelastic, then, by definition,

asses stick together after impact, moving with a velocity, v’.
ng Conservation of Linear Momentum, we find
例题答案
 (D) L/2
(E) 3L/4
Conservation of Momentum-Collision：例题
9. A wooden block of mass M is moving at speed V in a straight line.

How fast would the bullet of mass m need to travel to stop the block
(assuming that the bullet became embedded inside)?
(A) mV/(m + M)
𝒑𝒊𝒏𝒊𝒕𝒊𝒂𝒍 = 𝒑𝒇𝒊𝒏𝒂𝒍 ;
𝒎𝒗 + 𝑴𝑽 = 𝟎;
So
𝑴𝑽
𝒗= ;
𝒎
Conservation
ction of Momentum-Collision：练习
II: Free Response
1. A steel ball of mass m is fastened to a light cord of length L and
released when the cord is horizontal. At the bottom of its path, the ball
strikes a hard plastic block of mass M = 4m, initially at rest on a
frictionless surface. The collision is elastic. (a) Find the tension in the cord when the ball’s height above its lowest
position is L. Write your answer in terms of m and g.

(b) Find the speed of the block immediately after the collision.
(c) To what height h will the ball rebound after the collision?

2. A ballistic pendulum is a device that may be used to measure the muzzle

speed of a bullet. It is composed of a wooden block suspended from a
horizontal support by cords attached at each end. A bullet is shot into
the block, and as a result of the perfectly inelastic impact, the block
swings upward. Consider a bullet (mass m) with velocity v as it enters
the block (mass M). The length of the cords supporting the block each
have length L. The maximum height to which the block swings upward
after impact is denoted by y, and the maximum horizontal displacement
(a) Find the tension in the cord when the ball’s height above its lowest
is denoted by x.
练习答案
 Conservation
ction of Momentum-Collision：练习
II: Free Response
. A steel ball of mass m is fastened to a light cord of length L and
(a) of
released when the cord is horizontal. At the bottom Find
its the tension
path, in the cord when the ball’s height above its lo
the ball
strikes a hard plastic block of mass M = 4m, initiallyposition is a L. Write your answer in terms of m and g.
at rest on
frictionless surface. The collision is elastic.
(b) Find the speed of the block immediately after the collision.
(c) To what height h will the ball rebound after the collision?

2. A ballistic pendulum is a device that may be used to measure the m

speed of a bullet. It is composed of a wooden block suspended from
horizontal support by cords attached at each end. A bullet is shot
the block, and as a result of the perfectly inelastic impact, the bloc
swings upward. Consider a bullet (mass m) with velocity v as it en
the block (mass M). The length of the cords supporting the block e
have length L. The maximum height to which the block swings up
after impact is denoted by y, and the maximum horizontal displace
is denoted by x.
(a) Find the tension in the cord when the ball’s height above its lowest
练习答案
练习答案
plementary
there material. The rocketforces
are no external moves and
with the v. Themomentum
speedtotal engine is will rockets
be conserved.
propel this space shuttle during its
ned on and gases leave the rocket with speed u relative to the rocket. launch.
Conservation of Momentum-Explosion
This means that the Earth moves up a bit as the ball falls!
initial mass of the rocket including the fuel is M. After a short time δt
rocket has ejected fuel of mass δm. The mass of the rocket is therefore
The rocket equation
uced to M − δm and its speed increased to v + δv (Figure 2.85).
The best example of motion with vvarying mass is, of course, the rocket
M
(Figure 2.84).
This is quite a complex topic and is included here only as Figure 2.84 Exhaust g
supplementary
u – (v + δv) material. The rocket moves v + δwith
v speed v. The engine is rockets propel this spac
turned on and δm gases leave the rocket with speed u relative to the rocket. launch.
M – δm
The initial mass of the rocket including the fuel is M. After a short time δt
re 2.85 Diagram for deriving the rocket equation. The velocities are relative to an
the
rver ‘at restrocket
Applying thehaslaw
on the ground’ ejected fuel of mass
. of conservation δm. The mass
of momentum gives of(in the rocket is therefore
the equation
below
reduced termstoshaded
M − δthe sameits
m and colour
speedcancel out): to v + δv (Figure 2.85).
increased
Mv = (M − δm)(v + δv) − δm (u − v − δv) v
M
speed relative to ground

Mv = Mv + Mδv − vδm − δmδv − uδm + vδm + δmδv

Mδv = uδum– (v + δv) v + δv

2 MECHANICS 107
 there are no external forces and the total momentum will be conserved.
Conservation of Momentum-Explosion
This means that the Earth moves up a bit as the ball falls!

The rocket equation

The best example of motion with varying mass is, of course, the rocket
(Figure 2.84).
This is quite a complex topic and is included here only as Figure 2.84 Exhaust g
supplementary material. The rocket moves with speed v. The engine is rockets propel this spac
turned on and gases leave the rocket with speed u relative to the rocket. launch.
The initial mass of the rocket including the fuel is M. After a short time δt
the rocket has ejected fuel of mass δm. The mass of the rocket is therefore
reduced to M − δm and its speed increased to v + δv (Figure 2.85).
v
M

u – (v + δv) v + δv
ervation of Linear Momentum. In equation form, for two
ts colliding, we have
Conservation of Momentum-Explosion：例题
Example 5 An astronaut is floating in space near her
shuttle when she realizes that the cord that’s supposed
to attach her to the ship has become disconnected.
Her total mass (body + suit + equipment) is 89 kg.
She reaches into her pocket, finds a 1 kg metal tool,
and throws it out into space with a velocity of 9 m/s
directly away from the ship. If the ship is 10 m away,
how long will it take her to reach it?

ion. Here, the astronaut + tool are the system. Because of

ervation of Linear Momentum,
例题答案
Conservation of Momentum-2D：例题
Conservation of Momentum-2D：例题答案
Conservation of Momentum-2D：例题答案
Conservation of Momentum：练习
2004真题
Conservation of Momentum：练习答案
2004真题

C
Conservation of Momentum：练习
2008真题
Conservation of Momentum：练习答案
2008真题

D
 10 revolutions per second when a frictional

Conservation of Momentum：练习
torque is applied to stop it. How much work
is done by the torque in stopping the wheel?
(A)
(B)
(C) 2013真题
(D) 19. An arrow of mass and speed strikes and
(E) sticks to one end of a meterstick of mass as
shown in the diagram above. The meterstick is
initially at rest on a horizontal surface and free to
move without friction. The speed of the center of
mass of the stick-arrow system after the arrow
strikes is given by which of the following
expressions?

(A)
19. An arrow of mass and speed strikes and
sticks to one end of a meterstick of mass as (B)
shown in the diagram above. The meterstick is
initially at rest on a horizontal surface and free to (C)
18. A uniform beam of weight W is attached to a
move without friction.
wall Theat speed
by a pivot one endofandtheiscenter of
held horizontal
mass of the stick-arrow system
by a cable attached after
to the theend
other arrow
of the beam (D)
strikes is given
andby which
to the wall,of
as the following
shown above. T is the tension
expressions?in the cable, which makes an angle with the (E) 0
beam. Which of the following is equal to T ?
(A)
(A)
 10 revolutions per second when a frictional

Conservation of Momentum：练习
torque is applied to stop it. How much work
is done by the torque in stopping the wheel?
(A)
(B)
(C) 2013真题
(D) 19. An arrow of mass and speed strikes and
(E) sticks to one end of a meterstick of mass as
shown in the diagram above. The meterstick is
initially at rest on a horizontal surface and free to
move without friction. The speed of the center of
mass of the stick-arrow system after the arrow
strikes is given by which of the following
expressions?

(A)
19. An arrow of mass and speed strikes and
sticks to one end of a meterstick of mass as (B)
shown in the diagram above. The meterstick is
initially at rest on a horizontal surface and free to (C)
18. A uniform beam of weight W is attached to a
move without friction.
wall Theat speed
by a pivot one endofandtheiscenter of
held horizontal C
mass of the stick-arrow system
by a cable attached after
to the theend
other arrow
of the beam (D)
strikes is given
andby which
to the wall,of
as the following
shown above. T is the tension
expressions?in the cable, which makes an angle with the (E) 0
beam. Which of the following is equal to T ?
(A)
(A)
Conservation of Momentum：练习
2004真题
Conservation of Momentum：练习答案
2004真题

B、C
 the case where the particles all lie on a straight line.
Center of Mass axis. Select some point to be the origin (x = 0) and
positions of each particle on the axis. Multiply each pos
the mass of the particle at that location, and get the s
particles. Divide this sum by the total mass, and the res
is the center of mass:
The system of particles behaves in many respects as if
= m1 + m2 +…+ mn, were concentrated at a single l

If the system consists of objects that are not confine

On the equation
straight line, usesheet for the free-response
the equation above to findsection, th
the x-coo
will
centerbe represented
of mass, andasthefollows:
corresponding equation,

to find the y-coordinate of their center of mass (

equation to calculate the z-coordinate, if they are no
The system of particles behaves in many respects as if all its mass, M
 or masses
quation is λ with
= uniform density.
, therefore M = ∫dmNow=we will learn how to
∫λdx.
center of the
(b) Let mass
timeof objects
it takes with
the man to non-uniform density.
walk across the barge be
Center oft; then
denoted by Mass：例题
t = L/v. In this amount of time, the barge
moves a distance of L/4 in the opposite direction, so the
velocity of the barge is
Example 12 A bar with of length of 30 cm has a
linear density λ = 10 + 6x, where x is in meters and
λ is in kg/m. Determine the mass of the bar and the
Socenter
far weof mass
have of with
dealt this objects
bar. that can be considered point
masses, or masses with uniform density. Now we will learn how to
e find
example ofofamass
the center barofthat becomes
objects denser density.
with non-uniform along its length. Here
deal with the linear density λ as a function of x, λ(x). Each
gmentWe
ion. of the
canbar, x, has athe
determine different
mass mass,
of the m.bar Weby treat eachthe
using
a point mass and then take the limit as x approaches zero.
ition
he of linear
formula fordensity,
calculatingλ =the center
. Therefore
of mass of point masses,
lacing
Take the m with ofdm,
example we
a bar get
that the integral
becomes shown
denser along below.
its length. Here
we will deal with the linear density λ as a function of x, λ(x). Each
small segment of the bar, x, has a different mass, m. We treat each
m as a point mass and then take the limit as x approaches zero.
Using the formula for calculating the center of mass of point masses,
and replacing m with dm, we get the integral shown below.

M is the total mass, x is the distance to each dm, and you can

te for dm in terms of x. Linear density is mass per length, so

Center of Mass：例题答案
 the mass of the particle at that location, and get the su
s will not accelerate. The system of particles
particles. Divide this sumbehaves
by the in many
total respects
mass, asres
and the if
Center of Mass：例题 =the
is m1 center
+ m2 +…+
of mass:mn, were concentrated at a single lo
Example 10 Two objects, one of mass m and onesystem
If the of consists of objects that are not confine
mass 2m, hang from light threads from thestraight
ends ofline,
a use the equation above to find the x-coor
uniform bar of length 3L and mass 3m. Thecenter massesof m
mass, and the corresponding equation,
and 2m are at distances L and 2L, respectively, below
On the equation sheet for the free-response section, thi
the bar. Find the center of mass of this system.
will be represented as follows:

to find the y-coordinate of their center of mass (a

equation to calculate the z-coordinate, if they are not
single plane).

From the equation

 ticular, if the
concentrated at itsnet external
midpoint. force aon
Constructing the system
coordinate system is zero, then the center
the mass of the particle at that location, and get the su
his point as the origin, we now have three objects: one of mass
mass
–3L/2,will
–L), not accelerate.
mass 2m at (3L/2, –2L), and one of mass 3m particles. Divide this sumbehaves
by the total mass, and theasres
Center of Mass：例题答案
one of
is=the
The system of particles in many respects if
0):
m1center
+ m2 of mass:mn, were concentrated at a single lo
+…+
Example 10 Two objects, one of mass m and one of
If the system consists of objects that are not confine
mass 2m, hang from light threads from the ends of a
straight line, use the equation above to find the x-coor
uniform bar of length 3L and mass 3m. The masses m
center of mass, and the corresponding equation,
and 2m are at distances L and 2L, respectively,
On the below
equation sheet for the free-response section, thi
the bar. Find the center of mass of this system.
will be represented as follows:
(because it is uniform), so we may treat the total mass of the bar as
being concentrated at its midpoint. Constructing a coordinate system
with this point as the origin, we now have three objects: one of mass
m at (–3L/2, –L), one of mass 2m at (3L/2, –2L), and one of mass 3m
at (0, 0):

to find the y-coordinate of their center of mass (a

原点的选取可以不
equation
一样～ to calculate the z-coordinate, if they are not
single plane).

figure out the x- and y-coordinates of the center of mass From the equation
ately:

We figure out the x- and y-coordinates of the center of mass

ore, the center of mass is at separately:
ate again and establish the following:
efore, the center of mass is at
Center of Mass：例题
(xcm, ycm) = (L/4, –5L/6)

Fnet = Ma cm
ve to the midpoint of the bar.

Example 11 A man of mass m is standing at one end

(external) force acting on the system causes the

of a stationary, floating barge of mass 3m. He then
walks to the other end of the barge, a distance of L

elerate according to Newton’s Second Law. In

meters. Ignore any frictional effects between the
barge and the water.

ternal forcewalks aton thevelocity

system is zero, then the cente
(a) How far will the barge move?
(b) If the man an average of v,

ate.
what is the average velocity of the barge?

ion.

a) Since there are no external forces acting on the man + barge

system, the center of mass of the system cannot accelerate. In
particular, since the system is originally at rest, the center of
 he center of mass is a distance of L/8 from the midpoint
Center of Mass：例题答案
he barge, and since the mass is originally at the left end,
center of mass is a distance of L/8 to the left of the barge’s
point.

n the man reaches the

Symmetry ! other end of the barge, the center
mass will, by symmetry, be L/8 to the right of the midpoint
he barge. But, since the position of the center of mass
not move, this means the barge itself must have moved a
ance of
Center of Mass：例题答案

Relative motion!
Center of Mass：练习

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Center of Mass：练习答案

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Momentum：大题练习
Momentum：大题练习