1

AP Physics Chapter 9

Center of Mass and Linear Momentum

2

AP Physics

Lecture: Chapter 9 Q&A

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Standards: (AP Physics C: Mechanics)

Newtonian Mechanics D. Systems of particles, linear

momentum (12%)1. Center of mass 2. Impulse and momentum 3. Conservation of linear momentum,

collisions

4

System

• A system is a collection of objects. • A system includes the objects, but

not the space in between them.• In the system

All objects that belong to the system Not necessary the space enclosed by the system

5

Center of Mass

Center of mass (of a body or a system of bodies): the point that moves as though all of the mass were concentrated there and all external forces were applied there.

Position of Center of Mass:

1D:

3D:

cmx

ˆˆ ˆ, ,cm cm cm cm cm cm cmr x y z x i y j z k

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Position of Center of Mass: xcm (2 objects, 1D)

1 1 2 2

1 2cm

m x m xx

m m

xcm: position of center of mass of two objects

m1: mass of Object 1

m2: mass of Object 2

x1: position of (center of mass of) Object 1

x2: position of (center of mass of) Object 2

m1

m2

x2

•x1

•

7

Position of Center of Mass: rcm

(n objects, 3D)

1

1 n

cm i ii

x m xM

1

1 n

cm i ii

r m rM

M 1 2 nm m m im Total Mass

1

1 n

cm i ii

y m yM

1

1 n

cm i ii

z m zM

8

Position of Center of Mass:(Continuous Mass Distribution, 3D)

1

1

1

cm

cm

cm

x xdmM

y ydmM

z zdmM

zdVV

z

ydVV

y

xdVV

x

cm

cm

cm

1

1

1

uniform

density

V: volume of object

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Let C = 1, O = 2, then m1 = 12.01115u, m2 = 15.9994u.

Also, let x = 0 at C, then x1 = 0, and x2 = 1.131 10-10m.

xcm = ?

cmx

The center of mass is 6.46 10-11 m from the carbon atom, in between the two atoms.

Center of mass is closer to the object with larger mass.

Example: Pg238-102The distance between the centers of the carbon (C) and oxygen (O) atoms in a carbon monoxide (CO) gas molecule is 1.131 10-10 m. Locate the center of mass of a CO molecule relative to the carbon atom. (Find the masses of C and O in Appendix D.)

1 1 2 2

1 2

m x m x

m m

1015.9994 1.131 10

12.01115 15.9994

u m

u u

C O

0x

116.46 10 m

10

x

ySet up coordinates, then

m1 = 14g at (0, 11cm)

m2 = 14g at (22cm, 11cm)

m3 = 42g at (11cm, 22cm)

cmx

•cm

Symmetric?

m3m2m1

11 ,17.6cmr cm cm

Example: Pg229-3In Fig. 9-37, three uniform thin rods, each of length L = 22 cm, form an inverted U. The vertical rods each have a mass of 14 g; the horizontal rod has a mass of 42 g. a) What are the x coordinate and b) the y coordinate of the system’s center of mass?

14 0 14 22 42 1111

14 14 42

g g cm g cmcm

g g g

cmy 14 11 14 11 42 2217.6

14 14 42

g cm g cm g cmcm

g g g

1 1 2 2 3 3

1 2 3

m x m x m x

m m m

1 1 2 2 3 3

1 2 3

m y m y m y

m m m

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Newton’s Second Law for a System of Particles

• Fext: external force, force from object not included in the system

• M: total mass of system

• acm: acceleration of center of mass

, ,ext x cm xF Ma

3D

ext cmF Ma

, ,ext y cm yF Ma

, ,ext z cm zF Ma

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Special Cases and Application

What if the total external force is zero?

acm = 0

If vcm, i = 0

vcm = constant

vcm remains 0 CM does not move.

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Velocity of Center of Mass, vcm

21

2211

mm

xmxmxcm

1 1 2 2

1 2cm

m v m vv

m m

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Example: Pg238-104

An old Chrysler with mass 2400 kg is moving along a straight stretch of road at 80 km/h. It is followed by a Ford with mass 1600 kg moving at 60 km/h. How fast is the center of mass of the two cars moving?

Let m1 = 2400kg, v1 = 80km/h, m2 = 1600kg, v2 = 60 km/h

vcm = ?

cmv 1 1 2 2

1 2

m v m v

m m

2400 80 1600 6072

2400 1600

km kmkg kg

kmh hkg kg h

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Let x1 = 0, then x2 = 1m. Also, m1 = 0.10 kg,

m2 = 0.30kg.

?cmx

a) Fext =

cmx

Practice: Pg240-138Two particles P and Q are initially at rest 1.0 m apart. P has a mass of 0.10 kg and Q a mass of 0.30 kg. P and Q attract each other with a constant force of 1.0 10-2 N. No external forces act on the system. a) Describe the motion of the center of mass.b) At what distance from P’s original position do the particles collide?

b) What are we actually looking for?

What happens to the center of mass?

CM stays at rest.

1 1 2 2

1 2

0.10 0 0.30 10.75

0.10 0.30

kg kg mm x m xm

m m kg kg

0

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Newton’s Second Law

F ma

Define: (Linear) Momentum:

P is a vector.

Unit:

P mv

Same direction as v.

P

dvmdt d mv

dtdP

dt dP

Fdt

m v m

kgs

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Momentum of System of Particles

cmP Mv

ext cm

dPF Ma

dt

1 1 2 2 ...m v m v

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Practice: Pg230-18A 0.70 kg ball is moving horizontally with a

speed of 5.0 m/s when it strikes a vertical wall. The ball rebounds with a speed of 2.0 m/s. What is the magnitude of the change in linear momentum of the mass?

Let toward the wall be the positive direction, then

m = 0.70kg, vi = 5.0m/s, vf = -2.0 m/s

|P| = ?

P

4.9m

P kgs

f iP P f imv mv f im v v

0.70 2.0 5.0 4.9m m m

kg kgs s s

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Closed, isolated System

Closed System: n = constant n = constant No particles leave or enter the system.

Isolated System: Fext = 0 The sum of the external forces acting on the system

of particles is zero. There could be external forces acting on the system,

but the net external force is zero.

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Law of Conservation of Linear Momentum

In closed and isolated system:

If a component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change.

P: total momentum of system

P = constant Pi = Pf

P1i + P2i = P1f + P2f

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1 2 1 2 2

1

91 , 68 0.068 , 0, 4.0

?

i i f

f

mm kg m g kg v v v

sv

i fP P

Notice the negative sign. When we define v2 = 4.0 m/s, we already define the direction of v2f to be the positive direction.

Example: Pg232-35A 91 kg man lying on a surface of negligible friction shoves a 68 g stone away from himself, giving it a speed of 4.0 m/s. What speed does the man acquire as a result?

1 2 1 2i i f fP P P P 0

2 21

1

ff

m vv

m 3

0.068 4.02.9 10

91

mkg mskg s

1 1 2 2f fm v m v

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Motor = 1, Module = 2, v1i =v2i= 4300 km/h, m1 = 4m2,

v1f – v2f = -82 km/h v1f = v2f – 82 km/h = v2f - vR,

v2f = ?

i fP P

Practice: Pg232-37A space vehicle is traveling at 4300 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 82 km/h relative to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to earth after the separation?

1 2 1 1 2 2i f fm m v m v m v

1 2 1 2 2 2i f R fm m v m v v m v

1 2 2 1 2 1f i Rm m v m m v m v

1 2 12

1 2

i Rf

m m v m vv

m m

1 2 2 1f Rm m v m v

2 2 2

2 2

4 4

4i Rm m v m v

m m

5 4

5i Rv v

2 2

2

5 4

5i Rm v m v

m

35 4300 / 4 82 /4.4 10 /

5

km h km hkm h

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Practice

A railroad flatcar of mass M can roll without friction along a straight horizontal track. Initially, a man of mass m is standing on the car, which is moving to the right with speed v0. What is the change in velocity of the car if the man runs to the left so that his speed relative to the car is vrel?

Flatcar’s motion

Man’s motion

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Solution

m1 = M, m2 = m, v1i = v2i = vo, v2f = v1f - vrel,

v1=v1f - v1i = ?

i fP P

1 2 1 1 2o fm m v m v m

1 2 21

1 2

o relf

m m v m vv

m m

rel

o

mvv

M m

1 1 1f iv v v

1 2 1 1 2 2i f fm m v m v m v

1 f relv v 1 2 1 2f relm m v m v

o relM m v m v

M m

0rel

o

mvv v

M m

relmv

M m

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Practice: Pg236-85

The last stage of a rocket is traveling at a speed of 7600 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 290.0 kg and a payload capsule with a mass of 150.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 910.0 m/s.

a) What are the speeds of the two parts after they have separated? Assume that all velocities are along the same direction.

b) Find the total kinetic energy of the two parts before and after they separate; account for any difference.

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Solution: Pg236-85case 1, capsule 2, m1 = 290.0kg, m2 = 150.0kg,

v1i=v2i=vi = 7600 m/s, v2f – v1f = vR = 910.0 m/s

1 2) ?, ?f fa v v

i fP P

1 2 1 1 2 2i f fm m v m v m v

1 2 21

1 2

i Rf

m m v m vv

m m

290.0 150.0 7600 150.0 910.07290

290.0 150.0

m mkg kg kg

ms skg kg s

2 fv 1 f Rv v 7290 910.0 8200m m m

s s s

1 1 2 1f f Rm v m v v 1 2 1 2f Rm m v m v

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Solution: Pg236-85 (2)

b) Ki = ?, Kf = ?

iK

1 2f f fK K K

The increase in the KE comes from the spring, when the spring does positive work to separate the two parts.

21 2

1

2 im m v 2

101290.0 150.0 7600 1.271 10

2

mkg kg J

s

2 21 1 2 2

1 1

2 2f fm v m v

2 2

101 1290.0 7290 150.0 8200 1.275 10

2 2

m mkg kg J

s s

f iK K

28

System with Varying Mass: Rocket

Rocket accelerates forward by burning and ejecting fuel backward.

Mass of rocket keeps changing.Newton’s Second Law and

Conservation of Momentum still apply, though more complicated.

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First Rocket Equation

R = |dM/dt|, rate at which rocket losing mass (fuel)vrel: exhaust speed, backward speed of fuel relative to the rocket M: mass of rocket at that momenta: acceleration of rocket at that moment

relRv Ma

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Second Rocket Equation

vf: final speed of rocket after some fuel consumption

vi: initial speed of rocket

vrel: exhaust speed of fuel

Mi: initial mass of rocket

Mf: final mass of rocket after ejecting some fuel

ln if i rel

f

Mv v v

M

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vi = 6.0 103 m/s, vrel = 3.0 103 m/s, M = 4.0 104 kg,

a = 2.0 m/s2.

) ?b R

) ?a F

Example: Pg238-112A rocket is moving away from the solar system at a speed of 6.0 103 m/s. It fires its engine, which ejects exhaust with a velocity of 3.0 103 m/s relative to the rocket. The mass of the rocket at this time is 4.0 104 kg, and its acceleration is 2.0 m/s2. a) What is the thrust of the engine?b) At what rate, in kilograms per second, was exhaust ejected during the firing?

F 4 42

4.0 10 2.0 8.0 10m

ma kg Ns

relRv Ma

R

42

32

4.0 10 2.027

3.0 10rel

mkg

Ma kgsmv ss

32

vi = 400 m/s, v = 2.2 m/s,vrel = 1000 m/s

|M/Mi| = ?

f iv v v

rel

v

ve

rel

v

v

Practice: P238-114During a lunar mission, it is necessary to increase the speed of a spacecraft by 2.2 m/s when it is moving at 400 m/s. The exhaust speed of rocket engine is 1000 m/s. What fraction of the initial mass of the spacecraft must be burned and ejected for the increase?

i

M

M

2.2 /

1000 / 1.0022m s

m se

f i

i

M M

M

1.0022 0.00220.0022

1.0022 1.0022f f f

f f

M M M

M M

0.0022i

M

M

ln irel

f

Mv

Mln i

f

M

M

i

f

M

M 1.0022i fM M

33

Impulse

F

Define Impulse:

F t

J P Impulse-Momentum Theorem

Newton’s Second Law

Area under curve of F-t

ma v

mt

mv

t

mv

t

P

t

F t P

J F t

f iP P

tF dt

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Example:A cue stick strikes a stationary pool ball, exerting an average force of 50 N over a time of 10 ms. If the ball has mass 0.20 kg, what speed does it have after impact?

?

0,20.0,1010,50 3

f

i

v

vkgmstNF

J P f iP P

F t

fv

f imv mv fmv

350 10 102.5

0.20

N sF t m

m kg s

35

, ,i fm v v v v

) , ?a t F

3) 0.140 , 7.8 / , 3.8 10

?

b m kg v m s t s

F

Practice:A ball of mass m and speed v strikes a wall perpendicularly and rebounds in the opposite direction with the same speed. a) If the time of collision is t, what is the average force exerted by the ball on the wall?b) Evaluate this average force numerically for a rubber ball with mass 140 g moving at 7.8 m/s; the duration of the collision is 3.8 ms.

f iF t P P

2mvF

t

f imv mv 2m v mv mv

2mv

t

3

2 0.140 7.82

5703.8 10

mkg

mv sF N

t s

36

, 1.2 , 25 , 10i f

m mUpward m kg v v

s s

) ?a J

) 0.020 , ?b t s F

Practice: Pg231-26A 1.2 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an initial speed of 10 m/s. a) What impulse acts on the ball during the contact?b) If the ball is in contact with the floor for 0.020 s, what is the average force exerted on the floor?

f i f i f iJ P P P mv mv m v v

1.2 10 25 42m m m

kg kgs s s

J F t 42

21000.020

mkgJ sF N

t s

37

Series of Collisions

mF v

t

:F Average force on body being hit

Rate at which mass collides with the body

Change in velocity of hitting bodies

Impulse-Momentum Theorem

:m

t

:v

38

Example: Pg237-99

A pellet gun fires ten 2.0 g pellets per second with a speed of 500 m/s. The pellets are stopped by a rigid wall.

a) What is the momentum of each pellet?b) What is the kinetic energy of each pellet?c) What is the average force exerted by the stream

of pellets on the wall?d) If each pellet is in contact with the wall for 0.6

ms, what is the average force exerted on the wall by each pellet during contact? Why is this average force so different from the average force calculated in (c)?

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Solution: Pg237-990,/500,1,100.2,10 3

fi vsmvstkgmn

3) 2.0 10 500 1.0m m

a P mv kg kgs s

2

2 31 1) 2.0 10 500 250

2 2

mb K mv kg J

s

)m

c F vt

3) 0.6 10 , ?d t s F

Most of the time no bullet is in contact with the wall, average is much smaller.

f i

nmv v

t

310 2.0 10 50010

1i

mkg

nmv sN

t s

J P

imvF

t

f i f i iF t P P mv mv mv

3

33

2.0 10 5001.7 10

0.6 10

mkg

sN

s

31.7 10 N

40

Practice:A movie set machine gun fires 50 g bullets at a speed of 1000 m/s. An actor, holding the machine gun in his hands, can exert an average force of 180 N against the gun. Determine the maximum number of bullets he can fire per minute while still holding the gun steady?

0.050 , 0, 1000 , 180 , 60

?

i f

mm kg v v F N t s

sn

mF v

t

f i

nmv v

t

n

f

nmvt

180 60216

0.050 1000f

N sF tmmv

kgs

41

Collisions

Total momentum is always conserved.

Total kinetic energy:

Conserved Elastic Collision: Ki = Kf

Not conserved Inelastic Collision

Extreme case: Stick and move together Completely Inelastic Collision (No other collision loses more portion of the kinetic energy than this.)

Pi = Pf

42

Elastic Collisions in 1D

1 1 2 2 1 1 2 2i i f fm v m v m v m v

1 21 1

1 2

12 1

1 2

2

f i

f i

m mv v

m m

mv v

m m

if v2i = 0

1 2 21 1 2

1 2 1 2

1 2 12 1 2

1 2 1 2

2

2

f i i

f i i

m m mv v v

m m m m

m m mv v v

m m m m

Conservation of P:

Conservation of K: 2 2 2 21 1 2 2 1 1 2 2

1 1 1 1

2 2 2 2i i f fm v m v m v m v

43

Special Cases (of v2i = 0)

Equal masses: m1 = m21

2

f

f

v

v

1 1 2 2i i

cm

m v m vv

1

11 2 1 2

i

mv

m m m m

Massive target: m2 >> m1

Light target: m2 << m1

Motion of CM:

1 21 1

1 2

12 1

1 2

2

f i

f i

m mv v

m m

mv v

m m

0

1iv

1

2

f

f

v

v

0

1iv

1iv

12 iv1

2

f

f

v

v

cm

Pv

m

44

Example:A hovering fly is approached by an enraged elephant charging at 2.1 m/s. Assuming that the collision is elastic, at what speed does the fly rebound? Note that the projectile (the elephant) is much more massive than the stationary target (the fly).

1 1 2

2

1, 2, 2.1 ,

?

i

f

mElephant Fly v m m

sv

2 12 2 2.1 4.2f i

m mv v

s s

Notice that the elephant is chosen as mass 1 because of the formula we have.

45

Practice: Pg234-55

A cart with mass 343 g moving on a frictionless linear air track at an initial speed of 1.2 m/s strikes a second cart of unknown mass at rest. The collision between the carts is elastic. After the collision, the first cart continues in its original direction at 0.66 m/s.

a) What is the mass of the second cart?b) What is its speed after impact?c) What is the speed of the two-cart center of

mass?

46

Solution: Pg234-551 1 2 10.340 , 1.2 , 0, 0.66i i f

m mm kg v v v

s s

2) ?a m

2) fb v

) cmc v

1 21 1

1 2f i

m mv v

m m

2 1 2 1 1 1 1 1f i i fm v m v m v m v

2m

1 1 2 1 1 1 2 1f f i im v m v m v m v

2 1 1 1 1 1i f i fm v v m v v

1 1 1

1 1

0.340 1.2 0.660.099 99

1.2 0.66

i f

i f

m mkgm v v s s

kg gm mv vs s

11

1 2

2 2 0.3401.2 1.9

0.340 0.099i

m kg m mv

m m kg kg s s

1 1

1 2

0.340 1.20.93

0.340 0.099i

mkg

m v msm m kg kg s

47

Practice: Pg234-59

A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed.

a) What is the mass of the struck body?b) What is the speed of the two-body

center of mass if the initial speed of the 2.0 kg body was 4.0 m/s?

48

Solution: Pg234-590,4/,,0.2 2111 ifi vvvvvkgm

2) ?a m

1) 4.0 , ?i cm

mb v v

s

1 21 1

1 2f i

m mv v

m m

1 2 1 24 m m m m

1 1 2 24 4m m m m

1 1 2

1 1 2

f

i

v m m

v m m

/ 4 1

4

v

v

1 2 1 24 4m m m m

1 23 5m m 2 1

3 32.0 1.2

5 5m m kg kg

11

1 2cm i

mv v

m m

2.0

4.0 2.52.0 1.2

kg m m

kg kg s s

49

Practice

A platform scale is calibrated to indicated the mass in kilograms of an object placed on it. Particles fall from a height of 3.5 m and collide with the platform of the scale. The collisions are elastic; the particles rebound upward with the same speed they had before hitting the pan. If each particle has a mass of 110 g and collisions occur at the rate of 42 s-1, what is the average scale reading?

50

Solution:1

2, 0, 3.5 , , 0, 0.110 , 42 ,

?

ground f i fnup y y m v v v m kg st

F

21

2 imgy mv

mF v

t

But that is force on scale. By Newton’s Third Law, the force of scale on particle is 76.5N, and this is a normal force. Remember that scale reading (apparent weight) is the normal force.

readm

y=0

22 2 9.8 3.5 8.28i

m mv gy m

s s

8.28f

mv

s

8.28 8.28 16.56f i

m m mv v v

s s s

nmv

t

nm vt

142 0.110 16.56 76.5

mkg N

s s

2

76.57.8

9.8

N Nkg

mgs

51

Inelastic Collision

P is conserved.

K is not conserved (lost). Where does the energy go?

– Sound– Heat– Deformation, change in shape

52

Completely Inelastic Collision in 1D

If m1, initially moving with v1i, collides with m2, initially moving at v2i, and the two stick and move together with a combined velocity of Vf, theni fP P

Reverse CIC: Decay and explosion

1 2 1 2i i f fP P P P

1 1 2 2 1 2i i fm v m v m m V 1 1 2 2

1 2

i if

m v m vV

m m

Similar analysis

53

Example:

A 6.0 kg box sled is coasting across frictionless ice at a speed of 9.0 m/s when a 12 kg package is dropped into it from above. What is the new speed of the sled?

Motion on horizontal direction only.

1 1 2 26.0 , 9.0 , 12 , 0

?

i i

f

mm kg v m kg v

sv

1 1

1 2

if

m vv

m m

6.0 9.03.0

6.0 12

mkg

mskg kg s

54

1 1 2 2, 3000 , 6.0 , 500 , 0, 0.030 ,

?

idown m kg y m m kg v d m

F

1 2E E

Practice: Pg240-136A 3000 kg weight falls vertically through 6.0 m and then collides with a 500 kg pile, driving it 3.0 cm into bedrock. Assuming that the weight-pile collision is completely inelastic, find the magnitude of the average force on the pile by the bedrock during the 3.0 cm descent.

21 2

1

2mgy mv

1 13

1 2

3000 10.89.26

3000 500i

mkg

m v msV

m m kg kg s

4 3 3W K K K K

2

1 2 31 22

m m VF m m g

d

2 1 22 2 9.8 6.0 10.8

m mv gy m

s s

21 2 1 2 3

1

2m m g d F d m m V

2

62

3000 500 9.263000 500 9.8 5.0 10

2 0.030

mkg kg

mskg kg N

m s

1

2

4

3y = 0

55

Practice: Pg234-54

In Fig. 9-62, block 2 (mass 1.0 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 200 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 2.0 kg), traveling at speed v1 = 4.0 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

1 2v1

56

Solution: Pg234-54

1 2 1 2

max

2.0 , 1.0 , 200 / , 4.0 , 0

?

i i

mm kg m kg k N m v v

sx

1 1

1 2

2.0 4.02.67

1.0 2.0f

mkg

m v msv

m m kg kg s

i fE E

CIC:

Conservation of E during compression of spring:

2 2max

1 1

2 2fMv kx

max

1.0 2.02.67 0.33

200 /f

kg kgM mx v m

k N m s

57

Decay

Reverse process of completely inelastic collision.

Momentum is conserved.

Kinetic energy is not conserved, but total mass-energy is conserved. Gain in KE is given by loss in mass.– Energy released: Q = ΔK = -Δm c2

– atomic mass unit: u = 1.66 10-27 kg– 1 eV = 1.60 10-19 J

– Same analysis using conservation of momentum.

58

Example: Pg237-75E

A particle called - (sigma minus) is initially at rest and decays spontaneously into two other particles according to

- - + n.The masses are

m = 2340.5 me, m = 273.2me, mn = 1838.65me,

Where me (9.1110-31 kg or 0.511 MeV/c2) is the electron mass.

a) How much energy is transferred to kinetic energy in this process?

b) How do the linear momenta of the decay products (- and n) compare?

c) Which product gets the larger share of the kinetic energy?

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Solution: Pg237-75E

enee mmmmmm 65.1838,2.273,5.2340

) ?a K

b) The two particles should have momentum equal in magnitude and opposite in direction.

)c

Smaller particle has larger kinetic energy.

K

2 2 2228.65 228.65 0.511 / 117em c MeV c c MeV

2mc 2nm m m c 22340.5 273.2 1838.65e e em m m c

21

2K mv

2

2

mv

m

2

2

P

m

n

K

K

1838656.730

273.2n e

e

m m

m m

60

Collision in 2D

Glancing collision (as compared to head-on collision.)

Scattering (Not really hitting each other)

61

Collision in 2D (only for v2i = 0)

Conservation of linear momentum

1 1 1 1 1 2 2 2cos cosi f fm v m v m v

222

211

211 2

1

2

1

2

1ffi vmvmvm

If kinetic energy is also conserved.

1 2 90o m2

m2

v2f

m1 v1f

m1

v1i2

1

x

y

If m1 = m2:

1 1 1 2 2 20 sin sinf fm v m v

Px:

Py:

62

?

ˆ60,ˆ40,3600,2700,,

f

BiAiBA

v

jmphvimphvlbmlbmyNorthxEast

ix fxP P

x

y

Example:Two vehicles A and B are traveling west and south, respectively, toward the same intersection, where they collide and lock together. Before the collision, A (total weight 2700 lb) is moving with a speed of 40 MPH and B (total weight 3600 lb) has a speed of 60 MPH. Find the magnitude and direction of the velocity of the (interlocked) vehicles immediately after the collision.

A

B

iy fyP P

fv

A Aix A B f xm v m m v 2700 40

17.12700 3600

A Aixf x

A B

lb mphm vv mph

m m lb lb

B Biy A B f ym v m m v 3600 6034.3

2700 3600B Biy

f yA B

m v lb mphv mph

m m lb lb

1 1 34.3

tan tan 64 ,South of West17.1

f y o

f x

v mph

v mph

2 22 2 17.1 34.3 38fx fyv v mph mph mph

fv

63

External Force and Internal Energy Change

Eint: change in internal energy of system that causes the external force

Fext: external force that is caused by system

dcm: displacement of center of mass of system

: angle between Fext and dcm

cosint cmextdFE

In this case, instead of thinking Fextdcmcos as the work done by the external force, it is better to understand it as the work done by the system. If this work is positive, the Eint is negative, and the system loses internal energy.

64

then kinetic energy of CM is changed:

If there is also potential energy of any object within the system involved, then the mechanical energy of the system (which is expanded to include the earth and/or spring) will be changed:

coscmextcm dFK

coscmextcmcm dFUKE

65

Collision

An isolated event in which two or more bodies exert relatively strong forces on each other for a relatively short time.

If not isolated, not applicable, except when

Fext << Fint

Fext = 0 at the direction of consideration