4.1. The Center of Mass, Newton’s Second

Law for a System of Particles

4.2. Linear Momentum and Its Conservation

4.3. Collision and Impulse

4.4. Momentum and Kinetic Energy in Collisions

Chapter 4: Linear Momentum and Collisions

4.1. The Center of Mass. Newton’s Second Law for a System of Particles

4.1.1. The center of mass (COM)

4.1.1. The center of mass

•Consider a system of 2 particles of masses m1

and m2 separated by distance d:

dmm

mxcom

21

2

•If m1 at x1 and m2 at x2:

M

xmxm

mm

xmxmxcom

2211

21

2211

where M is the total mass of the system

•If the system has n particles that are strung out along the x axis:

n

i

iinn

com xmMM

xmxmxmx

1

2211 1...

a. Systems of Particles

•If the n particles are distributed in three dimensions:

n

i

iicom

n

i

iicom

n

i

iicom zmM

zymM

yxmM

x111

1,

1,

1

kzjyixr iiiiˆˆˆ

•If the position of particle i is given by a vector:

kzjyixr co mxo mco mco mˆˆˆ

•The center of mass of the system is determined by:

n

i

iicom rmM

r1

1

b. Solid Bodies

zdmM

zydmM

yxdmM

x comcomcom

1,

1,

1

where M is the mass of the object

•For uniform objects, their density are:

V

M

dV

dm

dVV

Mdm

zdVV

zydVV

yxdVV

x comcomcom

1,

1,

1

Sample Problem (p. 204)

Determine the center of mass of the plate

0

PS

PPSSPS

mm

xmxmx

P

SSP

m

mxx

3

1

)2( 22

2

RR

R

area

area

thickness

thickness

m

m

P

S

P

S

P

S

P

S

Rx S

RxP3

1

A system consists of three particles located as shown inthe figure. Find the center of mass of the system.

Sample Problem

4.1.2. Newton’s Second Law for a System of Particles

)1(co mn et aMF

netF

coma

zco mzn e tyco myn e txco mxn e t M aFM aFM aF ,,,,,,

: the net force for all external forces

: the acceleration of the center of mass of the system.

M : the total mass of the system.

4.2. Linear Momentum and Its Conservation

The linear momentum of a particle is a vector quantity defined as:p

vmp

ly .re s p e c tiv e p a r tic le , th eo f v e lo c ity th ea n d m a s s th ea re a n d w h e re vm

Newton’s second law is expressed in terms of momentum:

dt

pdFnet

p artic le . o n the fo rce externalnet theis w here n etF

a. Linear Momentum

(Unit: kg m/s)

•For a system of particles:

nnn vmvmvmpppP

...... 221121

comvMP

com

com aMdt

vdM

dt

Pd

The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass.

Question: Why do we need momentum?

Because momentum provides us a tool for studying collision of 2 or more objects.

b. Conservation of Linear Momentum:If the net external force acting on a system of particles is zero,

0netF

co nstantP

:z)o r y, x,X(0 If , Xn etFco nstantXP

14. Two particles are launched from the origin of the coordinatessystem at time t=0. m1=5.0 g is shot directly along the x axis with aconstant speed of 10 m/s. m2=3.0 g is shot with a velocity ofmagnitude 20.0 m/s, at an upward angle such that it always staysdirectly above particle 1 during its flight.(a) What is the maximum height Hmax reached by the COM of the two

particle system?(b) In unit-vector notation, what are the (b) velocity and (c)

acceleration of the com when the com reaches Hmax?

gyvv yy 22

0,22

,2

(a) At the maximum height:

max2

0,2 2gyv y

Particle 2 always stays directly above P.1:

xx vv ,1,2

(m/s) 17.32

,1

22

2

,2

220,2

xxy vvvvv

(m) 3.15max y (m) 74.521

max2max

mm

ymH

(b)

21

2211

1

1

mm

vmvmvm

Mv

n

iiicom

n

i

iicom amaM1

At the maximum height, v2,y=0:

x

xx

xcomycom vmm

vmvmvv ,1

21

,22,11

,, ;0

im/s) 10(comv

(c)

21

2211

mm

amamacom

)(m/s 68.3 2

21

2

mm

gmacom

j )m/s 68.3( 2coma

:hence downward, iscoma

4.3. Collision and Impulse• Consider a collision between a bat and a ball:

The change in the ball’s momentum is:

d ttFpd )(

from a time ti to a time tf:

f

i

f

i

t

t

t

tdttFpd )(

• The impulse of the collision is defined by:

f

i

t

tdttFJ )(

Jp

the change in the object’s momentum

the impulse ofthe object

(Unit: kg m/s)

If F(t) function is unknown: tFJ a vg

Examples:

1. A 0.70 kg ball is moving horizontally with a speed of 5.0 m/s whenit 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 ball?

vmpvmp

;

Since the ball is moving horizontally, therefore, this is one dimensional motion:

xx vmp

)( ifxx vvmvmp

m/s) (kg 4.95)-(-20.7 :m/s 5 m/s; 2 xif pvv

ivmp ˆkg.m/s)9.4(

x

iv

fv

2. A 1500-kg car travelling at a speed of 5.0 m/s makes a 900 turn in a time of 3.0 s and emerges from this turn with a speed of 3.0 m/s: (a) What is the magnitude of the impulse that acts on the car during this turn? Draw the impulse vector. (b) What is the magnitude of the average force on the car during this turn? (Final exam, June 2014)

iv

fv

ip

fppJ

(a)

if pppJ

)m/s kg(874622 ifppJ

(b)

(N)29153

8746avg

t

JF

4.4. Momentum and Kinetic Energy in Collisions

Three types of collisions: We consider a system of 2 bodies

1. Inelastic collision:

constantP :momentum total

ffii pppp 2121

constantKE

Some energy (KE) is transferred to other forms, e.g. heat, sound.

constant21

21

21

21

21

mm

pp

mm

pp

mm

Pv

ffiicom

2. Elastic collision: conserved. are and KEp

ffii pppp 2121

ffii KKKK 2121

• In one dimension: ffii vmvmvmvm 22112211

2

222

112

222

112

1

2

1

2

1

2

1ffii

vmvmvmvm

Special cases:

:02 iv

if

if

vmm

mv

vmm

mmv

1

21

12

1

21

211

2

:21 mm iff vvv 121 ;0

:12 mm ifif v

m

mvvv 1

2

1211

2;

:21 mm ifif vvvv 1211 2;

3. Perfectly inelastic collision: two bodies stick together after collision:

KE.not but conserved p

:21 fff vvv fii vmmvmvm )( 212211

Case 3

3.1. In one dimension:

3.2. In two dimensions:

fii vmmvmvm

)( 212211

Example: (Perfectly inelastic collision)

A 1000-kg car travelling east at 80.0 km/h collides with a 3000 kg car traveling south at 50.0 km/h. The two cars stick together after the collision. What is the speed of the cars after the collision?(Final exam, June 2014)

fii vmmvmvm

)( 212211

80 km/h

50 km/hfii ppp

21

ip1

ip2

fp

)km/h kg(170000

0.5030000.801000 2222

21

f

iif

p

ppp

m/s 11.8or )km/h(5.42)( 21

mm

pv

ff

Homework: 2, 5, 13, 14, 22, 25, 38, 49, 56, 67, 60, 64, 74 (pages 230-237)

1. Center of mass:

n

iiicom rm

Mr

1

1

3. Impulse:tFJ avg theorem)impulse-momentum(linear Jp

4. Conservation of Linear Momentum:

( a closed, isolated system)

constantP

•Inelastic Collisions: ffii pppp 2121

constantKE

•Elastic Collisions: ffii pppp 2121

ffii KKKK 2121

Review:

5. Momentum and Kinetic Energy in Collisions

Unit: kg m/s2. Linear Momentum vmp

dt

PdFnet

f

i

t

tdttFJ )(