The center of mass (com)Motion of a system of particles (Newton’s 2nd

law)Linear momentumImpulseConservation of linear momentumInelastic Collision in one dimensionElastic collision in 1 and 2 dimensions

Ch09: Center Of Mass And Linear Momentum 9.1: What is physics?

Analyzing complicated motion of any sort requires simplification via an understanding of physics. The complicated motion of a system of objects, such as a car or a ballerina or collection of particles, can be simplified if we determine a special point of the system—the center of mass of (com) that system.We can apply equations of motion to the

center of mass to represent the motion of that system

An example for complicate motion is a baseball bat when flipped into air points of bat moves along different path shapes except the center of mass moves a long a simple parabolic path (projectile motion)

9.2: The Center Of MassIf we consider a collection of particles or an extended object with a total mass, M.We can consider the system as a single particle with the total mass, M, concentrated at a single point called the center of mass (com).If a net force external is applied to the system, then you can apply Newton’s second law to the center of mass.

The center of mass of a system of particles is the point that moves as though:

all of the system’s mass were concentrated there all external forces were applied there.

M

xmx

n

iii

com

∑== 1

The position of center of mass xcom is the average position of the system’s masses on x-axis

n

nncom mmm

xmxmxmx++++++

=................

21

2211

Total mass of the system =M

; closer to bigger

9.2: The Center Of Mass: system of particles

For the two particle shown, the xcom is

For n particles along x-axis

For group of particles in space (more than one dimension), we need to find position vector for the center of mass comrr

kzjyixM

rmr CMCMCM

iii

comˆˆˆ ++==

∑ r

r

M

xmx i

ii

com

∑=

M

ymy i

ii

com

∑=

M

zmz i

ii

com

∑=

∑∑

==i

iii

ii

com rmMM

rmr r

r

r 1

9.2: The Center Of Mass: system of particles

A group of particles in two dimensions is shown in the figure. Find the center of mass if m1=1kg, m2=1kg, m3=2kg

m

mmmymymymycom

144

211)2)(2()0)(1()0)(1(

321

332211

==++

++=

++++

=

we have x and y dimensions

mjircom )ˆ1ˆ75.0( +=r

m

mmmxmxmxmxcom

75.043

211)0)(2()2)(1()1)(1(

321

332211

==++

++=

++++

=

9.2: The Center Of Mass: Example

∫= dmrM

rCMrr 1

∫= xdmM

xCM1

∫= ydmM

yCM1

∫= zdmM

zCM1

For an extended solid object (continuous distribution of matter)

M

rmr i

ii

CM

∑∆=

r

r

For ∆mi 0 dm

9.2: The Center Of Mass: Solid Bodiesa) show that the center of mass for the rode of length L lies at the middle b) if the rode is not uniform with λ = αx find center of mass

m=ρV dm= ρdV

m= σA dm=σdA

m=λL dm=λdL

a) Rode lies on x-axis, rode is uniform λ is constant

λdL=λdx because L lies on x-axis

b) λ is variable

9.2: The Center Of Mass: Example

For a system of particles in motion, we usually interesting in the motion of the center of mass that represents the whole system:

9.3: Newton’s Second Law For a System Of Particles

the velocity of the center of mass ( ) of the system is found from

comvr

∑=i

iicom vmM

v rr 1

nni

iicom vm.........vmvmvmvmvM rrrrrr++++== ∑ 332211

Hence, the total mass (M) multiplied by the velocity of center of mass ( ) represents the all system; it is equivalent to multiply each particle velocity with it’s mass and sum them

comvr

9.3: Newton’s Second Law For a System Of Particles

∑=i

iicom amM

a rr 1

netncom

nni

iicom

FFFFFaM

am.........amamamvmaMrrrrrr

rrrrrr

=++++=⇒

++++== ∑...........321

332211

taking the derivative of the , you can get the acceleration of the center of mass ( )

comvr

comar

Newton’s 2nd law

is the net force of all external forces that act on the system; it is equivalent to the sum of all forces act on all particles in the system

in components

9.3: Newton’s Second Law For a System Of Particles: Example

Consider the figure shown. what is the acceleration of the center of mass of the system, and in what direction does it move?

Since we have a system moving under external forces shown, the forces and the masses can be transferred to the center of mass (com) as shown

The com of the system will move as if all the mass were there and the net force acted there

9.3: Newton’s Second Law For a System Of Particles: Example: continued from previous slide

Newton’s 2nd law

x-component

y-component

and

It is a vector quantity has the direction of the velocity For a particle, any change in momentum due to velocity change is related to some external net force

The linear momentum of a particle with mass m and velocity is:

vrvmp rr

= SI unit is kg m/s ≡ N.s

9.4: Linear Momentum

pr

vr

netFr

From Newton’s 2nd law

Since Linear momentum is a vector quantity it may has components in the x-, y-, and z- direction

vmp rr=

zzyyxx mvpmvpmvp ===

dtpdFrr

=∑

kpjpipp zyxˆˆˆ ++=

r

the time rate of change of particle’s linear momentum is equal to the net force acting on the particle (other form of Newton’s 2nd law).

9.4: Linear Momentum

Momentum vector

Momentum components

where

The linear momentum for the center of mass that represents the system particles can be obtained from above equation:

9.5: Linear Momentum For a system of particles

For a system of particles in motion each has its own velocity and mass the velocity of the center of mass ( ) iscomvr

∑=i

iicom vmM

v rr 1

∑

∑++++===⇒

++++===

inicomcom

nni

iicomcom

p.........ppppvMP

vm.........vmvmvmvmvMP

rrrrrrr

rrrrrrr

321

332211wherecomcom vMP rr

= Momentum for system of particles

Hence, the momentum for the system is equivalent to the sum of all individual momentum of particles

9.6: Collision and impulse: Single CollisionMomentum of particle can only changes by a net external force

Consider a collision between a ball and a bat in baseball gameThe ball will experience a force that

varies during collision ball slows down, stop, and reverse direction changes due to that force From Newton’s 2nd law

we can have the change in momentum by integrating both sides over time during collision

A measure for magnitude and duration of a collision force (impulse )

9.6: Collision and impulse: Single Collision

whereIn components (x-components for instant)

the Impulse on the particle (ball in this case):

Since it is difficult to know during collisions we use the average force magnitude and ∆t:

Due to Newton’s 3rd law, the ball make an impulse on the bat of same magnitude but opposite in direction.

Impulse-momentum theorem Impulse is the net change in momentum due to F.

A car made an accident with a wall. Its velocity was 15 m/s to left beforthe accident. After the accident, it is recoil from the wall to the right with velocity 2.6 m/s. If the accident time is 0.15 s, find the impulse caused by the collision and the average force exerted on the car.m = 1500 kg, vi = -15.0 m/svf = 2.6 m/s, Dt = 0.150 sJ = ? Favg = ?

9.6: Collision and impulse: Example

Note: for x-axis

If v to the right v is +ve

If v to the left v is -ve

The impulse J = ∆pJ = pf - pi = mvf – mvi = m(vf - vi)J = 1500(2.6-(-15))J = 1500 (17.6) = 2.64×104 î N.s

Favg = ∆p/∆t = 2.64×104 / 0.15 = 1.76×105 î N

9.6: Collision and impulse: Example: continued from previous slide 9.6: Collision and impulse: Example

Figure below shows the path taken by a race car driver as he collides with the racetrack wall. If driver mass m is 80 kg finda) The impulse on the driver due to collisionb) Average force magnitude on driver if the collision lasts for 14 ms

x-component

y-component

From –ve x-axis

(a) (b)

Consider two particles interact together We can apply Newton’s Third law

constant 21 ==+ ∑ ppp rrr

9.7: Conservation of Linear Momentum

0)( 21 =+ ppdtd rr

Apply Newton’s 2rd law iiff

if

pppp 2121

pprrrr

rr

+=+

= ∑∑

∑∑∑∑∑∑

=

=

=

izfz

iyfy

ixfx

pp

pp

pp

Momentum is vector quantity it has components

conservedismomentumLinear⇒

Take care of velocity v sign (+ve or –ve)

9.7: Conservation of Linear Momentum

m1 = 100 kg, v1 = 5 m/s, m2 = 50 kg

Two skaters recoil each other from rest. After recoil Find a) the total momentum b) the momentum for heavy skater1 and c) the velocity for light skater2

a) ptot = ?, b) p1 = ?, c) v2 = ?Solution: momentum is conserved in x-axis

a) Σpi = ptot = 0 ; they start from rest

b) p1= m1v1= 100(-5)= - 500 kg m/s

c) Σpf = Σpi = ptot

p1f + p2f = 0 p2f = -p1f m2v2 = -p1f

v2=-p1f / m2 v2= -(-500)/50=10 m/s

9.7: Conservation of Linear Momentum: Example

Same with coordinates

9.7: Conservation of Linear Momentum: Example

Two-dimensional explosion: the figure shows a firecracker placed inside a coconut of mass M, initially at rest on a frictionless floor, blows the coconut into three pieces that slide across the floor. Piece C, with mass0.3M, has final speed vfC = 5.0 m/s. (a) What is the speed of piece B, with mass 0.20M? (b) What is the speed of piece A?

0.3M

0.2M

0.5M

= 5m/s

smv

.v.

Mv.Mv.

vmvmvmp

fB

fB

fCfB

fCyCfByBfAyAfy

/67.9

04811530

008sin3005sin200

00

=⇒

=+−

=°+°−

=++⇒=∑)1(0260129050

008cos3005cos2050

00

=++−

=°+°+−

=++⇒=∑

.v.v.

Mv.Mv.Mv.

vmvmvmp

fBfA

fCfBfA

fCxCfBxBfAxAfx

9.7: Conservation of Linear Momentum: Example: continued from previous slide

(a) What is the speed of piece B? (b) What is the speed of piece A?

∑∑∑∑∑∑ ==⇒= iyfyixfxif pppppp andrr

Since coconut was at rest before explosion 0== ∑∑ iyix pp

Sub. in (1) smv fA /3=⇒

p is conserved Inelastic collision

Particles don’t stick together but lose some energy

Completely inelastic collisionParticles stick together and lose some energy (deformation)

Elastic collisionParticles bounce off each other, no loss of energy

Mechanical Energy is conserved ONLY in elastic collisionsMomentum is conserved in both elastic and inelastic collisions

9.8: Momentum and Kinetic Energy in Collisions

ffii vmvmvmvm 22112211rrrr

+=+

( )222

2112

1iii vmvmK += ( )2

222

1121

fff vmvmK +=

ifloss KKE −=

Inelastic collisionMomentum is conserved

Mechanical energy is not conserved

The Energy loss =

∑∑ = fi pp rr

9.9: Inelastic Collision in One dimensions: Inelastic Collision

and

fii vmmvmvm rrr )( 212211 +=+

( )222

2112

1iii vmvmK += ( ) 2

2121

ff vmmK +=

ifloss KKE −=

Completely inelastic collisionMomentum is conserved

Mechanical energy is not conserved

The Energy loss =

∑∑ = fi pp rr

9.9: Inelastic Collision in One dimensions: Completely Inelastic Collision

and

Solution:

Momentum is conserved

9.9: Example: A 900 kg car had a velocity of 20 m/s to the right collides with other 1800kg car that was stopped. If the two car stuck together, find the final velocity after collision

∑∑ = fi pprr

Befor collision After collision

V2i = 0 Vf = ??V1i = 20 m/s

fii mmmm v)(vv 212211rrr

+=+

smmmmm ii

f /67.61800900

0)20(900vvv21

2211 =+

+=

++

=⇒rr

r

9.9: Example: Ballistic pendulum: A bullet (m1) is fired into a large block of wood (m2) suspended from some light wires. The bullet embeds in the block, and the entire system swings through a height h. How can we determine the speed of the projectile from a measurement of h?

Befor collision Immidiatlyafter collision

To find the speed of bullet v1A weneed to find vB

From conservation of energy(only for system after collision from B to C)

righttheto2

221

1

211

2

ghm

mmv

ghvmvmgh

A

BB

+=⇒

=⇒=

rWe have inelastic collision

∑∑ = fi pprr

BA mmm v)(0v 2111rr

+=+

1

211

v)(vmmm B

A

rr +

=⇒

∑∑ = fi pprr

∑∑ = fi KK

)1(vvvv 22112211 ffii mmmm rrrr+=+

=

Momentum is conserved

)2(21

21

21

21 2

222

112

222

11 ffii vmvmvmvm +=+

Energy is also conserved

From equations 1 & 2 )v-v(v-v 2121 ffiirrrr

−=

9.10: Elastic Collision in One dimensions9.10: Elastic Collision in One dimensions: Example

Two metal spheres, suspended by vertical cords, initially just touch. Sphere 1, with mass m1 = 30 g, is pulled to the left to height h1 = 8 cm, and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2, whose mass m2 = 75 g. What is the velocity v1f of sphere 1 just after the collision?

Sphere 1 speed before collision

Ef = Ei

)1(075003000470 21

22112211

ff

ffii

v.v..

vmvmvmvmrr

rrrr

+=+

+=+

)2(02521

)(

21

2121

ff

ffii

vv.

v-vv-vrr

rrrr

+−=−

−= Solve the two equations

9.10 : Example: A block (m1=1.6 kg) initially moving to the right at 4 m/s collides with a spring attached to a second block (m2=2.1 kg) moving to the left with at 2.5 m/s. The spring constant is 600 N/m. Find the velocities of the two blocks after the collision.

Befor collision After collision

Momentum is conserved

(1)

For elastic collision, we have

(2)

From eqn’s (1) and (2)

Conservation of momentum still stands:

Analyze each component separately:

Use conservation of kinetic energy if the collision is elastic:

ffii vmvmvmvm 22112211vrrr

+=+

fyfyiyiyfyiy

fxfxixixfxix

vmvmvmvmpp

vmvmvmvmpp

22112211

22112211

+=+⇒=

+=+⇒=

∑ ∑∑ ∑

( ) ( )222

211

222

211 2

121

iiff

if

vmvmvmvm

KK

+=+

=

∑∑ = fi pp rr

9.11: Collisions In Two Dimensions

9.11:Example: If before collision, m2 was at rest, find magnitude and direction of speed for m2 after collision (m1=1kg, m2=2kg)

v1i = 10 m/s, v1f =5 m/sθ = 30°

v2f = ?φ = ?

momentum is conserved in the two dimensions x and y

smvvv

vmvmvm

vmvmvmvmpp

fx

iff

ixixfxfxixfx

/83.2cos10cos233.4

0cos30cos

222

112211

22112211

==⇒=+

+=+⇒

+=+⇒=∑ ∑

φφ

φ

2-variables 2-eqn’s

m1

m2

x-component smvv

vmvm

vmvmvmvmpp

fyf

ff

iyiyfyfyiyfy

/25.1sin

0sin30sin

22

2211

22112211

−==−⇒

=−⇒

+=+⇒=∑ ∑

φ

φ

°−=−

==+= − 8.2383.225.1tan,/1.3

,

122

222 φandsmvvv

hence

fyfxf

y-component

m1

m2

9.11:Example: continued from previous slide:

Example: Rocket fired vertically, explode ataltitude of 1000 m into three fragements, as shown.The 3 fragements are of equal masses. Find the velocity of the third part after collision

Collision for system of particles

fcomicom vv ,,rr

=fcomicomfcomicom vMvMPP ,,,,rrrr

=⇒=

For isolated system of particles Momentum is conserved

m/sv

vM)(Mvmvmvm

)Mvm

Mvm

MvmM(MvMv

Ppp

Pppconservedismomentum

fx

fxxxx

xxxcom,xix

xcomfxix

comfi

2403

2403

00

0

3

3332211

332211

,

−=⇒

++=++=⇒

++=⇒=

==⇒

==⇒

∑∑∑∑

rrr

rrr

Rocket fired vertically

m/sv

vM)(MM

vmvmvmM

)Mvm

Mvm

Mvm

M(M

MvMv

Ppp

fy

fy

yyy

yyy

ycomiy

ycomfyiy

4503

03

4503

300

300

3

3

332211

332211

,

,

=⇒

++=

++=

++=

=

== ∑∑rrr

Same for momentum in the y-direction

smjiv fy /)ˆ450ˆ240(3 +−=⇒r

Collision for system of particles: Example: continued from previous slide