Chapter 9

Linear Momentum and Collisions

Linear momentum

• Linear momentum (or, simply momentum) of a point-like object (particle) is

• SI unit of linear momentum is kg*m/s

• Momentum is a vector, its direction coincides with the direction of velocity

vmp

p

Newton’s Second Law revisited

• Originally, Newton formulated his Second Law in a more general form

• The rate of change of the momentum of an object is equal to the net force acting on the object

• For a constant mass

dt

pdFnet

dt

pdFnet

dt

vmd )(

dt

vdm

am

Center of mass

• In a certain reference frame we consider a system of particles, each of which can be described by a mass and a position vector

• For this system we can define a center of mass:

M

rm

m

rmr i

ii

ii

iii

CM

Center of mass of two particles

• A system consists of two particles on the x axis

• Then the center of mass is

21

2211

mm

xmxmxCM

M

rm

m

rmr i

ii

ii

iii

CM

21

21 00

mm

mmyCM

0

Newton’s Second Law for a system of particles

• For a system of particles, the center of mass is

• ThenM

rm

m

rmr i

ii

ii

iii

CM

i

iiCM rmrM

2

2

2

2 )(

dt

rMd

dt

rdMaM CMCM

CM

2

2

dt

rmdi

ii

i

ii dt

rdm

2

2

iiiam

i

iF

Newton’s Second Law for a system of particles

• From the previous slide:

• Here is a resultant force on particle i

• According to the Newton’s Third Law, the forces that particles of the system exert on each other (internal forces) should cancel:

• Here is the net force of all external forces that act on the system (assuming the mass of the system does not change)

i

iCM FaM

netCM FaM

iF

netF

Newton’s Second Law for a system of particles

netCM FaM

Linear momentum for a system of particles

• We define a total momentum of a system as:

• Using the definition of the center of mass

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

i

iii

i vmpP

i

iivmP

dt

rmd ii

i

dt

rdM CM

CMvM

i

ii dt

rdm

M

rm

m

rmr i

ii

ii

iii

CM

Linear momentum for a system of particles

• Total momentum of a system:

• Taking a time derivative

• Alternative form of the Newton’s Second Law for a system of particles

CMvMP

dt

vdM

dt

Pd CM

CMaM

netF

netFdt

Pd

Conservation of linear momentum

• From the Newton’s Second Law

• If the net force acting on a system is zero, then

• If no net external force acts on a system of particles, the total linear momentum of the system is conserved (constant)

• This rule applies independently to all components

netFdt

Pd

0dt

Pd

constP

constPF xxnet 0,

Center of mass of a rigid body

• For a system of individual particles we have

• For a rigid body (continuous assembly of matter) with volume V and density ρ(V) we generalize a definition of a center of mass:

ii

iii

CM m

rmr

M

dmr

dV

dVrr

volume

volumeCM

Chapter 9Problem 37

A uniform piece of steel sheet is shaped as shown. Compute the x and y coordinates of the center of mass of the piece.

Impulse

• During a collision, an object is acted upon by a force exerted on it by other objects participating in the collision

• We define impulse as:

• Then (momentum-impulse theorem)

dttFpd net )(

)(tFdt

pdnet

f

i

f

i

t

t net

t

tdttFpd )(

f

i

t

t net dttFI )(

Ipp if

Elastic and inelastic collisions

• During a collision, the total linear momentum is always conserved if the system is isolated (no external force)

• It may not necessarily apply to the total kinetic energy

• If the total kinetic energy is conserved during the collision, then such a collision is called elastic

• If the total kinetic energy is not conserved during the collision, then such a collision is called inelastic

• If the total kinetic energy loss during the collision is a maximum (the objects stick together), then such a collision is called perfectly inelastic

Elastic collision in 1D

constK constP

2222

222

211

222

211 ffii

vmvmvmvm ffii vmvmvmvm 22112211

)()( 222111 fifi vvmvvm ))(( 11111 fifi vvvvm

))(( 22222 fifi vvvvm

iif vmm

mv

mm

mmv 2

21

21

21

211 )(

2

)(

)(

iif vmm

mmv

mm

mv 2

21

121

21

12 )(

)(

)(

2

Elastic collision in 1D: stationary target

• Stationary target: v2i = 0

• Then

iif vmm

mv

mm

mmv 2

21

21

21

211 )(

2

)(

)(

iif vmm

mmv

mm

mv 2

21

121

21

12 )(

)(

)(

2

if vmm

mmv 1

21

211 )(

)(

if vmm

mv 1

21

12 )(

2

Chapter 9Problem 18

A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle?

Perfectly inelastic collision in 1D

constP

fii vmmvmvm )( 212211

21

2211

mm

vmvmv ii

f

Collisions in 2D

constP constPx

constPy

Chapter 9Problem 57

A bullet of mass m is fired into a block of mass M initially at rest at the edge of a frictionless table of height h. The bullet remains in the block, and after impact the block lands a distance d from the bottom of the table. Determine the initial speed of the bullet.

Answers to the even-numbered problems

Chapter 9

Problem 81.39 kg m/s upward⋅

Answers to the even-numbered problems

Chapter 9

Problem 16(a) 2.50 m/s; (b) 3.75 × 104 J

Answers to the even-numbered problems

Chapter 9

Problem 20(a) 4.85 m/s; (b) 8.41 m

Answers to the even-numbered problems

Chapter 9

Problem 227.94 cm

Answers to the even-numbered problems

Chapter 9

Problem 58(a) – 0.667 m/s; (b) 0.952 m