Chapter 7

Linear Momentum

Collisions in One Dimension

The total linear momentum is conserved when two objects collide, provided they constitute an isolated system.

Elastic collision -- One in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision. Inelastic collision -- One in which the total kinetic energy of the system after the collision is not equal to the total kinetic energy before the collision; if the objects stick together after colliding, the collision is said to be completely inelastic.

Elastic collision of an isolated system à both momentum and kinetic energy are conserved:

Initial Final m1 m2 m1 m2

v01v02

v f 1v f 2

m1v01 +m2

v02 =m1v f 1 +m2

v f 2 ⇒ m1v01 +m2v02 =m1vf 1 +m2vf 212m1v

201 + 1

2m2v202 = 1

2m1v2f 1 + 1

2m2v2f 2

head-on collision

à can be shown by combining these equations for head-on collisions (see book):

v01 − v02 = −(vf 1 − vf 2 )

Example: Head-on elastic collision of an isolated system. A softball of mass 0.200 kg that is moving with a speed 8.3 m/s collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of 3.2 m/s. a) Calculate the velocity of the target ball after the collision, and b) calculate the mass of the target ball.

a) b)

v01 − v02 = −(vf 1 − vf 2 )v01 = 8.3 m/s v02 = 0 vf 1 = −3.2 m/s vf 2 = ?vf 2 = vf 1 + v01 − v02 = −3.2+8.3− 0 = 5.1 m/s

m1v01 +m2v02 =m1vf 1 +m2vf 2 m1 = 0.200 kg m2 = ?

m2 vf 2 − v02( ) =m1 v01 − vf 1( ) ⇒ m2 =m1v01 − vf 1vf 2 − v02

#

$%%

&

'((

∴m2 = 0.2008.3− −3.2( )

5.1− 0#

$%

&

'(= 0.45 kg

Collisions in One Dimension

Example: Inelastic collision -- Ballistic Pendulum The mass of the block of wood is 2.50 kg and the mass of the bullet is 0.0100 kg. The block swings to a maximum height of 0.650 m above the initial position. Find the initial speed of the bullet.

Collisions in One Dimension

22112211 ooff vmvmvmvm +=+

Apply conservation of momentum to the collision:

m1 +m2( )vf =m1vo1

vo1 =m1 +m2( )vf

m1

Collisions in One Dimension

Applying conservation of energy to the swinging motion:

221 mvmgh =

m1 +m2( )ghf = 12 m1 +m2( )vf2

221

ff vgh =

vf = 2ghf

Collisions in One Dimension

vo1 =m1 +m2( )vf

m1

vo1 =0.0100 kg+ 2.50 kg

0.0100 kg!

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$

%& 2 9.80m s2( ) 0.650 m( ) = +896m s

vf = 2ghf

vo1 =m1 +m2( )m1

2ghf

Center of Mass

The center of mass is a point that represents the average location for the total mass of a system.

21

2211

mmxmxm

xcm +

+=

Center of Mass

21

2211

mmxmxm

xcm +

Δ+Δ=Δ

21

2211

mmvmvmvcm +

+=

The motion of the center-of-mass is related to conservation of momentum. Consider the change in the positions of the masses in some short time Δt:

Divide by Δt

Center of Mass

vcm =m1v1 +m2v2m1 +m2

=PM

In an isolated system, the total linear momentum does not change, therefore the velocity of the center of mass does not change.

P

M

vcm =m1v1 +m2

v2m1 +m2

=

PM

In vector form,

P =MvCM

ΔPΔt

=Δ MvCM( )

Δt=M Δ

vCMΔt

=MaCM

∴FEXT =∑ M

aCM

F∑ EXT( )Δt =

Pf −P0 = Δ

P

Divide both sides by ΔtF∑ EXT

=ΔPΔt

Newton’s 2nd Law for a system of particles which act like a single particle of mass M concentrated at the center of mass

The Principle of Conservation of Linear Momentum

Example: Ice Skaters Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54 kg woman and one is a 88 kg man. The woman moves away with a velocity of +2.5 m/s. Find the recoil velocity of the man.

The Principle of Conservation of Linear Momentum

of PP

=

02211 =+ ff vmvm

2

112 m

vmv ff −=

( )( ) sm5.1kg 88

sm5.2kg 542 −=

+−=fv

Center of Mass

021

2211 =+

+=

mmvmvmvcm

BEFORE

AFTER

( )( ) ( )( ) 0002.0kg 54kg 88

sm5.2kg 54sm5.1kg 88≈=

+

++−=cmv

(54 kg)(+2.5 m/s) + (88 kg)(-1.5 m/s)

54 kg + 88 kg 0.02

Consider the “Ice Skater” example which was an isolated system: