Chapter 7

Impulse and Momentum

7.2 The Principle of Conservation of Linear Momentum

( ) of PPrr

!="tforces external average of sum

If the sum of the external forces is zero, then

of PPrr

!=0 of PPrr

=

PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM

The total linear momentum of an isolated system is constant(conserved). An isolated system is one for which the sum ofthe average external forces acting on the system is zero.

7.2 The Principle of Conservation of Linear Momentum

Applying the Principle of Conservation of Linear Momentum

1. Decide which objects are included in the system.

2. Relative to the system, identify the internal and external forces.

3. Verify that the system is isolated.

4. Set the final momentum of the system equal to its initial momentum.Remember that momentum is a vector.

7.3 Collisions in One Dimension

The total linear momentum is conserved when two objectscollide, 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.

7.3 Collisions in One Dimension

Example 8 A Ballistic Pendulim

The mass of the block of woodis 2.50-kg and the mass of the bullet is 0.0100-kg. The blockswings to a maximum height of0.650 m above the initial position.

Find the initial speed of the bullet.

7.3 Collisions in One Dimension

22112211 ooff vmvmvmvm +=+

Apply conservation of momentum to the collision:

( ) 1121 of vmvmm =+

( )1

211 m

vmmv fo

+=

7.3 Collisions in One Dimension

Applying conservation of energyto the swinging motion:

221 mvmgh =

( ) ( ) 2212

121 ff vmmghmm +=+

221

ff vgh =

( )( )m 650.0sm80.922 2== ff ghv

7.3 Collisions in One Dimension

( )1

211 m

vmmv fo

+=

( )( )m 650.0sm80.92 2=fv

( )( ) sm896m 650.0sm80.92kg 0.0100

kg 50.2kg 0100.0 21 +=!!

"

#$$%

& +=ov

7.5 Center of Mass

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

21

2211

mmxmxm

xcm +

+=

7.5 Center of Mass

21

2211

mmxmxm

xcm +

!+!=!

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

7.5 Center of Mass

21

2211

mmvmvmvcm +

+=

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

}P21

2211

mmvmvmvcm +

+=

P

7.2 The Principle of Conservation of Linear Momentum

Example: Ice Skaters

Starting from rest, two skaterspush off against each other onice where friction is negligible.

One is a 54-kg woman and one is a 88-kg man. The womanmoves away with a speed of +2.5 m/s. Find the recoil velocityof the man.

7.2 The Principle of Conservation of Linear Momentum

of PPrr

=

02211 =+ ff vmvm

2

112 m

vmv ff !=

( )( ) sm5.1kg 88

sm5.2kg 542 !=

+!=fv

7.5 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 kg0.02

Consider the “Ice Skater” example which was anisolated system: