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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!
"#
$
%& 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: