IMPULSE AND MOMENTUM
The impulse F
 t is a vector quantity equal in magnitude to the product of the force and the time interval in which it acts. Its direction is the same as that of the force.
F
 t Units: newton.second (N.s)
The momentum p of a particle is a vector quantity equal in magnitude to the product of its mass m and its velocity v .
p = m v Units: (kg.m/s)

Impulse (F

t) = change in momentum (m

v)

Spreading impulse out over a longer time means that the force will be less; either way, the change in momentum of the boxing glove, fist, and arm will be the same.

Contact time is reduced if arm's deceleration is kept as small as possible.
This is done by using "followthrough", which means to continue to push during the contact period.

CONSERVATION OF LINEAR MOMENTUM
According to the law of conservation of linear momentum, when the vector sum of the external forces that act on a system of bodies equals zero, the total linear momentum of the system remains constant no matter what momentum changes occur within the system.
Although interactions within the system may change the distribution of the total momentum among the various bodies in the system, the total momentum does not change. Such interactions can give rise to two general classes of events:

a. explosions, in which an original single body flies apart into separate bodies

b. collisions, in which two or more bodies collide and either stick together or move apart, in each case with a redistribution of the original linear momentum.

For two objects interacting with one another, the conservation of momentum can be expressed as: m v
1 1
 m v
2 2
 m v
1 1
'  m v
'
2 2 v
1 and v
2 v
' v
' are initial velocities, and are final velocities

7.4
Assume two objects have masses of 8 and 6 kg respectively. The 8 kg mass moves initially to the right with a velocity of 4 m/s and collides with the 6 kg mass moving to the left at 5 m/s. What is the total momentum before and after the collision?
m
1 m
2 v
1
= 8 kg
= 6 kg
= 4 m/s v
2
= -5 m/s
Initial momentum m
1 v
1
+ m
2 v
2
= 8(4) + 6(-5)
= - 2 kg m/s initial momentum = final momentum
Final momentum = - 2 kg m/s

ELASTIC AND INELASTIC COLLISIONS
If the Kinetic energy remains constant in a collision, the collision is said to be completely elastic.
If the colliding bodies stick together and move off as a unit afterward, the collision is said to be completely inelastic.
In inelastic collisions only the momentum is conserved.

In elastic collisions no permanent deformation occurs; objects elastically rebound from each other.
In head-on elastic collisions between equal masses,velocities are exchanged.

Inelastic collisions are characterized by objects sticking together and permanent deformation.

7.5
A 2 kg ball traveling to the left with a speed of 24 m/s collides head-on with a 4 kg ball traveling to the right at 16 m/s. Find the resulting velocity if the two balls stick together after impact. m
1 m
2 v
1
= 2 kg
= 4 kg
= -24 m/s v
2
= 16 m/s m
1 v
1
+ m
2 v
2
=( m
1
+m
2
)V
V
 m v
1 1
 m v
2 2 m
1
 m
2

2( 24)

4(16)
6
= 2.67 m/s to the right

7.6
The nucleus of an oxygen atom has a mass of 3.8x10
-25 kg and is at rest.
The nucleus is radioactive and suddenly ejects a particle of mass 6.6x10
-27 kg and speed 1.5x10
7 m/s. Find the recoil speed of the nucleus that is left behind.
m
1 m
2
= 3.8x10
-25 kg
= 6.6x10
-27 kg
0
 m v
'  m v
' v
2
'
= 1.5x10
7 m/s v
1
' 
 m v
2 2
' m
1

 x

27 x
7
(6.6 10 )(1.5 10 )

25
= -2.6x10
5 m/s

7.7
A 7500-kg truck traveling at 5 m/s east collides with a 1500-kg car moving at 20 m/s in a direction 210

. After the collision, the two vehicles remain tangled together. With what speed and in what direction does the wreckage begin to move?
m
1 m
2
= 7500 kg v
1
= 1500 kg v
2
= 5 m/s, 0º
= 20 m/s, 210º m
1 v
1
+ m
2 v
2
=( m
1
+m
2
)V

m
1 v
1
+ m
2 v
2
=( m
1
+m
2
)V x-comp y-comp
7500 (5) 0
1500 (20 cos 210º) 1500 (20 sin 210º)
Σx = 11,519 kg m/s Σy = - 15,000 kg m/s
Initial Momentum 
(11519)
2 
(15000)
2
= 18,912.7 kg m/s

initial momentum = final momentum
= (m
1
+ m
2
) V
V
 initial m
1
 m
2

18912.7
= 2.1 m/s
  tan
-1
15000
11519
= 52.5º
IV quadrant
V (2.1 m/s, 307.5º)