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CHAPTER 7
MOMENTUM
AND COLLISIONS
MOMENTUM
Momentum is the product of
the mass and velocity of a
body. Momentum is a vector
quantity that has the same
direction as the velocity of
the object.
MOMENTUM
p = mv
m = mass (kg)
v = velocity (m/s)
p = momentum (kg•m/s)
PROBLEM 1
A 2250 kg pickup truck has a velocity
of 25 m/s to the east. What is the
momentum of the truck?
m = 2250 kg
v = 25 m/s
p = mv = 2250(25)
= 56250 kg•m/s East
IMPULSE-MOMENTUM
THEOREM
Impulse is the product of the
average force and the time
interval during which the force
is exerted.
F = ma
Δv
F=m
Δt
F×Δt = m×Δv
Impulse = change in momentum
Applications of Impulse Momentum
Theorem
 Used
to determine stopping distances
and safe following distances of cars and
trucks.
 Used in designing safety equipment that
reduces force exerted on the object
during collisions. Example-nets and giant
air mattresses for fire-fighters.
 Also used in sports equipment and
games.
PROBLEM 3
A baseball of mass 0.14 kg is
moving at 35 m/s.
A) Find the momentum of the
baseball.
B) Find the velocity of a bowling
ball, mass 7.26 kg, if its
momentum is the same as the
baseball.
m = 0.14 kg
v = 35 m/s
p = mv
= 0.14(35)
= 4.9 kg•m/s
p  4.9 kg  m / s
m  7.26 kg
p 4.9
v 
m 7.26
 0.67 m / s
PROBLEM 6
The brakes exert a 640 N force on a car
weighing 15689 N and moving at 20
m/s. The car finally stops.
A) What is the car’s mass?
B) What is the car’s initial momentum?
C) What is the change in the car’s
momentum?
D) How long does the braking force act
on the car to bring it to a halt?
F  640 N
W  15689 N
t  ?
vi  20 m / s
vf  0 m / s
m?
pi  ?
p  ?
t  ?
W  15689 N
W 15689
m 
g
9.8
vi  20 m / s
 1600.92 kg
m  1600.92 kg
pi  mvi
 1600.92  20
 32018.4 kg  m / s
pi  32018 kg  m/s
pf  0 kg  m/s
Δp  pf  pi
 32018.4 kg  m/s
F  t  p
32018.4 kg  m/s
t 
640 N
 50 s
CONSERVATION OF
MOMENTUM
States that the momentum of any
closed, isolated system does not
change.
pinitial  pfinal
Isolated System has no net force
acting on it.
Closed System is a system in which
objects neither enter nor leave .
PROBLEM 11
Glider A of mass 0.355 kg moves
along a frictionless air track with a
velocity of 0.095 m/s. It collides with
a glider B of mass 0.710 kg moving
in the same direction at a speed of
0.045 m/s. After collision glider A
continues in the same direction with
a velocity of 0.035 m/s. What is the
velocity of glider B after collision?
Glider A
m A = 0.355 kg
Glider B
m B = 0.710 kg
v Ai = 0.095 m/s
v Bi = 0.045 m/s
v Af = 0.035 m/s
v Bf = ? m/s
Conservation of momentum means:
pinitial = pfinal
p Ai + pBi = pAf + pBf
m A v Ai + mB v Bi = mA v Af + mB vBf
m A v Ai + m B v Bi = m A v Af + m B v Bf
0.355(0.095) + (0.710)(0.045) = 0.355(0.035) + (0.710)v Bf
0.033725 + 0.03195 = 0.012425 + (0.710)vBf
0.065675 - 0.012425 = (0.710)vBf
0.05325 = (0.710)vBf
v Bf = 0.075 m/s
INTERNAL FORCES
Internal forces are forces between
objects within a system.
Example: If we consider a single car
as our system, forces are exerted on
objects within the car during a
collision (i.e. a crash dummy)
INTERNAL FORCES
EXTERNAL FORCES
External force is a force exerted by
an object outside the system.
Example: Our single car from the
previous example is still considered
our system. If the car collides with a
tree, then the force the tree exerts
on the car is an external force.
EXTERNAL FORCES
PROBLEM 17
Two campers dock a canoe. One
camper steps on the dock. This
camper has a mass of 80 kg and
moves forward at 4 m/s. With what
speed and direction do the canoe
and the other camper move if their
combined mass is 110 kg?
Camper
m a = 80 kg
v ai = 0 m/s
Camper and Canoe
v af = 4 m/s
m b = 110 kg
v bi = 0 m/s
v bf = ? m/s
p ai + p bi = paf + p bf
0 = m a v af + m b v bf
v bf
-m a v af
=
mb
v bf = -2.9m/s
Types of Collisions
– objects are apart
after the collision
ELASTIC
– objects “stick”
together after the collision
INELASTIC
COLLISIONS
INELASTIC
 Momentum is
conserved
 Some KE is
changed into
other forms
TOTALLY
ELASTIC
 Momentum is
conserved.
 KE is
conserved.
ELASTIC COLLISIONS

http://www.youtube.com/watch?NR=1&v=SBesU12g58I
INELASTIC COLLISIONS

http://www.youtube.com/watch?v=wFoPawE0LxA
SKATER/MEDICINE BALL
EXAMPLE
A 15 kg medicine ball is thrown at a
velocity of 20 km/hr to a 60 kg
skater who is at rest on ice. The
skater catches the ball and slides
with the ball across the ice.
Determine the velocity of the
skater and the ball after the
collision.
The collision occurs between a skater and a medicine
ball. Before the collision, the ball has momentum and the
person does not. The collision causes the ball to lose
momentum and the skater to gain momentum. After the
collision, the ball and the skater travel with the same
velocity ("v") across the ice.
SKATER/MEDICINE BALL EX
Before
Collision
Skater
0
Medicine ball 300
Total
300
After
Collision
60 * v
15 * v
300
p before = pafter
60v  15v  300
75v  300
v  4 km/hr
GRANNY/RALPH EXAMPLE
GRANNY/RALPH EXAMPLE
Before Collision
After Collision
Granny
80 * 6 = 480
80 * v
Ralph
0
40 * v
Total
480
480
p before = pafter
80v  40v  480
120v  480
v  4m/s
CAR/TRUCK EXAMPLE
CAR/TRUCK EXAMPLE
Before Collision
Truck
Car
Total
After Collision
3000 * 10 = 30 000 3000 * v
0
1000 * 15 = 15 000
30 000
30 000
pbefore = pafter
30000 = 3000v +15000
15000 = 3000v
v = 5m/s
Elastic
Elastic
Elastic
Above is a representation of 3 bullets with equal mass running into 3 blocks of
wood with equal mass. The first bullet passes through the block and maintains
much of its original momentum As a result, very little momentum gets
transferred to the block. The second bullet, expands as it enters the block of
wood which prevents it from passing all the way through it. As a result, most of
its momentum gets transferred to the block. (This is an inelastic collision.) The
third bullet (a rubber bullet) bounces off the block transferring all of it's own
momentum and then borrowing some more from the block. This has the most
momentum transferred to the block. (This is an elastic collision.)
KE and Collisions
Totally (Perfectly) Elastic Collision
KE is conserved: no sound or heat produced
Example: sub atomic particles (electrons),
attracting fields http://www.youtube.com/watch?v=x6n-QgjM4Ss
(Partially) Elastic Collision
some KE is lost as heat or sound
Is a broad range containing most collisions
Example: billiard balls http://www.youtube.com/watch?v=CgDDiDa3Kzk
http://www.youtube.com/watch?v=wFoPawE0LxA
KE and Collisions
Completely Inelastic Collision
KE is conserved: no sound or heat produced
Example: sub atomic particles (electrons), repelling
fields http://www.youtube.com/watch?v=NN_wwbx6Bew
Inelastic Collision
more KE is lost as heat or sound the colliding
objects stick together
Example: Coupling railroad cars
http://www.youtube.com/watch?v=qzV8ovAobfE
PROBLEM 20
A 2575 kg van runs into the
back of a 825 kg compact
car at rest. They move off
together at 8.5 m/s.
Assuming no friction with
the ground, find the initial
speed of the van.
m a  2575kg
m b  825kg
v ai  ? m / s
v bi  0
v af  8.5m / s
v bf  8.5m / s
m a v ai  m bv bi  m av af  m bv bf
2575v ai  825(0)  2575(8.5)  825(8.5)
2575v ai  21887.5  7012.5
2575v ai  28900
v ai  11.2m / s
PROBLEM 21
A 5g bullet is fired with a
velocity of 100 m/s toward a
10 kg stationary solid block
resting on a frictionless
surface. What is the change
in momentum of the bullet if
it becomes embedded in the
block?
m a  .005kg
m b  10kg
v ai  100m / s
v bi  0
v af  x
v bf  x
m av ai  m bv bi  m av af  m bv bf
.005(100)  10(0)  .005( x )  10( x )
.5  10.005x
x  .05m / s
Change in Momentum of Bullet
 ma (v af  v ai )
 .005(.05  100)
 0.5kgm / s