Download Physics 106P: Lecture 1 Notes

Chapter 7:
Linear Momentum (p)
(Linear) Momentum (p)
 Linear
Momentum (p) is defined as the
product of mass (m) and velocity (v):
p=mv
 SI
Units of linear momentum = kg.m/s
p
is a vector quantity: specified by
magnitude and direction.
 The
direction of p is the same as the
direction of velocity v.
 Example: A car of mass 1,500 kg is
traveling at a constant speed of 12 m/s due
north. What is its linear momentum?
p = mv= (1,500 kg) x (12 m/s)
= 18,000kg.m/s due north
 Constant momentum: Means both the mass
(m) and velocity (v) is not changing.
Usually, mass stays constant during motion.
 Some exceptions:
1. A rocket traveling: total mass changes as it
burns fuel.
2. An open freight car being loaded while in
motion. Its mass will change as it moves
Example:
1. A car of mass 1,500 kg is traveling at a
constant speed of 12 m/s in a circular
path. Is its linear momentum constant?
No: Because direction of velocity in a circular
motion changes.
2. A ball of mass 0.2 kg is thrown horizontally at
a wall. If it hits the wall at 25 m/s and
bounces back with the same speed, what is
the change in its momentum?
(A) 0 kg-m/s
(B) 5 kg.m/s
(C) -5 kg.m/s
(D) 10 kg.m/s (E) –10 kg.m/s
Example:
3. A ball of mass 0.2 kg is thrown at an
angle of 30o above the horizontal with a
speed of 5 m/s and travels as a
projectile.
(a)What is the y-component of its
momentum at the instant it is thrown?
(b)At its maximum height, what is its
momentum?
4. True/False? An object traveling with
larger velocity must have larger
momentum than another one traveling
with smaller velocity.
TOTAL MOMENTUM
Example 1:
Car (1) is moving due east (+) at 30 m/s.
Another car (2) is moving due east (+)
at 30 m/s. For these two cars each of
mass 1,500 kg, find their
(a) total kinetic energy.
(b) total momentum.
Example 2:
Car (1) is moving due east (+) at 30 m/s.
Another car (2) is moving due west (-)
at 30 m/s. For these two cars each of
mass 1,500 kg, find their
(a) total kinetic energy.
(b) total momentum.
Example 3:
Yes/No?
Is it possible to have a system of objects
where the total momentum is zero but
the total kinetic energy is not zero?
Conservation of Linear Momentum
Consider two objects with mass mA and mB
moving towards each other with initial
velocities v1(A) and v1(B)
V1(A)
V1(B)
mA
mB
Total initial momentum:
pi = mA V1(A) + mB V1(B)
If these two objects collide and later their
velocities after impact are V2(A) and V2(B):
V2(A)
V2(B)
mA
Total final momentum:
pf = mA V2(A) + mB V2(B)
mB
It has been established that on
condition that no net external forces
act on any system of colliding objects,
the total momentum of the system will
always remain conserved.
ie, pi = pf
OR:
mAV1(A) + mBV1(B) mA = mAV2(A)
+mBV2(B)
Law of Conservation of Linear Momentum:
Momentum is “Conserved” means it can not be
created nor destroyed
Can be transferred from one object to
another
 Total
Momentum does not change with time.
 Total momentum “before” = total momentum
“after.”
 This is a BIG deal!
 In
science, any law of conservation is a
very powerful tool in understanding the
physical universe.
Law of Conservation of Energy

In any natural process, total energy is
always “conserved”, i.e. energy can not be
created nor destroyed.
Can be transformed from one form to
another.
Can be transferred from one system to
another.
In science, any law of conservation is a
very powerful tool in understanding the
physical universe.
Example:
Jane and Fred are on skates facing
each other. Jane then pushes Fred so
he is going 2.0 m/s. If Fred is twice as
heavy as Jane, how fast does Jane
end up moving?
pinitial = pfinal
0 = mFred VFred + mJane VJane
VJane = mFred VFred / mJane = 4 m/s
Example:
Car (1) is moving due east at 30 m/s.
Another car (2) is moving due west at
30 m/s. For these two cars each of
mass 1,500 kg, find
(a) Total kinetic energy.
(b) Total momentum.
(c) Yes/No? Is it possible to have a
system of objects where the total
momentum is zero but the total kinetic
energy is not zero?
A bullet of mass 20 grams initially traveling
at a speed of 200 m/s lodges in a block of
wood of mass 2 kg at rest on a frictionless
floor. What is the velocity with which the
bullet and block of wood travel after
impact?
A 15,000 kg open box-car is moving at 7 m/s
on a level road. 3,000 kg of water then falls
straight down into the box-car. The speed
of the box-car now with the water in it is
Example 7.1: A car w/ mass 1200 kg is driving
north at 30 m/s, and turns east driving 13.6
m/s. What is the magnitude of the car’s
change in momentum?
pinitial = m vinitial = (1200 Kg) x 30 m/s = 36000 kg m/s North
pfinal = m vfinal = (1200 Kg) x 13.6 m/s = 16320 kg m/s East
North-South:
pfinal – pinitial = (0 – 36000) = -36000 kg m/s
East-West:
pfinal – pinitial = (16320 - 0) = +16320 kg m/s
Magnitude :
Sqrt(p2North – p2East ) = 39526 kg m/s
v2 = 13.6 m/s
v1 = 30
m/s
A ball is projected straight up. Which
graph shows the linear momentum of
the ball as a function of time?
(B)
(A)
t
(E)
t
t
(F)
t
(D)
(C)
t
t
IMPULSE
Objects A and B colliding: The force of impact A
exerts on B = FBA. This causes velocity of B to
change from v1(B) to v2(B)
FBA = maB = m[v2(B) – v1(B)]/t
OR FBA t = m(v2(B) – v1(B)) = p
• The quantity FBA t is called impulse (of a force).
Impulse = F t = p Unit = N.s
• Change in momentum requires force acting over
a time duration.
Force F (N)
IMPULSE
Time t (s)
Impulse = area under the graph
Force F (N)
200
0
0.2 0.4
0.6
Time t (s)
(a) Calculate the impulse
(b) If this impulse was applied on a 3 kg
mass at rest, what would its final
velocity be?
Ft = p
Example:
p = (mv) = m(v) and v = p/m
F = ma and a = F/m
A force of 30 N is applied for 5 s to
each of two objects of mass m
and M (m < M). Which of the
masses experience the greater
(a) Momentum change?
(b) Velocity change?
(c) Acceleration?
Why do we flex our knees when when
jumping?
• Increases the time of contact for the
ground to bring you to rest.
• In turn reduces force exerted on your
body.
A 160-gram baseball with a velocity of 20
m/s is hit by a bat and leaves at 25 m/s
in the opposite direction. If the contact
lasted for only 0.012 s, what was the
magnitude of the average force on the
ball? [1,000 grams = 1 kg]
Example on conservation of momentum:
A bullet of mass 200 g traveling at
a speed of 150 m/s hits a 3 kg
block of wood at rest on a
frictionless table. If the bullet
lodges inside the block, with what
speed will the bullet-block
composite travel after impact?
Elastic and Inelastic Collisions
1. Elastic Collisions:
Collisions in which the total kinetic
energy is conserved.
Kinitial = Kfinal
2. Inelastic Collisions:
Collisions in which the total kinetic
energy is NOT conserved.
Kinitial  Kfinal
Before:
V1(A)
mA
V1(B)
mB
Ki = ½ mAv21(A) + ½ mBv21(B)
After:
V2(A)
mA
V2(B)
mB
Kf = ½ mAv22(A) + ½ mBv22(B)
For Elastic Collision: Ki = Kf
OR ½ mAv21(A) + ½ mBv21(B) = ½ mAv22 (A) + ½ mBv22(B)
• In most cases, collisions occur inelastically.
• Part of the total initial kinetic energy is
converted to other forms of energy such
as light, heat, sound, etc.
• Ki = Kf + heat + sound + light, etc
• However, total energy and total linear
momentum are still conserved even in
inelastic collisions.
• Elastic collision is an ideal case. Collision of
billiard balls when no heat is produced is
the closest approximation to elastic
collision.
Center of Mass
Center of Mass = Balance point of a large object
= Balance point of a number of discrete objects
For an object with a regular shape (sphere,
cylinder, cube etc, CM is located at its
geometric center.
xcm
m1 x1  m2 x2  ....

m1  m2  ...
ycm
m1 y1  m2 y2  ....

m1  m2  ...
Example
m
m
x
xCM = (0 + mL)/2m = L/2
m
X=0
5m
X=L
x
xCM = (0 + 5mL)/6m = 5L/6
Example
A 55-kg man walks his 5-kg dog using
a 3 meter long lease. Where is the
center of mass of the man-dog
system?
Example: Find the center of mass.
y
7 kg
8
1 kg
6
4
3 kg
2
0
2
4
6
8
x
Collisions in Two Dimensions
When dealing with collisions in 2-D,
momentum conservation is applied
separately to the x and y components of
the total momentum:
px (Before) = px (After) -----[x-components]
And
py (Before) = py (After)
-----[y-components]
After
Before
8 m/s
m1
0.1 kg
y
At rest
m2
0.4 kg
v1
m1
60o
30o
v2
x
m2
(a) Find the speeds v1 and v2 after the collision.
(b) Is the collision elastic or inelastic?
A large seed initially at rest explodes
into two pieces which move off. Which
of these could be possible paths the
two pieces would take?
(II)
(I)
(III)
Two objects with different masses
(m and M with m < M) have the
same kinetic energy. Which has
the larger magnitude of
momentum?
[Hint: K = ½ p2/m]
A ball is projected straight up. Which
graph shows the total energy of the
ball as a function of time?
(B)
(A)
t
(E)
t
t
(F)
t
(D)
(C)
t
t
P7.35
A BMW of mass 2.0 x 103 kg is
traveling at 42 m/s. It approaches a
1.0 x 103 kg VW going 25 m/s in the
same direction and strikes it in the
rear. Neither driver applies the
brakes. Neglect frictional forces due
to the road and air resistance. If the
collision slows the BMW down to 33
m/s, what is the speed of the VW
after collision?
A 75-kg person jumps off a table and
lands on the ground with a speed of
3.5 m/s. By flexing his knees, he
comes to rest in 0.40 seconds.
Determine the average force exerted
on his body in this process.