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CHAPTER
CHAPTER 99
LINEAR
LINEAR MOMENTUM
MOMENTUM ,,
IMPULSE
IMPULSE AND
AND
COLLISIONS
COLLISIONS
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9.1 Linear Momentum
The principle of energy conservation can be used to solve problems that are
harder to solve just using Newton’s laws. It is used to describe motion of
an object or a system of objects.
A new concept of linear momentum can also be used to solve physical
problems, especially the problems involving collisions of objects.
Linear momentum of an object whose mass is m and is moving at a velocity
of v is defined as
ur
r
p = mv
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Consider two particles interact with each other
By Newton’s 3rd law:
r
r
F21 = − F12
r
r
F21 + F12 = 0
r
r
m 1a1 + m 2a 2 = 0
r
r
dv
dv
m1 1 + m 2 2 = 0
dt
dt
r
r
d (m 1v 1 ) d (m 2v 2 )
+
=0
dt
dt
r
r
d (m 1v 1 + m 2v 2 )
=0
dt
ur
r
Linear momentum p ≡ mv
(
d p1 + p 2
dt
)
=0
r
r
r
r
i.e., Total momentum p tot = ∑ p = p 1 + p 2
remains constant
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What can you tell from this definition about momentum?
Momentum is a vector quantity.
The heavier the object the higher the momentum
The higher the velocity the higher the momentum
Its unit is kg.m/s
What else can use see from the definition? Do you see force?
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ur
r
p = mv
(kg·m/sec.)
p x = mv x , p y = mv y , p z = mv z
VECTOR
Relationship to force:
dp d (mv)
=
= m d v + v dm
dt
dt
dt
dt
= ma
Mass usually doesn’t
change with time
= ∑F
Improved form of
2nd
Law:
r d pr
F
∑ = dt
Probably was Newton’s original form of 2nd Law.
Works even when mass does change, e.g., rockets.
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Momentum is conserved in the absence of a net external force.
r
r
r
p tot = p 1 + p 2 : Constant ⇒ Conservation of Total Momentum
r
r
r
r
∴ p 1 i + p 2 i = p 1f + p 2 f
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Momentum Conservation
r
r
∆P
FEXT =
∆t
r
∆P
=0
∆t
r
FEXT = 0
¾ The concept of momentum conservation is one of the most fundamental principles
in physics.
¾ This is a component (vector) equation.
- We can apply it to any direction in which there is no external force applied.
¾ You will see that we often have momentum conservation even when energy is not
conserved.
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Example 9.1
An archer standing at rest on frictionless ice. He
fires an arrow horizontally.
m1 = 60 kg , m2 = 0.5 kg , v2 = 50 m/s
What is the velocity of the archer after firing the
arrow?
r
r
m 1v 1f + m 2v 2f = 0
m r
r
v 1f = − 2 v 2f = − 0.42 m/s
m1
If the arrow were shot at an angle θ with theh
horizontal, what is the recoil velocity
m 1v 1f + m 2v 2f cos θ = 0
v 1f = −
m2
v cos θ
m 1 2f
READ Examples 9.2
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Example:
Figure
Figure below
below shows
shows aa 2.0
2.0 kg
kg toy
toy race
race car
car before
before and
and after
after taking
taking aa turn
turn on
on aa track.
track.
Its
Itsrspeed
speed isis 0.50
0.50 m/s
m/s before
before the
the turn
turn and
and 0.40
0.40 m/s
m/s after
after the
the turn.
turn. What
What isis the
the change
change
∆
∆ P in
in the
the linear
linear momentum
momentum of
of the
the car
car due
due to
to the
the turn?
turn?
r
v i = − (0.50 m / s ) jˆ and
r
v f = (0.40 m / s ) iˆ
r
r
Pi = M v i = (2.0 kg )( −0.50 m / s ) jˆ = ( −1 kg ⋅ m / s ) jˆ
r
r
Pf = M v f = (2.0 kg )(0.40 m / s ) iˆ = (0.80 kg ⋅ m / s ) iˆ
r r
r
∆ P = Pf − Pi
r
∆ P = (0.80 kg ⋅ m / s ) iˆ − ( −1.0kg ⋅ m / s ) jˆ
= (0.8 iˆ + 1.0 jˆ ) kg ⋅ m / s
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Example 9.2
A type of particle, neutral kaon (K0) decays (breaks up) into a pair of particles called
pions (π+ and π-) that are oppositely charged but equal mass. Assuming K0 is
initially produced at rest, prove that the two pions must have momenta that are equal
in magnitude and opposite in direction.
This reaction can be written as
K0
K0 → π + +π −
Since this system consists of a K0 in the initial state
which results in two pions in the final state, the
momentum must be conserved. So we can write
Since K0 is produced at rest its
momentum is 0.
pπ+
π+
π−
pπ−
p K 0 = pπ + + pπ −
p K 0 = pπ + + pπ − = 0
pπ + = − pπ −
Therefore, the two pions from this kaon decay have the
momenta with same magnitude but in opposite direction.
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9.2 Impulse and Momentum
The change of momentum in a given time interval
vi
vf
F
fk = 0
a =
∴
∆t
vf -vi
v -vi
, Q F = ma = m( f
)
∆t
∆t
F ∆t = m v f − m v i = p f − p i = ∆ p
Impulse = force times time=change in momentum
G G
JG
G
G
G
G
JG
∆ p mv − mv 0 m v − v 0
∆v
=
=
=m
= ma = ∑ F
∆t
∆t
∆t
∆t
(
)
r r
r
I = F ∆t = ∆ p
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Impulse and Momentum
A classic photograph taken by
Arthur Edgerton at MIT in 1935
showing the moment of impact
between bat and softball. The huge
force exerted by the bat on the ball
causes severe distortion of the ball
as it is hit.
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Impulse and Momentum
•
Newton’s 2nd Law:
r ∆ r
F=
( p)
∆t
r r
∆p = F ∆t
r r r r
∆p = pf − pi = F ∆t
A classic photograph taken by
Arthur Edgerton at MIT in 1935
showing the moment of impact
between bat and softball. The huge
force exerted by the bat on the ball
causes severe distortion of the ball
as it is hit.
r r
I = F ∆t
Impulse
ÆImpulse-momentum theorem
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Impulse and Momentum
r tf r
I = ∫ Fdt
ti
r
r
In general, F = F ( t )
r
ÆImpulse = area under F, t curve
r 1 tf r
F = ∫ Fdt
∆t ti
r r
I = F∆t
Simple case: constant Force
r r
I = F∆t
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Example;
(a) Calculate the impulse experienced when a 70 kg person lands on firm ground
after jumping from a height of 3.0 m. Then estimate the average force exerted on
the person’s feet by the ground, if the landing is (b) stiff-legged and (c) with bent
legs. In the former case, assume the body moves 1.0cm during the impact, and in
the second case, when the legs are bent, about 50 cm.
We don’t know the force. How do we do this?
Obtain velocity of the person before striking the ground.
K E = − ∆ PE
1 2
mv = − mg ( y − yi ) = mgyi
2
Solving the above for velocity v, we obtain
v = 2 gyi = 2 ⋅ 9.8 ⋅ 3 = 7.7 m / s
Then as the person strikes the ground, the
momentum becomes 0 quickly giving the impulse
I = F ∆t = ∆p = p f − pi = 0 − mv =
= −70 kg ⋅ 7.7 m / s = −540 N ⋅ s
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Example; cont’d
In coming to rest, the body decelerates from 7.7m/s to 0m/s in a distance
d=1.0cm=0.01m.
0 + vi
7.7
=
= 3.8 m / s
2
2
d
0.01m
The time period the collision lasts is
= 2.6 × 10 − 3 s
∆t = =
3.8 m / s
v
I = F ∆t = 540 N ⋅ s
Since the magnitude of impulse is
I
540
The average force on the feet during
=
= 2.1× 105 N
F=
this landing is
∆t 2.6 × 10 −3
2
2
How large is this average force? Weight = 70 kg ⋅ 9.8m / s = 6.9 × 10 N
The average speed during this period is
v=
F = 2.1× 105 N = 304 × 6.9 × 10 2 N = 304 × Weight
If landed in stiff legged, the feet must sustain 300 times the body weight. The person
will likely break his leg.
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Example ; cont’d
What if the knees are bent in coming to rest? The body decelerates from 7.7 m/s to 0
m/s in a distance d=50 cm=0.5 m.
The average speed during this period is still the same v =
0 + vi
7.7
=
= 3.8 m / s
2
2
d
0.5 m
=
= 1.3 × 10 − 1 s
3.8 m / s
v
I = F ∆t = 540 N ⋅ s
Since the magnitude of impulse is
I
540
The average force on the feet during
=
= 4.1× 103 N
F=
−1
this landing is
∆t 1.3 × 10
2
2
How large is this average force? Weight = 70 kg ⋅ 9.8m / s = 6.9 × 10 N
The time period the collision lasts
changes to
∆t =
F = 4.1× 103 N = 5.9 × 6.9 × 10 2 N = 5.9 × Weight
It’s only 6 times the weight that the feet have to sustain! So by bending the knee you
increase the time of collision, reducing the average force exerted on the knee, and will
avoid injury!
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Example 9.3
A golf ball of mass 50 g is struck by a club. The force exerted on the ball by the club
varies from 0, at the instant before contact, up to some maximum value at which the
ball is deformed and then back to 0 when the ball leaves the club. Assuming the ball
travels 200m, estimate the magnitude of the impulse caused by the collision.
vB
θB
m
The range R of a projectile is
v B2 sin 2θ B
R =
= 200m
g
Let’s assume that launch angle θi=45o.
Then the speed becomes:
Note the deformation of
the ball due to the large
force from the club
v B = 200 × g = 1960 = 44m / s
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Considering the time interval for the collision, ti and tf , initial speed and the
final speed are
v i = 0 (immediately before the collision)
v f = 44 m / s(immediately after the collision)
Therefore the magnitude of the impulse on the ball due to the force of the club is
ur
I = ∆ p = m v Bf − m v Bi = 0 + 0.05 × 44 = 2.2 kg ⋅ m / s
What is the average force on the ball during the collision with the club?
F =
I
= 200 N
∆t
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