Download Lecture 7 Energy (CH7)

Momentum & Center of ma
ass
Topics:
•
•
•
•
•
•
Impulse
Momentum
The impulse-momentum
Th
i
l
t
theorem
Center of mass
Conservation of momentum
Inelastic collisions
Sample question:
Male rams butt heads at high spe
eeds in a ritual to assert their
dominance. How can the force off this collision be minimized so as
to avoid damage to their brains?
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Slide 9-1
Impulse
The force of the foot
on the ball is an
impulsive force.
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Slide 9-8
Graphical Interpretation of Impulse
J = Impulse = area under
the force curve = Favg Δt
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Slide 9-9
Momentum
Momentum is the product of an
n object’s mass and its velocity:
→
→
p = mv
m
The unit of momentum is kg m/s; there is no
derived unit for momentum.
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Slide 9-10
The Impulse-Momentum Theorem
Impulse causes a change in momentum:
→
→
→
→
J =pf - pi = ∆p
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Slide 9-11
Example
A 0.5 kg hockey puck slides to the right at 10 m/s. It is hit with
a hockey stick that exerts the force
f
shown. What is its
approximate final speed?
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Slide 9-12
Two 1-kg stationary cue balls are struck by cue sticks. The cues
exert the forces shown. Which ba
all has the greater final speed?
A. Ball 1
B. Ball 2
C Both balls have the same final speed
C.
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Slide 9-13
Answer
Two 1-kg stationary cue balls are struck by cue sticks. The cues
exert the forces shown. Which ba
all has the greater final speed?
C Both balls have the same final speed
C.
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Slide 9-14
Definition of
impulse:
G
G G
J = Δp = FΔt
One can also define an average impulse when
the force is variable.
v
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Example
A car traveling at 20 m/s crashes into a bridge abutment.
Estimate the force on the driver iff the driver is stopped by
A. a 20-m-long
20
row off water-fille
f ed barrels
B. the crumple zone of her car (~1
( m). Assume a constant
acceleration.
l ti
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Slide 9-16
Example
A 500 kg rocket sled is coasting
g at 20 m/s. It then turns
on its rocket engines for 5.0 s, with a thrust of 1000 N.
What is its final speed?
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Slide 9-17
Example: A force of 30 N is applied for 5 sec to
each of two bodies of
o different masses.
30 N
Take m1<m2
m1 or m2
(a) Which mass has tthe greater moment
momentum
m
change?
Si
Since
th same force
the
f
is
i
Δp = FΔt
applied to each mass for the
same inte
erval, Δp is the same
for both masses.
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Example continued:
(b) Which mass has th
he greatest velocity
chang
ge?
Since bo
oth masses have the
Δp
Δv =
same Δp,
Δ the smaller mass
Δp
m
(mass 1) will have the larger
cha
hange in
i velocity.
l it
((c)) Which mass ha
as the greatest
g
accelerattion?
Δv
Since
a∝
∝Δv
the
mass
with
the
a=
greater ve
elocity change will have
Δt
the gre
eatest acceleration
acceleration.
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The Law of Conservation of
o Momentum
In terms of the initial and final total
t
momenta:
→
→
Pf = Pi
In terms of components:
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Slide 9-18
Example
A curling stone, with a mass off 20.0 kg, slides across the ice
at 1.50 m/s. It collides head on
n with a stationary 0.160-kg
hockey puck
puck. After the collision
n the puck’s
n,
puck s speed is 2
2.50
50
m/s. What is the stone’s final velocity?
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Slide 9-20
Example
p A rifle has a mas
ss of 4.5 kg
g and it fires a
bullet of 10.0 grams at a muzzle
m
speed of 820 m/s.
What is the recoil speed
p
of the rifle as the bullet
leaves the
e barrel?
Ass long
o g as the
e rifle
e is
s horizontal,
ho o a , there
e e will be
no net external force acting
a
on the rifle-bullet
system and momentu
um will be conserved.
pi = p f
0 = mb vb + mr vr
⎛ 0.01 kgg ⎞
mb
⎟⎟820 m/s = −1.82 m/s
∴ vr = −
vb = −⎜⎜
mr
g ⎠
⎝ 44.55 kg
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Center of
o Mass
The center of mass (CM)) is the point representing
the mean (average) po
osition of the matter in a
body This point need n
body.
not be located within the
bo
ody.
Th center
The
t off mass (of
( f a two
t
body
b d system)
t ) iis
found from:
x cm
m1 x1 + m 2 x 2
=
m1 + m 2
This is a “weighted” avera
age of the positions of the
particles that compose a body
body. (A larger mass is
more im
mportant.)
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In 2-dimensions
write
∑m x
=
∑m
i i
xcm
i
i
i
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∑m y
=
∑m
i
ycm
i
i
i
i
18
Example:
p Particle A is at the
t origin
g and has a mass
of 30.0 grams. Particle B has a mass of
grams. Where mus
st p
particle B be located so
10.0 g
that the center of mass (marked with a red x) is
p t ((2.0 cm,, 5.0 cm)?
)
located at the point
y
x
A
xcmm
ma xa + mb xb
mb xb
=
=
ma + mb
ma + mb
ycm
c
ma ya + mb yb
mb yb
=
=
ma + mb
ma + mb
x
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Example continued:
xcm =
(10 g )xb
10 g + 30g
xb = 8 cm
= 2 ccm
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ycm =
(10 g ) yb
30 g + 10 g
yb = 20 cm
= 5 cm
20
Example
p The p
positions of
o three p
particles are ((4.0
m, 0.0 m), (2.0 m, 4.0 m),
m and (-1.0 m, -2.0 m).
The masses are 4.0 kg,
kg, 6.0 kg,
g, and 3.0 kg
g
respectively. What is the
e location of the center of
masss?
y
2
1
x
3
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21
Example continued:
xcm
m1 x1 + m2 x2 + m3 x3
=
m1 + m2 + m3
(
4 kg )(4 m ) + (6 kg )(2 m ) + (3 kg )(− 1 m )
=
(4 + 6 + 3) kg
k
= 1.92 m
ycm
m1 y1 + m2 y2 + m3 y3
=
m1 + m2 + m3
(
4 kg )(0 m ) + (6 kg )(4 m ) + (3 kg )(− 2 m )
=
(4 + 6 + 3) kg
k
= 1.38 m
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22
Motion of the Center
C
of Mass
F an extended
For
t d db
body,
d it can be
b shown
h
th
thatt
p = mv
vcm.
F
From
this
thi it ffo
ollows
ll
th
thatt
ΣFext = ma
m cm.
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Collisions in One
O and Two
Dimension
Di
i
When there are no exte
ernal forces present
present, the
momentum of a system
m will remain unchanged.
(pi = pf)
If the kinetic energy
gy beforre and after an interaction
is the same, the “collisio
on” is said to be perfectly
elastic. If the kinetic energy changes, the collision
is inelastic.
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Inelastic Collisions
If, after a collision, the
e bodies remain stuck
together the loss off kinetic energy is a
together,
maximum. This type
e of collision is called
perfectly inelastic.
inelastic
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Slide 9-21
Example
p ((text p
problem 7.41):
) In a railroad freight
g
yard, an empty freight carr of mass m rolls along a
straight
g level track at 1.0 m/s and collides with an
initially stationary, fully loaded, boxcar of mass
p e together
g
upon
p collision.
4.0m. The two cars coupl
(a) What is the speed of the two cars after the
collisio
on?
pi = p f
p1i + p2i = p1 f + p2 f
m1v1 + 0 = m1v + m2 v = (m1 + m2 )v
⎛ m1 ⎞
⎟⎟v1 = 0.2 m/s
∴ v = ⎜⎜
⎝ m1 + m2 ⎠
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26
Example continued:
( ) Suppose
(b)
pp
instead tha
at both cars are at rest
after the collision. With
h what speed was the
loaded boxcar moving be
efore the collision if the
empty one had v1i = 1.0 m/s.
pi = p f
p1i + p2i = p1 f + p2 f
m1v1i + m2 v2i = 0 + 0
⎛ m1 ⎞
∴ v2i = −⎜⎜ ⎟⎟v1i = −0.25 m/s
⎝ m2 ⎠
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27
Example:
p Ap
projectile
j
of 1.0 kg
g mass approaches
pp
a
stationary body of 5.0 kg
g mass at 10.0 m/s and,
after colliding,
g, rebounds
s in the reverse direction
along the same line with a speed of 5.0 m/s. What
is the speed
p
of the 5.0 kg
g mass after the collision?
pi = p f
p1i + p2i = p1 f + p2 f
m1v1i + 0 = m1v1 f + m2 v2 f
v2 f
m1
(v1i − v1 f )
=
m2
1.0 kg
(10 m/ss − (− 5.0 m/s )) = 3.0 m/s
=
5.0 kg
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28
Example
Jack stands at rest on a skateb
board. The mass of Jack and
the skateboard together is 75 kg.
k Ryan throws a 3.0 kg ball
horizontally to the right at 4
4.0
0m
m/s to Jack,
Jack who catches itit.
What is the final speed of Jack
k and the skateboard?
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Slide 9-22
Example
A 10 g bullet is fired into a 1.0 kg
k wood block, where it
lodges. Subsequently, the bloc
ck slides 4.0 m across a floor
(µk = 0
0.20
20 for wood on wood)
wood). W
What was the bullet’s
bullet s speed?
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Slide 9-23
Perfectly elas
stic collision
Ap
perfectly
y elastic co
ollision is defined as
one in which there is no loss of kinetic
gy in th
he collision.
energy
http://ww
ww.walterfendt.de/ph14
p 4e/collision.htm
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One - Dimensionall Elastic Collisions
Consider two collidding objects with masses m1 and m2 ,
G
G
G
G
initial velocities v1i and v2i , and final velocities v1 f and v2 f ,
respectively.
Both linear momentum and kinetic enerrgy are conserved.
Li
Linear
momentum conservation:
i
m1v1i + m1v2i = m1v1 f + m1v2 f
2
2
m1v12i m1v22i m1v1 f m2 v2 f
Kinetic energy conservation:
+
+
=
2
2
2
2
wns, v1 f and v1 f .
We have two equations and two unknow
( 1)
(eq.
(eq. 2)
If we solve
l equations
ti
1 andd 2 for
f v1 f an
ndd v1 f we gett the
th following
f ll i solutions:
l ti
v1 f =
v2 f
m1 − m2
2m2
v1i +
v2i
m1 + m2
m1 + m2
2m1
m2 − m1
=
v1i +
v2i
m1 + m2
m1 + m2
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(9-
o Dimensions :
Collisions in Two
In this section we will remove the restriction that the
colliding objects move
m
along one axis. Instead we assume
that the two bodiees that participate in the collision
move in the xy -plaane. Their masses are m1 and m2 .
G
G
G
G
The linear momentum of the system is co
onserved: p1i + p2i = p1 f + p2 f .
If the system is elastic the kinetic energy is also conserved: K1i + K 2i = K1 f + K 2 f .
We assume that m2 is stationary and that after the collision particle 1 and
particle 2 move at angles θ1 and θ 2 with the
t initial direction of motion of m1.
In this case the conservation of momentu
um and kinetic energy take the form:
x − axis: m1v1i = m1v1 f cos θ1 + m2 v2 f cos θ 2 (eq. 1)
y − axis: 0 = −m1v1 f sin θ1 + m2 v2 f sin θ 2 (eq. 2)
1
1
1
m1v12i = m1v22 f + m2 v22 f (eq. 3) We have three equations and seven variables:
2
2
2
Two masses: m1 , m2 ; three speeds: v1i , v1 f , v2 f ; and two angles: θ1 , θ 2 . If we know
the values of four of these parameters wee can calculate the remaining three.
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Example:
p Body
y A of mass M has an original
g
velocity
y
of 6.0 m/s in the +x-direc
ction toward a stationary
bodyy (B)
( ) of the same mass.
m
After the collision,,
body A has vx=+1.0 m/s and
a vy=+2.0 m/s. What is
the
e magnitude
ag ude o
of bodyy B’ss velocity
e oc y a
after
e the
e
collis
sion?
Fi l
Final
Initial
A
A
vAi
B
B
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Example continued:
x
momentum
pix = p fx
p + p := p + p
1ix
2 ix
1 fx
y
momentum
piy = p fy
:
2 fx
m1v1ix + 0 = m1v1 fx + m2 v2 fx
Solve for v2fx:
v2 fx =
m1v1ix − m1v1 fx
m2
= v1ix − v1 fx
= 5.00 m/s
The mag. of v2 is
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p1iy + p2iy = p1 fy + p2 fy
0 + 0 = m1v1 fy + m2 v2 fy
Solve for
v2fy:
v2 fy =
− m1v1 fy
m2
= −v1 fy
= −2.00 m/s
/
v2 f = v
2
2 fy
+v
2
2 fx
= 5.40 m/s
Fig. 07.20
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