Download Welcome back to Physics 211

Welcome back to Physics 211
Today s
agenda:
•  Energy
•  Elastic
collisions
•  Work
Physics 211– Fall 2012
Lecture 08-1
1
Current assignments
•  HW#8 due this Friday at 5 pm.
–  Solutions will be posted by Saturday to help you
study
•  Midterm 2 in ONE week
–  on Tuesday, Oct 23:
–  Chapters 5-9
–  Forces, circular motion, momentum
–  Self-study problems from textbook posted online
–  Videos working through sample problems and
clicker questions from lectures available online
–  Extra office hours tomorrow (Friday) 3:30-5pm
Physics 211– Fall 2012
Lecture 08-1
2
Total Energy E = Ug + K
Gravitational Potential energy: for an object of
mass m: Ug = mgh
•  h is height above arbitrary reference line
Kinetic energy: For an object of mass m moving
with speed v: K = (1/2)mv2
Total Energy is conserved: Ei = Ef
Physics 211– Fall 2012
Lecture 08-1
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8-2.1 If the velocity at B is v,
then what is the velocity at C?
1. 
2. 
3. 
4. 
5. 
2v
v
sqrt(2) v
v/sqrt(2)
None of the above
Physics 211– Fall 2012
A
B
h
h/2
C
Lecture 08-1
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8-2.2: Starting from a height h, a ball rolls down a
frictionless shallow ramp of length l1 = h/sin(30) with an
angle 30 degrees, and then up a steep ramp of height h
with angle 60 degrees and length l2 = h/sin(60). How far up
the steep ramp does the ball go before turning around?
1.  ½ l2
2.  l1
3.  l1 sin(60)/sin(30)
4.  l2
5.  None of the above
Physics 211– Fall 2012
l1
l2
h
60
h
30
Lecture 08-1
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Demo: the “nose” pendulum
Physics 211– Fall 2012
Lecture 08-1
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Collisions
If two objects collide and the net force exerted on the system
(consisting of the two objects) is zero, the sum of their
momenta is constant.




pA,initial + pB,initial = pA,final + pB,final
The sum of their kinetic energies
may or may not be constant.
Physics 211– Fall 2012
Lecture 08-1
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Elastic and inelastic collisions
•  If K is conserved – collision is said to be elastic
e.g., cue balls on a pool table
KA,i + KB,i
KA,f + KB,f
•  Otherwise termed inelastic
e.g., lump of putty thrown against wall
KA,i + KB,i
KA,f + KB,f
•  Extreme case = completely inelastic -- objects
stick together after collision
Physics 211– Fall 2012
Lecture 08-1
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8-2.3 Cart A moving to the right at speed v collides with an
identical stationary cart (cart B) on a low-friction track. The
collision is elastic (i.e., there is no loss of kinetic energy of
the system).
What is each cart’s velocity after colliding (considering
velocities to the right as positive)?
Cart A
Cart B
1
-v
2v
2
- 1/3 v
0
3
4
Physics 211– Fall 2012
1
/3 v
4
/3 v
v
2
/3 v
Lecture 08-1
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Check conservation of momentum and energy
1
2
3
4
Cart A
(m)
-v
- 1/3 v
0
1
/3 v
Physics 211– Fall 2012
Cart B
Final
Final
(m)
momentum kin. energy
2v
4
/3 v
v
2
/3 v
Lecture 08-1
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Demo
•  Experiment –> zero sensors
Physics 211– Fall 2012
Lecture 08-1
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Elastic collision of two masses
v1i
m1
v2i = 0
v1f
v2f
m2
m1
m2
Momentum à m1v1i + 0 = m1v1f + m2v2f
Energy à (1/2)m1v1i2 + 0 = (1/2)m1v1f2 + (1/2)m2v2f2
Physics 211– Fall 2012
Lecture 08-1
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Special cases: (i) m1 = m2
v1i
m1
Physics 211– Fall 2012
v2i = 0
m2
v1f
m1
v2f
m2
Lecture 08-1
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Special cases: (ii) m1 << m2
v1i
v2i = 0
v1f
v2f
m1
Physics 211– Fall 2012
Lecture 08-1
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Midterm sample problem 1: In Sydney Australia there is a ride at an
amusement park called the rotor. The Rotor is a large, upright barrel, rotated
at 33 revolutions per minute. Once the barrel has attained full speed, the floor
is retracted, leaving the riders stuck to the wall of the drum. What is the static
coefficient of friction needed to keep a 60 kg man from slipping off the ride?
Physics 211– Fall 2012
Lecture 08-1
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Midterm sample problem 2: A 20 g ball of clay traveling east at 2.0 m/s
collides with a 30 g ball of clay travelling 30 degrees south of west at 1.0 m/s.
What are the speed and direction of the resulting 50g blob of clay?
Physics 211– Fall 2012
Lecture 08-1
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Springs -- Elastic potential energy
frictionless table
Force F = -kx (Hooke s law)
F
Area of triangle lying under
straight line graph of F vs. x
= (1/2)(+/-x)(-/+kx)
Us =
(1/2)kx2
Physics 211– Fall 2012
x
F = -kx
Lecture 08-1
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(Horizontal) Spring
frictionles
s table
•  x = displacement from relaxed state of spring
•  Elastic potential energy stored in spring:
Us = (1/2)kx2
(1/2)kx2 + (1/2)mv2 = constant
Physics 211– Fall 2012
Lecture 08-1
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8.2-4 A 0.5 kg mass is attached to a spring on a
horizontal frictionless table. The mass is pulled to stretch
the spring 5.0 cm and is released from rest. When the
mass crosses the point at which the spring is not
stretched, x = 0, its speed is 20 cm/s. If the experiment is
repeated with a 10.0 cm initial stretch, what speed will
the mass have when it crosses x = 0 ?
1.  40 cm/s
2.  0 cm/s
3.  20 cm/s
4.  10 cm/s Physics 211– Fall 2012
Lecture 08-1
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Work, Energy
•  Newton s Laws are vector equations
•  Sometimes easier to relate speed of a
particle to how far it moves under a
force – a single equation can be used –
need to introduce concept of work
Physics 211– Fall 2012
Lecture 08-1
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What is work?
•  Assume constant force in 1D
•  Consider:
vF2 = vI2 + 2a s
•  Multiply by m/2 à
(1/2)mvF2 - (1/2)mvI2 = m a s
•  But: F = ma
à (1/2)mvF2 - (1/2)mvI2 = F s
Physics 211– Fall 2012
Lecture 08-1
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Work-Kinetic Energy theorem (1)
(1/2)mvF2 - (1/2)mvI2 = F s
Points:
•  W = Fs à defines work done on
particle
= force times displacement
•  K = (1/2)mv2 à defines kinetic energy
=1/2 mass times square of v
Physics 211– Fall 2012
Lecture 08-1
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Improved definition of work
•  For forces, write F à FAB
•  Thus W = F s à WAB = FAB ΔsA is
work done on A by B as A undergoes
displacement ΔsA
•  Work-kinetic energy theorem:
Wnet,A = ΣBWAB = ΔK
Physics 211– Fall 2012
Lecture 08-1
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The Work - Kinetic Energy Theorem
Wnet = ΔK = Kf - Ki
The net work done on an object is equal to
the change in kinetic energy of the object.
Physics 211– Fall 2012
Lecture 08-1
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8.2-5 Suppose a tennis ball and a bowling ball
are rolling toward you. The tennis ball is moving
much faster, but both have the same momentum
(mv), and you exert the same force to stop each.
Which of the following statements is correct?
1. It takes equal distances to stop each ball.
2. It takes equal time intervals to stop each ball.
3. Both of the above.
4. Neither of the above.
Physics 211– Fall 2012
Lecture 08-1
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8.2-6 Suppose a tennis ball and a bowling ball
are rolling toward you. Both have the same
momentum (mv), and you exert the same force to
stop each.
It takes equal time intervals to stop each ball.
The distance taken for the bowling ball to stop is
1. less than.
2. equal to
3. greater than
the distance taken for the tennis ball to stop.
Physics 211– Fall 2012
Lecture 08-1
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8.2-7 Two carts of different mass are accelerated from
rest on a low-friction track by the same force for the
same time interval.
Cart B has greater mass than cart A (mB > mA). The final
speed of cart A is greater than that of cart B (vA > vB).
After the force has stopped acting on the carts, the
kinetic energy of cart B is
1.
2.
3.
4.
less than the kinetic energy of cart A (KB < KA).
equal to the kinetic energy of cart A (KB = KA).
greater than the kinetic energy of cart A (KB > KA).
Can t tell.
Physics 211– Fall 2012
Lecture 08-1
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