Survey of Physics

A frictionless, dynamic, calculus enabled, lecture oriented, descriptive, physics-motivated survey course.

Jim Branson Mayer Hall 3326 858-534-2978

1 Grades

• 30% Questions in lecture o Half credit for wrong answers (easy points) • 30% Biweekly Quizzes (4 or 5) • 16% Labs (2) • 24% Final

2 Proposed Syllabus Part I Part II

• Motion in one dimension • Relativity • Newton’s First Law o Three dimensions o "Every object persists in its state of rest or uniform o Muons and time dilation motion in a straight line unless it is compelled to o Laws of physics independent of inertial frame change that state by forces impressed on it” o Speed of light same in every frame o Newton’s Second Law o "Force is equal to the change im momentum (mv) per o Lorentz transformation change in time. For a constant mass, force is mass o Parable, relativity, 4-dimensions times acceleration.” o rotations, 4-vectors, rotations in 4D o F=ma (definition of force!) (not much physics) o general relativity Gravity near the earth is a constant • • The magic (ElectroMagnetic) field o "Force of gravity" is really an acceleration o Vector field Newton’s Third Law No • o Electric, Magnetic o "For every action, there is an equal and opposite reaction." o Charges generate E • Balance of forces Book o Currents generate B • Motion in 2 dimensions and vectors o Changing fields, generator Book The field in 4D o maximum range on the moon covers • • Energy is conserved material in • EM waves and light • Momentum is conserved o Transverse fields o Polarization o rocket more • Newton's Gravity detail than o Reflection and refraction o Rotational motion and orbits • Atoms, nuclei, quarks, electrons, photons o General Principle of Equivalence we will. • Quantum Mechanics, wave particle duality, o gravitational potential uncertainty principle • Angular Momentum is conserved • Cosmology, dark matter, dark energy, 11D • Simple Harmonic motion o traveling waves Mainly number problems. Mainly questions on concepts. 3 Example Problem: Position of a Car

• A car moves with constant velocity 100 m/s along a straight road. Assume it is at A) 1 mile x=0 when t=0. What is the position (x) of the car at t=20 B) 100 m/s seconds? C) 2000 m/s We will learn how to solve a limited set of problems. Numbers can change; objects can change; but the D) 2000 m problems on the quizzes will be the same ones we solve in class. E) 200 m

4 The Student Response System

• Turn it on • To enter ID • * to stop scan o A • * to get menu o Scroll to letter A o Finish putting in • Scroll to ID number • ↳ o There is a delete • Enter your button student ID • ↳ • Turn off • Turn on • Join ph11F07 class

5 Grades in 2006

• Any UCSD student can pass this class. B o Answer all the questions in class A (19/30) o Answer all the quiz C and final questions (14/54) o Hand in the labs D (8/16) • 41% just for trying

6 Fiat Lux And God said, Let there be light: and there was light. • A good description of the Big Bang (creation?). • We will take a scientific approach " F = j • Improved: “and God said "x µ! µ and there was light” !

We see that once the Laws of Nature were set, the universe developed all its complexity by just following those rules.

7 Laws of Nature in 2007

• Explain everything we see on (or near) earth and much more. o Complex systems often too difficult to compute but are in full agreement with basic laws. • Don’t explain some things about cosmos. • Don’t form a completely consistent theory of nature. • Have too many input parameters to be the real basic laws. o Expect to be able to derive these parameters from simpler Theory of Everything.

8 What can we measure about space-time?

• How many physical dimensions are there? o How many coordinates to we need to describe an event? • Do the experiment! • Are there hidden dimensions? o We think there are.

→ Perhaps the particles we are made of are confined to a “surface” in these extra dimensions.

→ Perhaps the size of the universe in extra dimensions is so small that we cannot notice them. • For now, concentrate on 4D = 3 space + 1 time. • What should be the units of space and time? o Feet and nanoseconds?

9 Moving in Space-Time

• Consider a particle in otherwise empty space. • Laws of Physics have important symmetries! o Independent of position o Independent of time o Independent of velocity

→ Inertial frame of reference o Do depend on acceleration (car example)

→ Acceleration and gravity are equivalent!

10 Symmetries Have Consequences

• Position symmetry implies conservation of momentum. o Newton’s laws are a consequence o Rotation symmetry implies conservation of angular momentum. • Time independence implies conservation of energy. • Laws of Physics are the same in any inertial frame of reference: Special relativity. • Acceleration and gravity are equivalent: General relativity.

11 The Student Response System

12 Practice Question 1

• What is Fiat Lux? A) An Italian Luxury car B) A car and a dishwashing liquid C) Latin for “Let there be light” D) The Big Bang

E) Motto of Revelle College

13 Practice Question 2:

• How many dimensions do 0) 0 5) 5 we live in? 1) 1 6) 10

2) 2 7) 11

3) 3 8) 26

4) 4 9) unknown

14 Practice Question 3:

• Why is the sky blue? A) Reflected light from the vast oceans B) Light from space is blue C) Blue light from the sun scatters more in air D) The air glows blue E) Just an illusion

15 Motion in One Dimension

Back to the 1600s

16 Air Track

• Air track reduces friction • Allows motion in one dimension (out of 3) • We set up a coordinate system (ruler)

1 2 3 4 5 6 7 8 9

17 Motion in One Dimension • Assume an object is moving along one direction. (We live in 3D). o Coasting through empty space o Ball dropped on earth o Object moving along a straight track o Driving your car along a straight road • Define coordinate x along that direction. • Position is a function of time: x(t) o x measured in meters (m) o t measured in seconds (s)

18 Position

• Origin: x=0 • Position of object can be measure with a meter stick: +3 meters from origin. • Position changes with time if object is moving. x(t)

19 Velocity • x(t) is the position. • Velocity is the rate of change of the position. o Your speedometer measures velocity o (positive or negative)

o (a vector in 3D (vx,vy,vz) ) • Units of velocity are m/s Definition of velocity! dx !x v(t) = = lim dt !t"0 !t dx = vdt x t x # dx = # vdt x0 0 t x(t) $ x = vdt 0 # 0 t x(t) $ x = vt for constant velocity 20 0 Simple Calculus

• Derivatives dx v = definition of velocity o Velocity dt dx o Acceleration ! vdt = ! dt + C indefinite integral • Integrals to derive dt solutions ! vdt = x + C contant of integration t x(t ) o Anti-derivative dx ! vdt = ! dt definite integral o Definite integral 0 xo dt t x(t ) • You will usually [vt]0 = [x]x0 pick limits to match problem

get the formula vt " 0 = x(t) " x0 evaluate at limits

derived this way. vt = x(t) " x0

x(t) = vt + x0 solve for x(t)

21 Recall Derivatives

d • Assume a is a (axn ) = anxn1 constant dx

d 3x2 = 6x • You remember this dx ( ) formula d (5x) = 5 dx

d (5) = 0 dx

22 Recall Integration

Integration is just the • n+1 opposite of n x differentiation ax dx = a n + 1 • Easy to remember x4 5x3dx = 5 4

23 Practice Problem:

• If you are in 0) not in Revelle 5) Fifth Choice Revelle College, 1) First choice 6)Sixth Choice where was Revelle College in 2) Second 7) Don’t your list of choice remember choices when you 3) Third Choice 8) applied to UCSD? 4) Fourth Choice 9)

24 Practice Question:

• In what year were you born?

25 Practice Problem: 1 Constant Velocity

dx !x v(t) = = lim 1) 150 m • A car is moving at dt !t"0 !t 30 m/s. How far dx = vdt x t 2) 150m/s does it move in 5 # dx = # vdt seconds? x0 0 3) 30 m t x(t) $ x = vdt 0 # 4) 30 m/s 0 x(t) $ x = vt 0 5)60 m

26 Velocity Changes with Time

27 You n know Derivative of ct from calculus • Assume x(t)= ctn o Where c is a constant o t is time • Differentiate to get v

dx d d v(t) = = (ct n ) = c (t n ) = c(nt n!1 ) = cnt n!1 dt dt dt

We will limit ourselves to these simple derivatives.

28 Practice Problem: 2 Compute Velocity

dx !x v(t) = = lim 1) 0.3t m t 0 If the position of the dt ! " !t • dx vdt = 2)0.3t m/s cart on an air track is x t given by x(t)=0.3 t # dx = # vdt x 0 3) 0.3 m/s meters, what is the 0 cart’s velocity? t x(t) $ x = vdt 4)0.15 m/s 0 # 0 x(t) $ x = vt 5) 0.6t m 0

29 2 Problem: Compute Velocity(2)

• If the position of the cart on an air 1)2 m/s track is given by x(t)=0.5t2+2t+20 2)5 m/s meters, what is the cart’s instantaneous velocity at t=5 3)7 m/s seconds? 4)7.5 m/s 5)10 m/s

(meters per second) 30 Newton’s First Law of Motion

31 Newton’s Three Laws of Motion

• “Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed on it.” • “Force is equal to the change in momentum (mv) per change in time. For a constant mass, force is mass times acceleration.” • “For every action, there is an equal and opposite reaction.”

32 Newton’s First Law

• “Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces impressed on it.” o Objects move with constant velocity o Applying a “force” can change the velocity

33 Air Track

• Air track reduces friction (but doesn’t eliminate it) o Could nearly eliminate it (space, superconductor…) • Velocity is constant (except for small friction) o (need to keep track horizontal in gravity)

1 2 3 4 5 6 7 8 9

34 Choice of Coordinate System • Pick any origin • Coordinate System can move at a constant velocity

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9 35 “Inertial” Coordinate System

• An inertial coordinate system is one that is moving at a constant velocity. o The coordinate system cannot be accelerating or changing its direction of motion. • The laws of Physics are the same in any inertial coordinate system. o Based on Newton’s first law o Extended by Einstein to include the fact that the speed of light is the same in any inertial coordinate system.

36 C Question: Inertial Systems

A) A coordinate system at rest.

• Which coordinate B) A coordinate system moving system is not an at a velocity of 10 m/s. inertial coordinate system? C) A coordinate system moving at a velocity of -10 m/s. D) A coordinate system with the x axis pointing down. E) A coordinate system moving

at a velocity of 2t m/s. 37 Integrate Velocity to get Position

• If we are given the position, we dx can differentiate to get the v = velocity. dt • If we are given the velocity we can dx = vdt integrate to get the position. x t • Assume a constant velocity v0 and dx = v0dt that we start at x=x0 at t=0. ! ! • (Learning to use integration is x0 0 much simpler than methods used x = v0t + x0 in book.)

38 You n know Integral of ct from calculus n • Assume v(t)= ct dx v(t) = = ct n o Where c is a dt constant dx = ct ndt o t is time t n+1 ct ndt = c indefinite integral • Integrate to get x(t) ! n + 1 x t ! dx = ! ct ndt integral with limits x0 0 t # t n+1 & t n+1 x " x0 = %c ( = c We will use only n + 1 n + 1 $ '0 these simple integrals

39 Example: Position from Velocity

• The velocity of a car increases linearly with time: v(t)=5t m/s. How far does the cat go between t=0 and t=10 seconds? •velocity is given dx v = = 5t dt •Put x on one side and v on t on other dx = vdt = 5tdt x 10 •Integrate both sides between dx 5tdt desired limits. ' = ' •(start at x=0 for simplicity) 0 0 10 •Compute integrals 10 ! t 2 " x = 5tdt = 5 = 250m ' # 2 $ 0 % &0 40 1 Problem: Position of a Car

A) 1 mile • A car moves with constant velocity 100 m/s along a B) 100 m/s straight road. Assume it is at x=0 when t=0. What is the C) 2000 m/s position (x) of the car at t=20 seconds? D) 2000 m

E) 200 m

41 Constant Acceleration

42 Constant a Acceleration

• x(t) is the position. • Velocity is the rate of change of the position. v increases linearly o Your speedometer measures velocity • Acceleration is the rate of change of the velocity.

x increases quadraticly dv v a(t) = = lim dt t0 t 43 Acceleration

• You undergo positive acceleration when you step on the gas in your car. o Velocity increases • Negative acceleration when you step on the brake. o Velocity decreases • (Accelerate to the side when you turn but we are still in one dimension.)

44 Constant a Constant Acceleration

• Constant acceleration is an important case for gravity near the earth. v increases linearly dx v = = at + v0 dv dt a = dx = vdt = at + v dt dt ( 0 ) dv = adt x t dx = at + v dt v t ! ! ( 0 ) x0 0 x increases quadraticly dv = adt t ! ! 2 v0 0 # t & 1 2 x " x0 = %a + v0t( = at + v0t v = at + v 2 2 0 $ '0 1 2 x = at + v0t + x0 2 45 Lab Schedule

• Week 3 o M 2:00 o T 4:00 o W 2:00 3:30 • Week 4 o M 2:00 3:30 o T W 2:00 3:30 • Week 5 o M 2:00 3:30 o T 4:00 o W 3:30

46 3 Problem: Constant Acceleration

dx Assume a car undergoes a v = • dt constant acceleration of 1)6 m/s/s x = v t + x a=12 m/s/s and that it starts 0 0 dv 2)6t m/s/s at zero velocity (v=0) and a = dt zero position (x=0) at t=0. 3)12t m/s/s v = a t + v What is the velocity at a 0 0 1 4)12t m/s later time t? x = a t 2 + v t + x 2 0 0 0 5)6t2 m/s/s

47 4 Problem: Constant Acceleration(2)

• Assume a car dx undergoes a constant v = dt acceleration of a=12 A)6 m x = v t + x m/s/s and that it starts 0 0 B)6t m dv at zero velocity (v=0) a = C)12t m and zero position dt D)12t2 m (x=0) at t=0. What is v = a0t + v0 2 the position at a later 1 2 E)6t m x = a t + v t + x time t? 2 0 0 0

48 Gravity Near the Earth

49 Gravity Near the Earth

• Gravity attracts two objects toward each other. • It causes an acceleration which is independent of the size or material of the objects. o (We can’t measure any difference between gravity and acceleration.) • Near the surface of the earth, the gravitational acceleration is approximately constant. o (It actually decreases with the square of the distance to the center of the earth but r+h≈r.) • Therefore gravity near the earth gives a constant acceleration g≈9.8 m/s/s.

50 Gravity Near the Earth

g h<

• Friction of the air resists fall of feather. • Remove (most of) the air from the tube and feather falls nearly as fast as washer.

52 C Question: Feather

A) Feather weighs less • Why does a feather fall B) Coin is metal much slower than a coin does? C) Friction D) Coin has more gravity E) Feather is for flying

53 C Question: Feather

A) cannonball • If we eliminate friction, B) hammer which falls fastest; a hammer, a cannonball, a C) baseball baseball, a feather? D) Cannonball and hammer E) All fall the same 54 g=9.8 m/s/s

• g is the acceleration of gravity (near the earth) o It depends a bit on position but we will ignore that. • The velocity decreases by 9.8 m/s in every second for an object falling freely near the earth.

o v=-9.8t+v0

55 C Question: Constant Gravity

• Why is gravity A) Gravity is always constant at small heights above B) Because the earth is round the the earth nearly a C) The weight of an object cannot constant depend on height acceleration D) The heights involved are much (independent of the less than the radius of the earth height)? E) The masses of the objects are much less than the mass of the earth 56 Dropping a Ball onto the Earth • Constant acceleration 1 due to gravity of a=-g x = at 2 + v t + x = 0 2 0 0 Drop the ball from a • 1 height x =h ! gt 2 + 0t + h = 0 0 2 • Drop the ball with no 1 initial velocity v =0 h = gt 2 0 2 • How long does it take 2h = t 2 to reach the ground? g 2h t = g

57 Air Track

• Slightly tilted air track gives constant (small) gravitational acceleration.

58 4 Problem: Dropping a Coin

dx A coin is dropped v = • dt from rest into a deep A)18m x = v t + x dark well. It falls for 6 0 0 dv B)66m seconds before a a = splash is heard. dt C)176m v = at + v About how deep is 0 D)332m the well? 1 2 x = at + v0t + x0 2 E)508m g=-9.8 m/s 59 4a Problem: Probe Drop

dx v = dt • A probe is dropped x = v t + x A)9 s from a plane at a 0 0 dv B)14 s height of 2000m. a = About how long does dt C)20 s v at v it take to reach the = + 0 D)28 s ground? 1 2 x = at + v0t + x0 2 E)40 s g=-9.8 m/s 60 Throwing a Ball into the Air

• Constant acceleration We have derived: due to gravity of a=-g dx • Throw the ball from a v = dt height x0≈0 x = v t + x • Throw the ball with an 0 0 dv initial velocity up of v0 a = • How high does it go? dt v = at + v • How long does it take 0 1 to reach the ground? 2 x = at + v0t + x0 2

61 Throwing a Ball into the Air

v = !gt + v0 = 0 velocity 0 at max height gt v • Constant acceleration = 0 v due to gravity of a=-g t = 0 time at max height g Throw the ball from a 1 • x = ! gt 2 + v t + x 2 0 0 height x0≈0 1 v2 v2 1 v2 x = ! 0 + 0 + 0 = 0 max height • Throw the ball with an 2 g g 2 g 1 initial velocity up of v0 x = ! gt 2 + v t + 0 = 0 on the ground 2 0 • How high does it go? " 1 # $ ! gt + v0 %t = 0 How long does it take & 2 ' • 1 gt = v to reach the ground? 2 0 2v t = 0 same time down as up g 62 5 Problem: Throwing a Ball

• A ball is thrown dx v = straight up into the dt A) 157 m air with an initial x = v t + x 0 0 B) 78 m velocity of 29.4 dv m/s. How high a = C) 44 m dt does it go? D) 20 m v = at + v0

1 2 E) 10 m x = at + v0t + x0 2 63 5a Problem: Bullets come down

A) 10 s A bullet is shot • dx straight up into the v = dt B) 20 s air with an initial x = v t + x velocity of 392 m/s. 0 0 dv C) 40 s How long does it a = dt take before it hits v = at + v D) 80 s the ground? 0 1 (assume initial x = at 2 + v t + x 2 0 0 E) 160 s height is 0)

64 Newton’s Second Law of Motion

65 Newton’s Three Laws of Motion

66 Newton’s Second Law

• “Force is equal to the change in momentum (mv) per change in time. For a constant mass, force is mass times acceleration.” • Definition of Force. Really no physics in addition to first law. (But a lot of use in engineering…) F = ma

67 Force of a spring

• Force increases as spring stretches from its normal (unstretched) length. • Wall microscopically behaves like a spring

with a much bigger F=-k(x-x0) spring constant k.

68 Force: Electric or Magnetic Field

• Fields are used to drive motors… • They can apply a force without two objects touching. F

69 Units

• We have already used some units o Length in meters o Time in seconds • Now we will add units F = ma of mass o Mass in kilograms

70 Units of Force

• From the equation F=ma, we can see that the units of force are kg⋅m/s/s. • This is called a Newton o N=kg⋅m/s/s • We will simply use kg⋅m/s/s as the units of force.

71 Problem: Constant Force

• The two front tires apply a constant force f to the car. • Neglect air resistance (friction). • Assume the car starts F = ma second law at x=0 at t=0. f a = f given force • What is the position m as function of time. 1 f x = at 2 = t 2 use equation 2 2m

72 6 Problem: 0 to 60 mph≈30 m/s

• A car has a mass dx of 1000 kg. It can v = generate a dt A) 3 s x = v t + x constant force of 0 0 B) 4 s dv 3000 kg⋅m/s/s. a = How quickly can it dt C) 5 s v = a0t + v0 reach a speed of D) 8 s 1 2 30 m/s? x = a0t + v0t + x0 2 E) 10 s F = ma 73 6 Problem: 0 to 60 mph≈30 m/s

• A car has a mass of 1000 kg. It can dx v = generate a dt A) 1 s x = v t + x constant force of 0 0 B) 2 s dv 12000 kg⋅m/s/s. a = How quickly can it dt C) 2.5 s v = at + v reach a speed of 0 D) 3 s 1 30 m/s? x = at 2 + v t + x 2 0 0 E) 6 s F = ma 74 Force of Gravity

• For a falling object, gravity applies a constant Force of gravity acceleration. (weight) • For an object being held up, the gravitational force (weight) is proportional to mass. o Weight is a force, mg (kg⋅ Force from table m/s/s) o Mass is m (kg) • The table exerts a contact F = !mg force to exactly counter the g force of gravity. o Statics not covered here 75 Units

• Mass has units of kilograms • Force has units of kg⋅m/s/s

Force from table

Fg = !mg

76 7 Problem: Weight

A) 0.5 kg

• A pool ball has x = v0t + x0 a mass of 0.5 B) -4.9 kg m/s/s v = a t + v kg. What is its 0 0 weight? 1 2 C) 4.9 kg m/s/s x = a t + v t + x 2 0 0 0 D) -4.9 kg F = ma

Fg = !mg E) 4.9 kg

77 7 Problem: Weight

A) 1000 kg A car has a • x = v0t + x0 mass of 1000 B) 4900 kg m/s/s v = at + v kg. What is its 0 weight? 1 C) 9800 kg m/s/s x = at 2 + v t + x 2 0 0 D) -9800 kg m/s/s F = ma F = !mg E) 9.8 m/s/s g 78 Newton’s third Law of Motion

79 Newton’s Three Laws of Motion

80 Newton’s Third Law

• “For every action, there is an equal and opposite reaction.” • If object A exerts a force on object B, then object B must exert an exactly opposite force on object A.

81 Air Track

• Demonstrate reaction on air track with different masses.

82 Example: Equal Reaction

F (t) = !F (t) equal and opposite reaction • Two carts on an 1 2 dv1 dv2 air track are m1 = !m2 F = ma pushed apart by a dt dt dv dv spring. m 1 dt = !m 2 dt integate over time of spring 1 " 2 " • The cart of mass dt dt v1 v2 m2 has a final m dv = !m dv zero initial velocity 1 " 1 2 " 2 velocity v2. 0 0

• What is the m1v1 = !m2v2 mv conserved velocity of the cart m v 1 v compute v with mass m ? 2 = ! 1 2 1 m2

m1 m2

83 Newton’s Three Laws of Motion

All three laws work in any coordinate system

moving with a constant velocity. 84 Conservation of Momentum p=mv

In Quantum Mechanics, we can show that momentum is conserved if the laws of physics are independent of position (translation symmetry).

85 Conservation of Momentum

• Newton’s third law implies conservation of momentum. F1 = !F2

m1a1 = !m2a2 initial momentum pinit = 0 m a dt = !m a dt 1 " 1 2 " 2 p final = m1v1 + m2v2 = 0 m1v1 = !m2v2 p = mv conserved • A very useful principle for problem solving.

86 Momentum on the Air Track

• Equal mass collision. • Inelastic Collision.

87 8 Problem: Air Track Carts

• Two air-track x = v0t + x0 carts (m1=1kg, v = a0t + v0 m2=2kg) are A) 6 m/s initially at rest. 1 2 x = a t + v t + x They are pushed 2 0 0 0 B) 5 m/s apart with a F = ma spring with cart-1 C) 4 m/s F = !mg getting a final g velocity of 6 m/s. F = !F D) 3 m/s What is the 1 2 velocity of cart-2? p = mv E) 2 m/s

88 8 Problem: Air Track Carts

Two air-track carts • x = v0t + x0 (m1=2kg, m2=5kg) v = a t + v are initially at rest. 0 0 A) 6 m/s They are pushed 1 apart with a spring x = a t 2 + v t + x with cart-1 getting a 2 0 0 0 B) 5 m/s final velocity of 10 F = ma m/s. What is the C) 4 m/s velocity of cart-2? Fg = !mg

F1 = !F2 D) 3 m/s p = mv E) 2 m/s

89 Rocket Equation (hard)

Assume fuel is dp fuel dp • ! = rocket third law ejected at a dt dt

constant velocity dm dvrocket ! v fuel = m vfuel relative to dt dt

rocket. !v fuel dm = mdvrocket o Rate of ejection is m1 dm v dm/dt !v = dv m(t) on left side fuel , m , rocket • Ignore gravity. m0 0 m 1 " m0 # v = !v fuel $ln (m)% = v fuel ln ( )m0 & ' * m1 +

Maximum velocity increases slowly with mass of fuel. (Multi-stage rocket useful) 90 Water Rocket Demo

91 9 Question: Single Stage Rockets • Assume a water rocket’s total mass is 90% water. Which improvement A)Increase water by a would increase a factor of 2 rocket’s maximum height B)Increase velocity of the most? water ejection by factor m 1 & m ) of 2 v = !v "ln m $ = v ln 0 fuel # ( )%m fuel ( + 0 ' m1 * C)Decrease mass of rocket (without water) by ln(2x) = ln(2) + ln(x) factor of 2 ln(2) = 0.693 D)All do the same ln 10 = 2.30 E)Can’t tell from info ( ) given 92 Air Track

• Equal mass collisions. • Demonstrate inelastic collisions with different masses.

93 13 Problem: Ballistics

• A bullet of mass x = v0t + x0 0.01 kg is fired into a A) 5 m/s v = a0t + v0 block of wood of 1 x = a t 2 + v t + x B) 10 m/s mass 0.99 kg. The 2 0 0 0 initial velocity of the F = ma C) 25 m/s bullet is 500 m/s. F = !mg What is the final g 1 2 D) 50 m/s velocity of the wood E = mgh + mv 2 with the bullet p = mv E) 200 m/s lodged inside of it? g = 9.8 m/s2 94 13a Trick Problem: Ballistics

• A bullet of mass 0.01 kg is fired into a x = v t + x block of wood of 0 0 v = a t + v A) 5 m/s mass 0.99 kg. The 0 0 1 final velocity of the x = a t 2 + v t + x B) 10 m/s wood with the bullet 2 0 0 0 lodged inside of it F = ma C) 25 m/s bullet is measured to Fg = !mg be 2 m/s. What was 1 2 D) 50 m/s the initial velocity of E = mgh + mv 2 the bullet? p = mv E) 200 m/s g = 9.8 m/s2 95 End Quiz 1

dx v ! dt x = vt + x0 constant velocity dv a ! dt v = at + v0 constant acceleration 1 x = at 2 + v t + x 2 0 0 F = ma

Fg = "mg g = 9.8 m/s2 p ! mv p = p conservation of momentum initial final 96 Newton’s Mechanics

• Laws of Physics the same in any inertial frame and for any position and any time. • Forces cause objects to accelerate. • Forces are equal and opposite. o Conservation of momentum. • Gravity near the earth causes a constant acceleration “down”. • Laws of mechanics can be simply written as mathematical formulas.

97 Projectiles in Three Dimensions

98 Three Space Dimensions

• The position of (the center of mass) of any object can be described by three coordinates (x,y,z). o We seem to live in a 3D world • We can describe the three coordinates! conveniently with a vector r (x, y,z) = • The vectors can also be thought of geometrically as having a length and a direction. o We will not make much use of this for simplicity. o However, we will think about the properties of vectors under rotations of the coordinate system.

99 Motion in Two Dimensions

• Most simple motion can be described in two dimensions if the coordinates are chosen wisely. o Projectile motion o Orbits of planets • For projectile motion near the earth, we can simply treat x and y as independent functions of time.

100 Apply Newton’s Laws in Each Dimension Separately

• Time is common to each spatial dimension. o x(t) satisfies same equations as in 1D. o y(t) satisfies same equations as in 1D. o t is the same • Choose x and y direction to make the problem easy.

101 x(t) and y(t) • Choose y direction to be anti-parallel to gravitational acceleration (up). o Constant acceleration -g. • x is then unaffected by gravity. o Constant velocity. Treat x and y as independent functions of time

102 Projectile Motion in 2D

• Initial vertical velocity of v0y • Initial horizontal velocity of v0x • To solve the problem o Compute y(t) as if throwing a ball into the air o Compute x(t) for constant velocity. 1 y = ! gt 2 + v t + y vertical motion with gravity. 2 0 y 0

x = v0 xt + x0 same t

103 Ballistic Cart Demo

• Projectile motion in 2D • Motion in the horizontal direction is not affected by vertical motion.

104 Range of a Projectile

• Initial vertical velocity of v0y • Initial horizontal velocity of v0x • To solve the problem v0 y ! gt = 0 v=0 at max ht. o Compute time in the air v o Compute how far 0 y tup = projectile moves g horizontally in that time. tup = tdown 2v t = 0 y time in air g 2v v d = v t = 0 x 0 y 0 x g 105 Range depends on Initial Angle

v0 y = v0 sin!

v0 x = v0 cos! 2 v0 2v0 xv0 y 2v sin! cos! d = = 0 g g v0y maximum for ! = 45! 1 θ sin! = cos! = 2 v 2 2 0x 2v0 sin! cos! v0 dmax = = g g If air resistance is significant, use a smaller angle. 106 10 Problem: Moon Golf

• Tiger Woods can hit 2v v A) 300 m a golf ball hard d = 0 x 0 y g enough to give it an B) 600 m initial velocity of 140 2v2 sin! cos! d = 0 mph or about 63 m/s. g C) 1200 m How far could he hit v2 the ball on the moon 0 D) 2400 m dmax = where g is smaller by g a factor of 6 and E) 3600 m there is no air? 107 10 Problem:

• The gun on a 2v v A) 500 m/s battleship can send a d = 0 x 0 y g shell to target 25 km B) 700 m/s away. What is the 2v2 sin! cos! d = 0 minimum velocity of g C) 1300 m/s the shell as it leaves v2 the gun? d = 0 D) 2400 m/s max g E) 3600 m/s

108 4a Problem: Probe Drop

• A probe is dropped from a plane flying 5000 m above A) 4 the ocean at a speed of 200 m/s. How many B) 8 seconds before the plane C) 16 passes over the target point on the water should D) 32 it drop the probe? E) 64

109 4a Problem: Probe Drop

• A probe is dropped from a plane flying 19600 m A) 4 above the ocean at a speed of 675 m/s. How B) 8 many seconds before the C) 16 plane passes over the target point on the water D) 31 should it drop the probe? E) 63

110 Shoot the Monkey Demo

• Projectile motion in 2D

111 Watermelon Drop

112 Watermelon Physics

• Watermelon is fairly aerodynamic o Ignore friction in fall • Hits sidewalk like a bag of water o Water nearly incompressible o Big problem as it hits sidewalk o Water at bottom must flow out so water at top can keep coming down • Fragments can fly out at full melon velocity or even faster • Treat fragment as projectile

113 Watermelon Drop 1 gt 2 = h 2 2h t = time to reach ground g v = gt = 2gh velocity on impact Assume some small fragment maintains its velocity and bounces out at 45 degrees. v2 d = = 2h range for 45 degree angle g

Friction will reduce range of small fragments, but, shock wave could produce greater velocity fragment. 114 11 Problem: Watermelon drop

• If a watermelon is dropped from a height of A) 10 m 35m, how far does the 1 2 x = at + v0t + x0 farthest piece fly from 2 B) 20 m the point of impact? v = at + v0 C) 35 m (While the collision is v2 d = very inelastic, assume max g D) 70 m that the lucky fragment keeps its velocity and E) 100 m flies at an optimal angle!) 115 11 Prob: Watermelon on Moon

• If a watermelon is dropped from a height of A) 10 m 35m on the moon, how 1 x = at 2 + v t + x far does the farthest 2 0 0 B) 20 m piece fly from the point v = at + v0 of impact? (While the C) 35 m collision is very inelastic, v2 d = assume that the lucky max g D) 70 m fragment keeps its velocity and flies at an E) 100 m optimal angle!) 116 Conservation of Energy

117 Conservation of Energy

• We know that momentum is conserved because the laws of physics are independent of where we are in space. • Quantum mechanics also tells us that a quantity called Energy is conserved because the laws of physics are independent of time. • We will consider an object in constant gravity to learn what Energy is.

118 Dropping a Ball • As I hold the ball above the floor, it has potential energy due to gravity. • When I drop the ball, gravity does work on the ball and it gains kinetic energy (energy inherent in motion). • If I define the potential energy to be zero at the height of the floor, all the potential energy has been converted into kinetic energy when the ball reaches the floor. • The sum of potential and kinetic energy is a constant since energy is conserved.

E = PE + KE conserved 119 (Bowling Ball?) Pendulum

• Ball oscillates between potential energy and kinetic energy. Never goes higher • V=0 than it initial position.

• Some transfer of v energy due to friction. o Energy goes into air Maximum PE flow and heat. Maximum KE

120 Energy in a Pendulum • Pull pendulum up to max height o Maximum potential energy mgh • Release at rest • Pendulum gets to lowest point o Maximum kinetic energy o Minimum potential energy • Pendulum goes back to max height • Energy oscillates back and forth between potential and kinetic

121 Potential Energy Proportional to Mass

Same PE for each • m m mass. PE>0 PE>0 y=h • If we attach the two masses together to make a single mass 2m, we se that PE is proportional to mass.

PE mf (h) = PE=0 m m Floor y=0 122 Potential Energy Proportional to h

• Assume PE=mf(h) o Same energy independent o how y=2h m the mass was raised to height. • If I drop the two masses they will reach the same velocities in constant gravity. o Kinetic energies must be the same. m y=h

PE(2h) ! PE(h) = PE(h) ! PE(0) Table height h PE(2h) = 2PE(h) PE = Cmh we can choose the constant PE = mgh clearly proportional to g Floor y=0 123 Kinetic Energy

If I drop a ball from 1 2 • h = gt constant acceleration height h, it converts 2 2h its PE into KE by the t = time falling in constant g time it reaches the g floor. v = gt = 2gh velocity after time t 2 • We can compute the v = 2gh square equation 1 KE which must be mgh = mv2 solve for mgh equal to the original 2 1 PE=mgh. KE = mv2 identify kinetic energy 2

124 Total Energy is Conserved

• Energy is conserved E = PE + KE total energy is conserved globally. 1 E = PE + mv2 for moving object in general • We have 2 learned how 1 E = mgh + mv2 moving object in constant gravity to compute 2 the energy of a single object in constant gravity.

125 Units of Energy

! m$ kg ' m2 mgh has units of (kg) (m) = "# s2 %& s2 1 kg ' m2 mv2 also has units of 2 s2 This is defined in SI units to be a Joule.

126 Energy

• Total energy is conserved. • There are other forms of energy than mechanical energy. o Gravitational energy, EM energy, nuclear energy o The energy we use ultimately comes from nuclear energy from the sun or a bit from nuclear energy on earth.

→ Hydrogen from Big Bang releases energy when it is fused into heavier nuclei • Energy which is used (dissipated into friction for example) does not disappear but becomes impossible to use for most purposes.

127 Energy of a Car

Ei = E f

PEi + KEi = PE f + KE f 1 1 mgh + mv2 = mgh + mv2 i 2 i f 2 f 1 mgh = mgh + mv2 i f 2 f

• A car is at rest at the top of a 1 2 mg h ! h = mv 100 m high hill. It rolls down ( i f ) 2 f the hill without friction. How 2g h ! h = v2 fast will the car be going once ( i f ) f

it reaches the bottom of the v f = 2g(hi ! hf ) = 2(9.8)(100) hill? 128 12 Problem: Car velocity

x = v t + x • A car is at rest at 0 0 v a t v A)30 m/s the top of a 510 m = 0 + 0

high mountain. It 1 2 x = a t + v t + x B)50 m/s rolls down without 2 0 0 0 F = ma friction. How fast C)100 m/s will the car be going Fg = !mg

once it reaches the 1 2 D)150 m/s E = mgh + mv bottom of the 2 mountain? p = mv E)200 m/s g = 9.8 m/s2 129 12a Problem: Car velocity

x = v t + x • A car is moving with 0 0 v a t v A)62 m a velocity of 35 m/s = 0 + 0

as it approaches a 1 2 x = a t + v t + x B)75 m hill. The driver lets 2 0 0 0 F = ma the car coast up the C)100 m hill. How high up Fg = !mg

the hill can it coast 1 2 D)124 m E = mgh + mv before it stops? 2 p = mv E)148 m g = 9.8 m/s2 130 13 Trick Problem: Ballistics

• A bullet of mass x = v0t + x0 0.01 kg is fired into a A) 5 m/s v = a0t + v0 block of wood of 1 x = a t 2 + v t + x B) 10 m/s mass 0.99 kg. The 2 0 0 0 initial velocity of the F = ma C) 25 m/s bullet is 500 m/s. F = !mg What is the final g 1 2 D) 50 m/s velocity of the wood E = mgh + mv 2 with the bullet p = mv E) 500 m/s lodged inside of it? g = 9.8 m/s2 131 Power

• A powerful car delivers a lot of energy per second to accelerate. • Power is Energy per second. • SI Unit is Watt=Joule/second. o W=kg⋅m2/s3

132 14 Problem: Electric Car

x = v0t + x0 How much power A)30 kW • v = a0t + v0 must a 1000 kg 1 x = a t 2 + v t + x B)60 kW electric car put out 2 0 0 0 to travel at 30 m/s F = ma up a hill with at C)90 kW Fg = !mg 10% slope? 1 2 D)120 kW 2 3 E = mgh + mv 1 kilowatt=1000 kg m /s 2 p = mv E)30 W g = 9.8 m/s2 133 14 Problem: Electric Hoist

x = v0t + x0 How much power A)2 kW • v = a0t + v0 must an electric 1 x = a t 2 + v t + x B)5 kW hoist put out to 2 0 0 0 raise a 1020 kg F = ma car at 2 m/s C)10 kW Fg = !mg straight up? 1 2 D)20 kW 2 3 E = mgh + mv 1 kilowatt=1000 kg m /s 2 p = mv E)500 W g = 9.8 m/s2 134 Force is -derivative of PE (skip derivation of this)

Remember this for later. d(PE) F = ! force from PE dx x2 PE2 " Fdx = ! " d(PE) integrate equation x1 PE1 x2 PE(x ) ! PE(x ) = ! Fdx work 2 1 " x1 PE = mgx example: gravity PE d(mgx) F = ! = -mg gives force of gravity dx

135 Ethanol Rocket

• Energy (from what) • Fire

136 Spent Energy

• The laws of Thermodynamics and Statistical Physics limit what can done with energy which has be used and converted into heat due to friction. • For heat to be useful to do work we need to have objects a (quite) different temperatures. • Once everything reaches thermal equilibrium, the energy is still there but we can’t use it for anything.

137 Energy from the Sun • The sun was formed mainly from hydrogen from the big bang. • Gravity attracts gas together o High pressure inside sun balances gravity • At high pressure, hydrogen fuses into helium and releases energy. o Later, helium fuses to make heavier nuclei o Nuclear fusion Surface temp 5000° C

• Ultimate source of energy is Radiates energy hydrogen from big bang.

138 Solar Energy

• Most sources of energy on earth come from the sun o Plants and animals o Oil and coal o Hydroelectric o Wind o solar

139 Nuclear Energy • Energy stored in heavy elements produced in a supernova before the formation of our solar system. • Fission in reactors (mainly Uranium) o Historically very safe except in USSR o Almost no pollution or greenhouse gas o Releases less radioactivity than burning coal o Has nuclear waste disposal problem

→ Yucca mountain • Geothermal o Mainly heavy elements in earth’s core producing heat o Some gravitational heating 140 C Question: Not from the Sun

• Which type of A) oil energy does not B) nuclear have its origin in solar radiation C) wind hitting the earth? D) hydroelectric

E) All of the above 141 C Question: Where did it go

A) pollution • We burn oil to drive around in our cars. B) CO Energy is conserved. 2 Where did the energy go C) Mainly heat once we arrive and turn the car off? D) Mainly wind

E) Mainly Sound

142 Pollution

• Burning fuels poorly can put unnatural chemicals in the air o Soot (global cooling) o Hydrocarbons (ozone) o Nitrous oxide (ozone) o Sulfur dioxide (acid rain) o All bad for breathing • “Clean burning” produces water and CO2

143 Greenhouse Effect

• High frequency (visible) light from the sun passes through the atmosphere. • Low frequency (infrared) light radiated by the earth is absorbed by complex gasses (CO2, H2O…) in the atmosphere. • Earth does not cool as well and the average temperature goes up (until energy in = energy out).

144 Roger Revelle's Discovery

Roger Revelle, seen here studying seawater chemistry, ca. 1936, and as a leading administrator as well as scientist, ca. 1958.

• Before scientists would take greenhouse effect warming seriously, they had to get past a counter- argument of long standing. It seemed certain that the immense mass of the oceans would quickly absorb whatever excess carbon dioxide might come from human activities. Roger Revelle discovered that the peculiar chemistry of sea water prevents that from happening. His 1957 paper with Hans Suess is now widely regarded as the opening shot in the global warming debates.

145 Global Warming

• Carbon, long stored in petroleum and coal in the earth, is burned and put in the atmosphere as CO2. • CO2 is natural o Emitted by animals o Used by plants to grow • But plants (and oceans) are not able to keep up. o Deforestation also bad • Less energy flows out than in

146 Temperature Follows CO2 Temperature Proxy

147 Industrial Revolution CO2 is outside geological bounds!

148 I don’t need to see this graph to know that we have big problems.

149 Global Cooling

• Soot, SO2, other light About 5 absorbing chemicals in the volcanoes upper atmosphere stop per century sunlight and cool earth. • Volcanoes a big source of cooling o Very large events are common in recent centuries. • Pollution contributed to cooling but has been greatly reduced. • Cooling can be very harmful.

150 C Question: Global Warming

• Which of these is a A) Sulfur dioxide major cause of global warming? B) Carbon monoxide C) Carbon dioxide

D) Hydrocarbons

E) Ozone

151 C Question: global cooling

• Which can cause A) volcanoes global cooling? B) pollution

C) Nuclear weapons

D) Asteroid impacts

E) All of the above

152 C Question: global warming

• How does CO2 A) It absorbs infrared radiation cause global warming? B) It increases plant growth rate C) It is absorbed in the ocean

D) It reflects visible light

E) It destroys the ozone

153 G Question: global warming

• By what factor is CO2 in the atmosphere projected to A) 1.4 increase between 1750 and 2050? B) 2

C) 100 ppm

D) 1.4%

E) 1.6

154 Global Dimming

• Pollution leads to more clouds! • Effect of global warming has been mitigated by pollution. • Reducing sulfur now will make things worse.

• Maybe we can put more benign “soot” in upper atmosphere.

155 Simple Harmonic Motion

And Waves

156 Simple Harmonic Motion

• Simple examples are spring oscillator or pendulum. • Almost anything will oscillate some way. • Restoring force • Mass (inertia) • Initial conditions

157 Spring and Mass Oscillator

• Spring provides restoring force F=-kx • Mass provides inertia m • Displace mass to start oscillator

Example of Harmonic Oscillator

158 The Oscillator Equation

F = !kx restoring force F = ma Newton's law dv d 2 x a = = acceleration is 2nd derivative of x dt dt 2 d 2 x F k = = ! x equate accelerations dt 2 m m x(t) = x0 cos("t) guess oscillatory solution

159 Sine and Cosine

• Periodic functions • Argument is angle o Radians for us o 2π radians per cycle • Radians o Needed for differentiation o Actually dimensionless

d sin(!t) = ! cos(!t) dt d cos(!t) = "! sin(!t) x(t) = sin(t / T ) dt 160 Check Solution x(t) = x0 cos(!t) guessed solution d sin(!t) = ! cos(!t) recall derivatve of sine function dt d cos(!t) = "! sin(!t) recall derivative of cosine too dt dx = "x ! sin(!t) take first derivatve of solution dt 0 d 2 x = "x ! 2 cos(!t) take second derivative dt 2 0 d 2 x k = " x oscillator equation dt 2 m k "x ! 2 cos(!t) = " x cos(!t) fill in both sides of eq. : solution verified 0 m 0 k ! 2 = remove common factors m k note that x0 can be anything but ! must be m 161 Examine Solution

x(t) = x0 cos(!t) solution verified k ! = condition on the constant ! m x0 can be anything initial condition: amplitude cos(0)=cos(2" )=1 !T=2" T is the period of the oscillation 1 ! f= = frequency is periods per second T 2" !=2"f ! is angular frequency

Called Harmonic Oscillator because frequency is constant. 162 Example: Spring Oscillator

• A spring with a spring constant k=2 kg/s2 is attached to a mass m=0.5 kg. The equilibrium position is at x=0. • The mass is released from x=0.5m at t=0. • What is the position x(t) as a function of time? k ! = = 4 / s2 = 2 / s m x(t) = 0.5cos(2t)

! 1 f = = Hz (1/s) 2" " 163 16 Problem: Spring Oscillator

A spring with a spring A) x(t) = 0.2sin(9t) • x(t) = x0 cos(!t) constant k=9 kg/s2 is ______k attached to a mass B) x(t) = 0.2cos(9t) m=1 kg. The mass is ! = m ______released from x=+0.2m at t=0. What !T=2" C) x(t) = 0.5cos(3t) is the position x(t) as a 1 ! ______function of time? (The f= = T 2" D) x(t) = 0.2cos(3t) equilibrium position is ______at x=0) !=2"f E) x(t) = 0.5cos(9t)

164 16 Problem: Spring Oscillator

A spring with a A) x(t) = 0.3cos(0.5t) • x(t) = x0 cos(!t) spring constant k=1 ______2 k kg/s is attached to a B) x(t) = 0.3cos(0.25t) mass m=4 kg. The ! = m ______mass is released from x=+0.3m at t=0. !T=2" C) x(t) = 0.6cos(.25t) What is the position 1 ! ______x(t) as a function of f= = T 2" D) x(t) = 0.3cos(2t) time? (The ______equilibrium position !=2"f is at x=0) E) x(t) = 0.6cos(0.5t) 165 17 Problem: Spring Oscillator 2

A) 4 • A spring with a spring x(t) = x cos(!t) constant k=12 kg/s2 is 0 attached to a mass k B) 2 m=3 kg. The mass is ! = m released from C) 0.32 x=+0.2m at t=0. What T=2 is the frequency f of ! " the oscillator? (The 1 ! D) 0.64 equilibrium position is f= = T 2" at x=0) E) 0.2 !=2"f Per second 166 17 Problem: Spring Oscillator 2

A) 4.5/s • A spring with a spring x(t) = x cos(!t) constant k=20 kg/s2 is 0 attached to a mass k B) 2/s m=2.5 kg. The mass is ! = m released from C) 0.9/s x=+0.4m at t=0. What T=2 is the frequency f of ! " the oscillator? (The 1 ! D) 0.64/s equilibrium position is f= = T 2" at x=0) E) 0.45/s !=2"f Per second 167 Velocity of Oscillator

Velocity can be computed from position

x(t) = x0 cos(!t) dx v = = "x0! sin(!t) dt

168 18 Problem: Spring Oscillator v

A) v(t) = !0.2sin(2t) • A spring with a spring x(t) = x cos(!t) constant k=12 kg/s2 is 0 ______attached to a mass m=3 k kg. The mass is released ! = B) v(t) = !0.4 cos(2t) from x=+0.2m at t=0. m ______What is the velocity of the oscillator as a !T=2" C) v(t) = !0.4sin(2t) function of time? (The equilibrium position is at 1 ! ______x=0) f= = T 2" D) v(t) = 0.2cos(2t) !=2"f ______E) v(t) = 0.4sin(2t) v(t) = #x ! sin(!t) 0 169 18 Problem: Spring Oscillator v

A) v(t) = 0.2sin(2t) • A spring with a x(t) = x0 cos(!t) spring constant k=1 ______kg/s2 is attached to a k B) v(t) = 0.1sin(2t) mass m=4 kg. The ! = mass is released m ______from x=-0.2m at t=0. !T=2" C) v(t) = !0.1sin(2t) What is the velocity 1 ! ______of the oscillator as a f= = function of time? T 2" D) v(t) = 0.2sin(t / 2) (The equilibrium ______position is at x=0) !=2"f E) v(t) = 0.1sin(t / 2) v(t) = #x ! sin(!t) 0 170 Energy in the Spring Oscillator

1 KE = mv2 standard KE 2 x x 1 PE = !" Fdx = " kxdx = kx2 calculate from F 0 0 2 x(t) = x0 cos(#t) v(t) = !x0# sin(#t) 1 1 Energy is time E=KE+PE= mv2 + kx2 sum 2 2 independent, 1 2 2 2 1 2 2 E = mx0# sin (#t) + kx0 cos (#t) moving back 2 2 and forth k # = m between kinetic 1 1 1 and potential E = kx2 sin2 (#t) + kx2 cos2 (#t) = kx2 2 0 2 0 2 0 energy 171 End Quiz 2

172 Quiz2 Formulas x = v0t + x0 constant velocity v = a t + v constant acceleration 0 0 1 1 E = mgh + mv2 Energy x = a t 2 + v t + x 2 2 0 0 0 E = E Conservation of Energy F = ma Newton's 2nd initial final F = !kx H.O. Force Fg = !mg force of gravity near earth x(t) = x0 cos("t) H. O. position g = 9.8 m/s2 k p = mv momentum " = Angular Frequency m p = p conservation of momentum initial final "T=2# T is the period 2v0 xv0 y 1 " d = range of projectile f= = Frequency g T 2# 2v2 sin" cos" "=2#f d = 0 g v(t) = !x0" sin("t) H.O. velocity v2 d = 0 max g 173 Traveling Waves

• Traveling waves are very common o Water waves o Waves on a rope o Sound waves o Radio waves o Light waves • They can carry energy from one place to another • We will not derive wave equations but will go right to the solution

174 Wave Machine, slinky and Rope • Transverse and longitudinal waves

175 Waves on a String

• Imagine string with some tension and no displacement. o No waves • String at an angle still the same. o No waves • String with some curvature o Restoring force

o Waves F a !2w(x,t) !2w(x,t) • Restoring force from 2nd space v2 = derivative of string position. !x2 !t 2 Tension • Acceleration 2nd time derivative v = mass / length of position. 176 Traveling Wave Solution

w(x,t) = Acos(kx ! "t) wave " v = phase velocity phase k 2# f = " frequency k$ = 2# wavelength

177 Check Solution !2w(x,t) !2w(x,t) v2 = wave equation !x2 !t 2 Tension v = velocity from physics of wave eq. mass / length w(x,t) = Acos(kx " #t) guess traveling wave solution # v = velocity from solution phase k !2w(x,t) = "Ak 2 cos(kx " #t) !x2 !2w(x,t) = "A# 2 cos(kx " #t) !t 2 "v2 Ak 2 cos(kx " #t) = "A# 2 cos(kx " #t) plug in calculations v2k 2 = # 2 remove common factors # v = ± solution works, requirement on v k 178 Waves on rope

• Transverse displacement wave • Produced by some initial displacement (plucking)

Can move in either 2 2 • F kg ! m / s m m v = = = = direction µ kg / m s2 s F is tension in rope µ is mass per unit length

179 19 Problem: Waves

• A wave on a rope has a displacement of d(x,t)=0.3cos(6x-3t) A) 0.6 m/s meters. What is the velocity of a peak in this B) -2 m/s wave? w(x,t) = Acos(kx ! "t) wave C) -0.5 m/s " v = phase velocity phase k D) 2 m/s 2# f = " frequency E) 0.5 m/s k$ = 2# wavelength

180 19 Problem: Waves

• A wave on a rope has a displacement of d(x,t)=0.5cos(5x+7t) A) 0.5 m/s meters. What is the velocity of a peak in this B) -0.5 m/s wave? w(x,t) = Acos(kx ! "t) wave C) 1.4 m/s " v = phase velocity phase k D) -1.4 m/s 2# f = " frequency E) 5 m/s k$ = 2# wavelength

181 Sound

• Longitudinal pressure wave • Produced by vibrating object

E kg / m / s2 m2 m v = = = = ! kg / m3 s2 s E is the Elastic modulus; pressure/(stretching factor) ! is density; mass per unit volume gas is similar but some thermodynamic effects v " 344 m/s in air, 1500 m/s in sea water, 5100 m/s in steel rod

182 Water Waves

• Deep water for λ20d

g m/s2 v = 2v = for deep water (gravity) waves phase group k 1/m velocity depends on wavelength k! = 2" vphase = vgroup = gd for shallow water (d is depth)

183 Light Waves

• Transverse Electric and Magnetic waves • Any wavelength • v=c=3X108 m/s o Phase and group

184 Electromagnetic Spectrum

• Electromagnetic waves o Move at the velocity of light c o Can have any wavelength

→ From low frequency radio waves

→ To Gamma rays

185 20 Problem: Waves 2

A sound wave has a pressure • A) 628 Hz of P(x,t)=20cos(1.95x-628t) in some units. What is the B) 200 Hz frequency of this sound? C) 100 Hz w(x,t) = Acos(kx ! "t) wave " v = phase velocity D) 344 m/s phase k 2# f = " frequency E) 20 Hz k$ = 2# wavelength Hz is 1/s 186 20 Problem: Waves 2

• A sound wave has a pressure of P(x,t)=20cos(1.95x-628t) in A) 6.3 m some units. What is the B) 200 Hz wavelength of this sound? C) 1.95 m w(x,t) = Acos(kx ! "t) wave " v = phase velocity D) 344 m phase k 2# f = " frequency E) 3.2 m k$ = 2# wavelength Hz is 1/s 187 21 Problem: Waves

A) 6.28 • The Electric field in the light wave of my green laser is mm E(x,t)=5cos((1.26X107)x-(3.78X1015)t) B) 6.28 µm in some units. What is the wavelength of this light? C) 6.28 nm w(x,t) Acos(kx t) wave = ! " D) 0.5 µm " vphase = phase velocity k E) 0.5 nm 2# f = " frequency k$ = 2# wavelength 1 µm=10-6 m -9 1 nm=10 m 188 Principle of Superposition

• Principle seen from wave equation.

o If w1(x,t) is a solution of the wave equation

o And w2(x,t) is a solution

o Then a(w1(x,t))+b(w2(x,t)) is also a solution

→ Because equation is linear in w • Waves can co-exist on same string…

189 Wave Group

• The waves we surf on are not cosine waves. • They are like the single pulse I put on the rope. • We call this a wave group (or packet) o It is a superposition of many cosine waves o It travels at the group velocity d! vgroup = o There may be dispersion dk

190 Interference

• Waves can add together constructively or destructively. • Diffraction of light

Two slit diffraction of light 191 Diffraction of Light

• Pass light beam through two slits. • Path length to screen differs. • Light is detected as individual photons o E=hf • Interference of waves o Even if intensity is turned down so there only one photon at a time 192 x(t) x cos( t) x = v t + x 1 = 0 ! 0 0 E = mgh + mv2 v = a t + v k 0 0 2 ! = m 1 2 p = mv x = a t + v t + x !T=2" 2 0 0 0 m v + m v 1 ! F = ma v = 1 1 2 2 f= = CM T 2" F mg m1 + m2 g = ! !=2"f 2 v1 ! v1 " vCM g = 9.8 m/s v(t) = #x0! sin(!t) 2v v v ! v " v w(x,t) = Acos(kx # !t) d 0x 0 y 2 2 CM = ! g v = "v v = 1 final 1initial phase k 2v2 sin" cos" d = 0 v = "v 2" f = ! g 2 final 2initial k$ = 2" 2 v0 dmax = g 193 Newton’s Universal Law of Gravity

194 Newton’s Universal Gravity Newton’s Constant Gm m • Newton related the 1 2 F = 2 gravity observed on r12 earth to the force that m3 G = 6.673! 10-11 causes planets and 2 kg s comets to follow elliptical orbits. • Inverse square • The attraction • Attractive force between spheres had • Direction is along the line been measured in the connecting the two lab prior to his masses. papers. • Equal and opposite forces on the two objects.

195 Acceleration in Circular Motion

• If we turn our car hard to the left, we are pushed to the right. • This means there is an acceleration to the left. • In circular motion, there is an acceleration toward the center of the circle which can be easily

calculated. v a v2 a = r

196 Ball on a String

• String pulls directly inward. • Ball moves in a circle.

197 Example: Mass of the Earth

• The radius of the earth is about 6367 kilometers. Estimate the mass of the earth.

GMm F = = mg r2 gr2 (9.8m / s2 )(6367000m)2 M = = = 6 ! 1024 kg G m3 6.673 ! 10-11 kg s2

198 22 Problem: Mass of the moon

Gm m F = 1 2 r 2 22 • The radius of the 12 A) 7.3X10 kg v2 moon is about 1737 a = kilometers and the r B) 4.2X1016 kg acceleration of GM v = gravity on the r C) 4.2X1018 kg surface is about 2"r 2"r 3/ 2 ! = = 1.62 m/s/s. v Estimate the mass GM D) 2.2X1020 kg m3 of the moon. -11 G = 6.673# 10 2 kg s E) 6.0X1023 kg GMm PE = $ r 199 Orbits of Planets

• Newton showed his laws of motion and gravity implied elliptical orbits. o Circle is just an ellipse where the two axes are the same length. o (Unbound objects follow hyperbola.) o Most of the planets have nearly circular orbits. • Gravitational force accelerates planets ma = Fgravity inward toward sun, keeping the orbit v2 GMm circular. m = r r 2 Period of orbit of planets just depends GM on 3/2 power of radius of orbit. v = (and mass of sun) r 2"r 2"r 3/ 2 ! = = v GM 200 Period of Orbits: Kepler’s Laws

ma = F • Orbits of planets are gravity v2 GMm m = ellipses with sun at r r 2 GM one focus. v = r • Orbits sweep out 2"r 2"r 3/ 2 ! = = equal area per unit v GM time. • Square of the Period is proportional to cube of semi-major axis.

201 23 Problem: Mass of the sun

Gm m F = 1 2 r 2 24 12 A) 7.3X10 kg • The radius of v2 a = the earth’s r B) 6.0X1024 kg orbit is about GM v = 150 billion r C) 2.0X1030 kg 2"r 2"r 3/ 2 meters. What ! = = v GM D) 2.2X1032 kg is the mass of 3 -11 m the sun? G = 6.673# 10 2 kg s E) 6.0X1032 kg GMm PE = $ r 202 A) B) Mass of sun C) D) E)

! earth = 1 yr = 365.25(24)(60)(60) = 31557600 s m3 G = 6.673 " 10-11 kg s2 2#r 3/2 ! = GM 2#r 3/2 GM = ! 3 2 3 4 2 1.5 1011 m 2 3 4# r # ( " ) 30 kg s m M = 2 = 3 = 2.00 " 10 3 2 G! $ m ' 2 m s 6.673 " 10-11 (31557600s) %& kg s2 () M = 2.00 " 1030 kg 203 Ganymede Statistics

Discovered by Simon Marius & Galileo Galilei Date of discovery 1610 Mass (kg) 1.48e+23 Mass (Earth = 1) 2.4766e-02 Equatorial radius (km) 2,631 Equatorial radius (Earth = 1) 4.1251e-01 Mean density (gm/cm^3) 1.94 Mean distance from Jupiter (km) 1,070,000 Rotational period (days) 7.154553 Orbital period (days) 7.154553 Mean orbital velocity (km/sec) 10.88 Orbital eccentricity 0.002 Orbital inclination (degrees) 0.195 Escape velocity (km/sec) 2.74

204 23 Problem: Mass of Jupiter Discovered by Simon Marius & Galileo Galilei Date of discovery 1610 Mass (kg) 1.48e+23 Mass (Earth = 1) 2.4766e-02 Equatorial radius (km) 2,631 Equatorial radius (Earth = 1) 4.1251e-01 Mean density (gm/cm^3) 1.94 Mean distance from Jupiter (km) 1,070,000 Rotational period (days) 7.154553 24 Orbital period (days) 7.154553 A) 7.3X10 kg Mean orbital velocity (km/sec) 10.88 Orbital eccentricity 0.002 Gm m F 1 2 Orbital inclination (degrees) 0.195 = 2 27 Escape velocity (km/sec) 2.74 r12 B) 1.9X10 kg v2 a = r C) 2.0X1029 kg GM What is the v = • r mass of 2"r 2"r 3/ 2 32 ! = = D) 2.2X10 kg Jupiter? v GM m3 G = 6.673# 10-11 32 kg s2 E) 6.0X10 kg GMm PE = $ r 205 Calculate Mass of Jupiter Discovered by Simon Marius & Galileo Galilei

Gm1m2 Date of discovery 1610 F = 2 Mass (kg) 1.48e+23 r12 Mass (Earth = 1) 2.4766e-02 v2 Equatorial radius (km) 2,631 a = Equatorial radius (Earth = 1) 4.1251e-01 r Mean density (gm/cm^3) 1.94 Mean distance from Jupiter (km) 1,070,000 GM v = Rotational period (days) 7.154553 r Orbital period (days) 7.154553 Mean orbital velocity (km/sec) 10.88 3/ 2 2"r 2"r Orbital eccentricity 0.002 ! = = v Orbital inclination (degrees) 0.195 GM Escape velocity (km/sec) 2.74 m3 G = 6.673# 10-11 3/ 2 2 2"r 2"r kg s ! = = GMm v GM PE = $ 3 r -11 m G = 6.673# 10 kg s2 2"r 3/ 2 GM = ! 4" 2r 3 4" 2 (1.07 # 109 )3 M = = = 1.9 # 1027 G! 2 (6.67 # 10-11)[(7.15)(24)(3600)]2 206 Liquid Accelerometer

• Shows acceleration of circular motion • Height proportional to PE • We can compute shape v = !r v2 a = = ! 2r r 2 Feffective = ma = m! r 1 PE = m! 2r2 effective 2

207 Example: Earth’s Velocity

• The earth orbits the sun in an approximately circular orbit with a radius of 150 billion meters. The sun has a mass of about 2X1030 kg. What must the earth’s velocity be?

GMm v2 F = = m r2 r " m3 % 6.673 ! 10-11 2 ! 1030kg GM #$ kg s2 &' ( ) v = = = 30000m / s r 1.5 ! 1011m

208 23 Problem: Orbit of Saturn

GMm F = r 2 7 12 A) 5.8X10 m v2 a = r B) 3.84X108 m GM v = r 3/ 2 6 2"r 2"r C) 6.4X10 m ! = = • Saturn takes 29.5 v GM m3 11 years to circle the G = 6.673# 10-11 D) 1.5X10 m kg s2 sun. What is the GMm PE = $ 12 radius of its orbit? r E) 1.4X10 m 11 rearth orbit = 1.5X10 m $ 209 23 Problem: Orbit of Mars • Mars takes 1.88

years to circle the GMm F = sun. What is the 2 r12 radius of its orbit? v2 A) 1.7X1011 m a = r GM 11 v = B) 1.9X10 m r 2"r 2"r 3/ 2 ! = = C) 2.1X1011 m v GM m3 G = 6.673# 10-11 11 kg s2 D) 2.3X10 m GMm PE = $ 11 r E) 2.7X10 m 11 rearth orbit = 1.5X10 m $ 210 Phobos & Diemos: Moons of Mars

From the surface of Mars, the motions of Phobos and Deimos appear very different from that of our own moon. Phobos rises in the west, sets in the east, and rises again in just 11 hours. Deimos, being only just outside synchronous orbit—where the orbital period would match the planet's period of rotation—rises as expected in the east but very slowly. Despite the 30 hour orbit of Deimos, it takes 2.7 days to set in the west as it slowly falls behind the rotation of Mars, then just as long again to rise.[55]

Because Phobos' orbit is below synchronous altitude, the tidal forces from the planet Mars are gradually lowering its orbit. In about 50 million years it will either crash into Mars’ surface or break up into a ring structure around the planet. Mars has two tiny natural moons, Phobos and Deimos, which orbit very close to the planet and are thought to be It is not well understood how or when Mars came to capture its captured asteroids. two moons. Both have circular orbits, very near the equator, which is very unusual in itself for captured objects. Phobos's Both satellites were discovered in 1877 by Asaph Hall, and unstable orbit would seem to point towards a relatively recent are named after the characters Phobos (panic/fear) and capture. There is no known mechanism for an airless Mars to Deimos (terror/dread) who, in Greek mythology, capture a lone asteroid, so it is likely that a third body was accompanied their father Ares, god of war, into battle. Ares involved—however, asteroids as large as Phobos and Deimos was known as Mars to the Romans. are rare, and binaries rarer still, outside of the asteroid belt.

211 23 Synchronous Orbit of Earth

A synchronous orbit has the GMm F = same orbital period as the r 2 rotational period (1 day for the 12 v2 earth). A satellite stays above a = 7 one spot on the earth. (at r A) 4.2X10 m equator)2"r 2"r 3/ 2 GM ! = = v = 8 v GM r B) 2.1X10 m 3 3/ 2 -11 m 2 r 2 r G = 6.673# 10 " " What is the2 radius ! = = • kg s v GM C) 4.2X109 m 3 ! GofM a 3/synchronous2 = r -11 m 2 G = 6.673# 10 "orbit of the earth? kg s2 10 ! 2GM D) 2.3X10 m r 3 = GMm 4" 2 PE = $ r ! 2GM [(24)(3600)]2 (6.67 # 10-11)(6.0 # 1024 ) 11 r 3 3 4.2 11107 m E) 2.7X10 m = 2 = 2 r = 1.5= X10# m 4" 4" earth$orbit 24 m = 6.0 # 10 kg 212 r = 42000 km earth Gravitational Potential Energy

• In constant gravity, PE=mgy • The force of gravity is given by F=-mg • We can compute the force from the PE d(PE) F = ! y dy • We can compute the PE from the force PE = ! F dy " y GMm GMm PE = ! F dr = dr = ! " r " r2 r 213 Gravitational Potential Energy

GMm F = ! 2 force on object of mass m falling toward moon Derive potential r • dv GMm energy in inverse m = ! apply second law dt r 2 square gravity. dv GM = ! acceleration independent of mass • Integral is a bit dt r 2 tricky but you don’t dv dt GM = ! divide by v need to do it dt dr vr 2 GM yourself. vdv = ! dr put v on left side and r on right r 2 • PE is negative r v 1 GM going to zero at vdv = ! dr do the definite integrals " " 2 0 r r infinity. 0 r v2 # 1& 1 ) 1 1 , = !GM %! ( = GM + ! . 2 r * r1 r0 - $ 'r0 mv2 GMm GMm = ! KE = PE0 ! PE1 2 r1 r0 GMm PE = ! PE goes to zero at infinity r 214 Example: Escaping Earth’s Gravity

1 GM m E = KE + PE = mv2 ! earth • What on!earth 2 r earth velocity is 1 E = mv2 # 0 needed to " 2 1 GM m escape the mv2 # earth 2 r earth’s earth 2GM gravity? v # earth rearth • Need m3 G = 6.673 $ 10-11 enough kg s2

energy to 2GM earth vescape = go r=infinity rearth 3 to escape. % -11 m ( 24 2 6.673 $ 10 6 $ 10 kg &' kg s2 )* ( ) m = = 11200 6.4 $ 106 m s ( ) 215 24 Problem: Moon Escape

Gm m • The moon has a F = 1 2 r 2 mass of 7.35X1022 12 A) 190 m/s v2 kg and a radius of a = r 1,738,000 m. What B) 1950 m/s GM is the escape v = r C) 2380 m/s velocity of the 2"r 2"r 3/ 2 ! = = moon? v GM D) 3450 m/s 3 -11 m G = 6.673# 10 2 kg s E) 4370 m/s GMm PE = $ r 216 A) Escape from the Moon B)

1 GM m E = KE + PE = mv2 ! moon C) on! moon 2 r moon D) 1 2 E" = mv # 0 2 E) 1 GM m mv2 # moon 2 rmoon 2GM v # moon rmoon m3 G = 6.673 $ 10-11 kg s2

2GM moon vescape = rmoon % m3 ( 2 6.673 $ 10-11 7.35 $ 1022 kg &' kg s2 )* ( ) m = = 2378 (1.738 $ 106 m) s 217 Space Shuttle in Orbit

• Shuttle said to be in “free fall” o No forces other than gravity • Astronaut floating inside is also in free fall • Acceleration of gravity is independent of mass o Shuttle and Astronaut have the same acceleration and free fall together o Astronaut just floats freely inside shuttle o No apparent effect of gravity

218 General Principle of Equivalence

• Newton’s Theory of gravity gives very accurate calculations of orbits… • Einstein’s theory of Gravity, General Relativity, gives some small corrections. • General Relativity says that a mass, like the sun, curves space-time. o Objects follow the shortest path in the curved space-time, seeming to accelerate to us. o In fact, astronauts experience no gravitational force or acceleration. They are in free fall. • GR tells us we cannot distinguish between gravity and acceleration (if we are inside a closed elevator for example.)

219 Free Fall and Equivalence

• In free fall, all objects just follow the shortest path is curved space-time. Objects inside a space ship follow the same path as the space ship. • If the ship fires its rockets, there is an acceleration not due to gravity and the astronauts will feel this as a force.

220 Vectors and Scalars

Properties when we rotate the coordinates system.

221 Vectors •We live in 3 space Euclidian Space dimensions. Cartesian Coordinate System •Any point in 3D can be y described with 3 coordinates. •The coordinates depend on the orientation of the coordinate system. z=1 X=3 ! •Position is a vector. v = (3,2,1) •Other physical quantities are also 3D vectors (velocity, acceleration, y=2 force…) x ! ! d x v = dt ! ! dv z a = dt 222 Lengths of Vectors

• Vector coordinates y depend on orientation of the coordinate system. ! v = 3,2,1 ( ) • Length of a vector does not depend on x coordinate system. Length of vector is • z related to dot product. ! ! v ! v = v2 length v independent of system 223 3D Coordinate system and Rotations

224 Dot Product of Vectors

Dot product of vectors in ! ! • r ! r = x2 + y2 + z2 = r2 3D are easily calculated. ! "! r ! p = xp + yp + zp = rpcos" • Dot product of vector x y z with itself gives length squared. • Dot products of vectors do not change when coordinate axes are rotated. o Dot product of vectors is a scalar.

225 C Question: Common Sense?

A) Translation of the • Which changes of origin coordinate systems B) Change of inertial change the length of frame velocity vectors? C) Rotation of coordinate axes D) All of the above

E) None of the above 226 C Question: Vectors

A) Inversion of the • Which changes of axes coordinate systems B) Change of inertial change velocity vectors? frame C) Rotation of coordinate axes D) All of the above

E) None of the above 227 C Question: Scalars

A) ! ! • Which of these does not x ! x change under any rotation ! ! B) v ! x of the coordinate axes? C) ! ! v ! v D) All of the above

E) None of the above

228 Rotational Symmetry

• The laws of Physics are the same if we rotate the coordinate system. • Rotation symmetry implies!" conservation" !" of angular momentum. L = r ! p o Remember: Translation symmetry implies conservation of momentum p=mv. • Cross product of vector is a vector perpendicular to each original vector. !" " !" | L |=| r ! p |= rpsin"

229 C Question: Conservation

• Which of the following A) Energy are conserved in nature? B) Momentum C) Angular momentum

D) All of the above

E) None of the above

230 Laws of Physics not Independent of…

• Accelerating frame • Parity: inversion of coordinate system • Almost but not quite independent of o Time reversal o Charge conjugation

231 Conservation of Angular Momentum

232 Conservation of Angular Momentum

• Independence of the laws of physics under rotations implies Angular Momentum is conserved. • Angular momentum is momentum times distance from center of rotation. ! ! ! ! ! L = r ! p a vector perpendicular to both r and p ! ! length is rpsin" where " is the angle between r and p

233 Bike Wheel Gyro and Spinning Stool

• Angular momentum conserved • Angular momentum points along axis of wheel • Turning the wheel changes its angular momentum • Other things must get the angular momentum

234 Why Didn’t Planets Fall into Sun?

• Solar system forms from cloud of gas and solids. o Gravity pulls it together to form sun • Angular momentum conserved o As matter moves to smaller radius, it must move faster to conserve angular momentum. • Some matter (1%) stays at a distance from sun to maintain angular momentum. • Sun spins so there is significant angular momentum there, but r is small so it can’t be too much.

235 Gyroscope

• Gyroscope has a good deal of angular momentum • It can’t just fall over because angular momentum (vector) is conserved • Force of gravity causes it to precess o Small orbit of angular momentum about vertical

236 Rotating Galaxies • Galaxies form due to small density fluctuations in the universe growing. • Collapse is countered by angular momentum which keeps everything from falling into the center. • Galaxy develops into a disk of some kind with angular momentum vector perpendicular to the disk. 237 Dark Matter • The existence of dark matter was first seen in the period of orbits of stars on the periphery of galactic clusters. • Period did not decrease with radius as expected. • Indicates some invisible mass spread over region larger than galaxy. • Dark matter. 238 C Question: Solar System

• If conservation A)The orbits of the planets should all of angular be in nearly the same plane. momentum is an important B)Most of the mass should be in factor in the planets. formation of the C)The plane of our solar system solar system, should be parallel to the planes of what properties other nearby systems. should the solar D) Both A and B. system have. E) All of the above.

239 C Question: Dark Matter

• The rotation curves of A)the same as visible stars galactic B)in a sphere instead of a disk clusters C)more concentrated than the visible indicate that stars the dark D)spread out to larger radius than the matter visible stars distribution is: E)mainly in the black hole at the center of the galaxy 240 Dropped

241 Inelastic Collisions

• Momentum is conserved in all collisions. • Elastic collisions also conserve energy. • Inelastic Collisions have some energy loss.

242 Example: Inelastic Collision

• A 1000 kg space ship (m + M )v f = mvi moves at a relative m velocity of 2 m/s as it v = v docks with a 100,000 f m + M i kg space station. After v f ! 0.02 m/s the docking, what velocity does the space station have in the frame in which it was initially at rest? 243 Problem: Docking Space Ships

• A 1000 kg space ship moves at a relative velocity of 6 m/s Numerical Answer! as it docks with another ship with a mass of 2000 kg. After the docking, what velocity (in m/s) do the docked ships have in the frame in which ship-2 was initially at rest?

244 Equal Mass Elastic Collisions

• In elastic collisions, Energy is also conserved. o E=mv2/2 • For equal mass, both momentum and energy are conserved if the incoming cart stops and the target cart gets the same velocity as the incoming cart had.

245 Elastic Collisions • In an Elastic Collision (of two masses) no kinetic energy is lost (to friction, heat, sound…) o KE is conserved o Momentum is always conserved. • Solution easy in Center of Mass Frame o Frame where total momentum is zero o Otherwise arithmetic is long

246 Solution in rest frame of m2

Even harder if m is moving! 2 2 ! m2 $ ! m2 $ ! m2 $ ! m2 $ p final = pinitial 2 1 v ± 2 1 v ' 4 m + 1 'm v2 + 1 v2 "# m initial %& "# m initial %& "# 1 m %& "# 1 initial m initial %& m v + m v = m v v = 2 2 2 2 1 1 2 2 1 initial 1 ! m2 $ 2 m 1 v is known; solve for v # 1 + & initial 2 " m2 % m v ! m v m 2 1 initial 1 1 1 ! m $ ! m $ ! m $ ! m $ v2 = +m2v2 = = (vinitial ! v1 ) 1 1 1 1 m m # & ± # & ' #1+ & # '1+ & 2 2 " m2 % " m2 % " m2 % " m2 % v1 = vinitial E final = Einitial ! m1 $ #1+ & " m2 % 1 2 1 2 1 2 m1v1 + m2v2 = m1vinitial 2 2 2 2 2 ! m $ ! m $ ! m $ 1 ± 1 + 1' 1 # m & # m & # m & have 2 equ. in 2 variables v1 and v2 " 2 % " 2 % " 2 % v1 = vinitial 2 ! m $ 1 1 " m % 1 1+ 1 2 1 2 "# m %& m1v1 + m2 $ (vinitial ! v1 )' = m1vinitial 2 2 2 # m2 & 2 ! m1 $ 2 # & ± 1 m 2 " m2 % m ± m 2 1 2 v = v = 1 2 v m1v1 + (vinitial ! v1 ) = m1vinitial 1 initial initial ! m1 $ m1 + m2 m2 1+ "# m %& m2 2 m v2 + 1 v2 ! 2v v + v2 = m v2 two solutions (+)=no collision 1 1 m ( initial initial 1 1 ) 1 initial 2 m1 ' m2 2 2 2 v1 = vinitial m m m m1 + m2 m v2 + 1 v2 ! 2 1 v v ! m v2 + 1 v2 = 0 1 1 1 initial 1 1 initial initial m m ! m ' m $ m2 m2 m2 1 1 1 2 v2 = (vinitial ' v1 ) = #1' & vinitial 2 2 2 m2 m2 " m1 + m2 % " m1 % 2 " m1 % " 2 m1 2 % m + v + !2 v v + !m v + v = 0 m ! m m m m $ m ! 2m $ ! 2m $ $ 1 ' 1 $ initial ' 1 $ 1 initial initial ' 1 1 + 2 1 ' 2 1 2 1 m m m v2 = ' vinitial = vinitial = vinitial # 2 & # 2 & # 2 & m # m + m m + m & m # m + m & # m + m & 2 " 1 2 1 2 % 2 " 1 2 % " 1 2 % 247 Center of Mass

• Center of mass is m1x1 + m2 x2 xCM = the average position m1 + m2 of mass. dx dx m 1 m 2 dx 1 + 2 • In CM frame, CM sits CM = dt dt = 0 dt m + m at origin. 1 2 dx1 dx2 • So, total momentum m1 + m2 = 0 dt dt is zero m1v1 + m2v2 = 0

p1 + p2 = 0

248 Solution in Center of Mass frame

• In CM frame total Transformation to CM frame m1v1 + m2v2 momentum is zero. vCM = m1 + m2

• Energy conserving v1 ! v1 " vCM

solution is simple. v2 ! v2 " vCM

v1 final = !v1initial Transformation back

v1 ! v1 + vCM v2 final = !v2initial v2 ! v2 + vCM

249 15 Problem: Center of Mass

1 E = mgh + mv2 • Consider the 2 collision of two p = mv A) 6.25 m/s masses. m v + m v 1 1 2 2 B) 1.25 m/s vCM = o m1=3 kg, v1=10 m/s m1 + m2 o m =5 kg, v =-4 m/s 2 2 v ! v " v C) 6 m/s • What is the velocity 1 1 CM v ! v " v of the center of 2 2 CM D) 14 m/s mass? v1 final = !v1initial E) -6.25 m/s v2 final = !v2initial 250 15 Problem: Elastic Collision

1 E = mgh + mv2 • Consider the same 2 collision of two p = mv A) 4 m/s masses. m v + m v 1 1 2 2 B) -1.25 m/s vCM = o m1=3 kg, v1=10 m/s m1 + m2 o m =5 kg, v =-4 m/s 2 2 v ! v " v C) 1.25 m/s • What is the velocity 1 1 CM v ! v " v of mass2 in the 2 2 CM D) -5.25 m/s center of mass after v1 final = !v1initial the collision? E) 5.25 m/s v2 final = !v2initial 251