### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# EssentialPhysics PHYS200

DU

GPA 3.82

### View Full Document

## 60

## 0

## Popular in Course

## Popular in Physics 2

This 276 page Class Notes was uploaded by Milton O'Hara on Sunday October 11, 2015. The Class Notes belongs to PHYS200 at Duquesne University taught by Staff in Fall. Since its upload, it has received 60 views. For similar materials see /class/221280/phys200-duquesne-university in Physics 2 at Duquesne University.

## Reviews for EssentialPhysics

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/11/15

CHAPTER 21 Chapter 211 Magnetic Fields If you ever played with magnets as a kid you ll know the following 1 A11 magnets have two ends called poles These poles are given the labels of North and South s 2 Like poles attract and unlike poles repel f A quot n jl 7 1 J 3 I J r L 739 y z xquot 7 3 y i yquot 31 V 5 5 4 5 La Kb Like poles repel Unlike poles attract Sources of magnetic fields 1 magnetic moments amp magnetic domains Magnetic d omains M ii 12 Permanent magnet b Induced magnetism a Unmagnetized iron 2 electric currents Chapter 212 Force on Moving Charges Due to Magnetic Fields As we ve seen from Chapter 18 a charged particle feels a force from an electric field Under certain conditions charged particles can also feel forces from magnetic fields Those conditions are l The charged particle is moving 2 The movement of the charged particle is in a direction that is not along or opposite the direction of the magnetic field If these conditions are met then the force on the charged particle is found from the following formula ITBqi7x Here q is the charge on the particle 7 is the velocity of the particle and B is the magnetic field The SI unit for the magnetic field is the tesla T Where 1T 2 W coulomb meter The multiplication symbol on the right hand side of this formula is a special kind of multiplication called a cross product The cross product is similar to the dot product which you ll see in Chapter 6 in that there is a dependence on the angle between the velocity and magnetic field vectors The main difference is that the result of the cross product is a vector quantity whereas the result of the dot product is a scalar quantity Let s focus on how to handle the cross product For any two vectors A and E A x 3 W E sine However the result of a cross product is a vector so we need a direction The direction of the cross product follows a righthand rule There are a few different rules you can follow just make sure you are using your right hand From the book The Old School version T x E E 6 a A R H R l Right I hand r i E A Also notice that in the cross product the order of the vectors matters that is to say that A x B B x A Other conventions you will see The direction of the magnetic field or any other vector may point into or out of the page This information is conveyed in the following manner B into paper B out of paper X X X X X e xxxxx 39 You can think of these directions in the following way Vectors can be represented by arrows If the arrow is pointing toward you out of the paper you will see the point thus the dot If the arrow is pointing away from you into the paper you will see the etching thus the X c 9 a o In summary The force experienced by a charged particle moving in a magnetic field is given by the formula FBqi7xB Where the magnitude of the force is given by FB q g sine and the direction of the force is found using the righthand rule Right hand rule gives the direction of the force for a positive charge If the charge is negative use the right hand rule but the force will point in the opposite direction Example 1 A uniform magnetic field points to the right A positively charged particle is moving upwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 2 A uniform magnetic field points into the page A positively charged particle is moving upwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 3 A uniform magnetic field points out of the page A negatively charged particle is moving upwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 4 A uniform magnetic field points to the left A positively charged particle is moving to the right in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 5 A uniform magnetic field points to the right A negatively charged particle is moving downwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 6 A uniform magnetic field points to the left A positively charged particle is moving downwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 7 A uniform magnetic field points to the right A negatively charged particle is moving upwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 8 A uniform magnetic field points downward A negatively charged particle is moving to the right in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field CHAPTER 4 Forces amp Newton s Laws of Motion Force push or pull that affects an object s motion There are two types 1 contact forces frictional normal 2 action at a distance forces gravitational electric Inertia natural tendency of an object to remain at rest or in motion at a constant speed along a straight line The mass of an object is a quantitative measure of its inertia It s the resistance to the change in an object s motion caused by a net force Newton s 1St Law of Motion An object continues in a state of rest or in a state of motion at a constant speed along a straight line unless compelled to change that state by a net force Net force vector sum of all forces 217 SF gt EFX EFy Limitations Inertial reference frame Where the 1st law is valid a frame of reference that is not accelerating Newton s Second Law of Motion When a net external force 217 acts on an object of mass m the acceleration 5 that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass The direction of the acceleration is the same as the direction of the net force The unit of force is the Newton N kgigm s 215 m51 2F ma EFy ma X x y Limitations 1 Very small for small things such as atoms amp molecules need to use quantum mechanics 2 Very fast for speeds close to the speed of light need to use relativity Newton s Third Law Whenever one body exerts a force on a second body the second body exerts an oppositely directed force of equal magnitude on the first body AKA the action reaction law Suppose that the mass of the spacecraft is ms 11000 kg and that the mass of the astronaut is m A 92 kg In addition assume that the astronaut exerts a force of P 36 N on the spacecraft Find the accelerations of the spacecraft and the astronaut 46 Types of Forces An Overview Fundamental Forces 1 Gravitational Force 2 Strong Nuclear Force stability of nuclei 3 Electroweak Force charged particles amp radioactive decay of some nuclei Beam 39 7m ml 39 whim Newton s Second Law 44 The Vector Nature of Newton s Second Law of Motion E m EFx mazC EFy ma y FreeBodv Diagram A sketch representing an object and all of the forces acting on that object They are used with Newton s second law of motion to help determine the acceleration of the object Make sure to include a coordinate system y 560 N Opposing force 560 N a b Freebody diagram of the car Example 1 Two forces act on an object mass 300 kg 131 400 N 300 and 132 600 N 900 Find the magnitude and direction relative to the XaXis of the acceleration of the object Make sure you include a freebody diagram in your solution Example 2 Chapter 4 4 During a circus performance a 72kg human cannonball is shot out of a 18mlong cannon If the human cannonball spends 095 s in the cannon determine the average net force exerted on him in the barrel of the cannon Chapter 4 8 47 Gravitational Force Mass amp Weight Gmm A F 12quot g r2 2 G6673x10 11N7quot21 kg Chapter 4 CHAPTER 6 61 Work Done by a Constant Force The Principle of Conservation of Energy Energy is neither created nor destroyed Energy is only transferred from one type to another Transfer kind of misleading No material is owing to or from the object such as water ows but rather like the electronic transfer of money One account increases while the other account decrease without any material thing passing between the accounts Work is energy transferred to or from an object by means of a force acting on that object The SI unit of work is the Joule l Joule N m The mathematical de nition of work W done by a force lE over a tiny displacement d is given by the formula Wf d restriction assumption the applied force is constant over the entire path Chapter 6 WI7 A The right hand side of this equation is a special type of multiplication called a dot product For any two vectors the dot product is defined as follows A 9 B A E AHE COSB Applying this de nition to work gives us W AS A cos9 removing the bulky notation but remembering we are dealing with magnitudes WFAScos9 Where 9 is the angle between the force and the displacement vectors Chapter 6 Special Angles 00 90 180 So is work always done on an object if there is an applied force Conceptual question Are there forces that never do work on an object even if that object is moving Chapter 6 Focusing on the sign l x 39 assume the block moves down the ramp Energy transferred to an object is positive work energy transferred from an object is negative Chapter 6 4 Example 1 Chapter 6 7 A person pushes a 160 kg shopping cart at a constant velocity for a distance of 220 m She pushes in a direction of 290 below the horizontal A 480 N frictional force opposes the motion of the cart a What is the magnitude of the force that the shopper exerts Determine the work done by b the pushing force c the frictional force and d the gravitational force Chapter 6 5 Example 2 Chapter 6 71 Suppose that in the figure that 110gtlt103 J of work are done by the force I7w magnitude 300 N in moving the suitcase a distance of 500 m At What angle 6 is the force oriented with respect to the ground Chapter 6 6 Example 3 Chapter 6 1 The brakes of a truck cause it to slow down by applying a retarding force of 30103 N to the truck over a distance of 850 m What is the work done by this force on the truck Is the work positive or negative Why Chapter 6 7 Example 4 Chapter 6 6 The drawing shows a plane diving toward the ground and then climbing back upward During each of these motions the lift force L acts perpendicular to the displacement s which has the same magnitude of 17gtlt103 m in each case The engines of the plane exert a thrust T which points in the direction of the displacement and has the same magnitude during the dive and climb The weight W of the plane has a magnitude of 59104 N In both motions net work is preformed due to the combined action of the forces L T and W a Is more net work done during the dive or climb Explain b Find the difference between the net work done during the dive and climb w a Dive W s b Climb V Chapter 6 8 64 Conservative and Nonconservative Forces Forces can be broken into two categories using the concept of work Forces can be said to be either conservative or nonconservative The distinction between these types of forces is made based on how the force acts over a given path Conservative Force De nition 1 A force is conservative when the work it does on a moving object is independent of the path between the obj ect s initial and final positions Chapter 6 21 Example 11 Work Done by the Gravitational Force Suppose a 500 kg mass is moved from A to B along the three different paths shown below Let gl 100 ms2 and find the work done by gravity as the mass is moved horizontally and then vertically along the red path 500m B 500m Chapter 6 22 Example 12 Work Done by the Gravitational Force Suppose this same 500 kg mass is moved from A to B along the black path shown below Let gl 100 ms2 and find the work done by gravity along this path 500m B 500m Chapter 6 23 Example 13 Work Done by a Frictional Force Suppose a 500 kg mass is moved from A to B on a level surface along the three paths shown below Let the magnitude of the frictional force be 100 N and find the work done by the frictional force as the mass is moved along the red path 500m B 500m Chapter 6 24 Example 14 Work Done by a Frictional Force Suppose the same 500 kg mass is moved from A to B on a level surface along the black path shown below Let the magnitude of the frictional force be 100 N and find the work done by the frictional force along this path 500m B 500m Chapter 6 25 Conservative Force De nition 2 A force is conservative when it does no net work on an object moving around a closed path A closed path is one in which the starting and ending point is the same Example 15 Work Done by the Gravitational Force Closed Path Suppose a 500 kg mass is moved from A along the red path to B and then along the green path back to A Find the work done on the mass by the gravitational force Let 100 ms2 500m B 500m Chapter 6 26 Example 16 Work Done by a Frictional Force Closed Path Suppose a 500 kg mass is placed on a level surface The mass is then moved from A along the red path to B and then along the green path back to A Find the work done on the mass by a frictional force if its magnitude is 200 N 500m B 500m Chapter 6 27 Table 62 page 158 Conservative Forces Nonconservative Forces Gravitational Force Static and kinetic frictional forces Elastic spring force Air resistance Electric force Tension Normal force With two types of forces that means we can talk about two types of work the work done by conservative and nonconservative forces We can then modify the WorkEnergy Theorem a bit Wm AKE WC Wm AKE We won t worry about the electric force right now We would have to go back into the later chapters to talk about the energy contained in a given arrangement of charges This means the gravitational force is the only conservative force we ve studied so far W W APE c gravtty W W AKE grawty no W AKE W no gravtty Wm AKE APE Chapter 6 28 Example 17 Chapter 6 27 A bicyclist rides 50 km due east while the resistive force from the air has a magnitude of 30 N and points due west The rider then turns around and rides 50 km due west back to her starting point The resistive force from air on the return trip has a magnitude of 30 N and points due east a Find the work done by the resistive force during the round trip b Based on your answer to part a is the resistive force a conservative force Explain Chapter 6 29 Q rm 5 j392 VI V23YM2 M V2VMIWZ 1 9min 4 1 w w 12 r a rwmmJMi Mal pd n g 5 quot4 LvaE O JtccHWS VzF W WEKQ 39 39zyuf39s0 27y N C q 0300 k5 0275 A H r zf L 0quot WS ax Hm quotg u Amvn a M rrJ U0 72 9 71F v0 vaF v3 tW 2r v TZI v1F A75 02gmlggt 1L C039139h15 39 LrSUMEE 71quot7 mg 39 k urnw 3 H 6435 Converting Units Whether you are working with base or derived units you must be comfortable converting between different standards of measurement The key to converting units is understanding you are just multiplying by a well chosen 1 Example 1 How many minutes are in 23 hours kMODV JO W39V ul39 39s 5 1 Leaf cw dPS CA Fo cl 0 S d m A Q in ll 0 w A T 145 yr Cldj o O 1311 1 w 39l Niels use llamas TO be minis cat 3 bag MA Q yg 80 wLM l L T lllOmmwlCS 1 2 6 5 95 66 Nonconservative Forces and the WorkEnergy Theorem mesh mesh Wm E f E Wm AKE APE 67 Power The average power done on or by an object is the work done on or by the object over a given amount of time Mathematically this is given by where PW is the average power Wis the work done and I is the time it takes to do the work Power is a scalar quantity with units ofWatts W W Js This is not the only formula for average power Average power can also be found from P Fv 11V 11V where F is the applied force and VW is the average speed Chapter 6 37 Where does PM F vavcome from Is it magic Nope p K av PW F AS2COSB P FASJCOSB 1V t PM F vav cosB so if IE and 17 are in the same direction 8 0 P Fv 11V 11V Chapter 6 3 8 Example 23 Chapter 6 56 Bicyclists in the Tour de France do enormous amounts of work during a race For example the average power per kilogram generated by Lance Armstrong m 750 kg is 650 W per kilogram of his body mass a How much work does he do during a l35km race in which his average speed is 120 ms b Often the work done is eXpressed in nutritional Calories rather than in joules EXpress the work done in part a in terms of nutritional Calories noting that l joule 2389 X 10394 nutritional Calories Chapter 6 39 Example 24 Chapter 6 57 A car accelerates uniformly from rest to 200 ms in 56 s along a level stretch of road Ignoring friction determine the average power required to accelerate the car if a the weight of the car is 900 X 103 N and b the weight of the car is 14 X 104 N Chapter 6 40 68 Other Forms Types of Energy amp Conservation of Energy Energy is neither created nor destroyed Energy is only transferred from one type to another This statement can be summed up mathematically in a simple equation EiEf However this statement takes into account all types of energy an object possesses at two instances making it a tough equation to work with EiEf KEI PEl Eil gh Ei50 quotdKEfPEfEjjgh E d Chapter 6 41 Problem solving techniques tips Forces amp Work What is always true of work done by the following forces gravitational magnetic kinetic frictional Analysis of Linear Motion Kinematics Forces Work amp Energy Chapter 6 42 Example 25 A basketball player makes a jump shot The 0600 kg ball is released at a height of 200 111 above the oor with a speed of 720 ms The ball goes through the net 310 111 above the oor at a speed of 420 ms What is the work done on the ball by air resistance Chapter 6 43 Example 26 A 75 gram Frisbee is thrown from a point 11 m above the ground with a speed of 12 ms When it has reached a height of 21 m its speed is 105 ms a How much work was done on the Frisbee by its weight b How much work was done by air drag Chapter 6 44 Example 27 A 25 kg bear slides from rest 12 n1 down a lodgepole pine tree moving with a speed of 56 ms just before hitting the ground What is the average frictional force that acts on the bear Chapter 6 45 Example 28 A 30 g bullet with a horizontal velocity of 500 ms stops 12 cm Within a solid wall a What is the change in its mechanical energy b What is the magnitude of the average force exerted by the wall that stops the bullet Chapter 6 46 Example 29 During a rock slide a 520 kg rock slides down from rest down a hillside that is 500 m long and 300 m high The coefficient of kinetic friction between the rock and hill surface is 025 a How much work is done by kinetic friction b What is the speed of the rock at the bottom of the hill Chapter 6 47 Example 30 An out elder throws a baseball with an initial speed of 366 ms Just before an infielder catches the ball at the same level its speed is 335 ms How much of the ball s mechanical energy is lost due to air drag The mass of the baseball is 0255 kg Chapter 6 48 Example 31 A 94 kg projectile is red vertically upward Air drag dissipates 68 kJ during its ascent How much higher would it have gone were air drag negligible Chapter 6 49 Concept question Chapter 6 1 Two forces F1 and F2 are acting on the box as shown in the drawing causing the box to move across the oor The box moves in the same direction as F1 The two forces are drawn to scale Which force does more work Justify your answer Concept question Chapter 6 2 A box is being moved with a velocity v by a force P parallel to v along a level horizontal oor The normal force is FN the kinetic frictional force is fk and the weight of the box is mg Decide which forces do positive zero or negative work Provide a reason for each answer Chapter 6 50 Concept question Chapter 6 3 A force does positive work on a particle that has a displacement pointing in the X direction This same forced does negative work on a particle that has a displacement pointing in the y direction In what quadrant does the force lie Account for your answer Concept question Chapter 6 4 A sailboat is moving at a constant velocity a Is work being done by a net external force acting on the boat EXplain b Recognizing that the wind propels the boat forward and the water resists the boat s motion what does your answer in part a imply about the work done by the wind s force compared to the work done by the water s resistive force Chapter 6 51 Concept question Chapter 6 5 A ball has a speed of 15 ms Only one external force acts on the ball After this force acts the speed of the ball is 17 ms Has the force done positive or negative work Explain Concept question Chapter 6 6 A slowmoving car may have more kinetic energy than a fast moving motorcycle How is this possible Chapter 6 52 Concept question Chapter 6 10 In a simulation on earth an astronaut in his space suit climbs up a vertical ladder On the moon the same astronaut makes the same climb In which case does the gravitational potential energy of the astronaut change by a greater amount Account for your answer Concept question Chapter 6 13 The roller coaster called the Steel Dragon starts with a speed of 30 ms at the top of the drop and attains a speed of 429 ms when it reaches the bottom If the roller coaster were to then start up an identical hill would its speed be 30 ms at the top of this hill Ignore friction Explain your answer in terms of energy concepts Chapter 6 53 Concept question Chapter 6 14 A person is riding on a Ferris Wheel When the Wheel makes one complete turn is the net work done by the gravitational force positive negative or zero Justify your answer Chapter 6 54 49 Static and Kinetic Frictional Forces Static Friction static means no movement Static Friction has two main characteristics restriction assumption macroscopic can see objects Characteristic 1 Static friction is independent of the area of contact between the surface and object restriction assumption hard and nondeformable surfaces Characteristic 2 The magnitude of the static frictional force is proportional to the normal force The constant of proportionality is called the coefficient of static friction denoted by MS and is a unitless quantity The coefficient of static friction is not surprisingly material dependent among other things temperature material s condition rough polished The magnitude of the static frictional force can be zero to a maximum amount of fsmax depending on the applied force The maximum value of this force is just before the object starts to move This is stated mathematically by the very important and often used condition fsmax MSFN As the book mentions this only relates the magnitudes of the normal and static forces and in this case the lack of arrows indicates magnitude Forces are vectors and these vectors do not act in the same direction However this relationship ties components together and goes back to the vector nature of EF mil Chapter 4 28 Kinetic Friction kinetic means motion is present Kinetic Friction has three main characteristics restriction assumption macroscopic objects Characteristic 1 Kinetic friction is independent of the area of contact between the surface and object restriction assumption hard and nondeformable surfaces Characteristic 2 Kinetic friction is independent of the speed of sliding motion restriction assumption the speed is small Characteristic 3 The magnitude of the kinetic frictional force is proportional to the normal force The constant of proportionality is called the coefficient of kinetic friction denoted by Mk and is a unitless quantity The magnitude of the kinetic frictional force is given by f k lukF N The same warning applies this relates the magnitudes of the two forces Both static and kinetic friction act in a direction to oppose motion whether that motion is occurring or could potentially occur Chapter 4 29 Example 15 Chapter 4 37 A block Whose weight is 450 N rests on a horizontal table A horizontal force of 360 N is applied to the block The coefficients of static and kinetic friction are 0650 and 0420 respectively Will the block move under the in uence of the force and if so What will be the block s acceleration Chapter 4 30 Example 16 Chapter 4 41 A 81kg baseball player slides into second base The coefficient of kinetic friction between the player and the ground is pk 049 a What is the magnitude of the frictional force b If the player comes to rest after 16 s What is his initial speed Chapter 4 31 Example 17 Chapter 4 44 An ice skater is gliding horizontally across the ice with an initial velocity of 63 ms The coefficient of kinetic friction between the ice and the skate blades is 0081 and air resistance is negligible How much time elapses before her velocity is reduced to 28 ms Chapter 4 32 Example 18 Chapter 4 51 A stuntInan is being pulled along a rough road at a constant velocity by a cable attached to a moving truck The cable is parallel to the ground The mass of the stuntInan is 109 kg and the coefficient of kinetic friction between the road and him is 0870 Find the tension in the cable Chapter 4 33 CHAPTER 1 Understanding Units In Physics almost every number comes with a unit The units of a number explain to everyone paying attention exactly What standard the observer has used in the measurement The units are part of the number and must be retained Base Units and Derived Units 0To create a system of measurement a small set of base units must be defined Only these need to be defined in terms of outside standards We will be using the MKS base system Not surprising the three base units are Meter length Kilogram mass Second time By contrast derived units are defined as combinations of the base units needing no other outside standard Examples include meters per second ms speed and velocity kgT m momentum aka the Newton force m s2 Watch your units Many times you must make sure they are in the same base SI vs British for example Also you must watch your prefixes convert centimeters to meters for example Chapter 1 l Converting Units Whether you are working with base or derived units you must be comfortable converting between different standards of measurement The key to converting units is understanding that you are just multiplying by a well chosen one What does this mean Example 1 How many minutes are in 23 hours Chapter 1 Not all conversions will be of the simple onestep variety However the process is the same Example 2 Chapter 1 2 Vensa Vuloic survived the longest fall on record Without a parachute when her plane eXploded and she fell 6 miles 551 yards What is this distance in meters Chapter 1 3 Example 3 Chapter 1 6 A bottle of Wine known as a magnum contains a volume of 15 liters A bottle known as a jeroboam contains 0792 US gallons How many magnums are there in one jeroboam From the back of the text 1 US gallon 3785 liters Chapter 1 4 Dimensional Analysis Each unit measures a certain dimension Following the book s example we will let certain letters represent different dimensions L dimensions of length meters feet M dimensions of mass grams slugs T dimensions of time seconds The game here is to take an equation and replace variables with the letters above If an equation is dimensionally correct both sides of the equation will give the same dimension after some algebra is done During addition or subtraction dimensions are like units LL LT LT L M During multiplication or division dimensions act like numbers MLMT MTL LT LM Chapter 1 5 Who cares Will tell you if an equation is correct avoids transcription errors of variables Will give confidence in numerical answers does not help with constants however Example 4 Which of these equations is dimensionally correct In this example x denotes position dimensions of length v denotes speed dimensions of length divided by time and Idenotes time V x nal xiniml Vt x nal xiniml g 139 xfinal xiniml Vt X V 39 inal xiniml i f t D neither equation 4 both equations Chapter 1 Example 5 Which of these equations is dimensionally correct In this example x denotes position dimensions of length v denotes speed dimensions of length divided by time a denotes acceleration dimensions of length divided by time squared and I denotes time 2 xfinal xiniml Vinitalt a t x nal xiniml Vinitalt at 2 39 xfinal xiniml Vinitalt a t 39 x nal xiniml Vinitalt at U neither equation 4 both equations Chapter 1 7 Example 6 Given the basic dimensions of Length L Mass M and Time 3 2 T the combination of units T3 reduces to M3TL2 LT3 M3L T2 2 32 3MT L 2 4ML T Chapter 1 Example 7 Given the basic dimensions of Length L Mass M and Time WM L2 1 T the combination of units ML2 TZMZ reduces to 1 Hey trick question The end result is dimensionless 1 2 2L 2 1 3 ML2 2 ML 4MUL L2 Chapter 1 Significant Figures Concept Not all numbers your calculations give are necessarily meaningful Your answer cannot be more precise than the least precise value in your calculation Who cares Homework and exam solutions will require the correct number of significant figures for full credit Actually exam solutions have one point reserved for sig figs Rules 1 nonzero digits are significant 2 zeros as the last digit AND to the right of a decimal point are significant 3 zeros to the right of the decimal point AND not in between significant figures are NOT significant they are just placeholders 4 zeros in between signi cant figures ARE significant 5 round your answer to the place where the last significant figure occurs in the numbers you use 6 any constants or conversion factors have an infinite number of significant figures 7 e Chapter 1 10 Example 8 How many significant figures do the following numbers have 1 100 2 15 3 107 4 000452 5 1000 6 800452 Chapter 1 Example 9 A circle has a of radius 21 meters What s the area of the circle 1 14 m2 2 139 m2 3 1385 m2 4 10 1112 Chapter 1 Example 10 A rectangle has sides of 225 meters and 110 centimeters The area of this rectangle is 1 2475 m2 5 2475 m2 2 247 m2 6 248 m2 3 24 m2 7 24 m2 4 25 m2 8 25 1112 Chapter 1 13 Example 11 How many days are in 100 hour and 1800 minutes Answer Chapter 1 Example 12 A vehicle is traveling at 5500 ms How many miles per hour is this Helpful conversions 1 ft 03048 m 1 mile 5280 feet Answer Chapter 1 Example 13 A car is traveling at 650 mileshr How many meters per second is this Helpful conversions 1 ft 03048 n1 1 mile 5280 feet Answer Chapter 1 l6 Trigonometric Functions The trigonometric functions are the ratio of sides of a right triangle The definition of the three most commonly used trig functions are opposite side sme hypotenuse adjacent side cosine hypotenuse opposite side tangent adjacent side Which leads to a common mnemonic device SOH CAH TOA How does this work with respect to triangles and specific angles Chapter 1 l7 Trigonometric Functions cu E a adjacent side C b opposite side or A s1nA hyp0tenuse Sln C 81113 C adjacent side or b B comm W cosA E Similarly cos C opposite side or t A E 9 tanA adjacent side an b tanB a Chapter 1 18 Example 14 In the right triangle below angle A is 53 degrees and side 0 is 41 meters long How long is side a Answer How long is side b Answer Chapter 1 l9 Example 15 In the right triangle below angle B is 35 degrees and side 0 is 14 meters long How long is side a Answer How long is side b Answer Chapter 1 20 Trigonometric Functions A Calculator Check There are two main units used to measure angles degrees and radians You must be able to switch between these units on your calculator You must also be able to tell which mode your calculator is operating in Which is a good way to test your calculator s mode 1 sin0 2 cos0 3 tan45 Other useful stuff 360 27 radians The Pythagorean theorem 12 b2 2 Quadrant 11 1 Quadrant I Sine is Sine is Cosine is quot Cosine is I I I I I I 1 1 Quadrant III Quadrant IV Sine is Sine is Cosine is 1 Cosine is Chapter 1 21 Example 16 Chapter 1 problem 13 A highway is to be built between two towns one of which lies 350 km south and 720 km west of the other What is the shortest length of highway that can be built between the two towns and at what angle would this highway be directed with respect from due west Answer Chapter 1 22 Example 17 Chapter 1 16 amp 17 The drawing shows sodium and chlorine ions positioned at the corners of a cube that is part of the crystal structure of sodium chloride common table salt The edge of the cube is 0281 nm 1 nm l nanometer 1x10quot9 m in length Find the distance between the sodium ion located at one corner of the cube and the chloride ion located on the diagonal at the opposite comer What is 6 Sodium ion Chloride ion 4 J V N I 0281 nanometers 9 Chapter 1 23 HORIZONTAL ACCELERATION CARD a am no 848 03753 Chapter 1 24 Example 18 A very curious student 150 meters tall from feet to eye stands 340 meters away from a tall agpole This student just happens to have a triangulation sextant a device for measuring angles from the horizontal and measures an angle of 225 degrees How tall is the agpole Answer Chapter 1 25 Example 19 The same curious student 150 m tall meters from feet to eye stands on the 10th street bridge and looks toward Duquesne University An angle of 150 degrees from the horizontal to the top of the bluff is measured using a triangulation sextant If this student is standing 217 meters away from the bluff how tall it Answer Chapter 1 27 Scalar a quantity that can be described by a single number and units Vector a quantity described by a strengt its magnitude and a direction Scalar Quantities Vector Quantities Distance Displacement Speed Velocity Energy Force Mass Acceleration Temperature Electric Field Having to keep track of a direction may seem to have little consequence but it can mean vector answers differ wildly compared to scalar answers As a result if you are required to supply an answer that is a vector you must include a magnitude and direction For the direction to make sense it is very helpful to include a coordinate aXis Your choice of a coordinate axis will be required for quiz and exam problems dealing with vector quantities If a problem dealing with a vector quantity asks for just the magnitude of the vector only a positive number with appropriate units is required Giving a direction in this case is wrong Similarly if you are working with a scalar quantity and you give a direction in your answer you are giving an incorrect answer Chapter 1 28 Vector Addition amp Subtraction Graphical Method Required Tools 1 2 3 Two or more vectors and their directions Pencil and eraser pens are not a good idea Ruler or graph paper and protractor Method head to tail 1 2 3 Draw a Cartesian coordinate system amp decide on a scale Draw 1St vector s magnitude to scale and along correct direction From the head of the 1St vector draw in the 2nd vector s magnitude to scale and along correct direction Repeat step 3 for additional vectors Draw in the resultant vector connect the start of the first vector to the head of the last vector drawn The length of this vector is the resultant vector s magnitude to scale The resultant s direction can be specified in a variety of ways You will get practice using this method in lab Pros no math needed Cons very tedious and time consuming not always practical Chapter 1 y Baa x 29 Vector Addition amp Subtraction An Introduction When dealing with vector addition and subtraction using algebra it is helpful to introduce some concepts Vector components In this class we will work with two dimensions most of the time Using a standard coordinate aXis these dimensions are referred to as the x and y dimensions A two dimensional vector magnitude and direction can be decomposed into its vector components These vector components are found using trigonometric functions and describe how much of the vector lies in each dimension Notation of Vector Components Algebraically a decomposed vector will be written in the following form called component form AAx Ay Here AK and Ay are the magnitudes of the vector components 22 and i tell you in which direction these magnitudes point Examples Chapter 1 30 Vector Addition amp Subtraction Algebraic Method Method 1 J U Uib Take every vector written in magnitudedirection form and break each vector into its components X and y Do this in an organized manner to avoid confusion Write each vector in terms of its components using 35 and 9 notation This is called the component form of the vector Add like components For vector subtraction add the negative of the components you got from step 2 Find the resulting vector s magnitude using the Pythagorean theorem Find the resulting vector s direction from the definition of the tangent function Be very careful here the tangent function is restricted to angles between 0 and 90 on most calculators You must be aware your calculator s angle may only be a reference ang Pros much quicker than the graphical method Cons not very visual Chapter 1 31 Steps 1 amp 2 Decomposition of a Vector Into Its Components i Take every vector written in magnitudedirection form and break each vector into its components X and y Do this in an organized manner to avoid confusion Write each vector in terms of its components using 35 and y notation This is called the component form of the vector J Decomposing a vector means finding that vector s two components using trig functions These components are always perpendicular to each other It can be helpful to draw a rough sketch to start its decomposition This sketch includes a coordinate aXis and the original vector The original vector is the hypotenuse of a right triangle and its components are the sides of that triangle You must determine what angle direction is given to you and then use the appropriate trigonometric function to find the components Make sure to be aware of what quadrant the vector is in This may C C help you assign the correct or signs to your components Chapter 1 32 Example 20 Vector A has a magnimge of 5 km and is oriented 30 from the horizontal Write vector A in component form Example 21 Vector 13 has a magnitude of 6 ms and is oriented 40 from the vertical Write vector B in component form Chapter 1 33 Example 22 Vector E has a magnitude of 7 km and is oriented 135 from the horizontal Write vector C in component form Example 23 Vector 15 has a magnitude of ms and is oriented 240 from the positive yaXis Write vector D in component form Chapter 1 34 Example 24 Chapter 1 23 A circus performer begins his act walking out along a nearly horizontal high wire He slips and falls to the safety net 250 feet below The magnitude of his displacement from the beginning of the walk to the net is 267 ft a How far out along the high wire did he walk b Find the angle that his displacement vector makes below the horizontal Answer Chapter 1 35 Example 25 On takeoff an airplane climbs with a speed of 180 ms at an angle of 34 above the horizontal The speed and direction of the airplane constitute a vector quantity known as the velocity The sun is shining directly overhead How fast is the shadow of the plane moving along the ground That is what is the magnitude of the horizontal component of the plane s velocity Answer Chapter 1 36 Steps 3 5 Vector Addition amp Subtraction 3 Add like components For vector subtraction add the negative of the components you got from step 2 4 Find the resulting vector s magnitude using the Pythagorean theorem 5 Find the resulting vector s direction from the definition of the tangent function Be very careful here the tangent function is restricted to angles between 0 and 90 on most calculators You must be aware your calculator s angle may only be a reference angle 31 Chapter 1 37 Example 26 Chapter 1 55ish You are on a treasure hunt and your map says quotWalk due west for 52 paces then walk 300quot north of west for 42 paces and finally walk due north for 25 pacesquot What is your displacement at the end of your journey with respect to your starting position Give your direction with respect to due east and your magnitude in paces Answer Chapter 1 38 Example 27 Chapter 1 problem 60 In wandering a grizzly bear makes a displacement of 1563 m due west followed by a displacement of 3348 m in a direction 320 north of west What are a the magnitude and b the direction of the displacement needed for the bear to return to its starting point Specify the direction relative to due east Answer Chapter 1 39 Example 28 Chapter 1 42 The magnitudes of the four displacement vectors shown in the drawing areA 160 m B 110 m C 120 m and D 260 m Determine the a magnitude and b direction for the resultant that occurs When these vectors are added together Specify the direction as a positive counterclockwise angle from the x axis Answer Chapter 1 40 29 Chapter 1 27 A car is being pulled out of the mud by two forces that are applied by the two ropes shown in the drawing The dashed line in the drawing bisects the 300 angle The magnitude of the force applied by each rope is 2900 newtons Arrange the force vectors tail to head and use the graphical technique to answer the following questions a How much force would a single rope need to apply to accomplish the same effect as the two forces added together b How would the single rope be directed relative to the dashed line 2900 newtonsl lt 3000 2900 newtons Answer Chapter 1 41 Example 30 Chapter 1 56 Displacement vector 11 points due east and has a magnitude of 200 km Displacement vector l3 points due north and has a magnitude of 375 km Displacement vector 6 points due west and has a magnitude of 250 km Displacement vector D points due south and has a magnitude of 300 km Find a the magnitude of the resultant vector 11 DDD and b its direction as a positive angle relative to due west Chapter 1 42 Example 31 Chapter 1 63 What are the a x and b y components of the vector that must be added to the following three vectors so that the sum of the four vectors is zero Due east is the x direction and due north is the 32 direction 11 113 units 600 south of west 13 222 units 350quot south of east 6 177 units 230quot north of east Chapter 1 43 CHAPTER 21 Chapter 211 Magnetic Fields If you ever played with magnets as a kid you ll know the following 1 A11 magnets have two ends called poles These poles are given the labels of North and South s 2 Like poles attract and unlike poles repel f A quot n jl 7 1 J 3 I J r L 739 y z xquot 7 3 y i yquot 31 V 5 5 4 5 La Kb Like poles repel Unlike poles attract Sources of magnetic fields 1 magnetic moments amp magnetic domains Magnetic d omains M ii 12 Permanent magnet b Induced magnetism a Unmagnetized iron 2 electric currents Chapter 212 Force on Moving Charges Due to Magnetic Fields As we ve seen from Chapter 18 a charged particle feels a force from an electric field Under certain conditions charged particles can also feel forces from magnetic fields Those conditions are l The charged particle is moving 2 The movement of the charged particle is in a direction that is not along or opposite the direction of the magnetic field If these conditions are met then the force on the charged particle is found from the following formula ITBqi7x Here q is the charge on the particle 7 is the velocity of the particle and B is the magnetic field The SI unit for the magnetic field is the tesla T Where 1T 2 W coulomb meter The multiplication symbol on the right hand side of this formula is a special kind of multiplication called a cross product The cross product is similar to the dot product which you ll see in Chapter 6 in that there is a dependence on the angle between the velocity and magnetic field vectors The main difference is that the result of the cross product is a vector quantity whereas the result of the dot product is a scalar quantity Let s focus on how to handle the cross product For any two vectors A and E A x 3 W E sine However the result of a cross product is a vector so we need a direction The direction of the cross product follows a righthand rule There are a few different rules you can follow just make sure you are using your right hand From the book The Old School version T x E E 6 a A R H R l Right I hand r i E A Also notice that in the cross product the order of the vectors matters that is to say that A x B B x A Other conventions you will see The direction of the magnetic field or any other vector may point into or out of the page This information is conveyed in the following manner B into paper B out of paper X X X X X e xxxxx 39 You can think of these directions in the following way Vectors can be represented by arrows If the arrow is pointing toward you out of the paper you will see the point thus the dot If the arrow is pointing away from you into the paper you will see the etching thus the X c 9 a o In summary The force experienced by a charged particle moving in a magnetic field is given by the formula FBqi7xB Where the magnitude of the force is given by FB q g sine and the direction of the force is found using the righthand rule Right hand rule gives the direction of the force for a positive charge If the charge is negative use the right hand rule but the force will point in the opposite direction Example 1 A uniform magnetic field points to the right A positively charged particle is moving upwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 2 A uniform magnetic field points into the page A positively charged particle is moving upwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 3 A uniform magnetic field points out of the page A negatively charged particle is moving upwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 4 A uniform magnetic field points to the left A positively charged particle is moving to the right in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 5 A uniform magnetic field points to the right A negatively charged particle is moving downwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 6 A uniform magnetic field points to the left A positively charged particle is moving downwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 7 A uniform magnetic field points to the right A negatively charged particle is moving upwards in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 8 A uniform magnetic field points downward A negatively charged particle is moving to the right in this magnetic field The force the charged particle feels is l to the right 2 to the left 3 upwards 4 downwards 5 into the page 6 out of the page 7 the particle feels no force from the magnetic field Example 9 Chapter 21 69 Due to friction with the air an airplane has acquired a net charge of 170x10 5 C The plane moves with a speed of 280x102 ms at an angle 9 with respect to the earth s magnetic field the magnitude of which is 500x10 5 T The magnetic force on the airplane has a magnitude of 230x10 7 N Find the angle 9 There are two possible angles Example 10 Chapter 21 64 In New England the horizontal component of the earth s magnetic field has a magnitude of 1610 5 T An electron is shot vertically straight up from the ground with a speed of 21106 ms What is the magnitude of the acceleration caused by the magnetic force Ignore the gravitational force acting on the electron Example 11 Chapter 21 10 ish A magnetic eld has a magnitude of 1210 3 T and points into the page A positive 18 C charge moves at a speed of 31106 ms to the right Determine force that acts on the charge Example 12 Chapter 21 2 A charge of 83 C is traveling at a speed of 74 X 106 ms in a region of space Where there is a magnetic field The angle between the velocity of the charge and the field is 52 A force of magnitude 54 X 10393 N acts on the charge What is the magnitude of the magnetic field Chapter 213 Motion of Charged Particles in Magnetic Fields From the previous section if there is a charged particle moving perpendicular to a magnetic field that particle will feel a force How does this force affect its motion X X X X 391 X X gt X X gt X X X X X X X X X X X X X X Thus if the entire velocity of the charged particle is perpendicular to the magnetic eld the motion of the particle is circular Now back when we were talking about acceleration I mentioned there were three different ways an object can accelerate because acceleration is a vector 1 the object s speed can change the magnitude changes 2 the object s direction can change with the speed staying constant 3 both of the above can change not going to happen in this class In the second case acceleration due to changing direction with a constant speed is called centripetal acceleration and found using acentripe ml a5 v2 7 If only a component of the velocity is perpendicular to the magnetic field then the motion is no longer circular The motion is helical because the component of the velocity parallel or anti parallel to the eld does not affect the particle s trajectory Entire V perpendicular to B Part of V perpendicular to B Chapter 21 4 The Mass Spectrometer Using Newton s Second Law for charged particles moving in a magnetic field and ignoring other forces except for the magnetic force you find that the radius of the circular path depends on the mass of the particle 215 m FB q I 17sin6 Let 6 90 B out of paper This behavior is extremely useful Machines called mass spectrometers exploit this behavior for different purposes As the book points out physicists use mass spectrometers to determine the relative abundances of isotopes as well as their masses Chemists use these to help them identify unknown molecules and anesthesiologists use them to gather information on the gases in a patient s lungs 17 Example 13 An electron moves at a speed of 60106 ms perpendicular to a constant magnetic eld The path is a circle of radius 1310 3 m a Draw a sketch showing the magnetic field and electron s path b What is the magnitude of the field c Find the magnitude of the electron s acceleration Example 14 Chapter 21 11 A charged particle enters a uniform magnetic eld and follows the circular path shown in the drawing a Is the particle positively or negatively charged Why b The particle s speed is 140 ms the magnitude of the magnetic eld is 048 T and the radius of the given path is 960 m Determine the mass of the particle given that its charge has a magnitude of 82x10 4C B out of paper Example 15 Chapter 21 17 Two isotopes of carbon carbon12 and carbon13 have masses of 1993x10 27kg and 21 59x10 27kg respectively These two isotopes are singly ionized e and each is given a speed of 6667x105ms The ions then enter the bending region of a mass spectrometer where the magnetic field is 08500 T Determine the spatial separation between the two isotopes after they have traveled through a halfcircle 20 Example 16 Chapter 21 13 A beam of protons moves in a circle of radius 023 m The protons move perpendicular to a 025T magnetic field a What is the speed of each proton b Determine the magnitude of the centripetal force that acts on each proton 21 Example 17 Chapter 21 12 In the operating room anesthesiologists use mass spectrometers to monitor the respiratory gases of patients undergoing surgery One gas that is often monitored is the anesthetic iso urane molecular mass 306 X 103925 kg In a spectrometer a single ionized molecule of iso urane charge e moves at a speed of 741 X 103 ms on a circular path that has a radius of 0161 m What is the magnitude of the magnetic field that the spectrometer uses 22 Example 18 Chapter 21 23 A particle of mass 68 X 10398 kg and charge 72 C is traveling due east It enters perpendicularly a magnetic field Whose magnitude is 30 T After entering the field the particle completes onehalf of a circle and eXits the field traveling due west How much time does the particle spend in the magnetic field 23 71 The Impulse Momentum Theorem Where are we going in chapter 7 The goal is to be able to analyze and predict the outcome of simple onedimensional collisions To reach this goal we will concentrate on the concept of linear momentum Like energy linear momentum of an object is conserved under the right circumstances It is the conservation of linear momentum that will allows us to study collisions between objects and predict the outcome of such collisions Chapter 7 1 The linear momentum of an object is defined as p mv where f is the object s linear momentum In is the object s mass and i7 is the object s velocity The units for linear momentum has are N s Notice that linear momentum is a vector How the heck does this relate to What we ve been doing Actually it s at the heart of forces Newton s Second Law of Motion the uncensored version is This is the most general way to state the second law of motion Chapter 7 2 Using just a bit of general differential calculus reveals the censored fonn of the second law you ve been using 313 mi7 using the product rule for derivatives E i7dm m d dt dt d d usually 171 is constant over time m 0 so that a d A 2F 0mv which in turn gives the uncensored version of Newton s Second Law of Motion we ve been using E md This is nice but it s restrictive The term 17m is called thrust and is used for topics like rocket motion Now forgetting the calculus here s the algebra version a A i7 2 At Chapter 7 3 By rearranging the terms we arrive at the Impulse Momentum Theorem a 2FAt My E AtEf i 213Atmi7f mi7i Which in words simply states that an impulse the net force times the contact time results in a change of momentum Chapter 7 Example 1 Chapter 7 50 When jumping straight down you can be seriously injured if you land stifflegged One way to avoid injury is to bend your knees upon landing to reduce the force of impact A 75 kg man has a speed of 64 ms just before landing on the ground a In a stiff legged landing he comes to a halt in 20 ms Find the magnitude of the average net force that acts on him during this time b When he bends his knees he comes to a halt in 010 seconds Find the magnitude of the average net force now Chapter 7 5 Example 2 Two men pushing a stalled car generate a net force of 680 N for 72 seconds What is the nal linear momentum of the car But who cares about the car s linear momentum Me not so much Most often you will be asked to find final velocities If this car has a mass of 1500 kg What would be its final velocity Chapter 7 Example 3 Chapter 7 2 A 62kg person standing on a diving board dives straight down into the water Just before striking the water her speed is 550 ms At a time of 165 s after she enters the water her speed is reduced to 110 ms What is the average net force that acts on her when she is in the water Chapter 7 7 Example 4 Chapter 7 4 A baseball m 149 g approaches a bat horizontally at a speed of 402 ms 90 mih and is hit straight back at a speed of 456 ms 102 mih If the ball is in contact with the bat for a time of 110 ms What is the average force exerted on the ball by the bat Neglect the weight of the ball since it is so much less than the force of the bat Choose the direction of the incoming ball as the positive direction Chapter 7 Example 5 Chapter 7 11 A 0500kg ball is dropped from rest at a point 120 m above the oor The ball rebounds straight upward to a height of 0700 m What is the impulse of the net force applied to the ball during the collision with the oor Chapter 7 72 The Principle of Conservation of Linear Momentum When is linear momentum conserved To find out we have to start with the uncensored version of Newton s Second Law Al 3 2 At In a way similar to chapter 6 and the net work done on an object recall Wm was separated into two types of work Wm and WC we will split up the forces into two types internal and external forces a a EFIILI 217616 E Internal forces that objects within the system exert on each other External forces by agents external to the system on the objects Chapter 7 10 From Newton s Third Law of Motion the action reaction pairs are the internal forces and because they are always equal in magnitude but opposite in direction to each other their sum cancels out a a EFIILI 217616 E a Aa 0 3F T This leaves us only concerned with the external forces acting on the system 212m Pl If the vector sum of the external forces is also zero then we arrive at the conservation of linear momentum Al At 0Af7 13ipf Chapter 7 l l What is this system that was mentioned earlier Is this magic Not really but it can be confusing The system consists of the objects Whose motion you re studying Figure 79 on page 188 illustrates the concept of a system and how it can change to suit our needs For our purposes the system Will consist of the objects involved in a collision Chapter 7 12 Problem solving tips If you decide to use momentum to solve a problem I believe it is best to start with 213m Identify if there are any external forces If there are none you then can use conservation of linear momentum R pf If there are external forces but they sum to zero as in the pool table picture on the previous page you can still use the conservation of linear momentum Note This is very similar to solving problems using energy If you decide to use energy to solve a problem I believe one of the best ways to start is with Wm E 1 El h Identify if there are any nonconservative forces doing work If there are none you then use conservation of mechanical energy E I ElWCh Chapter 7 13 Example 6 Chapter 7 22 The lead female character in the movie Diamonds are Forever is standing at the edge of an offshore oil rig As she fires a gun she is driven back over the edge and into the sea Suppose the mass of a bullet is 0010 kg and its velocity is 720 ms Her mass including the gun is 51 kg a What recoil velocity does she acquire in response to a single shot from a stationary position assuming that no external force keeps her in place b Under the same assumption What would her recoil velocity be she shoots a blank cartridge that ejects a mass of 5010 4 kg at a velocity of 720 ms Chapter 7 14 Example 7 Chapter 7 25 Kevin has a mass of 87 kg and is skating with inline skates He sees his 22 kg younger brother up ahead standing on the sidewalk with his back turned Coming up from behind he grabs his brother and rolls off at a speed of 24 ms Ignoring friction find Kevin s speed just before he grabbed his brother Chapter 7 15 Example 8 Chapter 7 51 A twostage rocket moves in space at a constant velocity of 4900 ms The two stages are then separated by a small explosive charge placed between them Immediately after the eXplosion the velocity of the 1200 kg upper stage is 5700 ms in the same direction as before the eXplosion What is the velocity of the 2400 kg lower stage after the explosion Chapter 7 l6 Example 9 Chapter 7 16 A 55kg swimmer is standing on a stationary ZlOkg oating raft The swimmer then runs off the raft horizontally with a velocity of 46 ms relative to the shore Find the recoil velocity that the raft would have if there were no friction and resistance due to the water Chapter 7 Example 10 Chapter 7 19 A fireworks rocket is moving at a speed of 450 ms The rocket suddenly breaks into two pieces of equal mass which y off with velocities V1 and V2 as shown in the drawing What is the magnitude of a V1 and b V2 4 Chapter 7 Example 11 Chapter 7 53 By accident a large plate is dropped and breaks into three pieces The pieces y apart parallel to the oor As the plate falls its momentum has only a vertical component and no component parallel to the oor After the collision the component of the total momentum parallel to the oor must remain zero since the next external force acting on the plate has no component parallel to the oor Using the data shown in the drawing find the masses of pieces 1 and 2 300 ms 2500 179 ms m1 1307 ms Chapter 7 l9 73 Collisions in One Dimension Inelastic Collision A collision in which the total kinetic energy after the collision is not the same as it was before the collision If the objects stick together the collision is said to be completely inelastic Elastic Collision A collision in which the total kinetic energy after the collision is the same as it was before the collision In a typical elastic collision problem you are given two masses and their initial velocities and you want to find the final velocities of the masses To do this we need to use conservation of linear momentum and conservation of kinetic energy pi pf KEI KEf The goal here is to find a way to easily solve any onedimensional elastic collision involving two objects This is neat because collision problems are very intense algebraically but we re going to make it simple This method will only work for one dimensional elastic collisions Chapter 7 20 Conservation of Linear Momentum 1211 31i132i131f132f m11711m2172i m11711fm2172f ml lz m1l71 f m2l72 f m2 721 NOTE no direction has been specified yet Momentum is a vector we ll have to put direction in sometime To keep everything general we ll put the directions in at the end Chapter 7 21 Chapter 7 Conservation of Kinetic Energy KEI KEf KE KE KE KE 11 21 1f 2f 2 1 2 1 2 1 2 imv imv imv imv 1 2 1 2 1139 2 2139 2 1f 2 2f mlv2 m2v2 m1v2 m v2 11 21 Lf 2f mlv2 m1v2 m2v2 m2v2 11 1f 2f 21 22 Now for the mathematical technique mlv v v v m2v v v v 1139 1f 1139 1f 2f 2139 2f 2139 This gives another equation you can use for onedimensional elastic collisions For any onedimensional elastic collision you can use these two equations with out penalty Because we have not speci ed any directions you must do so when using these equations Chapter 7 23 Example 12 ignore air resistance Two kids are having a snowball fight One snowball of mass 0500 kg travels with a velocity of 450 ms to the right Another snowball of mass 0375 kg travels to with a velocity of 625 ms to the left Assuming these snowballs stick together when they meet a completely inelastic collision what is the nal velocity Is kinetic energy conserved Verify your answer mathematically Let s say these kids are chucking super balls at each other so that the collisions are elastic What are the velocities of the super balls after the collision Chapter 7 24 Example 13 ignore air resistance Two kids are having a snowball fight One snowball of mass 0450 kg travels with a velocity of 400 ms to the right Another snowball of mass 0250 kg travels to with a velocity of 750 ms to the right Assuming these snowballs stick together when they meet a completely inelastic collision what is the final velocity Is kinetic energy conserved Verify your answer mathematically Let s say these kids are chucking super balls at each other so that the collisions are elastic What are the velocities of the super balls after the collision Chapter 7 25 Example 14 Chapter 7 26 A 1055 kg van stopped at a traffic light is hit directly in the rear by a 715 kg car traveling With a velocity of 225 ms Assume that the transmission of the van is in neutral the brakes are not being applied and the collision is elastic What is the final velocity of a the car and b the van Chapter 7 26 Ballistic Pendulum a Completely Inelastic Collision A ballistic pendulum is used to determine speeds of projectiles and consists of a penetrable material usually wood or Styrofoam connected by a string as shown below If a projectile of mass 500 g undergoes an inelastic collision with the block of wood mass of 275 g and the combination of the bullet and block undergo a change of height of 150 cm what was the initial speed of the projectile a b hf 0650 m Chapter 7 27 Example 15 Chapter 7 49 In a football game a receiver is standing still having just caught a pass Before he can move a tackler running with a velocity of 45 ms grabs him The tackler holds onto the receiver and the two move off together with a velocity of 26 ms The mass of the tackler is 115 kg Assuming that momentum is conserved find the mass of the receiver Chapter 7 28 5 K 39 quotr Cfl 3 Emurucc NOLJCY70391 Lynquot H 31 L39Wlpl w w Wm VeinW X gtLJ T IC 1 ft df 39ffcj39lf DEL351 a cYE 1A YR Lr439 7 TI I I I I c 3 7 x bu w I 5 r why j gt 74 Am I I I 397 I I 4 if Fk L quot Dm I 39 Cum wry dy 0 Lin 1 p I 1 gt VF VIA1 v 02 cf iv 1 1 L 3HFIMIIS 4 f 1 90L 5395 1151 URVC 55ng 7 v37 QG we rhrwww r gt1 t 1 39w39 quot 1ka 39 via 39 Viz 39 A quot fair w BM w 3977 IJU Ml 6 s 674 A I V 3 H f39vaJam 25565154 TLVS 6quot 3 90039 3quot 30quot Projectile Motion further analysis Let s now apply these ideas to projectile motion problems Chapter 3 Example 3 Chapter 3 21 The drawing shows two planes each dropping an empty fuel tank At the moment of release each plane has the same speed of 135 ms and each tank is at the same height of 200 km above the ground Although the speeds are the same the velocities are different at the instant of release because one plane is ying at an angle of 150 above the horizontal and the other is ying at an angle of 150 below the horizontal Find the a magnitude and b direction of the velocity with which the fuel tank hits the ground if it is from plane A Find the c magnitude and 1 direction of the velocity with which the fuel tank hits the ground if it is from plane B In each part give the direction as a positive angle with respect to the horizontal 7 Fueltank 15 h Plane A plane B Chapter 3 1 1 Example 4 Chapter 3 17 A hotair balloon is rising straight up with a speed of 30 ms A ballast bag is released from rest relative to the balloon when it is 95 m above the ground How much time elapses before the ballast bag hits the ground Chapter 3 12 Example 5 Chapter 3 29 A diver runs horizontally with a speed of 120 ms off a platform that is 100 111 above the water What is his speed just before striking the water Chapter 3 Example 6 Chapter 3 text example 3 The drawing shows an airplane moving horizontally with a constant velocity of 115 ms at an altitude of 1050 m The direction to the right has been chosen as the x direction and upward is the y direction The plane releases a quotcare package that falls to the ground along a curved trajectory Ignoring air resistance determine the time required for the package to hit the ground Chapter 3 14 Example 7 Chapter 3 13 Suppose now that this plane is traveling with a horizontal velocity of 230 ms If all other factors remain the same determine the time required for the package to hit the ground Chapter 3 15 Example 8 Chapter 3 16 A golfer imparts a speed of 303 ms to a ball and it travels the maximum possible distance before landing on the green The tee and the green are at the same elevation a How much time does the ball spend in the air b What is the longest quothole in onequot that the golfer can make if the ball does not roll when it hits the green Chapter 3 l6 Example 9 Chapter 3 33 An Olympic long jumper leaves the ground at an angle of 23 and travels through the air for a horizontal distance of 87 m before landing What is the takeoff speed of the jumper Chapter 3 Example 10 Chapter 3 63 A golf ball rolls off a horizontal cliff with an initial speed of 114 ms The ball falls a vertical distance of 155 m into a lake below a How much time does the ball spend in the air b What is the speed v of the ball just before it strikes the water Chapter 3 18 Example 11 Chapter 3 24 A majorleague pitcher can throw a ball in excess of 410 ms If a ball is thrown horizontally at this speed how much will it drop by the time it reaches a catcher who is 170 111 away from the point of release Chapter 3 l9 Example 12 Chapter 3 61 A dolphin leaps out of the water at an angle of 35 above the horizontal The horizontal component of the dolphin39s velocity is 77 ms Find the magnitude of the vertical component of the velocity Chapter 3 20 Example 13 Chapter 3 62 A bullet is fired from a ri e that is held 16 m above the ground in a horizontal position The initial speed of the bullet is 1100 ms Find a the time it takes for the bullet to strike the ground and b the horizontal distance traveled by the bullet Chapter 3 21 Example 14 Chapter 3 66 The highest barrier that a projectile can clear is 135 m when the projectile is launched at an angle of 150 above the horizontal What is the projectile39s launch speed Chapter 3 22 Example 15 Chapter 3 20 A car drives horizontally off the edge of a cliff that is 54 m high The police at the scene of the accident note that the point of impact is 135 m from the base of the cliff How fast was the car traveling when it drove off the cliff Chapter 3 23 Example 8 Four charges labeled q 11 q and 14 are placed on the comers of a square of length 150 meters as shown in the picture What is the force on charge q located in the middle of the square ifq 300 pC q 100 pC q 100 11C q 500 11C and q4 500 pC 6 DaleNIL CMYclf a gr lm Farms Le39lbc repel 43 UkhkC H g c39l L H E Look of stMyc lx r 39 51W 5 0 Y3 dlcLlt 0M 13 6 45 2 LL 42 Sf 2 llwltwal WW var x look e4quot L15C FAJI39S Y quotLv hdH 265 uveoueu cmu l at Gw wueu s dJJ TLA L Lecmk m ml 1 ma mum mgr rsth L r 7 imam 2 t a r J FMJ use 4 P f ixajcll39np Titanem lbquot o7 om r1 mn ownmquot Mg r 2 orisomy22 GI plug r H s ivJ Solve 410 The Tension Force Tension Tension in a rope or cable acts to pull the rope or cable apart Newton s third law of motion is useful to understand this force when it is used to pull objects along level or inclined surfaces A force is applied to a rope r ng Ideally the tension in the rope exerts a force on the block However using Newton s 3rd law there must be an equal force but in the opposite direction acting on the rope from the block Thus the tension tends to pull the rope apart When working with tension in problems for this class we will assume an ideal rope or cable Restriction An ideal rope or cable has no mass Chapter 4 23 The tension force can seem even stranger when combined with a pulley When working with pulleys in this class we will assume the use of an ideal pulley Restriction An ideal pulley has no mass and is frictionless 422 N W 5 185 N Chapter 4 24 Example 12 Chapter 4 50 The drawing shows a wire tooth brace used by orthodontists The topmost tooth is protruding slightly and the tension in the wire exerts two forces T and T on this tooth in order to bring it back into alignment If the forces have the same magnitude of 21 0 N what is the magnitude of the net force exerted on the tooth by these forces Chapter 4 25 Example 13 Chapter 4 56 During a storm a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts The limb exerts a downward force of 151 N on the wire The left section of the wire makes an angle of 140 relative to the horizontal and sustains a tension of 447 N Find the a magnitude and b direction as an angle relative to horizontal of the tension that the right section of the wire sustains Chapter 4 26 Example 14 Chapter 4 73 A cable is lifting a construction worker and a crate as the drawing shows The weights of the worker and the crate are 965 N and 1510 N respectively The acceleration of the cable is 0620 msz upward What is the tension in the cable a below the worker and b above the worker a 0620 ms2 Chapter 4 27 71 The Impulse Momentum Theorem Where are we going in chapter 7 The goal is to be able to analyze and predict the outcome of simple onedimensional collisions To reach this goal we will concentrate on the concept of linear momentum Like energy linear momentum of an object is conserved under the right circumstances It is the conservation of linear momentum that will allows us to study collisions between objects and predict the outcome of such collisions Chapter 7 1 The linear momentum of an object is defined as p mv where f is the object s linear momentum In is the object s mass and i7 is the object s velocity The units for linear momentum has are N s Notice that linear momentum is a vector How the heck does this relate to What we ve been doing Actually it s at the heart of forces Newton s Second Law of Motion the uncensored version is This is the most general way to state the second law of motion Chapter 7 2 Using just a bit of general differential calculus reveals the censored fonn of the second law you ve been using 313 mi7 using the product rule for derivatives E i7dm m d dt dt d d usually 171 is constant over time m 0 so that a d A 2F 0mv which in turn gives the uncensored version of Newton s Second Law of Motion we ve been using E md This is nice but it s restrictive The term 17m is called thrust and is used for topics like rocket motion Now forgetting the calculus here s the algebra version a A i7 2 At Chapter 7 3 By rearranging the terms we arrive at the Impulse Momentum Theorem a 2FAt My E AtEf i 213Atmi7f mi7i Which in words simply states that an impulse the net force times the contact time results in a change of momentum Chapter 7 Example 1 Chapter 7 50 When jumping straight down you can be seriously injured if you land stifflegged One way to avoid injury is to bend your knees upon landing to reduce the force of impact A 75 kg man has a speed of 64 ms just before landing on the ground a In a stiff legged landing he comes to a halt in 20 ms Find the magnitude of the average net force that acts on him during this time b When he bends his knees he comes to a halt in 010 seconds Find the magnitude of the average net force now Chapter 7 5 Example 2 Two men pushing a stalled car generate a net force of 680 N for 72 seconds What is the nal linear momentum of the car But who cares about the car s linear momentum Me not so much Most often you will be asked to find final velocities If this car has a mass of 1500 kg What would be its final velocity Chapter 7 Example 3 Chapter 7 2 A 62kg person standing on a diving board dives straight down into the water Just before striking the water her speed is 550 ms At a time of 165 s after she enters the water her speed is reduced to 110 ms What is the average net force that acts on her when she is in the water Chapter 7 7 Example 4 Chapter 7 4 A baseball m 149 g approaches a bat horizontally at a speed of 402 ms 90 mih and is hit straight back at a speed of 456 ms 102 mih If the ball is in contact with the bat for a time of 110 ms What is the average force exerted on the ball by the bat Neglect the weight of the ball since it is so much less than the force of the bat Choose the direction of the incoming ball as the positive direction Chapter 7 Example 5 Chapter 7 11 A 0500kg ball is dropped from rest at a point 120 m above the oor The ball rebounds straight upward to a height of 0700 m What is the impulse of the net force applied to the ball during the collision with the oor Chapter 7 CHAPTER 4 Forces amp Newton s Laws of Motion Force push or pull that affects an object s motion There are two types 1 contact forces frictional normal 2 action at a distance forces gravitational electric Inertia natural tendency of an object to remain at rest or in motion at a constant speed along a straight line The mass of an object is a quantitative measure of its inertia It s the resistance to the change in an object s motion caused by a net force Newton s 1St Law of Motion An object continues in a state of rest or in a state of motion at a constant speed along a straight line unless compelled to change that state by a net force Net force vector sum of all forces 217 SF gt EFX EFy Limitations Inertial reference frame Where the 1st law is valid a frame of reference that is not accelerating Newton s Second Law of Motion When a net external force 217 acts on an object of mass m the acceleration 5 that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass The direction of the acceleration is the same as the direction of the net force The unit of force is the Newton N kgigm s 215 m51 2F ma EFy ma X x y Limitations 1 Very small for small things such as atoms amp molecules need to use quantum mechanics 2 Very fast for speeds close to the speed of light need to use relativity Newton s Third Law Whenever one body exerts a force on a second body the second body exerts an oppositely directed force of equal magnitude on the first body AKA the action reaction law Suppose that the mass of the spacecraft is ms 11000 kg and that the mass of the astronaut is m A 92 kg In addition assume that the astronaut exerts a force of P 36 N on the spacecraft Find the accelerations of the spacecraft and the astronaut 46 Types of Forces An Overview Fundamental Forces 1 Gravitational Force 2 Strong Nuclear Force stability of nuclei 3 Electroweak Force charged particles amp radioactive decay of some nuclei Beam 39 7m ml 39 whim Newton s Second Law 44 The Vector Nature of Newton s Second Law of Motion E m EFx mazC EFy ma y FreeBodv Diagram A sketch representing an object and all of the forces acting on that object They are used with Newton s second law of motion to help determine the acceleration of the object Make sure to include a coordinate system y 560 N Opposing force 560 N a b Freebody diagram of the car Example 1 Two forces act on an object mass 300 kg 131 400 N 300 and 132 600 N 900 Find the magnitude and direction relative to the XaXis of the acceleration of the object Make sure you include a freebody diagram in your solution Example 2 Chapter 4 4 During a circus performance a 72kg human cannonball is shot out of a 18mlong cannon If the human cannonball spends 095 s in the cannon determine the average net force exerted on him in the barrel of the cannon Chapter 4 8 47 Gravitational Force Mass amp Weight Gmm A F 12quot g r2 2 G6673x10 11N7quot21 kg Chapter 4 Example 3 Determine the magnitude of the acceleration due gravity on a 500 kg object on the surface of the earth The earth s radius is 638x106 in the earth s mass is 598x1024 kg and G667x1011Nm2kg2 Chapter 4 Example 4 Chapter 4 23 A satellites are placed in a circular orbit that is 359 X 107 m above the surface of the earth What is the magnitude of the acceleration due to gravity at this distance Chapter 4 l l Example 5 Chapter 4 21ish A planet has an equatorial radius of 600 X 107 m and a mass of 567 X 1026 kg Compute the acceleration of gravity at the equator of this planet Chapter 4 Weight One common force in every day use is weight To find the weight of an object you take the object s mass times the acceleration due to the gravitational force If you are far from the surface of a planet such as the earth you need to find this acceleration just like the previous examples ma I I w gravttatlomzlforce If you are on earth which is common we make use of the fact that the acceleration due to gravity has been measured accurately to simplify the calculation 15W mg Fw mg Close to the earth s surface g 981 ms2 Weight always acts downward or more correctly toward the center of the planet Mass The mass of an object is a measure of an object s inertia This does not change due to a change in gravitational fields ie going from the earth to the moon Chapter 4 l3 Example 6 Chapter 4 27 Mars has a mass of 646x1023 kg and a radius of 339x106 m a What is the acceleration due to gravity on Mars b How much would a 65kg person weigh on this planet magnitude only Chapter 4 14 Example 7 Chapter 4 19 A bowling ball mass 72 kg radius 011 m and a billiard ball mass 038 kg radius 0028 m may each be treated as uniform spheres What is the magnitude of the maximum gravitational force that each can exert on the other Chapter 4 Example 8 Chapter 4 28 Three uniform spheres are located at the corners of an equilateral triangle Each side of the triangle has a length of 120 m Two of the spheres have a mass of 280 kg each The third sphere mass unknown is released from rest Considering only the gravitational forces that the spheres exert on each other What is the magnitude of the initial acceleration of the third sphere Chapter 4 l6 Ch 48 The Normal Force and Apparent Weight The Normal Force Denoted JEN the normal force is the component of the force a surface exerts on an object with which it is in contact The direction of the normal force is always perpendicular to the surface Chapter 4 However the normal force is not always equal to the weight of the object Chapter 4 18 Example 9 use 100 ms2 A A 500 kg mass rests on a table What is the normal force that the table exerts on the mass B A 150 kg mass is placed on top of the previously mentioned mass What is the normal force that the table exerts on the 500 kg mass C A cable is attached to the 500 kg mass which exerts a 500 N force upwards away from the surface What is the normal force that the table exerts on the 500 kg mass Chapter 4 l9 Example 10 A 500 kg block rests on a wooden ramp with an angle of 300quot with respect to the horizontal What is the magnitude of the normal force on the block Chapter 4 20 Apparent weight Scales when functioning correctly give weights of objects When you use a scale you re measuring the normal force the scale is exerting on the object There are situations when this is not the case however Chapter 4 21 Example 11 A 900 kg person stands on a scale in an elevator What is the apparent weight when a the elevator is accelerating upward at 175 msz b the elevator is moving upward at a constant speed of 150 ms c the elevator is accelerating downward at 130 msz Chapter 4 22 410 The Tension Force Tension Tension in a rope or cable acts to pull the rope or cable apart Newton s third law of motion is useful to understand this force when it is used to pull objects along level or inclined surfaces A force is applied to a rope r ng Ideally the tension in the rope exerts a force on the block However using Newton s 3rd law there must be an equal force but in the opposite direction acting on the rope from the block Thus the tension tends to pull the rope apart When working with tension in problems for this class we will assume an ideal rope or cable Restriction An ideal rope or cable has no mass Chapter 4 23 The tension force can seem even stranger when combined with a pulley When working with pulleys in this class we will assume the use of an ideal pulley Restriction An ideal pulley has no mass and is frictionless 422 N W 5 185 N Chapter 4 24 Example 12 Chapter 4 50 The drawing shows a wire tooth brace used by orthodontists The topmost tooth is protruding slightly and the tension in the wire exerts two forces T and T on this tooth in order to bring it back into alignment If the forces have the same magnitude of 21 0 N what is the magnitude of the net force exerted on the tooth by these forces Chapter 4 25 Example 13 Chapter 4 56 During a storm a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts The limb exerts a downward force of 151 N on the wire The left section of the wire makes an angle of 140 relative to the horizontal and sustains a tension of 447 N Find the a magnitude and b direction as an angle relative to horizontal of the tension that the right section of the wire sustains Chapter 4 26 Example 14 Chapter 4 73 A cable is lifting a construction worker and a crate as the drawing shows The weights of the worker and the crate are 965 N and 1510 N respectively The acceleration of the cable is 0620 msz upward What is the tension in the cable a below the worker and b above the worker a 0620 ms2 Chapter 4 27 49 Static and Kinetic Frictional Forces Static Friction static means no movement Static Friction has two main characteristics restriction assumption macroscopic can see objects Characteristic 1 Static friction is independent of the area of contact between the surface and object restriction assumption hard and nondeformable surfaces Characteristic 2 The magnitude of the static frictional force is proportional to the normal force The constant of proportionality is called the coefficient of static friction denoted by MS and is a unitless quantity The coefficient of static friction is not surprisingly material dependent among other things temperature material s condition rough polished The magnitude of the static frictional force can be zero to a maximum amount of fsmax depending on the applied force The maximum value of this force is just before the object starts to move This is stated mathematically by the very important and often used condition fsmax MSFN As the book mentions this only relates the magnitudes of the normal and static forces and in this case the lack of arrows indicates magnitude Forces are vectors and these vectors do not act in the same direction However this relationship ties components together and goes back to the vector nature of EF mil Chapter 4 28 Kinetic Friction kinetic means motion is present Kinetic Friction has three main characteristics restriction assumption macroscopic objects Characteristic 1 Kinetic friction is independent of the area of contact between the surface and object restriction assumption hard and nondeformable surfaces Characteristic 2 Kinetic friction is independent of the speed of sliding motion restriction assumption the speed is small Characteristic 3 The magnitude of the kinetic frictional force is proportional to the normal force The constant of proportionality is called the coefficient of kinetic friction denoted by Mk and is a unitless quantity The magnitude of the kinetic frictional force is given by f k lukF N The same warning applies this relates the magnitudes of the two forces Both static and kinetic friction act in a direction to oppose motion whether that motion is occurring or could potentially occur Chapter 4 29 Example 15 Chapter 4 37 A block Whose weight is 450 N rests on a horizontal table A horizontal force of 360 N is applied to the block The coefficients of static and kinetic friction are 0650 and 0420 respectively Will the block move under the in uence of the force and if so What will be the block s acceleration Chapter 4 30 Example 16 Chapter 4 41 A 81kg baseball player slides into second base The coefficient of kinetic friction between the player and the ground is pk 049 a What is the magnitude of the frictional force b If the player comes to rest after 16 s What is his initial speed Chapter 4 31 Example 17 Chapter 4 44 An ice skater is gliding horizontally across the ice with an initial velocity of 63 ms The coefficient of kinetic friction between the ice and the skate blades is 0081 and air resistance is negligible How much time elapses before her velocity is reduced to 28 ms Chapter 4 32 Example 18 Chapter 4 51 A stuntInan is being pulled along a rough road at a constant velocity by a cable attached to a moving truck The cable is parallel to the ground The mass of the stuntInan is 109 kg and the coefficient of kinetic friction between the road and him is 0870 Find the tension in the cable Chapter 4 33 411 Equilibrium Applications of Newton s Laws of Motion Equilibrium a fancy way of saying an object is not accelerating Implications Well g ma Chapter 4 34 When is an object experiencing no acceleration You know about one and a half of the situations Not moving not accelerating Moving with constant speed not accelerating that s only about half right Acceleration is a change in velocity both are vectors amp vectors have a magnitude and direction You can get an acceleration by changing l the velocity s magnitude speeding up or slowing down 2 the velocity s direction 3 both the velocity s magnitude and direction Technically there is no acceleration when 1 an object does not move at rest 2 an object moves at a constant velocity without changing direction Chapter 4 3 5 Example 19 Chapter 4 47 A supertanker mass 170x108 kg is moving with a constant velocity Its engines generate a forward thrust of 740x105 N Determine a the magnitude of the resistive force exerted on the tanker by the water and b the magnitude of the upward buoyant force exerted on the tanker by the water Chapter 4 36 Example 20 Chapter 4 51 A stuntInan is being pulled along a rough road at a constant velocity by a cable attached to a moving truck The cable is parallel to the ground The mass of the stuntInan is 109 kg and the coefficient of kinetic friction between the road and him is 0870 Find the tension in the cable Chapter 4 37 Example 21 Chapter 4 49 Three forces act on a moving object One force has a magnitude of 800 N and is directed due north Another has a magnitude of 600 N and is directed due west What must the third force be such that the object continues to move with a constant velocity Chapter 4 3 8 Example 22 Chapter 4 59 A toboggan slides down a hill and has a constant velocity The angle of the hill is 800 with respect to the horizontal What is the coefficient of kinetic friction between the surface of the hill and the toboggan Chapter 4 39 Example 23 Chapter 18 23 A small spherical insulator ofmass 649 x 1039Z kg and charge 0600 C is hung by athin Wire ofnegligible mass A charge of 0900 C is held 0150 m away from the sphere and directly to the right of it so the Wire makes an angle With the vertical see the dmwing Find a the angle and b the tension in the Wire UlECrr J 0600 C 0 900 m Chapta4 412 Nonequilibrium Applications of Newton s Laws of Motion Nonequilibrium a fancy way of saying the net force on an object is not zero Implications Well g ma EFX max EFy may Chapter 4 41 Example 24 Chapter 4 69 A student is skateboarding down a ramp that is 600 meters long and inclined at 180 with respect to the horizontal The initial speed of the skateboarder at the top of the ramp is 260 ms Neglect friction and find the speed at the bottom of the ramp Chapter 4 42 Example 25 Chapter 4 76 A 205 kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp The ramp is inclined at 300 with respect to the horizontal The coefficient of kinetic friction between the log and the ramp is 0900 and the log has an acceleration of 0800 msz Find the tension in the rope Chapter 4 43 Example 26 Chapter 4 74 A crate is resting on a ramp that is inclined at an angle 9 above the horizontal As 9 is increased the crate remains in place until 9 reaches a value of 380 Then the crate begins to slide down the slope a Determine the coefficient of static friction between the crate and ramp surface b The coefficient of kinetic friction between the crate and ramp surface is 0600 Find the acceleration of the moving crate Chapter 4 44 Example 27 A girl is sledding down a slope that is inclined at 300 with respect to the horizontal A moderate Wind is aiding the motion by providing a steady force of 105 N that is parallel to the motion of the sled The combined mass of the girl and sled is 650 kg and the coefficient of kinetic friction between the runners of the sled and the snow is 0150 How much time is required for the sled to travel down a 175 m slope starting from rest Chapter 4 45 Example 28 Chapter 4 99 Two objects 450 and 210 kg are connected by a massless string that passes over a massless frictionless pulley The pulley hangs from the ceiling Find a the acceleration of the objects and b the tension in the string Chapter 4 46 Example 29 A rescue helicopter is lifting a man weight 822 N from a capsized boat by means of a cable and harness a What is the tension in the cable when the man is given an initial upward acceleration of 110 ms2 b What is the tension during the remainder of the rescue when he is being pulled upward at a constant velocity Chapter 4 47 Example 30 Chapter 4 79 A box is sliding up an incline that makes an angle of 150 with respect to the horizontal The coef cient of kinetic friction between the box and the surface of the incline is 0180 The initial speed of the box at the bottom of the incline is 150 ms How far does the box travel along the incline before coming to rest Chapter 4 48 The assumption used to derive the kinematics equations in section 24 and 25 is 1 constant acceleration 2 constant velocity 3 constant speed 4 none of the above Chapter 2 15 Chapter 2 Kinematics Equations the real ones X f Xcomponent 2 Axv tla t2 me 2 x x v tia t2 x me me x ycomponent 2 v tiat y y yhy Ly 2 y 1 2 Ay vigyt j ayt 17i7Zz t Xcomponent vf x vi x axt ycomponent V vat fay Ly Y From the MeriamWebster Dictionary Kinematics a science that deals with motion apart from considerations of mass and force Equation Number Equation 24 Vf viat 1 27 x vivft 1 2 28 xf xivil al 29 v vi22an Chapter 2 Using the Kinematics Equations How to Choose Equation Variables Number Equation X a vf t Vi 24 Vf vl 01 I I I I 27 x vvf2 I I I I 28 xfxiviz az 2 I I I I 29 v vi220Ax I I I I Other useful stuff Quadratic equations AX2BXC0 B IBZ 4AC 2A X Chapter 2 18 Apply the solution for quadratic equations to x xi vilalz AX2BXCO math Chapter 2 xf f 1 xi viz jazz phy51cs Example 8 Chapter 2 13 A motorcycle has a constant acceleration of 25 ms2 Both the velocity and acceleration of the motorcycle point in the same direction How much time is required for the motorcycle to change its speed from a 21 to 31 ms and b 51 to 61 ms Chapter 2 20 Example 9 Chapter 2 22 a What is the magnitude of the average acceleration of a skier who starting from rest reaches a speed of 80 ms when going down a slope for 50 s b How far does the skier travel in this time Answer Chapter 2 21 Example 10 Chapter 2 67 A jetliner traveling northward is landing with a speed of 690 ms Once the jet touches down it has 750 m of runway in which to reduce its speed to 610 ms Compute the average acceleration magnitude and direction of the plane during landing Answer Chapter 2 22 Example 11 Chapter 2 53 A cement block accidentally falls from rest from the ledge of a 530mhigh building When the block is 140 m above the ground a man 20 m tall looks up and notices that the block is directly above him How much time at most does the man have to get out of the way Answer Chapter 2 23 Example 12 Chapter 2 29 Suppose a car is traveling at 200 ms and the driver sees a traffic light turn red After 0530 s has elapsed the reaction time the driver applies the brakes and the car decelerates at 700 msz What is the stopping distance of the car as measured from the point Where the driver first notices the red light Chapter 2 24 Example 13 Chapter 2 20 A cart is driven by a large propeller or fan which can accelerate or decelerate the cart The cart starts out at the position x 0 m with an initial velocity of 50 ms and a constant acceleration due to the fan The direction to the right is positive The cart reaches a maximum position of x 125 m where it begins to travel in the negative direction Find the acceleration of the cart Chapter 2 25 Example 14 Chapter 2 23 The left ventricle of the heart accelerates blood from rest to a velocity of 26 cms a If the displacement of the blood during the acceleration is 20 cm determine its acceleration in cms2 b How much time does blood take to reach its nal velocity Chapter 2 26 Chopkr 8 Emmi 6 5 milch 5 461nm 51 10059 gzzs52wa a f Qf quX39U39WWCXZWELC3 5 em151 xmm gmfj 3454 9 156 2 of SOM339 F quota 0 07 F a 07711 0 1 L Lth Now for the good stuff Example 15 Chapter 2 24 A cheetah is hunting Its prey runs for 30 s at a constant velocity of 90 ms Starting from rest What constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time Chapter 2 27 Example 16 Chapter 2 17 A car is traveling along a straight road at a velocity of 360 ms when its engine cuts out For the next twelve seconds the car slows down and its average acceleration is 51 For the next siX seconds the car slows down further and its average acceleration is 52 The velocity of the car at the end of the eighteensecond period is 280 ms The ratio of the average acceleration values is 5152 150 Find the velocity of the car at the end of the initial twelvesecond interval Chapter 2 28 Example 17 Chapter 2 28 A race driver has made a pit stop to refuel After refueling he leaves the pit area with an acceleration Whose magnitude is 60 ms2 and after 40 s he enters the main speedway At the same instant another race car that is on the speedway and traveling at a constant speed of 700 ms overtakes and passes the entering car If the entering car maintains its acceleration how much time is required for it to catch the other car Chapter 2 29 Example 18 Chapter 2 31 A car is traveling at a constant speed of 33 ms on a highway At the instant this car passes an entrance ramp a second car enters the highway from the ramp The second car starts from rest and has a constant acceleration What acceleration must it maintain so that the two cars meet for the first time at the next eXit which is 25 km away Chapter 2 30 Example 19 Chapter 2 72 A drag racer starting from rest speeds up for 402 m with an acceleration of 170 msz A parachute then opens slowing the car down with an acceleration of 610 msz How fast is the racer moving 350 m after the parachute opens Chapter 2 31 Equation Variables Number Equation X a vf t Vi 24 Vf Vi 5 1 27 x vivft 1 2 28 xfxivit at 29 v vi2Zan Equation Variables Number Equation X a vf t Vi 24 Vf Vi 5 1 27 x vivft 2 28 xfxivit at 29 v vi2Zan Equation Variables Number Equation X a vf t Vi 24 vfviat 1 27 x vivft 1 2 28 xfxivit at 29 v vi220Ax Chapter 2 32 Section 26 Free Fall In section 26 two of the main assumptions are I ignore air resistance J velocity is constant U small drop distance 4 1amp2 52amp3 C 1amp3 712amp3 00 there are none Chapter 2 33 If you drop an object in the absence of air resistance it accelerates downward at 98 msz If instead you throw it downward its downward acceleration after release is 1 less than 98 ms2 2 98 ms2 3 more than 98 ms2 Chapter 2 34 Example 20 Chapter 2 38 From the top of a cliff a person uses a slingshot to fire a pebble straight downward which is in the negative direction The initial speed of the pebble is 90 ms a What is the acceleration magnitude and direction of the pebble during the downward motion b After 050 s how far beneath the cliff top is the pebble Chapter 2 35 Example 21 Chapter 2 45 From her bedroom Window a girl drops a waterfilled balloon to the ground 60 m below If the balloon is released from rest how long is it in the air Chapter 2 36 Objects being thrown up but not in the I m sick sense Problem solving help This goes for all problems but may be very helpful for objects being acted on by gravity only Give a rough sketch with a labeled axis This will help you assign the correct directions and signs to variables Include given information Put these on the object s trajectory in your sketch Label anything else that will be assumed given from rest special points those sorts of things Chapter 2 3 7 AssumptionsRestrictions thus far 0 Acceleration is constant This must be true to use any kinematics equations 0 For an object in free fall ignore air resistance 0 For an object in free fall the distance of the fall will be small compared to the earth s radius Chapter 2 3 8 65 The Conservation of Mechanical Energy The combination of kinetic and gravitational potential energy is called total mechanical energy or mechanical energy for short The WorkEnergy Theorem from section 64 can be written in terms of mechanical energy Wm AKE APE Wm KEf KEI PEf PEI Wm KEf PEf KEi PEI W KE PE HKE PE no f f l l mesh mesh Wnc El39 and if there is no work done by nonconservative forces then Emech Emech f 139 Which is called the conservation of mechanical energy Physically this equation means the following In the absence of nonconservative forces the mechanical energy of an object KE PE is the same for every point on its path Chapter 6 30 0 25 50 75 100 PB KE l l l I 0 25 50 75 100 PB KE 0 25 50 75100 33m PE KE 1 0 25 50 75100 PE HEW I 0 25 50 75100 Chapter 6 31 Example 18 Chapter 6 69 A polevaulter approaches the takeoff point at a speed of 900 ms Assuming that only this speed determines the height to which he can rise nd the maximum height at which the vaulter can clear the bar Chapter 6 32 Example 19 A runaway truck is moving at 80 mihr 358 ms when it encounters an emergency escape ramp as shown below What is the minimum length the ramp has to be to stop the truck if friction is ignored Why do real escape ramps have a thick covering of sand or gravel Chapter 6 3 3 Example 20 Chapter 6 40 A particle starting from point A in the drawing is projected down the curved runway Upon leaving the runway at point B the particle is traveling straight upward and reaches a height of 400 rn above the oor before falling back down Ignoring friction and air resistance nd the speed of the particle at point A 1 vo 400 m 300 m Chapter 6 3 4 Example 21 In the picture below a 500 kg block starting from rest slides along a frictionless surface Ifthe radius ofthe loop is 100 m and h 5R a What is the speed of the block at the bottom of the ramp b What is the speed of the block at point Q c What is the speed of the block at the top of the loop d ifthe surface is not frictionless and friction does 50 J of work by the time the block reaches the top of the loop What is the maximum height the loop can be Chapter 6 35 Example 22 Chapter 6 37 A 470g golf ball is driven from the tee with an initial speed of 520 ms and rises to a height of 246 m a Neglect air resistance and determine the kinetic energy of the ball at its highest point b What is its speed when it is 80 m below its highest point Chapter 6 36 72 The Principle of Conservation of Linear Momentum When is linear momentum conserved To find out we have to start with the uncensored version of Newton s Second Law Al 3 2 At In a way similar to chapter 6 and the net work done on an object recall Wm was separated into two types of work Wm and WC we will split up the forces into two types internal and external forces a a EFIILI 217616 E Internal forces that objects within the system exert on each other External forces by agents external to the system on the objects Chapter 7 10 From Newton s Third Law of Motion the action reaction pairs are the internal forces and because they are always equal in magnitude but opposite in direction to each other their sum cancels out a a EFIILI 217616 E a Aa 0 3F T This leaves us only concerned with the external forces acting on the system 212m Pl If the vector sum of the external forces is also zero then we arrive at the conservation of linear momentum Al At 0Af7 13ipf Chapter 7 l l What is this system that was mentioned earlier Is this magic Not really but it can be confusing The system consists of the objects Whose motion you re studying Figure 79 on page 188 illustrates the concept of a system and how it can change to suit our needs For our purposes the system Will consist of the objects involved in a collision Chapter 7 12 Problem solving tips If you decide to use momentum to solve a problem I believe it is best to start with 213m Identify if there are any external forces If there are none you then can use conservation of linear momentum R pf If there are external forces but they sum to zero as in the pool table picture on the previous page you can still use the conservation of linear momentum Note This is very similar to solving problems using energy If you decide to use energy to solve a problem I believe one of the best ways to start is with Wm E 1 El h Identify if there are any nonconservative forces doing work If there are none you then use conservation of mechanical energy E I ElWCh Chapter 7 13 Example 6 Chapter 7 22 The lead female character in the movie Diamonds are Forever is standing at the edge of an offshore oil rig As she fires a gun she is driven back over the edge and into the sea Suppose the mass of a bullet is 0010 kg and its velocity is 720 ms Her mass including the gun is 51 kg a What recoil velocity does she acquire in response to a single shot from a stationary position assuming that no external force keeps her in place b Under the same assumption What would her recoil velocity be she shoots a blank cartridge that ejects a mass of 5010 4 kg at a velocity of 720 ms Chapter 7 14 Example 7 Chapter 7 25 Kevin has a mass of 87 kg and is skating with inline skates He sees his 22 kg younger brother up ahead standing on the sidewalk with his back turned Coming up from behind he grabs his brother and rolls off at a speed of 24 ms Ignoring friction find Kevin s speed just before he grabbed his brother Chapter 7 15 Example 8 Chapter 7 51 A twostage rocket moves in space at a constant velocity of 4900 ms The two stages are then separated by a small explosive charge placed between them Immediately after the eXplosion the velocity of the 1200 kg upper stage is 5700 ms in the same direction as before the eXplosion What is the velocity of the 2400 kg lower stage after the explosion Chapter 7 l6 Example 9 Chapter 7 16 A 55kg swimmer is standing on a stationary ZlOkg oating raft The swimmer then runs off the raft horizontally with a velocity of 46 ms relative to the shore Find the recoil velocity that the raft would have if there were no friction and resistance due to the water Chapter 7 Example 10 Chapter 7 19 A fireworks rocket is moving at a speed of 450 ms The rocket suddenly breaks into two pieces of equal mass which y off with velocities V1 and V2 as shown in the drawing What is the magnitude of a V1 and b V2 4 Chapter 7 Example 11 Chapter 7 53 By accident a large plate is dropped and breaks into three pieces The pieces y apart parallel to the oor As the plate falls its momentum has only a vertical component and no component parallel to the oor After the collision the component of the total momentum parallel to the oor must remain zero since the next external force acting on the plate has no component parallel to the oor Using the data shown in the drawing find the masses of pieces 1 and 2 300 ms 2500 179 ms m1 1307 ms Chapter 7 l9

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.