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# Physics 101- 1-D Kinetics and Introduction AP Physics 101

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This 52 page Class Notes was uploaded by Kevin Kudah Vitalis on Monday August 15, 2016. The Class Notes belongs to AP Physics 101 at University of Pennsylvania taught by Dr. Kristen Pacada in Fall 2016. Since its upload, it has received 19 views. For similar materials see Phys 101 in PHYSICS (PHY) at University of Pennsylvania.

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Date Created: 08/15/16

Earth and moon from Mercury This week (8/28): •Read Ch. 1 on your own - especially section 1-8. •We’ll start Chapter 2 in class today (1D kinematics) •Begin the procedure in the document on homework - start a daily habit of working problems from Chapter 2 (all) and 3 (3-1 through 3-4)! •First in-class quiz will be Wednesday next week, 9/2, and will cover vectors (Giancoli 3-1 -> 3-4) (our only quiz without a recitation ﬁrst). •Please get in the habit of emailing me through the Canvas site - this is the fastest, easiest way for me to stay on top of your emails. Context-rich Problems: You and your friends have a 5 m length of sturdy bungee cord from moving day. You’re curious whether you could use it to bungee jump inside the atrium of the hospital, which is 20 m high - the cord is strong and it seems like you have plenty of room to spare. A quick recollection and test using Hooke’s law from class shows you that the spring constant for the cord is about 100 N/m, and you weigh 50 kg. Assuming you can get the rigging right, is this certain death? You have just had a brilliant idea for new startup business. You are going to deliver pizza directly to customers while they are traveling in their cars. You will accomplish this using remotely controlled ﬂying drones that will drop the pizza delivery directly into the car while passing overhead. (Only customers with a sunroof or convertible are eligible for your services.) The drone will ﬂy on a level trajectory parallel to, and directly above, the trajectory of the customer's car. You will have to release the pizza at the correct time in order for the pizza to land directly in the car. Unfortunately, the FAA has not yet decided on the rules governing how high and how fast privately owned drones may ﬂy. This means you do not yet know how fast or how high your drone will be ﬂying. Furthermore, you will not know in advance how fast each customer will be traveling. This means that you must do all your calculations without numeric values for these parameters. In terms of these parameters, how long before passing over the customer's car should your drone release the delivery? How far will the drone be from the car at this point in time? r b n m 5 f N S r ( v T T P o M T T o A g d d t k d d d e e R s s o r s s e n n b p p p d p p i c c c S l 4 l o k o o d s s . 4 E io i a r i i n i s S U r s , @ , t t bo o R L t o o y t o sn s r P lp .n . c n . a t T e r i u s r o O . 3 n m . rv en N T T o e M A m A A a a a ] t o d e e y y o a s s l l c c tc s n o t t H p i t a i pn a a A S o 2 m ip a a a r r P C e nn h a d h r ( s O S e r r d a e r . s A . r d i i b t , v H r e e m v r e . . . o . h 1 h a M h h l p p p s S p e o i i S c c i p i i n lf l A E t t lhm l l a t t P I n n a e a p t n n L FI fp n s 1 s dn a f . f . C PC a 0 m n , n s c r t T Y e nn a 0 a r o b e O S e r d 8 r r s m r N C T A P M A y M M S e p s o l e a a a a o o a i o o M e i e h e a d d MT a s a a a r r r . P TE cp c i i p c h o h O C- r o t t / cd e a e C A t i n n o s y e l D L p i o o ns a i r U e s r r a y p E h h a . h h o o S e s s o e r i i L b r f a u s g g O m n nt e n o s s R I a dn r f f l c c . G A s c u ar u r s . s R L o h o o o i l. s S c t n n n s , v I ll n s s s r e r N . s n n n . t . . . ) Core of this course: ! ! ∑ F = ma Kinematics: Mathematical description of motion Instantaneous velocity Acceleration Derivation of the “kinematic equations” Some deﬁnitions: Position - where something is on a ruler/tape measure/yardstick Symbol: x n Displacement - how far something moved on the tape measure x − x Math statement: f i Symbol: Δx Δx x x i f Model for kinematics: drop a random object: Displacement of ball: xi 1m xf− x i Δx 0m−1m = −1m Time for ball to drop (Δt) : tf−t i Δt 0.45s− 0s = Δ0.45s x 0m f x 1m i If I drop a heavier weight a distance of 1 m: 1. The heavier weight takes more time to fall the same distance. 2. The heavier weight takes less time to fall the same distance. 3. The heavier weight takes the same amount of time to fall the same distance. x f 0m Average velocity: displacement divided by the time it took v = Δx Δt xf− x i v = tf−t i v = vf+v i 2 (deﬁnition of “average” - applies for case of constant acceleration) Average velocity - notice units are meters per second (m/s): 1m x i v = x f x i t −t f i x f 0 What was the average velocity of the object while it fell? x − x v = f i v = 0m−1m = −2.22m /s tf−t i 0.45s−0s Is this how fast it was going when it started? How fast was it going when it hit the ﬂoor? Are these different? Can we ﬁgure out how different? Can we ﬁgure out the ﬁnal, instantaneous velocity of the ball knowing only the average velocity? x − x Instantaneous velocity: v = lim = f i t→0 t −t f i Was theﬁnal instantaneous velocity the same as the initial instantaneous velocity? Can we ﬁgure out the ﬁnal, instantaneous velocity of the ball knowing only the average velocity? x − x v +v v = f i = v = f i hint: what was the initial t −t 2 instantaneous velocity? f i xf− x i v fv i = tf−t i 2 2 ( f x i) vf+v =i t −t v ind t aie both zero f i 2Δx vf= We know Δx and Δt so we Δt can ﬁnd v! f Can we ﬁgure out the ﬁnal, instantaneous velocity of the ball knowing only the average velocity? 2Δx v f We know Δx and Δt so we can ﬁnd v ! f Δt 2×−1m vf= = −4.44m /s 0.45s The ball was going 4.44 m/s toward the ground when it hit the ground! If the velocity changed, it accelerated - what is acceleration? Δv a = deﬁnition of “acceleration” Δt v −v vf−v i a = f i a = limt→0 = t −t tf−t i f i instantaneous acceleration - but is always constant in this course... What are the units of acceleration? Δx Δv Δt Δx 1 Δx m a = = = × = 2 s2 Δt Δt Δt Δt Δt Do we have a way to ﬁgure out the acceleration of the ball? a = Δv deﬁnition of “acceleration” Δt v fv i v fv i a = a = lim Δt→0 = tf−t i t ft i We also know displacement and time... Do we have a way to ﬁgure out the acceleration of the ball? Δx Δv v = a = Δt Δt xf− x =ivΔt vf= v +iΔt v +v x − x = f i× Δt f i 2 vi+aΔt +v i 2v iaΔt x f x =i × Δt x f x =i × Δt 2 2 2 x f x +viΔt +i/2aΔt Now we can ﬁnd acceleration! Find acceleration (“g”) with new relationship: 2 Δx = v it +1/2aΔt Δx =1/2aΔt 2 2Δx 2⋅−1m a = = = −9.8m /s 2 Δt 2 (0.45s) 2 We just re-derived acceleration due to gravity on earth by the simple expedient of dropping a random object! Neat! Another useful relationship (no time info needed): v f v +ait 2a(x − x )= v −v 2 2 v −v f i f i Δt = f i 2 2 a vf= v +iaΔx x f x = iΔt x = x +vΔt f i ⎛ v +v ⎞⎛ v −v ⎞ x = x + f i f i f i ⎝ 2 ⎠⎝ a ⎠ v −v 2 x − x = f i f i 2a Kinematic equations - memorize by next week! Δx v = x f x +iΔt Δt v +v v = f i 2 Δv v = v +aΔt a = f i Δt 2 2 2 x f x +i Δt i1/2aΔt vf= v +iaΔx Free-fall kinematic problems Δx v = x f x +vit Δt v +v v = f i 2 Δv v = v +aΔt a = f i Δt 2 2 2 x f x +viΔt +i/2gΔt v f v +igΔx objects in free-fall: a = g = -9.8 m/s 2 when up is the positive-x direction. This week: •1D kinematics (Ch. 2) and vectors (Ch. 3.1 - 3.4) •Recitation problem on Mon/Tue will involve 1D kinematics •Quiz on Wednesday will be vector addition •Quiz will be one problem graded according to the problem solving rubric (set-up and organization count for points!) Vectors are a mathematical object that describes an amount (magnitude) with an associated direction: Θ (angle) Vectors: Scalars (objects Displacement with only Velocity magnitude): Acceleration Mass Force Energy Momentum Work Vectors add with other vectors, and multiply with scalars vector +vector = vector vector × scalar = vector vector + scalar “Tip to tail” method of vector addition: Vector addition Vector subtraction ! c b b! a a! ! c ! ! ! ! ! ! a +b = c a −b = c Vector addition ! ! c ! a = magnitude of a - only the length b ! a The vector that results from a vector sum is the “resultant” Any arbitrary number of vectors can be added to ﬁnd a given “resultant” We * could* work this problem with a ruler and protractor, but that would be tedious - need a better way... What if we just expressed all vectors as the sum of their x- and y-axis components? x-axis y-axis What if we just made all vectors the sum of their x- and y-axis components? Now, what is the magnitude of the x-component of the red vector? The magnitude of the y-component? x-axis y-axis ! ! ax= a cosΘ ! ! a y a sinΘ ! a a x-axis θ ! y ax y-axis ! ! ! ! ax= a cosΘ bx= b cosΦ ! ! ! ! a y a sinΘ by= b sinΦ ! ! ! b b a ! y x-axis ay bx θ a x y-axis ! ! ! ! a x a cosΘ ! x b c!sΦ ! ! ! a +b = c ! ay= a sinΘ b y b sinΦ ! ! ! ! c = a +b c c b b !x !x !x y a ! y c y a +y y x-axis ay bx θ a x! cx ! ! ! cx= a cosΘ+ b cosΦ ! ! ! cy= a sinΘ+ b sinΦ y-axis ! ! ! a +b = c ! ! ! c b ! cy ! Φ! by γ a ! bx x-axis θ ! ay ax ! ! ! ! cx= a cosΘ+ b cosΦ x ! 2 ! 2 ! 2 ! ! ! c = c + c cy= a sinΘ+ b sinΦ x y ! ! 2 ! 2 c = c x + c y y-axis −1⎛cy⎞ γ = tan ⎜ ⎟ ⎝cx⎠ That method is universal, even for things on the same axis, in different directions ! c = a cosΘ+ b cosΦ x ! ! ! Θ = 0º,Φ =1!0º a +b = c c = a − b !x ! ! c y a sinΘ+ b sinΦ x-axis b c a! y-axis vector!subtraction: a −b = c ! ! c = a cosΘ− b cosΦ ! x ! ! cy= a sinΘ− b sinΦ ⎛ c ⎞ ! γ = tan−⎜ y⎟ b ⎝ cx⎠ a Φ x-axis γθ a! ! x cyis negative, so γ is c also negative - directed below x-axis. y-axis ! Find the description of vector C ! ! a!+b = c a = 5 ! ! c b = 7 θ=40º ! Φ=-80º a Φ x-axis θ γ ! c = a cosΘ+ b cosΦ = 5cos40º+7cos(−80º)= 5.04 b x ! cy= a sinΘ+ b sinΦ = 5sin40+ 7sin(−80)= −3.68 ! ! 2 ! 2 2 2 c = cx + cy = 5.04 +−3.68 = 6.24 γ = tan−1 cy⎞ = tan1⎛−3.68 ⎞ = −36º y-axis ⎝ cx⎠ ⎝ 5.04 ⎠ r b n m 5 f N S r ( v T T P o M T T o A g d d t k d d d e e R s s o r s s e n n b p p p d p p i c c c S l 4 l o k o o d s s . 4 E io i a r i i n i s S U r s , @ , t t bo o R L t o o y t o sn s r P lp .n . c n . a t T e r i u s r o O . 3 n m . rv en N T T o e M A m A A a a a ] t o d e e y y o a s s l l c c tc s n o t t H p i t a i pn a a A S o 2 m ip a a a r r P C e nn h a d h r ( s O S e r r d a e r . s A . r d i i b t , v H r e e m v r e . . . o . h 1 h a M h h l p p p s S p e o i i S c c i p i i n lf l A E t t lhm l l a t t P I n n a e a p t n n L FI fp n s 1 s dn a f . f . C PC a 0 m n , n s c r t T Y e nn a 0 a r o b e O S e r d 8 r r s m r N C T A P M A y M M S e p s o l e a a a a o o a i o o M e i e h e a d d MT a s a a a r r r . P TE cp c i i p c h o h O C- r o t t / cd e a e C A t i n n o s y e l D L p i o o ns a i r U e s r r a y p E h h a . h h o o S e s s o e r i i L b r f a u s g g O m n nt e n o s s R I a dn r f f l c c . G A s c u ar u r s . s R L o h o o o i l. s S c t n n n s , v I ll n s s s r e r N . s n n n . t . . . ) ! ! Find the vector sum of R and G ! ! ! ! ! ! R+G = B Bx= R +xG x x ! ! ! By= R +yG y α 30º R = 3acos(−90º)= 0a R = 3 asin(−90º)= − 3 a a ! a x 2 y 2 2 ! ! 1 ! 3 R G G x acos(−60º)= a G y asin(−60º)= − 2 a 2 60º 60º ! ! ! 1 1 ! ! ! 3 3 a Bx= R x G =x0a+ a =2 2 a By= R y G =y− a+− a = − 3a ! 2 2 B ! B = B 2+ B 2= 1a + 3a = a 1+12 = a 13 ! x y 4 4 2 G By − 3a tanα = = = −2 3 Bx 1a 2 −1 α = tan (−2 3)≈ −74º Can we ﬁnd this bridge? The rock falls, hits the water, and then the sound takes some time to travel back to microphone. How tall is this bridge? The speed of sound in air is 340 m/s. The rock falls, hits the water, and then the sound takes some time to travel back to microphone. How tall is this bridge? The speed of sound in air is 340 m/s. sound a=-9.8 m/s xt? m vi0 m/s vs=340 m/s 1 +2 =T T=6 s t1=? t2=? xb=0 m rock The rock falls, hits the water, and then the sound takes some time to travel back to microphone. How tall is this bridge? The speed of sound in air vs=340 m/s. sound xt? m a=-9.8 m/s For constant velocity (sound): i=0 m/s vs⋅t 2 x − t b s =340 m/s 2 For free-fall (rocvi 11/2gt = x 1 x b t t1+2 =T T=6 t1+t =2T 1 =? t2=? t2= T −t 1 Set displacement for sound and rock equal: 2 −v st =2vt +1/i 1 1 xb=0 m 2 rock −v Ts(t = vt 1)/2gt i 1 1 1 gt +vt −v t +v T = 0 1 i 1 s 1 s 2 1 2 gt 1v t +vs 1= 0 s Can use quadratic equation to ﬁnd t ! 1 2 form is (ax +bx+c=0) The rock falls, hits the water, and then the sound takes some time to travel back to microphone. How tall is this bridge? The speed of sound in air s =340 m/s. xt? m T=6 1/2gt −v t +v T = 0 a=-9.8 m/s 1 s 1 s vi0 m/s (ax +bx+c=0) vs=340 m/s t1=? t2=? −b ± b − 4ac +v ± v − 4(1/2g)v T t = = s s s 1 2a 2⋅1/2⋅g t1= 5.5s xb=0 m 2 2 2 Δx =1/2at =1/2⋅−9.8m /s ⋅ 5.5s ( ) = −148m ⎛ 3.3ft ⎞ 148m⋅ ⎜ ⎟ = 488 ft ⎝ 1m ⎠ The bridge is ~490 feet from the water... r b n m 5 f N S r ( v T T P o M T T o A g d d t k d d d e e R s s o r s s e n n b p p p d p p i c c c S l 4 l o k o o d s s . 4 E io i a r i i n i s S U r s , @ , t t bo o R L t o o y t o sn s r P lp .n . c n . a t T e r i u s r o O . 3 n m . rv en N T T o e M A m A A a a a ] t o d e e y y o a s s l l c c tc s n o t t H p i t a i pn a a A S o 2 m ip a a a r r P C e nn h a d h r ( s O S e r r d a e r . s A . r d i i b t , v H r e e m v r e . . . o . h 1 h a M h h l p p p s S p e o i i S c c i p i i n lf l A E t t lhm l l a t t P I n n a e a p t n n L FI fp n s 1 s dn a f . f . C PC a 0 m n , n s c r t T Y e nn a 0 a r o b e O S e r d 8 r r s m r N C T A P M A y M M S e p s o l e a a a a o o a i o o M e i e h e a d d MT a s a a a r r r . P TE cp c i i p c h o h O C- r o t t / cd e a e C A t i n n o s y e l D L p i o o ns a i r U e s r r a y p E h h a . h h o o S e s s o e r i i L b r f a u s g g O m n nt e n o s s R I a dn r f f l c c . G A s c u ar u r s . s R L o h o o o i l. s S c t n n n s , v I ll n s s s r e r N . s n n n . t . . . ) The rock falls, hits the water, and then the sound takes some time to travel back to microphone. How tall is this bridge? The speed of sound in air v =34s m/s. xt? m T=6 For constant velocity (sound): v st =2x − x i f For free-fall (rock):t +1/2at = x − x a=-9.8 m/s i 1 1 f i vi0 m/s v =340 m/s 1 +t 2T s t = T −t 2 1 t =? t =? 2 1 2 −v st =2vt +i 1at 1 −v T(−t = v) +1/2at 2 s 1 i 1 1 2 1/2at +v1 −v ti 1 T =s 1 s 2 xb=0 m 1/2at −1 t +vs 1= 0 s Can use quadratic equation! (ax +bx+c=0) Grading rubric: −b ± b − 4ac −v s v − s(1/2a)v T s 1 = = Useful description 2a 2⋅1/2⋅a t = 5.5s Physics approach 1 Speciﬁc application of physics 2 2 2 Δx =1/2at =1/2⋅−9.8m / s ⋅ 5.5s ( )= −148m Mathematical procedures Logical progression ⎛ 3.3ft ⎞ 148m⋅ ⎝ ⎠ = 488 ft 1m A man hops a boxcar of a train moving at a constant velocity of 5 m/s. He starts moving when he is exactly even with the car door and accelerates at 1.4 m/s until he is sprinting at his top speed of 6 m/s. He sprints at this speed until he catches up with the door and hops in. •How long does this take? •How far did he run before he jumped in? v=0 m/s v =6 m/s Δx = v Δt +1/2aΔt t1 t2 v =5 m/s 2 2 Δx Δx a=1.4 m/s v f v i2aΔx 1 2 t1=?2 t =?B Δx =? vf= vi+aΔt Δx B 1 2 Δx B Δx +1Δx 2 v B(+1 2) = at1+v t f 2 1 2 Δx = at 2 ⎛ v ⎞ 1 ⎛v ⎞2 1 2 1 v f +t = a f +v t Δx = v t B ⎝ a 2 ⎠ 2 ⎝ a ⎠ f 2 Solve for2t ! 2 f 2 2 Δx B v t f( 1 2) 1 1 ⎛ m ⎞ m m v −v v ⋅⎜6 ⎟ −5 ⋅6 v f at 1 2 f B f 2 ⎝ s ⎠ s s v t2= = m m m = 8.6 s f = t a(v Bv f) 1.4 ⎜ 5 −6 ⎟ a 1 s2⎝ s s ⎠ A man hops a boxcar of a train moving at a constant velocity of 5 m/s. He starts moving when he is exactly even with the car door and accelerates at 1.4 m/s until he is sprinting at his top speed of 6 m/s. He sprints at this speed until he catches up with the door and hops in. •How long does this take? •How far did he run before he jumped in? vi0 m/s 2 t1 t2 vf=6 m/s Δx = viΔt +1/2aΔt vB=5 m/s vf= vi+2aΔx Δx 1 Δx 2 a=1.4 m/s2 t1=?,2t =?,BΔx =? vf= vi+aΔt Δx B 1 ⎛ m ⎞2 m m 1 v −v v ⋅⎜6 ⎟ −5 ⋅6 2 f B f 2 ⎝ s ⎠ s s t2= = m m m = 8.6 s t +t = 8.6 s + 4.3 s =12.9 s a(vB−v f) 1.4 2 ⎜5 −6 ⎟ 1 2 s ⎝ s s ⎠ m Δx = v t (t ) = 5m ⋅12.9 s = 64.4 m 6 B B 1 2 s t = vf = s = 4.3s 1 a m 1.4 2 s

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