Physics 111. Thursday, September 9, Ch 3: Kinematic Equations Structured Approach to Problem Solving. Ch 4: Projectile Motion EXAMPLES!

Transcription

1 ics Thursday, ember 9, 2004 Ch 3: Kinematic Equations Structured Approach to Problem Solving Ch 4: Projectile Motion EXAMPLES!

2 Announcements Help sessions meet Sunday, 6:30-8 pm in CCLIR 468 Wednesday, 8-9 pm in NSC 118/119

3 Announcements You must bring both your Lab Manual and Lab Notebook to Lab or you will not be allowed to begin this week. There will be another short quiz at the beginning of lab.

4 (uniform acceleration ONLY) v = v + a(δt) f i a = constant x = x + v (Δt) + 1 f i i 2 a(δt) 2 where Δt t f t i These equations are easy to derive using calculus.

5 (uniform acceleration ONLY) a = constant How do we interpret these relationships graphically? x f = x i + v i (Δt) + 1 a(δt) 2 v f = v i + a(δt) 2 velocity clock reading The area under the curve is the distance traveled by the object over the time interval.

6 (uniform acceleration ONLY) a = constant How do we interpret these relationships graphically? t f v f x f = x i + v i (Δt) Area of rectangle = v i t f a(δt) 2 velocity v i clock reading Δv=at f v f = v i + at f Δv = v f - v i = at f Area of triangle = at f2 /2

7 What if the acceleration is NOT constant? velocity The area under the curve is still the distance traveled by the object over the time interval. Δt clock reading Δx = v(t) Δt Split up the interval of interest into a bunch of little time intervals in which the velocity is approximately constant.

8 What if the acceleration is NOT constant? velocity The area under the curve is still the distance traveled by the object over the time interval. Δt clock reading d = x f x i = lim Δt 0 Δx = v(t) Δt

9 Let s set up a procedure for problem solving that is thoughtful and organized. Our procedure needs to be versatile enough to help us work through a wide range of problems of varied difficulty. Structured Approach to Problem

10 This approach involves 3 steps, two of which we already have covered: 1) ical Representation 2) Pictorial Representation (new!) 3) Mathematical Representation

11 1) ical Representation Motion Diagrams We ll learn other ical Representations in the near future

12 3) Mathematical Representation Definitions of position, displacement, velocity, and acceleration Kinematic equations More to come!

13 2) Pictorial Representation (New!) Draw pictures of the physical situation at all the key times in the problem Define the coordinate system Define the variables List knowns and unknowns

14 The approach is best demonstrated with a problem. Let s try this one together: George lives in Chicago and want to meet up with his friend Annabel for lunch, but Annabel lives in Pittsburg, 400 miles East of Chicago. If George leaves home at 9 am traveling 60 mph East and Annabel leaves at the same time traveling 40 mph West, where will they meet?

15 George lives in Chicago and want to meet up with his friend Annabel for lunch, but Annabel lives in Pittsburg, 400 miles East of Chicago. If George leaves home at 9 am traveling 60 mph East and Annabel leaves at the same time traveling 40 mph West, where will they meet? 1) ical Representation - Motion Diagram v George Chicago a George = 0 a Annabel = 0 v Annabel Pittsburg

16 George lives in Chicago and want to meet up with his friend Annabel for lunch, but Annabel lives in Pittsburg, 400 miles East of Chicago. If George leaves home at 9 am traveling 60 mph East and Annabel leaves at the same time traveling 40 mph West, where will they meet? 2) Pictorial Representation Unknown: t f, x f v G x = 0 x G,i t i a G = 0 v A Known: t i = 0, x A,f = x G,f = x f, x G,i = 0, x A,i = +400 miles v G = +60 mph, v A = - 40 mph, a G = 0, a A = 0 x A,f x G,f t f v G a A = 0 va x A,i t i +x

17 George lives in Chicago and want to meet up with his friend Annabel for lunch, but Annabel lives in Pittsburg, 400 miles East of Chicago. If George leaves home at 9 am traveling 60 mph East and Annabel leaves at the same time traveling 40 mph West, where will they meet? 3) Mathematical Representation Kinematic equation for George: Kinematic equation for Annabel: x G, f = x G,i + v G,i (Δt) a G (Δt)2 = 0 + (60mph)(t f t i ) + 0 = (60mph)t f x A, f = x A,i + v A,i (Δt) a A (Δt)2 = 400miles + ( 40mph)(t f t i ) + 0 = 400miles (40mph)t f

18 George lives in Chicago and want to meet up with his friend Annabel for lunch, but Annabel lives in Pittsburg, 400 miles East of Chicago. If George leaves home at 9 am traveling 60 mph East and Annabel leaves at the same time traveling 40 mph West, where will they meet? 3) Mathematical Representation The condition for meeting at time t f is that x G,f = x A,f, so we need to set the two kinematic equations from the last slide equal to one another. x A, f = 400miles (40mph)t f = (60mph)t f = x G, f (100mph)t f = 400miles t f = 4hours

19 George lives in Chicago and want to meet up with his friend Annabel for lunch, but Annabel lives in Pittsburg, 400 miles East of Chicago. If George leaves home at 9 am traveling 60 mph East and Annabel leaves at the same time traveling 40 mph West, where will they meet? 3) Mathematical Representation So, they ll meet up at 1 pm, but the question wants to know where. To find out, we can use either of the kinematic equations, plugging in our answer for t f. x A, f = 400miles (40mph)(4hours) = 240miles

20 George lives in Chicago and want to meet up with his friend Annabel for lunch, but Annabel lives in Pittsburg, 400 miles East of Chicago. If George leaves home at 9 am traveling 60 mph East and Annabel leaves at the same time traveling 40 mph West, where will they meet? 4) Sanity Check It s useful to ask yourself if the answer makes sense. We know that George was traveling with a higher speed. We also know that they both left at the same time. That means that when they meet up, they ll both have been traveling for the same amount of time. So, George must have gone further than Annabel. Indeed, our answer of 240 miles East of Chicago is farther from Chicago than Pittsburg. Good deal!

21 Becky & Tom are skiing on a dangerous, double black diamond slope in Vail. Today, the slope is completely ice, so there s no friction. The angle of the slope is 10 0 above horizontal. Tom starts from rest. If Becky starts down the slope 5 s after Tom, what s the minimum initial speed Becky must have in order to catch Tom before they reach the bottom of the slope, 500 m away? 1) ical Representation - Motion Diagram t = 5s t = 0 vtom atom v Becky Problem Solving Worksheet a Becky

22 Becky & Tom are skiing on a dangerous, double black diamond slope in Vail. Today, the slope is completely ice, so there s no friction. The angle of the slope is 10 0 above horizontal. Tom starts from rest. If Becky starts down the slope 5 s after Tom, what s the minimum initial speed Becky must have in order to catch Tom before they reach the bottom of the slope, 500 m away? Unknown: t f, x f, v T,f, v B,f, v B,i Known: t T,i = 0, t B,i = 5s θ = 10 0, x T,i = x B,i = 0 x T,f = x B,f = x f = 500m, v T, i = 0 a T = a B = 1.70 m/s 2 = +(9.81 m/s 2 ) sin (10 0 ) x = 0 x B,i, x B,i t B,i, t T,i v T,i = 0 vb,i 2) Pictorial Rep. a B = a T +x θ x T,f = x B,f = 500 m t f v T, f v B, f

23 Becky & Tom are skiing on a dangerous, double black diamond slope in Vail. Today, the slope is completely ice, so there s no friction. The angle of the slope is 10 0 above horizontal. Tom starts from rest. If Becky starts down the slope 5 s after Tom, what s the minimum initial speed Becky must have in order to catch Tom before they reach the bottom of the slope, 500 m away? 3) Mathematical Representation Kinematic equation for Tom: x T, f = x T,i + v T,i (Δt) a T (Δt)2 500m = 1 2 a T (t f t T,i ) 2 = 1 2 (1.70 m s 2 )(t f 0) 2 = (0.852 m s 2 )t f 2 t f = 24.3s

24 Becky & Tom are skiing on a dangerous, double black diamond slope in Vail. Today, the slope is completely ice, so there s no friction. The angle of the slope is 10 0 above horizontal. Tom starts from rest. If Becky starts down the slope 5 s after Tom, what s the minimum initial speed Becky must have in order to catch Tom before they reach the bottom of the slope, 500 m away? 3) Mathematical Representation Kinematic equation for Becky: x B, f = x B,i + v B,i (Δt) a B(Δt) 2 = v B,i (t f t B,i ) a T (t f t B,i ) 2 500m = v B,i (24.3s 5.0s) (1.70 m s 2 )(24.3s 5.0s) 2 v B,i = 9.50 m s

25 Becky & Tom are skiing on a dangerous, double black diamond slope in Vail. Today, the slope is completely ice, so there s no friction. The angle of the slope is 10 0 above horizontal. Tom starts from rest. If Becky starts down the slope 5 s after Tom, what s the minimum initial speed Becky must have in order to catch Tom before they reach the bottom of the slope, 500 m away? 4) Sanity Check 9.50 m/s translates to about 20 miles/hour. That doesn t seem unreasonable. Olympic downhill champs can reach speeds > 70 mph. Becky & Tom - St 4 - Sanity

26 Two-Dimensional Motion This chapter looks at motion in two dimensions. These problems are simply 2, one-dimensional problems. We ll look at a special case of 2-D motion: projectile motion.

27 Ch 4: Two-Dimensional Motion Class Worksheet #1 In this demo, which object should strike the floor first? (Assume no air resistance.) 1) Falling block 2) Flying block 3) Neither. They ll hit at the same time.

28 Ch 4: Two-Dimensional Motion Motion Diagrams v v a a Both blocks are falling under the influence of Earth s gravity, so they both have the same acceleration. Note that the x- components of velocity do not change while the y-components match in this case. Projectile motion demo (W#1)

29 Ch 4: Two-Dimensional Motion The block demo is an example of an object undergoing: Class of problems in which an acceleration is present in only one direction of motion The acceleration is often due to the Earth s gravity in experiments conducted at or near the surface of the Earth.

30 Ch 4: Two-Dimensional Motion An artillery shell is fired with an initial velocity of 300 m/s at 55.0 o above the horizontal. It explodes on a mountainside 42.0 s after firing. What are the x and y coordinates of the shell when it explodes? (0,0) = location of the launching point of the shell. 1) ical Representation - Motion Diagram v a Problem Solving Worksheet

31 Ch 4: Two-Dimensional Motion An artillery shell is fired with an initial velocity of 300 m/s at 55.0 o above the horizontal. It explodes on a mountainside 42.0 s after firing. What are the x and y coordinates of the shell when it explodes? (0,0) = location of the launching point of the shell. 2) Pictorial Representation Known: t i = 0, x i = 0, y i = 0, v i = 300 m s θ i = v x = v i cosθ = m s v y,i = v i sinθ = m s t f = 42.0 s, a x = 0, a y = -g +y v i θ i x i, y i, t i a Unknown: t f, x f, y f, v y,f x f, y f, t f v f +x

32 Ch 4: Two-Dimensional Motion An artillery shell is fired with an initial velocity of 300 m/s at 55.0 o above the horizontal. It explodes on a mountainside 42.0 s after firing. What are the x and y coordinates of the shell when it explodes? (0,0) = location of the launching point of the shell. 3) Mathematical Representation Let s do the x-coordinate first: x f = x i + v x (Δt) a x (Δt)2 = 0 + (172.1 m s )(42.0s 0) = 7.23km

33 Ch 4: Two-Dimensional Motion An artillery shell is fired with an initial velocity of 300 m/s at 55.0 o above the horizontal. It explodes on a mountainside 42.0 s after firing. What are the x and y coordinates of the shell when it explodes? (0,0) = location of the launching point of the shell. 3) Mathematical Representation Now for the y-coordinate: y f = y i + v y,i (Δt) a y (Δt)2 = 0 + (245.7 m s )(42.0s 0) ( 9.81m s 2 )(42.0s 0) 2 = 1.67km