buku physics11

e x a m p l e 5 Maximum height of a projectileFor the cannonball in Example 4, find its maximum height.Solution and Connection to TheoryGivenWe need the y direction only because the maximum height involves thedisplacement in the y direction only.y} v1y ϭ 20 m/s v2y ϭ 0 m/s ay ϭ Ϫ9.8 m/s2 ⌬dy ϭ ? ⌬t ϭ ?The final velocity is zero because at the maximum height, the cannonballstops moving up.The formula of choice for this case is v→22 ϭ v→12 ϩ 2a→⌬d→Then ⌬dy ϭ ᎏv22y Ϫᎏv12y ϭ ᎏ(0 m2/(sϪ)ᎏ29Ϫ.8 (m2/0ᎏs2m) /s)2 ϭ ᎏϪϪ14090.6ᎏmm2//ss22 ϭ 20 m 2ayThe cannonball will rise 20 m above its original location.Fig.3.18 Finding Maximum HeightGiven x } ⌬dx ϭ vx⌬t Is maximum YES Use kinematics Find ⌬t m ethovalues y } ⌬dy ϭ v1y⌬t ϩ 1/2ay⌬t2 height equations Find dy s of Set v2y ϭ 0 do ces pr required? 89 NO Solve projectile problem as in Fig. 3.16Finding Final VelocityTo find the final velocity of the projectile as it hits the ground, we need tofind the values of the two vector components that comprise it. In the x direc-tion, the final velocity is the same as the initial velocity because there is nohorizontal acceleration. Then we calculate the final velocity in the y direc-tion from the given data. We get the resultant final velocity by adding thetwo components using the head-to-tail method. With a scale diagram, youcan measure the magnitude and direction of the final velocity. For a moreprecise value, use Pythagoras’ theorem and the inverse tangent of the ratioof the magnitudes of the components. chapt e r 3: Motion in Two Dimensions

A circus cannon fired Emanuel e x a m p l e 6 Finding the final velocityZacchini over three Ferris wheels thatwere 18 m high. Launch velocity was What is the final velocity of the Great Projecto, if he is shot out of a can-37 m/s [R53°U] (Fig. 3.19). He non with a horizontal velocity component of 19 m/s and a vertical com-travelled horizontally 69 m. A net ponent of 23 m/s? For his stunt, he lands in a net 2.0 m above the pointwas placed at that location. Many of launch and 70 m away (Fig. 3.20A).performers black out momentarily asthey accelerate in the barrel. They Fig. 3.20Amust regain consciousness beforelanding, or risk breaking their necks. v1y vx vx v2yFig.3.19 Human cannonball ⌬dy Range For image Fig.3.20B see student vx vx text. ␪ v2y ៬vT v2y Solution and Connection to Theory For image Given see student x} vx ϭ 19 m/s. Because the velocity is constant in the x direction, this text. value is also the final x velocity. y} ay ϭ Ϫ9.8 m/s2 ⌬dy ϭ 2.0 m v1y ϭ 23 m/s v2y ϭ ? The formula of choice is v→22 ϭ v→12 ϩ 2a→⌬d→ v2y ϭ Ϯ͙(ෆ23 mෆ/s)2 ϩෆ2(Ϫ9ෆ.8 m/sෆ2)2.0 m ϭ Ϯ22 m/sThe final vector can be calculated We obtain two answers (a positive and a negative) because Projecto is two metres above the point of launch at two different times. Just afterfrom the given values using the for- leaving the cannon (up two metres vertically), his vertical velocity is 22 m/s [up]. As he reaches the net, he has a vertical velocity of 22 m/s [down]. Here, we are interested in the final velocity only.mulas Because Projecto is moving downward, the velocity is negative.v22 ϭ v22x ϩ v22y and ␪ ϭ tanϪ1 ᎏv2ᎏy Therefore, v2y ϭ Ϫ22 m/s. Now we combine the x and y components of v2x the final velocity from Fig. 3.20A by connecting them head to tail.90 u n i t a : M ot i o n a n d Forc es

In Fig. 3.20B, the final velocity vector was obtained by adding the twocomponents. The resulting vector diagram is a right-angle triangle. UsingPythagoras’ theorem,vT ϭ ͙ෆ(19 mෆ/s)2 ϩෆ(Ϫ22ෆm/s)2 ϭ 29 m/sIf you used a scale diagram, then measure the angle. If not, use the tan-gent function, substituting the magnitudes of the components.␪ ϭ tanϪ1 ᎏ22 mᎏ/s ϭ 49° 19 m/sTo find the compass directions, we check the signs of the components,vx (ϩ) and v2y (Ϫ). The signs tell us that Projecto was moving to the rightand down. His final velocity is 29 m/s [R49°D].Fig.3.21 Finding Final Projectile Velocityx } v1x ϭ constant v2x ϭ v1x v2x vf ϭ ͙vළ2ළxළ2ළϩළළv2ළyළ2ළ rocesp of HT v2y v2y sOUG v2x THuttin it ally } ay, v1y, ⌬dy v2y ϭ͙ළv1ළyළ2ළϩළළ2ළaළyළ⌬ළdළy ␪ ϭ tanϪ1 gethFig.3.22 Complete Projectile Overview Givens g To p er xy ax ϭ 0 y ay ϭ Ϫ9.8 m/s2⌬dx, v1x ϭ v2x ⌬dy, v1y⌬d៬ ϭ v៬1⌬t ϩ 1/2 a៬⌬t2 Can YES Complete the Calculate v2y vf ϭ ͙vළ2ළx2ළϩළළvළ2ළyළ2ළ you initial problem v2y calculate ␪ ϭ tanϪ1 v2x time in x ? NOCalculate time in y chapt e r 3: Motion in Two Dimensions 91

pplying 1. a) Calculate the time an object takes to fall 100 m if it starts from rest. theCo b) Calculate the time it takes the object to reach the ground if the ncepa object has an initial velocity of 10 m/s upward. ts c) Calculate the time it takes the object to reach the ground if theFig.3.23A object has an initial velocity of 10 m/s downward. d) Assuming that the object had a horizontal velocity of 5.0 m/s,92 find the range for a), b), and c). 2. A soccer ball is kicked at 25 m/s at an angle of 25° to the ground. Find a) the time it takes the ball to reach the maximum height in its trajectory. b) the maximum height of the ball. c) the time it takes to land on the ground again. d) the range the ball travels. e) the final velocity of the ball (with angle), using logic and calculation. Golf Golf is all about projectile motion. The angle of the face of a golf club head relative to the centreline (an imaginary line perpendicular to the ground) is referred to as the club head loft. The number of the club iron or wood indicates the size of the club head loft. The higher the number, the larger the loft angle. The club head loft also indicates the angle of the initial velocity of the ball. For the same force exerted on a ball, the nine-iron club will cause the golf ball to fly higher and travel a shorter distance than the three-iron club. In principle, a loft angle of 0° transfers all of the energy of the swing directly to the ball, causing the ball to travel the farthest. However, because of gravity pulling the ball down, a ball hit off a standard one-inch tee would travel only Fig.3.23B a few metres before hitting the ground. Thus, the bigger the loft angle, the greater the trajec- Centreline tory and the smaller the range. Loft angle 3. If a wood with loft 18° is used to hit a ball 90° that is in the air for 10.9 s, calculate Iron a) the distance the ball travels if its initial Loft velocity is 18.5 m/s. angle b) the maximum height of the ball. c) Repeat a) and b) for a loft angle of 8°. 90° 4. A pitching wedge causes the ball to travel 31 m for the same initial velocity as in Question 3 (18.5 m/s). What is the loft angle of the club if the ball spends 3.66 s in the air? Wood unit a: Motion and Forces

3.4 Relative VelocitiesTarzan and Tarzana, brawny ape people, decide to have a race across a river.Both are excellent swimmers and have the same ability. Tarzan swims straightacross, landing on shore directly across from where he started. Tarzana jumpsin and points herself across the river, but ends up farther downstream whenshe reaches the other side. The two are pictured in Fig. 3.24. Fig.3.24 Family race Who finished first? Then there was Cheetah, faithful brawny ape pet, 93who chose to swim across the same river, but at a time when the floods wereraging and the river flowed at five times the velocity it did when Tarzan andTarzana swam it. Cheetah too is a good swimmer and has the same ability asTarzan and Tarzana. Did Cheetah finish the crossing in a shorter time? Thesequestions will be answered in this section.Relative MotionIn trying to answer the posed problem, we need to define what relativemotion is. Suppose you are travelling in a car and are being observed bya highway patrolman sitting by the side of the road, as shown in Fig.3.25(a). The patrolman will measure your velocity relative to his position.In this case, the patrolman is not moving. Let’s assume the velocity hemeasured is 140 km/h. Your velocity is said to be 140 km/h [E], meas-ured relative to the ground. Now suppose you pass the patrol car with a velocity of 140 km/h while itis moving with a velocity of 100 km/h in the same direction. This time, thevelocity measured is 40 km/h relative to the patrol car. While the patrol car hasa velocity of zero in its own reference frame, its 100 km/h velocity must bemeasured relative to the ground, as in Fig. 3.25(b). Notice that your velocityis the same, but the two measurements done by the patrolman are different.The bottom line is that a velocity value is measured relative to a referenceframe. In everyday use, we are accustomed to using the ground as a station-ary reference frame. This frame is implied in most problem situations. chapt e r 3: Motion in Two Dimensions

Fig.3.25 Relative velocity POLICE ICE (a) (b) Now we can apply this knowledge to the swimmer in the river. If Tarzana floats with the current, it will appear to her that she is not moving relative to the water. To actually measure her velocity, someone must stand on shore (relative to ground) and watch her go by. In fact, to measure the river’s veloc- ity, we also must measure it relative to the shore (relative to the ground). In order to keep these velocities distinct, we use a series of subscripts. v→og is the velocity of the person or object relative to the ground. v→mg is the velocity of the medium the person or object is in relative to the ground. v→om is the velocity of the person or object relative to the medium he/she/it is in. This velocity is not readily observed because it is affected by the medium’s velocity. It can be considered the person’s or object’s ability with- out influence (such as swimming in water with no current). You can come up with your own subscripts; choose symbols that you’ll find easy to remember. The relationship between the three quantities is v→og ϭ v→om ϩ v→mg Since the vectors v→om and v→mg are being added, they must be joined head to tail. The vector v→og is only ever attached to other vectors tail to tail or head to head.94 u n i t a : M ot i o n a n d Forc es

e x a m p l e 7 Relative motion in one dimension Fig.3.26In one dimension, the river problem has the three possible cases shownin Fig. 3.26. When the river flows, its velocity is 1.0 m/s [east]. Tarzanacan swim at 2.0 m/s (in still water).Case 1. Tarzana swims in a dammed up river (no current).Case 2. Tarzana swims with the current (dam broke).Case 3. Tarzana swims against the current.Solution and Connection to TheoryCase 1 GivenAssuming east is positive, the givens are v→om ϭ 2.0 m/s and v→mg ϭ 0 m/sThen v→og ϭ v→om ϩ v→mg.Since the vectors are collinear, v→og ϭ 2.0 m/s ϩ 0 m/s ϭ 2.0 m/s.Tarzana swims 2.0 m/s [E]. Without anything affecting her velocity, wesee her swim at 2.0 m/s [E], as illustrated in Fig. 3.27.Fig.3.27 ϩ0 ϭ៬vom ϭ 2 m/s [E] v៬mg ϭ 0 ៬vog ϭ 2 m/s [E] chapt e r 3: Motion in Two Dimensions 95

Case 2 Given v→mg ϭ 1.0 m/s [E] v→om ϭ 2.0 m/s [E] Since the vectors are collinear, you can add them algebraically. v→og ϭ v→om ϩ v→mg v→og ϭ 2.0 m/s ϩ 1.0 m/s ϭ 3.0 m/s The positive value indicates that the velocity direction is to the east. Notice in Fig. 3.28 that Tarzana is capable of swimming at 2.0 m/s, but is seen to be swimming at 3.0 m/s. This is because the current is helping her. What we see is just the resultant, v→og. Fig.3.28 ៬vom v៬mg v៬og ϩϭ v៬om ϭ 2 m/s [E] ៬vmg ϭ 1 m/s [E] ៬vog ϭ 3 m/s [E] Case 3 Given Since v→om ϭ 2.0 m/s [W], we write v→om ϭ Ϫ2.0 m/s because we have des- ignated east as positive. v→mg ϭ 1.0 m/s v→og ϭ v→om ϩ v→mg v→og ϭ Ϫ2.0 m/s ϩ 1.0 m/s ϭ Ϫ1.0 m/s Notice in Fig. 3.29 that even though Tarzana is capable of swimming at 2.0 m/s, she is seen to swim only at 1.0 m/s [W] because she is now fight- ing the current. Fig.3.29 ϩϭ ៬vmg ៬vog v៬om v៬om ϭ 2 m/s [W] ៬vmgϭ 1 m/s [E] ៬vog ϭ 1 m/s [W] Fig.3.30 Calculating Relative Velocity in One Directionco nnecti ts the ng ncep Object’s velocity ϩ Velocity of medium ϭ Object’s velocityCo within the medium relative to ground relative to ground96 v៬om ៬vmg v៬og unit a: Motion and Forces

Two-dimensional Relative Velocity Fig.3.31 Determine the patternWe have learned that when we use vector equations in twodimensions, values cannot be substituted into the equationdirectly. This is because the vector quantities are no longercollinear. The vector equation tells you how to draw the vec- v៬og v៬og v៬og v៬ogtor diagram. The plus sign (؉) in the equation tells you to con-nect the vectors head to tail. By looking at the equation v→og ϭv→mg ϩ v→om, we can see that the vectors v→mg and v→om are the onesconnected head to tail. No matter how we rearrange theequation, v→og is never added to the other two vectors. Thismeans that v→og is never connected head to tail in any diagram.Figure 3.31 shows a series of vector additions. For each case,v→og is identified. You should notice that it is always connectedeither head to head or tail to tail to the other two vectors.As with other 2D vector problems such as displacement and projectilemotion, two methods can be used to solve them. The first is to create a scalediagram using the vector equation to guide your drawing. The final vectoris then measured along with its direction. The other method is to usetrigonometric formulas and Pythagoras’ theorem.We are now closer to answering the opening question.e x a m p l e 8 Relative motion in two dimensionsCase 1. Assume Tarzan can swim at 2.0 m/s, the current is 1.0 m/s [E],and Tarzan is seen to swim directly across (assume north).Case 2. Assume Tarzana can swim at 2.0 m/s, the current is 1.0 m/s [E],and Tarzana is seen to direct her body directly across (north).Case 3. Assume Cheetah can also swim at 2.0 m/s. The water current isnow 5.0 m/s [E]. Cheetah swims like Tarzana and points directly north.Find the final velocity of each.Solution and Connection to Theory Fig.3.32 Case 1 v៬mgCase 1 Given ៬vomv→ogϭ ? [N] v→om ϭ 2.0 m/s [?] v→mg ϭ 1.0 m/s [E] ␪From the vector equation v→og ϭ v→om ϩ v→mg, construct a vector diagram. v៬ogFrom Fig. 3.32, notice that in order for Tarzan to swim directly across theriver, he must angle his body into the current while still maintaining 97some motion towards the far shore.Vector v→og, pointing directly north, will be the resultant of v→mg and v→om.Note that v→om is the hypotenuse of the right-angle triangle. Then usePythagoras’ theorem to find v→og. chapt e r 3: Motion in Two Dimensions

vog ϭ ͙ෆv2om Ϫෆv2mg ϭ ͙ෆ(2.0 m/ෆs)2 Ϫෆ(1.0 mෆ/s)2 ϭ 1.7 m/s To find the angle ␪ as indicated in Fig. 3.32, we can use ␪ ϭ sinϪ1 ᎏvmᎏg ϭ sinϪ1 ᎏ1.0 mᎏ/s ϭ 30° vom 2.0 m/s This gives the direction as [N30°W] or [W60°N]. While Tarzan swims 2.0 m/s in that direction, his velocity relative to the ground is 1.7 m/s [N].Fig.3.33 Case 2 Case 2 Given v→mg ϭ 1.0 m/s [E] ៬vmg v→og ϭ ? [?] v→om ϭ 2.0 m/s [N] v៬om ៬vog Construct a vector diagram. From Fig. 3.33, you can see that Tarzana is concentrating only on swimming north. In still water, you would see her ␪ moving at 2.0 m/s directly north. However, she is being pushed east by the current as she tries to move north. These two velocities act at the same time to create the angled motion. Notice, though, that her body posi- tion stays pointing north as she approaches the far shore at an angle. Using Pythagoras’ theorem, vog ϭ ͙vෆ2om ϩෆv2mg ϭ ͙ෆ(2.0 m/ෆs)2 ϩෆ(1.0 m/ෆs)2 ϭ 2.2 m/s The angle can be obtained from ␪ ϭ tanϪ1 ᎏvmᎏg ϭ tanϪ1 ᎏ1.0 mᎏ/s ϭ 27° vom 2.0 m/s Thus, Tarzana’s velocity relative to the ground is 2.2 m/s [N27°E] or [E63°N]. Because of her angled motion, she travels farther than Tarzan at a higher speed.Fig.3.34 Case 3 Case 3 Given ៬vmg v→og ϭ ? [?] v→om ϭ 2.0 m/s [N] v→mg ϭ 5.0 m/s [E] In the vector diagram of Fig. 3.34, you can see that Cheetah’s velocity rel- ative to the ground is v→og. Using Pythagoras’ theorem,៬vom vog ϭ ͙vෆ2om ϩෆv2mg ␪ ៬vog ϭ ͙(ෆ2.0 mෆ/s)2 ϩෆ(5.0 mෆ/s)2 ϭ 5.4 m/s The angle can be obtained by ␪ ϭ tanϪ1 ᎏvmᎏg ϭ tanϪ1 ᎏ5.0 mᎏ/s ϭ 68° vom 2.0 m/s The direction is [N68°E] or [E22°N]. So Cheetah travels at 5.4 m/s in the direction 68° east of north.98 unit a: Motion and Forces

In all three cases, the diagrams could have been constructed by setting ascale, using the givens, and connecting them in a manner dictated by thevector equation. The final answers could have been obtained by measuring.Finding the Time to CrossSo who finishes first in this race? To find out, we need to look at the three Fig.3.35 Aligning displacement1.7 m/svector diagrams, which are shown together in Fig. 3.35. and velocity vectors2.0 m/s The equation v→ ϭ ᎏ⌬⌬dᎏ→t rearranged to find time, becomes ⌬t ϭ ᎏ⌬v→dᎏ→. We nowneed to find out which v to use. If you look at the diagrams in Fig. 3.35, the 1 m/s2.0 m/svelocity vector that points in the direction of the desired motion (north inour case) is the appropriate choice. The other vectors involve the motion of ⌬d៬ Scalethe current, which does nothing to help anyone cross to the other side.Tarzan swims at 1.7 m/s in the north direction. Notice that for Tarzana and 50mCheetah, the velocity directed north is 2.0 m/s. Thus, Tarzan finishes last, Nwith Tarzana and Cheetah tied for first. The lesson of this story is that ifyou wish to cross the river quickly, then direct all your energies to do so. ⌬d៬ W ETarzan used up some of his velocity in fighting the current in order to endup in a position directly across from where he started. S ⌬d៬Airplanes Magnitude DirectionFor airplanes and flight paths, the method of approaching the problem is the →vpg Ground speed Tracksame as for our jungle swimmers. The velocity of the plane relative to theground is the velocity we see. It is a combination of the plane’s capability, →vpa Air speed Headingcalled the velocity of the plane relative to air (v→pa), and the velocity of the airrelative to the ground (v→ag). These two velocities continually combinethroughout the flight to produce the final velocity we see.e x a m p l e 9 Relative velocity involving planesA pilot with a heading of [N30°E] and an airspeed of 400 km/h flies intoa wind coming from the north at 110 km/h. What is the plane’s velocityrelative to the ground and how long would it take to complete a journeyof 1000 km [N30°E]?Solution and Connection to TheoryGiven v→ag ϭ 110 km/h [S] v→pg ϭ ? [?]v→pa ϭ 400 km/h [N30°E]v→pg ϭ v→pa ϩ v→ag chapt e r 3: Motion in Two Dimensions 99

TRIGONOMETRIC EQUATIONS Fig.3.37A In Fig. 3.36, the angles of the trian- ៬vpa ϭ 400 km/h gle are labeled using capital letters, and the sides opposite them are vpayϭ 400 sin 60˚ v៬pg ϭ 310 km/h 236 km/h labeled using lowercase letters. ϭ 346 km/h Whenever the lengths of two sides 30˚ and the contained angle (the angle between them) are known, the v៬ag ϭ 110 km/h 60˚ ␪ cosine law is used. 200 km/h vpax ϭ 400 cos 60˚ c2 ϭ a2 ϩ b2 Ϫ 2ab cos C . ϭ 200 km/h Similarily, if we know the value of an Component method angle, the side opposite it, plus x} vpgx ϭ vpax ϩ vagx another angle or side, we can use the sine law. vpgx ϭ (400 km/h) cos 60° ϩ 0 km/h ϭ 200 km/h y} vpgy ϭ vpay ϩ vagy ᎏsinᎏA ϭ ᎏsinᎏB ϭ ᎏsinᎏC ab c vpgy ϭ (400 km/h) sin 60° Ϫ 110 km/h ϭ 236 km/hFig.3.36 ͙ෆෆvpg ϭ v2pgx ϩ v2pgy c B ϭ ͙ෆ(200 kෆm/h)2ෆϩ (236ෆkm/hෆ)2 ϭ 310 km/h a The angle for the ground velocity isA bC ␪ ϭ tanϪ1 ᎏvpᎏgy ϭ tanϪ1 ᎏ236 kᎏm/h ϭ 50°Fig.3.37B vpgx 200 km/h ␤ ϭ 30˚ Since both the x and y components are positive, the direction is [E50°N].v៬pa ϭ 400 km/h Therefore, v→pg ϭ 310 km/h [E50°N]. v៬ag ϭ 110 km/h v៬pg To find the time involved, we use the velocity of the plane relative to the ground. Since the distance covered is the actual displacement, the v→pg ␣ velocity vector is the one that lines up with the displacement direction. Therefore, the time it takes to reach the destination is ␪60˚ ⌬t ϭ ᎏ⌬ᎏd ϭ ᎏ1000ᎏkm = 3.23 h vpg 310 km/h Sine/Cosine Method First we construct the vector diagram, knowing that v→pa and v→ag are con- nected head to tail (Fig. 3.37B). Now we use the cosine law to find vpg: v2pg ϭ v2pa ϩ v2ag Ϫ 2vpavag cos␤ vpg ϭ ͙ෆ(400 kෆm/h)2ෆϩ (110ෆkm/hෆ)2 Ϫ ෆ2(400ෆkm/h)ෆ(110 ෆkm/h)ෆcos 30ෆ° ϭ 310 km/h100 u n i t a : M ot i o n a n d Fo rc es

To find the angle, we use the sine law: ᎏsinᎏ␣ ϭ ᎏsinᎏ␤ vag vpg sin␣ ϭ ᎏ131100 kkᎏmm//hh sin 30° ␣ ϭ 10°From the diagram, to find the angle from the horizontal,␪ ϭ 60° Ϫ 10° ϭ 50° (as before). Both methods lead to the same final result. The plane’s ground veloc-ity is 310 km/h [E50°N]. It will take 3.23 h to fly 1000 km.Fig.3.38 Calculating Time of Travel v៬om v៬mg Select velocity Calculate time guttin v៬og in direction of travel as it all of requested To geth Solve for displacement ⌬d p pplyin relative velocity vector v er the ncep ⌬t ϭ 1011. a) Clearly state the meaning of the vectors v→og, v→om, and v→mg. g b) Use the cases of a person swimming in a river, a plane flying in Co the air, and a person throwing a ball while running to illustrate a the meaning of the vectors in a). ts2. For the case of a wind blowing from the east at 80 km/h and a plane capable of flying at 200 km/h, find a) the velocity of the plane relative to the ground if the pilot points the plane north. b) the velocity of the plane relative to the ground and the heading the pilot must take in order to fly directly north. c) the velocity of the plane relative to the ground if the pilot points the plane [N20°W].Make sure you state the givens for each case.3.5 Average AccelerationConsider the equation a→ ϭ .ᎏ(v→2 Ϫᎏv→1) In two dimensions, the velocity vectors ⌬tare usually in different directions. We know that if the vectors are notcollinear, then we cannot substitute directly into the acceleration equation.We must use the component method of solving problems. chapt e r 3: Motion in Two Dimensions

In the equation a→ ϭ ᎏ(v→2 ⌬Ϫᎏt v→1) , time does not affect the direction of acceleration. Therefore, we only need to solve for ⌬v→ ϭ v→2 Ϫ v→1. Using the component method, we separate the horizontal and vertical compo- nents for v→1 and v→2, as we did in the previous example. We substitute the hori- zontal components into the equation ⌬v→ϭ v→2 Ϫ v→1, then we do the same for the vertical components. Finally, we use Pythagoras’ theorem to find the magni- tude of the resultant vector. We use the function tanϪ1 ᎏvᎏy to obtain the angle. After you obtain ⌬v→, divide it by the time to vx determine the acceleration. Fig.3.39A Solving a 2D Vector Problem ethom 2D Find Break vectors Solve equation in x Use Pythagoras’ ofs vector defining into x and y Solve equation in y o cesd problem equation components theorem andpr ␪ ϭ tanϪ1aodpjpacoesnitte102 e x a m p l e 1 0 Finding vector acceleration Find the acceleration of a car moving at 30 m/s [E60°N], which then veers off and moves directly east at 25 m/s in 2.5 s. Solution and Connection to Theory Given v→2 ϭ 25 m/s [E] ⌬t ϭ 2.5 s a→ ϭ ? v→1 ϭ 30 m/s [E60°N] The defining equation for the vector problem is ⌬v→ ϭ v→2 Ϫ v→1, which will be solved in both the x and y directions. Next we obtain the components of each velocity. v1x ϭ (30 m/s) cos 60° ϭ 15 m/s v1y ϭ (30 m/s) sin 60° ϭ 26 m/s v2x ϭ 25 m/s v2y ϭ 0 m/s Solving in the x direction, ⌬vx ϭ v2x Ϫ v1x ϭ 25 m/s Ϫ 15 m/s ϭ 10 m/s Solving in the y direction, ⌬vy ϭ v2y Ϫ v1y ϭ 0 m/s Ϫ 26 m/s ϭ Ϫ26 m/s unit a: Motion and Forces

Therefore, ⌬v ϭ ͙(ෆ10 m/ෆs)2 ϩෆ(Ϫ26ෆm/s)2 ϭ 28 m/s. Fig.3.39B v1x ϭ 10 m/s ␪Substituting the magnitudes of the velocities into the v1y v៬1equation, the angle ⌬៬v v1y ϭ Ϫ26 m/s␪ ϭ tanϪ1 ᎏ(26 mᎏ/s) ϭ 69°. 60˚ v1x etho 10 m/s ofFrom the signs of the x (ϩ) and y (Ϫ) values for ⌬v→, we o cescan see that the direction is [E69°S]. Dividing ⌬v→ by2.5 s, we obtain the acceleration, 11 m/s2 [E69°S]. v៬2Fig.3.39C Finding Vector Acceleration Find 2D Use Break v៬1 and v៬2 ⌬vx ϭ v2x Ϫ v1x ⌬v ϭ͙ළ⌬ළv2ළxළϩළළ⌬ළvළ2ළyළළ macceleration ⌬vy ϭ v2y Ϫ v1y s ⌬v៬ϭ v៬2 Ϫ v៬1 into components ␪ ϭ tanϪ1 ⌬vy d ⌬vx pr a៬ ϭ ⌬៬v ⌬tSubtracting Vectors Using a DiagramThe component method does not require the use of a vector diagram. As Fig.3.40 →vpa ϭ →vpg Ϫ →vaglong as you have a defining equation and the components of the vectorquantities involved, the problem can be solved algebraically. But what does A ៬vagthe subtraction of vectors look like? ϩ ៬vpg In order to visualize vector subtraction, consider the diagram represent- v៬paing a plane flying north with a wind blowing east. From Fig. 3.40, we can Wsee that the plane’s actual velocity is to the northeast. N E S Because the defining equation is v→pg ϭ v→pa ϩ v→ag, the vectors v→pa and v→agare connected head to tail. This is the meaning of the plus sign in vectoraddition. But if the equation is rearranged to solve for either v→pa or v→ag, a neg-ative sign appears because one of these quantities must be subtracted fromv→pg in order to solve for the missing quantity: v→pa ϭ v→pg Ϫ v→ag or v→ag ϭ v→pg Ϫ v→pa Notice from Fig. 3.40 that v→pg is connected either head to head or tail to Ϫ Btail to the other two vectors. This is the vector diagram representation forsubtraction. We will use the tail-to-tail method of connecting vectors to illus-trate subtraction. From Example 10, we used vector subtraction, ⌬v→ ϭ v→2 Ϫ v→1. Figure 3.41illustrates several examples of velocity vector subtraction. Notice that theresultant always points toward the head of v→2. chapt e r 3: Motion in Two Dimensions 103

v៬2Fig.3.41 ⌬→v ϭ →v2 Ϫ →v1 v៬1 v៬1 v៬2 v៬2 v៬1 v៬1 v៬2 v៬1ALTERNATIVE METHOD v៬1 ⌬៬v ⌬៬v ⌬៬vFOR SUBTRACTING VECTORS (a) v៬2 (c) v៬2 (b)→v2 Ϫ →v1 ϭ →v2 ϩ (Ϫ→v1)→v1 can be subtracted from →v2 by adding Ϫ→v1. (Ϫ→v1) is→v1 pointing in the opposite direction. Add →v2 ϩ (Ϫ→v1) in the examples in Fig. 3.41 to verify that the resultant is the same. Fig.3.42 General Method of Vector Subtractionnnectico ts the ngncepCo Using the Cosine and Sine Law Methods Because we used triangles in the last example, there is an alternate approach to solving the problem. After we create a vector diagram, we can treat it like a triangle and use trigonometric equations to find the missing information. e x a m p l e 1 1 Sine/cosine method for solving vector problems Find the acceleration of a car moving at 30 m/s [E60°N], which then veers off and moves directly east at 25 m/s in 2.5 s. Solve using either the sine law or the cosine law. Fig.3.43A104 u n i t a : M ot i o n a n d Fo rc es

Solution and Connection to TheoryGiven v→2 ϭ 25 m/s [E] ⌬tϭ 2.5 s a→ ϭ ?v→1 ϭ 30 m/s [E60°N]First we draw a vector diagram. Fig.3.43B E ␪Connect the vectors tail to tail, as illustrated in Fig. 3.43B. Notice that ⌬v→ 5 m/shas its head touching v→2. The angle between the two vectors is 60°. We ⌬៬vknow the size of the contained angle between two known sides. This v៬1means that we can use the cosine law.⌬v2 ϭ v12 ϩ v22 Ϫ2v1v2 cos 60°⌬v2 ϭ (30 m/s)2 ϩ (25 m/s)2 Ϫ2(30 m/s)(25 m/s) cos 60°⌬v2 ϭ 775 m2/s2 ⌬v ϭ 27.8 m/s. N ␪ 60°This number rounds off to 28 m/s due to significant digits. v៬2 W STo find the angle for this vector, we use the sine law to find ␪. ᎏsinᎏ␪ ϭ ᎏsin 6ᎏ0° 30 m/s 27.8 m/s sin␪ ϭ 0.933, ␪ ϭ 69°As before, the angle is [E69°S]. To obtain the acceleration magnitude, wedivide 28 m/s by 2.5 s. Thus, the acceleration is 11 m/s2 [E69°S].The following example is done using both the component and trigonomet-ric methods and is accompanied by a scale diagram.e x a m p l e 1 2 Comparing methods — change in velocityA puck is shot towards the boards at a velocity of 25.0 m/s at an angle of30° to the boards. If the puck ricochets off the boards at an angle of 20°relative to the boards and at a speed of 20.0 m/s, find the change in veloc-ity of the puck. Fig.3.44Solution and Connection to TheoryGiven v→2 ϭ 20.0 [R20°U] 5 m/sv→1 ϭ 25.0 m/s [R30°D] v៬1 ϭ 25 m/s v៬2 ϭ 20 m/s 30˚ 20˚ Boards chapt e r 3: Motion in Two Dimensions 105

Fig.3.45 v៬2 Trigonometric Method. We first construct a vector subtraction dia- gram, as in Fig. 3.45. Notice that the interior angle is 50°. Since we knowBoards 20° the lengths of two sides and the interior angle between them, we can use 30° the cosine law. ⌬៬v ⌬v2 ϭ (25.0 m/s)2 ϩ (20.0 m/s)2 Ϫ2(25.0 m/s)(20.0 m/s) cos 50° v៬1 ␪ ⌬v2 ϭ 382 m2/s2 30° ⌬v ϭ 19.5 m/s To find the angle for this vector, we first use the sine law to find ␪. ᎏ20s.0inᎏm␪ /s ϭ ᎏ1s9in.55ᎏm0°/s sin␪ ϭ 0.786, ␪ ϭ 52° Referring back to Fig. 3.45, we see that this angle is not measured from the boards. Because the angle must be given in relation to one of the car- dinal directions (U, D, R, L), add 30°, as shown. 30° ϩ 52°ϭ 82°. The change in velocity is 19.5 m/s [L82°U] (or [U8°L]). Component Method. Obtain the components of each velocity vector. v1x ϭ (25.0 m/s) cos 30° ϭ 21.7 m/s v1y ϭ (Ϫ25.0 m/s) sin 30° ϭ Ϫ12.5 m/s The negative sign occurs because the direction of the y component is down. v2x ϭ 20.0 m/s cos 20° ϭ 18.8 m/s v2y ϭ 20.0 m/s sin 20° ϭ 6.8 m/s Solving in the x direction, ⌬vx ϭ v2x Ϫ v1x ϭ 18.8 m/s Ϫ 21.7 m/s ϭ Ϫ2.9 m/s Solving in the y direction, ⌬vy ϭ v2y Ϫ v1y ϭ 6.8 m/s Ϫ (Ϫ12.5 m/s) ϭ 19.3 m/s Therefore, ⌬v ϭ ͙(ෆϪ2.9ෆm/s)2ෆϩ (19ෆ.3 m/sෆ)2 ϭ 19.5 m/s. ␪ ϭ tanϪ1 ᎏ19.3ᎏm/s ϭ 82° 2.9 m/s (We actually kept more decimal places in the calculation than are shown.) By checking the signs of the x (Ϫ) and y (ϩ) values for ⌬v (see Fig. 3.45), we can see that the direction is [L82°U]. In summary, we can solve relative velocity problems, total displacement problems (with two displacements), and average acceleration problems using the same two methods. The vector equations v→pg ϭ v→pa ϩ v→ag ⌬d→total ϭ → ϩ → a→ ϭ ᎏ⌬⌬ᎏv→t d1 d2106 u n i t a : M ot i o n a n d Fo rc es

tell you how to combine the given vectors in a vector diagram. Once the tri-angle is formed, the appropriate trigonometric law is used. For the compo-nent method, these are the defining equations into which the componentsof the given vectors are substituted.Fig.3.46 Solving Problems Using Vectors More guttin than two it all non-collinear To gethVector vectors? NO Trigonometric pproblem methods erpplyin the YES Triangle method ncep Set up equation and Component method draw vector triangle 107 Choose appropriate equation Set up and solve Solve triangle equations in x and with cosine and sine laws y directions Use Pythagoras’ theorem and ␪ ϭ tanϪ1 vy vx1. a) Given two vectors, describe the three possible methods of g solving a vector equation. Co a b) Describe the difference in meaning when two vectors are con- ts nected tail to tail and head to head.2. a) For a puck travelling at 120 km/h [E], then changing its direction to [N] in 0.5 s, calculate the puck’s acceleration. b) Now assume the puck changed to a direction given by [N25°W]. Calculate the acceleration using the component and cosine law methods. c) Now assume the puck changed to a new velocity of 100 km/h [N25°W]. Calculate the acceleration using the component and cosine law methods.3. Repeat Question 2 in Section 3.4 Applying the Concepts using the sine/cosine method. chapt e r 3: Motion in Two Dimensions

S T S c i e n c e — Te c h n o l o g y — S o c i ety —S E Environmental Interrelationships Global Positioning System The Global Positioning System (GPS), commissioned by the United States Department of Defense, is a navigation system that allows users to deter- mine their location anywhere on Earth. GPS uses communications between satellites as well as a series of monitoring stations and ground-based anten- nas (Fig. STSE.3.1). Triangulation calculations (Fig. STSE.3.2) between the satellites and the user make accurate results possible.Fig.STSE.3.1 The segments of the GPS Fig.STSE.3.2 Triangulation with satellitesFig.STSE.3.3 Wherify child watch Radio signals travelling at the speed of light are timed as they travel For image from the satellite to the hand unit and back to the satellite. The distances see student between the satellite and user can then be calculated using a simple kine- text. matics equation. dsatellite ϭ ᎏc2ᎏt For image see student where dsatellite is the distance between the satellite and the GPS user, c is the speed of light (3.0 ϫ 108 m/s), and t is the time it takes for the signal to text. descend to Earth and return to the satellite. The product, ct, is divided by two108 because we are only interested in the distance between Earth and the satellite. Receivers on Earth have a “line of sight” that allows them to communi- cate with at least five GPS satellites at any one time. First widely used during the Persian Gulf War in 1991, GPS systems are now sold in many stores for personal use, and are also offered as options in some new cars. Wherify Wireless of Redwood Shores, California, is marketing a Personal Location System that uses GPS satellites and cellular phone technology to determine the location of children (Fig. STSE.3.3). Children with this watch can press a button to be directly connected to the 911 emergency system, and police can be dispatched to their exact location. As navigators in the global village, we may never need paper maps again. unit a: Motion and Forces

Design a Study of Societal Impact 109 Research GPS. What is the resolution of the system? How many satel- lites does it use? Are there plans to add more satellites? Think of a new application for GPS. The evolution of GPS required the cooperation of government, sci- ence, business, and industry. Brainstorm the various fields of physics that are directly involved in developing new technology. Research other careers that are directly involved in the development of this technology. For each of these careers, examine the extent to which knowledge of physics is an asset. Does the widespread use of GPS raise privacy issues? How should these issues be addressed? There is only a limited amount of space “up there” in which to position satellites around Earth, and rich countries seem to have first pick. Should all countries have guaranteed access to the services the satellites provide? Which governing body would be capable of moni- toring accessibility to satellites?Design an Activity to Evaluate Conduct an activity to locate an object on a large, open floor or table- top (large lab bench) using the principle of triangulation. Use the cracks between floor tiles (other marks for a tabletop) to set up a two- dimensional grid axis map. Designate three positions in this area to be used as “satellites.” Using a sonic ranger (and computer interface) or tape measure, measure the distances between each “satellite” and the test object in order to locate the object on the area map. Your test object should be small and not easily discernible, such as one coin out of many that are laid out on the floor. Looking for one object among many similar objects will highlight the importance of accuracy for each measurement as well as the need for more positioning “satel- lites.” Extend this activity to three dimensions by lifting your three satellites to different heights using retort stands or tripods.Build a Structure Organize a class activity or competition similar to “capture the flag,” where teams hide their “flag” (a small marker) somewhere on the school grounds, or even in the town or city in which you live. Teams must find the flag using a map that gives at least three reference points, as well as the respective distances from these points to the flag. Add more reference points if necessary. The winning team is the one that finds the flag first using a minimum number of reference points. chapt e r 3: Motion in Two Dimensions

S U M M A RY S P E C I F I C E X P E C TAT I O N S You should be able to Understand Basic Concepts: Explain the difference between one-dimensional and two-dimensional vector equations. Add vectors in two dimensions graphically. Analyze uniform motion in two dimensions by creating scaled vector diagrams. Find the x and y components of any vector at any angle. Use the component method to find the resultant in a two-dimensional vector problem. Describe and solve projectile motion problems in two dimensions using kinematics equations. Identify the vectors involved in relative motion. Draw representations of problems involving relative motion. Solve relative velocity problems using graphical and trigonometric methods. Describe situations involving average acceleration in two dimensions. Solve average acceleration problems using graphical and trigonometric methods. Develop Skills of Inquiry and Communication: Analyze the motion of objects using vector diagrams and two-dimensional vector equations. Describe why projectiles undergo parabolic motion. Design and carry out experiments involving projectile motion. Describe situations involving relative motion. Relate Science to Technology, Society, and the Environment: Analyze the role of projectiles in sports. Explain how GPS works. Relate GPS to the transportation industry, present and future. Equations Projectile motion: x direction ⌬dx ϭ v1x⌬t y direction ⌬dy ϭ v1y⌬t ϩ ᎏ21ᎏay⌬t2 ⌬d→T ϭ → ϩ → ϩ → ϩ … d1 d2 d3 v→og ϭ v→om ϩ v→mg a→ ϭ ᎏ⌬⌬ᎏv→t110 u n i t a : M ot i o n a n d Fo rc es

EXERCISES Conceptual Questions 5. Does the acceleration of a projectile ever go to zero? 1. What motion would you expect if both the horizontal and vertical components of the 6. What other factors enter into projectile prob- velocity were constant? Provide instances lems that we haven’t discussed? where this can occur. 7. Find examples of sports that use projectile 2. Find examples of motion where there is an motion without spin and of sports that use acceleration in the horizontal direction as well. projectile motion with spin. What does spin do for the projectile? 3. Consider this possible demonstration: A stu- dent aims and shoots a peashooter at a honey 8. At what point in a projectile’s path is the pail suspended by an electromagnet (Fig. 3.47). speed a As the pea leaves the shooter, a sensor detects a) maximum? its exit and sends a signal to the electromag- b) minimum? net, shutting it off. This causes the pail to fall. As long as the student has aimed the shooter 9. For the diagram of baseball hits (Fig. 3.48), correctly (directed at the initial position of the rank the hits in order of when they land. pail), the projectile will hit the pail. If you change the distance between the shooter and Fig.3.48 the pail, and the initial velocity of the projec- tile, the result will always be the same. 123 4 Explain why this is so. 10. An airplane is moving with a constant veloc-Fig.3.47 ity. You are sitting inside the airplane and drop a ball. Describe what the ball’s path 4. A batter in baseball sometimes says that the looks like to you inside the plane and to approaching baseball rises. In fact, the effect someone standing still, observing the action of lift is not strong enough to create this on the ground. Assume Superman’s X-ray, action. What motion does the ball actually telescopic vision. undergo? 11. Describe a satellite’s orbit in terms of a pro- jectile launch. Problems 3.1 Vectors in Two Dimensions 12. Copy the vectors of Fig. 3.49 into your note- book. Add them and draw in the total dis- placement vector. Label the start and end of the net path. No calculations are required. chapt e r 3: Motion in Two Dimensions 111

Fig.3.49 b) Find the total displacement for the first two legs of the walk.(a) d៬2 d៬3 d៬1 c) What was the displacement of the final part of the walk home? d) What was the average speed for the whole walk? e) What was the average velocity for the first two parts of the walk?(b) d៬2 d៬3 d៬4 3.2 Parabolic Motion d៬1(c) d៬1 d៬3 d៬4 16. Copy the sketch of the soccer ball’s trajectory in Fig. 3.51 into your notebook. Complete the vector diagrams at the indicated points (draw the component vectors). d៬2 d៬5 Fig.3.51 vx13. Copy the velocity vectors of Fig. 3.50 into v៬o v៬f your notebook. Add them and find the result- ant velocity. No calculations are required.Fig.3.50 v៬2 17. Given the trajectory in Fig. 3.52 of a baseball thrown and a baseball dropped, use the vec-(a) (b) v៬1 v៬2 tors given to create vector diagrams at the ball v៬1 positions indicated.14. Given a velocity of 4 m/s [E] and a second Fig.3.52 velocity of magnitude 6 m/s, a) find the maximum and minimum values 18. For a projectile thrown horizontally at 50 km/h, for the final velocity and draw a vector a) calculate the velocities in the vertical direc- representation for each situation. tion at 1.0 s, 2.0 s, 3.0 s, and 4.0 s. b) Draw four more possible final velocity vec- b) Find the final velocity for each of these tor diagrams. times. c) Using either trigonometric techniques or scaled diagrams, find the final velocity for 19. For a projectile thrown at 50.0 km/h but at each of your vector diagrams. 45° to the horizontal, both the vertical and horizontal components are 35.4 km/h15. A student walking his pet duck walks 0.4 km (motion is up and out). [N], then 0.3 km [E], and then returns home. The whole walk took 0.5 hours. a) What was the total displacement for the whole walk?112 u n i t a : M ot i o n a n d Fo rc es

a) Calculate the vertical velocities at 1.0 s, 2.0 s, a) how far the skydiver falls. 3.0 s, and 4.0 s. b) how far the skydiver moves horizontally. c) the final vertical velocity of the skydiver. b) What are the horizontal velocities at these d) the final velocity of the skydiver. times? 28. A plane flying level at 80 m/s releases a pack- c) Find the final velocities for the four times. age from a height of 1000 m. Find d) Sketch the resultant trajectory. a) the time it takes for the package to hit the ground. 3.3 Projectile Motion Calculations b) the distance it travelled horizontally. c) the final velocity of the package.20. A movie scene has a car fall off a cliff. If the car took 5.5 s to reach the ground, how high 29. The cannon in Fig. 3.53 is on the edge of a was the cliff? cliff 500 m above ground. The ball leaves the cannon at 100 m/s. Calculate the range for21. If the car in Problem 20 had an initial hori- each of the three cases. zontal velocity of 26 m/s, how far from the a) Case 1, cannon shoots horizontally, cliff bottom did the car land? vx ϭ 100 m/s, v1y ϭ 0 b) Case 2, cannon is elevated 60°22. A bullet is shot horizontally from a gun. If the c) Case 3, cannon is depressed 60° bullet’s speed exiting the muzzle is 325 m/s and the height of the gun above the ground is 2.0 m, Fig. 3.53 a) how long was the bullet in the air? b) how far did the bullet travel horizontally ϩ before it hit the ground? 60˚23. A tennis player serves a tennis ball from a ϩ height of 2.5 m. If the ball leaves the racket horizontally at 160 km/h, how far away will the ball land?24. A pitcher throws a baseball at 140 km/h. If 60˚ the plate is 28.3 m away, how far does the ball drop if we assume the ball started travelling 500 m toward the plate horizontally? 30. After serenading your girlfriend under her25. Two pennies are sitting on a table 1.2 m high. balcony window, you toss a love letter to her Both fall off the table at the same time, except wrapped around a small rock (she has a base- one is given a significant push. If the pushed ball glove and knows how to use it). If the penny is moving at 4.1 m/s horizontally at the window is 15 m away and you toss the rock time it leaves the table, with a vertical velocity component of 13 m/s a) which penny lands first? and a horizontal component of 10 m/s, how b) how far from the table does the pushed high up the wall was the balcony, assuming penny land? she caught the rock without having to reach?26. For Problems 22 to 24, calculate the final velocity of each projectile.27. A plane is flying horizontally with a speed of 90 m/s. If a skydiver jumps out and free falls for 10.6 s, findchapt e r 3: Motion in Two Dimensions 113

31. Will a football, kicked at 14.0 m/s vertically velocity of the boat as seen by a person on shore. and 9.0 m/s horizontally, clear a bar 3.0 m Note: The term heading is used to describe the high and 20 m away from the kicker? Solve in direction the boat must point in order to success- two different ways. fully travel in its desired direction.32. Will a tennis ball served horizontally at 38. A plane is seen to travel in a direction [S30°W]. 100 km/h from a height of 2.2 m clear a net If its ground velocity was 300 km/h and the 0.9 m high and 10 m away? Solve in two wind speed is 150 km/h south, what is the different ways. plane’s velocity relative to the air?33. Emanual Zacchini was shot over three ferris 39. A boat wishes to cross a lake and end up wheels, landing in a net at the same height directly south of where it started. The boat is from which he was shot (described in Fig. capable of moving at 34.0 km/h. If there is a 3.19). Given his initial velocity of 27 m/s current of 8.0 km/h flowing to the west, find [R53°U] and range of 69 m, find a) the heading the boat must take in order to a) his maximum height reached. successfully complete the trip. b) the time spent in air, using two different b) the velocity relative to the ground. methods. c) the time (in seconds) it took the boat to c) his final velocity, using logic and computation. cross if the lake was 21 km in the direction the boat had to travel. 3.4 Relative Velocities 40. How much faster is it to point yourself and34. a) Create a series of arbitrary vector additions swim directly across instead of fighting the (two vectors) and identify the vector that current and swimming directly across if the could represent v→pg. width of the river is 1000 m, you are capable of swimming at 2.2 m/s, and there is a cur- b) Draw a vector addition triangle where no rent of 1.6 m/s? vector can represent v→pg. 41. A large cruise boat is moving at 15 km/h east35. A plane is flying at 100 km/h north. A passen- relative to the water. A person jogging on the ger walks along the length of the plane. What ship moves across the ship in a northerly are the maximum and minimum velocities of direction at 6 km/h. What is the velocity of the passenger relative to the ground? Assume the jogger relative to the water? her velocity relative to the plane is 1.5 km/h. 42. A shortstop running at 2.0 m/s toward third36. Given the following conditions, what is the rel- base catches and throws a ball toward home ative velocity of two cars if their ground speeds plate at 35 m/s. If the shortstop and catcher are 80 km/h for car A and 45 km/h for car B? are lined up in a direct line of sight when the a) Both cars are heading in the same direction, shortstop throws the ball, A behind B in the reference frame of A. a) at what angle to his body should he throw b) Use the reference frame of B for part a). in order for the ball to move in a straight c) Both cars are heading toward each other. line directly from him to the catcher? Take the reference frame of A. b) If the distance the ball travels is 20 m, how d) Use B’s reference frame for part c). long does the ball take to get to the catcher?37. A boat wishes to travel east. If there is a current 43. A plane flies a square route, with each side of 10 km/h flowing north and the boat is capable 1500 m in length (north, west, south, then of travelling at 30 km/h, find the heading and114 u n i t a : M ot i o n a n d Fo rc es

east). If there is a wind blowing from the east that point. Then redraw the vectors in pairsat 20 km/h during the whole trip, find the total (1 & 2, 2 & 3, etc.), combining each pair in atime it will take the plane to complete the jour- way that shows the relation of the two veloc-ney, given that the plane is flying at 80 km/h. ity vectors to the acceleration vector.3.5 Average Acceleration 46. Redraw the vectors in Fig. 3.56 into your notebook, and then draw the x and y compo-44. Sketch the vectors v→1 and v→2 in Fig. 3.54 in nents of each. Then calculate the x and y components. The vectors areyour notebook and complete the vector dia- a) v→1 ϭ 54 km/h [N33°E] b) v→2 ϭ 70 km/h [W71°N]grams for b) v→1 Ϫ v→2 c) v→3 ϭ 43 km/h [E18°N]a) v→2 Ϫ v→1 d) v→1 ϩ v→2 d) v→4 ϭ 50 km/h [S45°W]c) a→, given ⌬t ϭ 3.0 s e) v→5 ϭ 27 km/h [E40°S]Fig.3.54 E N Fig.3.56 W EWE v៬2 N S v៬2 S WE Scale v៬1 1 cm ϭ 5 m/s S v៬1 v៬3 v៬5 v៬445. Sketch the parabolic trajectory of Fig. 3.55 in 47. A boat sails at 8.0 km/h [S40°E]. What are your notebook. Vector v→1 is the initial hori- the components of its velocity in zontal velocity of 15 m/s. With a constant a) the southward direction? downward acceleration, the projectile follows b) the eastward direction? the shape of a parabola. The marked points c) a direction given by [N50°E]? show its location at subsequent 1 s intervals. At points 2, 3, 4, and 5, construct the vector 48. a) A car does a slow turn. If it moves at a that represents the velocity of the projectile at constant speed of 50 km/h, what is the change in velocity for the car as it movesFig.3.55 from a direction north to west? v៬1 2 b) If it took the car 5.0 s to do this, calculate its acceleration.1 49. If the car in Problem 48 moved from an initial 3 direction of [N] to one of [N20°E], what is its change in velocity and its acceleration? Scale 41 cm ϭ 10 m/s 50. Given that a ball moving at 30 m/s [S10°W] hits a wall and moves off at 5.0 m/s [S30°E], 5 115 chapt e r 3: Motion in Two Dimensions

find the change in the velocity of the ball. Use 56. For the following path of 50 m [N47ºW], component and trigonometric methods. 22 m [W43ºN], 30 m [E60ºS], 30 m [E], and 44 m [ N75ºE], find51. a) Calculate the speed of the tip of a second a) the total distance travelled. hand of length 5.0 cm. b) the total displacement. c) the direction of the most direct route back b) Calculate the average acceleration of the tip to the start. from the moment it hits the “12” position to the moment it hits the “3” position. 57. Remember that speed is distance divided by time and velocity is displacement divided by c) Calculate the average acceleration of the tip time. Given that the time for a complete trip of the second hand from the moment it is 0.15 h, find the speed and velocity in hits the “6” position to the moment it hits Problem 56. the “9” position. 58. a) An airplane has a heading east with an air52. a) Calculate the speed of the tip of a minute speed of 120 km/h. A wind is blowing to hand of length 25 cm. the south at 40 km/h. Calculate its velocity relative to the ground. b) How many degrees does the hand move through when its position changes from b) How long would it take to complete a flight the “12” position to the “2” position? of 1000 km? c) Find the average acceleration of the tip of 59. A boat is moving at 15 km/h in a direction the minute hand in b). given by [N29ºW]. If there is a water current of 5 km/h [S],53. What is the average acceleration of a car a) what is the heading and speed of the boat going clockwise around a circular track of relative to the water? radius 40 m at a constant speed if it takes the b) If the passengers wish to get to a point car 12.5 s to complete one lap? The angle 790 m away in the direction in which the between the positions of the car from t1 to t2 boat is actually moving, how long will it is 60° and the initial position of the car is at take them to cover the distance? the southernmost point on the circular track. 60. A large man inhales a quantity of helium inThe following questions involve the component order to fly east. If he is capable of flying atmethod and the trigonometric method for total 26 km/h by waving his arms and there is adisplacement and relative velocity problems. wind of 10 km/h [S20ºE], find his headingWhen solving problems with two vector and ground speed.quantities given, use the trigonometricmethod. When three or more vectors are 61. A motorized canoe is pointed [N20ºW] in agiven, use the component method. river. If it has the capability to move at 5 m/s and is filmed moving at 7.6 m/s, what is the54. Boris hops 120 km [E60ºN], then 60 km [N], velocity of the river? then 40 km [ W30°N]. Boris has big legs and is happy. Find his displacement. 62. A plane with ground speed 380 km/h flies [N30°E]. Given a wind velocity of 80 km/h55. Calculate the total displacement for a trip [S] during the entire flight, find the plane’s 12 km [N30ºE], 15 km [E], 5 km [N], and heading and air speed. 20 km [S70ºE].116 u n i t a : M ot i o n a n d Fo rc es

3.1 Initial Velocity of a Projectile LABORATORY EXERCISES Purpose Data To find the initial velocity of a projectile. 1. Create a chart for the 10 measured ranges. 2. Record the height of the table. Equipment Uncertainty (See Appendix A) Steel ball, Grooved ramp, Carbon paper Blank paper, Metre stick, Plumb line, Tape Assign a procedural and instrumental uncertainty for the measurements. Calculate the standard devi- Procedure ation of the mean, if that is part of your course. 1. Position the ramp a few centimetres behind Calculations the edge of the table and tape it in place. 1. Calculate the time the ball took to fall. 2. Roll the steel ball down the ramp so that the Use ay ϭ Ϫ9.8 m/s2 and ⌬dy ϭ v1y⌬t ϩ ᎏ21ᎏay⌬t2 ball leaves the ramp in a horizontal direction, (you know the value for v1y). and note where it hits the floor. 2. Use the time and the average value of the 3. Place a blank sheet of paper on the floor and range to find the initial velocity. Again, the tape it down. appropriate equation is ⌬d→ ϭ v→1⌬t ϩ ᎏ21ᎏa→⌬t2. 4. Use the plumb line to locate where the edge Discussion of the desk is projected onto the floor. Mark the point on the paper. You will be measuring 1. What were the three values that were implied the range from here. in doing this experiment? 5. Place a carbon sheet onto the paper, carbon 2. Why did we use 10 trials instead of one? side down. 3. If you calculated the standard deviation of the 6. Roll the ball down the ramp. mean, relate its value to the confidence level 7. Remove the carbon paper and measure the you have in the obtained value for v→1. 4. Why don’t we need the angle of the ramp? distance from the mark on the paper to the 5. If another steel ball was used and was point of first contact with the ball. dropped at the same time as the moving ball 8. Repeat nine more times. left the table, which one would land first? 9. Measure the height of the table. 6. If the ramp was moved to the edge of the table, how does that change the experiment?Fig.Lab.3.1 Is the answer to Question 4 still the same?Vertical height Length of roll Conclusion Summarize your result. Extension 1 Find the final velocity and compare its magni- tude to the initial velocity. Vertical height Extension 2 Redo the projectile lab, only this time put the ramp right beside the table edge. You will need to measure the angle of the ramp. chapt e r 3: Motion in Two Dimensions 117

4 Newton’s Fundamental Laws Chapter Outline 4.1 The World According to Newton 4.2 Newton’s Second Law, F→net ϭ ma→ 4.3 Free-body Diagrams 4.4 Free-body Diagrams in Two Dimensions 4.5 Newton’s Third Law and Free-body Diagrams ST S E Supplemental Restraint Systems 4.1 Newton’s Second Law For image see student text. By the end of this chapter, you will be able to • relate forces to motions • solve problems of force and motion using free-body diagrams • design and implement labs related to different types of motion118