Unit 2 Kinematics 2-D
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Drexel University**We aren't endorsed by this school
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FOOD AND H 435
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Physics
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Jan 7, 2025
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14
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Physics 20 Workbook Unit 2: 2-D Kinematics1 PHYSICS 20 UNIT 2: 2-D KINEMATICS PROJECTILE MOTION
Physics 20 Workbook Unit 2: 2-D Kinematics2 Lesson 1(2-D vectors; tail-to-tip addition)A. 1. For the following vectors, a) b) express the angle in NEWS and in RCS. 2. Sketch the resultant vector by adding each set of vectors. a) b) 3. Determine the resultant displacement (magnitude and direction) for the following: a) A person walked 8.50 km North, and then 5.70 km East b) A boat sailed 12.0 km West, and then 19.0 km South 1468V2 V1 V3 V1 V2
Physics 20 Workbook Unit 2: 2-D Kinematics3 4. A person walked 1.8 km West, and then an unknown distance South. If the magnitude of the resultant displacement is 3.0 km, then determine: a) the distance the person walked South b) the angle of the resultant displacement 5. Sketch the resultant displacement vector, using a tail-to-tip diagram. No calculations. 14 m at 20N of W, and then 25 m at 10W of S B. 6. Determine the resultant displacement (magnitude and direction) for the following: a) A person walked 8.0 blocks East, 5.0 blocks North, and then 14.0 blocks West b) A plane flew 140 km North, then 290 km East, and then 300 km South 7. A boat travelled an unknown distance North, and then 19.0 km East. If the magnitude of the resultant displacement is 23.0 km, then determine: a) the distance the boat travelled North b) the angle of the resultant displacement
Physics 20 Workbook Unit 2: 2-D Kinematics4 8. Three vectors are added together: 1V= 72.0 units South ; 2V= 55.0 units East ; 3VIf the resultant vector is zero, then determine the magnitude and direction of 3V. 9. Sketch the resultant displacement vector, using a tail-to-tip diagram. No calculations. 5 km at 30E of S, and then 11 km at 70N of E SOLUTIONS 1. a) 14W of N ; 104(−256) b) 68W of S ; −158(202) 2. a) 21VVR+=b) 321VVVR++=(see diagram on right) 3. a) 10.2 km at 33.8E of N (56.2N of E) b) 22.5 km at 57.7S of W (32.3W of S) 4. a) 2.4 km b) 53S of W (37W of S) 5. See diagram on right. 6. a) 7.8 blocks at 40N of W (50W of N) b) 331 km at 28.9S of E (61.1E of S) 7. a) 13.0 km b) 55.7E of N (34.3N of E) 8. 90.6 units at 52.6N of W (37.4W of N) 9. See diagram on right.
Physics 20 Workbook Unit 2: 2-D Kinematics5 Lesson 2(Components)A.1. Determine the x- and y-components for the following vectors: a) 2.50 km at 180b) 3.90 m South c) 27.0 km at 340d) 928 m at 43.0N of W 2. A vector has an x-component of 160 m East and a direction of 71.0N of E. a) What is the y-component of the vector? b) The magnitude of the vector? 3. You wish to add three vectors. Their components are shown below: Vector 1: x1= 15.0 km West y1= 46.0 km North Vector 2: x2= 67.0 km East y2= 93.0 km South Vector 3: x3= 80.0 km West y3= 12.0 km South What are the magnitude and direction of the resultant (sum) vector?
Physics 20 Workbook Unit 2: 2-D Kinematics6 4. Add the following vectors, using components. a) 45.0 m North, 60.0 m at −210b) 3.90 km at 62.0S of E, 5.80 km at 24.0S of W B. 5. A vector has a y-component of 3.10 km South. The x-component is directed East and the magnitude of the resultant vector is 5.00 km. a) What is the x-component? b) What is the angle of the resultant vector? 6. Add 7.00 km at 18.0W of N and 11.0 km at 69.0E of N using components. 7. Two vectors are added: 1V= 460 units at 81.0N of W and 2V. If the resultant vector is 270 units at 25.0N of E, then determine 2V. SOLUTIONS 1. a) x= 2.50 km W ; y= 0 b) x= 0 ; y= 3.90 m S c) x= 25.4 km E ; y= 9.23 km S d) = 137; x= 679 m W ; y= 633 m N 2. a) y= 465 m N b) R= 491 m 3. xT= 28 km W ; yT= 59 km S ; R= 65.3 km at 64.6S of W (25.3W of S) 4. a) Rx= −51.9615 m , Ry= 75 m ; R= 91.2 m at 55.3N of W (34.7W of N) b) Rx= −3.4677 km ; Ry= −5.8026 km; R= 6.76 km at 59.1S of W (30.9W of S) 5. a) x= 3.92 km E b) 38.3S of E (51.7E of S) 6. Rx= 8.1063 km , Ry= 10.5994 km ; R= 13.3 km at 52.6N of E (37.4E of N) 7. x2= 316.66 , y2= −340.23 ; V2= 465 units at 47.1S of E (42.9E of S)
Physics 20 Workbook Unit 2: 2-D Kinematics7 Lesson 3(Relative Velocity) A. 1. A boat can travel at 7.0 m/s in still water. If it is placed in a river that has a current of 3.0 m/s South, what would be the boat’s velocity with respect to the shore if the boat headed:a) upstream b) downstream 2. A boat that can travel at 11.0 m/s in still water is placed into a 135 m wide river that has a current of 8.0 m/s North. If it is aimed directly West, then calculate: a) the velocity of the boat with respect to the ground b) the time it takes to travel to the other side of the river c) how far downstream the boat travels 3. A river that is 57.0 m wide has a current that is 4.90 m/s West. A boat can travel at 6.50 m/s in still water. If the boat must travel directly North across the stream, a) at what angle must the boat be aimed? b) what would be the magnitude of boat’s velocity (with respect to the shore)?c) how much time would it take to reach the opposite shore? B. 4. A plane that can travel at 400 km/h in still air heads directly into a 50.0 km/h wind that blows directly West. How long would it take this plane to reach its destination, which is 1900 km due East?
Physics 20 Workbook Unit 2: 2-D Kinematics8 5. A plane that can travel at 200 km/h in still air heads directly South. However, a wind comes from the West (towards the East) and blows it off course. If the angle of the plane’s final velocity is 6.50E of S, then determine: a) the wind speed b) how far the plane travels in 3.00 hours 6. A plane must travel due East, but there is a wind that is blowing from the North (towards the South) at 45.0 km/h. If the plane’s actual speed is 170 km/h, then determine: a) the heading (angle) of the plane b) the speed of the plane if it was in still air 7. A plane that can travel at 120 km/h in still air heads towards the West. However, there is a 55.0 km/h wind that is directed at 70.0N of E. What is the actual velocity of the plane (with respect to the ground)? SOLUTIONS 1. a) 4.0 m/s upstream b) 10.0 m/s downstream 2. a) 14 m/s at 36N of W (54W of N) b) 12 s c) 98 m 3. a) 48.9E of N (41.1N of E) b) 4.27 m/s c) 13.3 s 4. v= 350 km/h East ; t= 5.43 h 5. a) 22.8 km/h b) v= 201.29 km/h ; d= 604 km 6. a) 14.8N of E (75.2E of N) b) 176 km/h 7. xT= −101.1889 km/h ; yT= 51.6831 km/h ; v= 114 km/h at 27.1N of W (62.9W of N)
Physics 20 Workbook Unit 2: 2-D Kinematics9 LESSON 4(1-D Projectiles) Assume 1-D motion and no air resistance. A.1. A rock is thrown upward at a speed of 15.0 m/s. a) What would its maximum height be (measured from its release height)? b) How much time would it take to go up and then return to its original height? 2. On Planet X, an object is launched upward with a speed of 28.0 m/s. If it takes a total time of 7.50 seconds for it to go up and return to its original (release height), a) what is the acceleration due to gravity on this planet? b) what is the maximum height of this object? 3. An object is launched upward at a speed of 26.0 m/s. What is this object’s velocity when it reaches a height of 30.0 m above its release height? Why are there two answers? 4. An object is launched upward at a speed v. If it is in the air for 3.80 s and its final position is 11.0 m above the original position, then what is the speed v? B. 5. An object is dropped from a height of 85.0 m. What is the object’s average velocity?
Physics 20 Workbook Unit 2: 2-D Kinematics10 6. An object is thrown upward at a speed of 17 m/s. How much time would it take for this object to reach a position that is 10 m below its original position? Hint: Find the final velocity first. 7. On Planet G, an object is thrown upward and it flies through the air for 16.0 seconds. If the object’s velocity is 13.0 m/s downward when it is located 24.0 m below its original position, then what is the acceleration due to gravity on this planet? 8. On Planet Y, an object that is dropped from a height of 12.0 m takes 3.20 s to reach the ground. If another object on this planet is thrown up in the air and it takes 8.50 s to reach its original position, then to what maximum height does it reach? SOLUTIONS 1. a) 11.5 m b) 3.06 s 2. a) 7.47 m/s2down b) 52.5 m 3. It reaches this height going up and going down. v= 9.35 m/s up and 9.35 m/s down 4. 21.5 m/s 5. vf= 40.837 m/s downwards ; vavg= 20.4 m/s downward 6. vf= 22.0273 m/s downwards ; t= 4.0 s 7. 1.44 m/s2downwards 8. a= 2.3438 m/s2downwards ; d= 21.2 m
Physics 20 Workbook Unit 2: 2-D Kinematics11 Lesson 5(Horizontal Projectiles) Assume no AR and a level surface. A.1. A rock is thrown horizontally off of a 10 m high building with a speed of 14 m/s. How far away from the base of the building will the rock hit? 2. A crate is dropped from a plane that is moving forward at a speed vat a height of 85.0 m. If the crate lands 120 m horizontally from the release position, then what is the speed v? 3. A bullet is projected horizontally from a height Hat a speed of 350 m/s. If it lands 225 m horizontally from the release position, than what is the initial height H? 4. A rock is thrown horizontally off of a cliff that is 35.0 m high with a speed of 18.0 m/s. What is the velocity of the rock just before it hits? Hint: Find both components of the final velocity first.
Physics 20 Workbook Unit 2: 2-D Kinematics12 B. 5. On Planet X, a rock is thrown horizontally at a speed of 29.0 m/s and at a height of 3.20 m. If the rock travels a horizontal distance of 52.0 m, then what is the acceleration due to gravity? 6. A rock is thrown horizontally from a height of 2.50 m. When it lands, the angle of its velocity vector is 36.0below the horizontal. What was the initial speed? 7. A car driver, travelling at 90.0 km/h, notices that he is heading directly for a cliff when he is 20.0 m away from the edge. He applies the brakes and experiences a backward acceleration of 8.00 m/s2, but it is not enough. The car goes over the cliff. If the cliff is 12.0 m high, then calculate how far the car travels horizontally while in the air (until the car lands). SOLUTIONS 1. t= 1.4278 s ; d= 20 m 2. t = 4.1628 s ; v= 28.8 m/s 3. t = 0.6429 s ; h= 2.03 m 4. vfx= 18 m/s forward ; vfy= 26.2 m/s down ; v= 31.8 m/s at 55.5below horizontal 5. t= 1.7931 s ; a= 1.99 m/s2down 6. vfy= 7.0036 m/s ; vi= vfx= = 9.64 m/s 7. Moving at 17.4642 m/s when it reaches the cliff’s edge.While in air: t= 1.5641 s ; d= 27.3 m
Physics 20 Workbook Unit 2: 2-D Kinematics13 Lesson 6(Diagonal Projectiles) Assume no AR and a level surface. A. 1. A golf ball is struck with an initial velocity of 47.0 m/s at an angle of 28.0above the horizontal. If this ball lands and rolls and additional 15.0 m, then determine the maximum height that this ball achieves. 2. A football is kicked with an initial velocity of 25 m/s at a projection angle of 34. Will this ball go over the 9.2 m uprights, if they are located 40 m away? Prove by calculation. B. 3. An object is thrown with an initial speed of 14.0 m/s. If the maximum height of this object is 6.00 m, then what was the projection angle?
Physics 20 Workbook Unit 2: 2-D Kinematics14 4. A soccer ball is projected at an angle of 31.0. If it stays in the air for a total time of 3.00 seconds, then what was its initial speed? 5. For the world record shot put in 1990, the shot was projected at a velocity of 14.5 m/s at 37above the horizontal. If the release height was 2.0 m, then how far (horizontally) was it thrown? Hint: Find the vertical component of the final velocity first. SOLUTIONS 1. a) t= 4.4985 s ; dair= 186.68 m ; dT= 202 m b) hmax= 24.8 m 2. t= 1.93 s ; h= 8.71 m Not high enough (h< 9.2 m) 3. viy= 10.8499 m/s ; = 50.8above the horizontal 4. viy= 14.715 m/s ; vi= 28.6 m/s 5. vfy= 10.7419 m/s downward ; t= 1.9845 s ; d= 23 m