Version 001 – Quest 2 Review Vectors &amp; 2d kinematics – tubman – (Phys1A2122p3) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 1 ~ =C ~ +D ~ 2. P ~ =C ~ +B ~ 3. P ~ =A ~ −B ~ correct 4. P 001 (part 1 of 2) 10.0 points Vector Addition 05 88051 ~ Express the vector R C R A P D B ~ ~ ~ ~ the edges of a in terms of A, B, C, and D, parallelogram. ~ =D ~ −A ~ 1. R ~ =A ~ −D ~ 2. R ~ =A ~ −B ~ 3. R ~ =C ~ +B ~ 4. R ~ =B ~ +D ~ 5. R ~ =B ~ +A ~ correct 6. R ~ =A ~ +D ~ 7. R ~ =A ~ −C ~ 8. R ~ =B ~ −A ~ 9. R ~ =C ~ +D ~ 10. R Explanation: Apply the parallelogram rule of addition: ~ and B; ~ the join the tails of the two vectors A resultant vector is the diagonal of a parallel~ and B ~ as two of its ogram formed with A sides. 002 (part 2 of 2) 10.0 points ~ in terms of A, ~ B, ~ C, ~ and Express the vector P ~ D, ~ =B ~ −A ~ 1. P ~ =C ~ −A ~ 5. P ~ =B ~ +D ~ 6. P ~ =B ~ +A ~ 7. P ~ =D ~ −A ~ 8. P ~ =A ~ −D ~ 9. P ~ =A ~ +D ~ 10. P Explanation: By the triangle method of addition ~ +P ~ =A ~ B ~ =A ~ −B ~ . P 003 10.0 points Resultant Displacement 05 109772 A person walks 34 m East and then walks 31 m at an angle 30◦ North of East. What is the magnitude of the total displacement? 1. 76.6011 2. 60.2479 3. 56.1532 4. 70.9834 5. 62.79 6. 67.4562 7. 58.459 8. 57.5253 9. 66.7191 10. 56.8649 Correct answer: 62.79 m. Explanation: Let : Ra = 34 m , Rb = 31 m , θ = 30◦ . and Version 001 – Quest 2 Review Vectors &amp; 2d kinematics – tubman – (Phys1A2122p3) The total displacement is the vector sum of the two displacements, so Rx = Ra + Rb cos θ = 34 m + (31 m) cos 30◦ = 60.8468 m and Ry = Rb sin θ = (31 m) sin 30◦ = 15.5 m . The final displacement is R= = q q Rx2 + Ry2 (60.8468 m)2 + (15.5 m)2 = 62.79 m . 004 (part 1 of 2) 10.0 points Blown Off Course 168934 A ship is expecting to travel to its home port 510 km due East. Before the ship starts to travel, a severe storm comes up and blows the ship 490 km due South. How far is the ship from its home port? 1. 776.981 2. 643.506 3. 672.012 4. 806.04 5. 514.782 6. 797.559 7. 477.074 8. 707.248 9. 766.094 10. 757.694 Correct answer: 707.248 km. Explanation: Let : dship = 510 km and dwind = 490 km . Choose a coordinate system with the positive x-axis representing 0◦ and the positive y-axis representing 90◦ . 2 N Ship W ind R θ E The displacements are at right angles to each other, so q ~ kRk = d2ship + d2wind q = (510 km)2 + (490 km)2 = 707.248 km . 005 (part 2 of 2) 10.0 points At what angle North of East must the ship travel to reach its destination? Let East be 0◦ and North 90◦ . 1. 32.7352 2. 25.8664 3. 28.8427 4. 23.9625 5. 30.1137 6. 25.0675 7. 26.5651 8. 21.6444 9. 43.8543 10. 34.8545 Correct answer: 43.8543 degrees. Explanation: The ship was blown to the South, so it must travel to the North and East to reach its destination. If θ is the angle its destination makes with East, then W is the side opposite θ and S is the side adjacent, so that W 490 km θ = arctan = arctan S 510 km = 43.8543◦ . 006 (part 1 of 2) 10.0 points Version 001 – Quest 2 Review Vectors &amp; 2d kinematics – tubman – (Phys1A2122p3) Big Bertha 36992 During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.15 km/s at an initial inclination of 22.7◦ to the horizontal. The acceleration of gravity is 9.8 m/s2 . How far away did the shell hit? 1. 559.786 2. 408.362 3. 283.289 4. 85.7919 5. 117.531 6. 186.477 7. 24.8054 8. 641.356 9. 483.685 10. 335.851 Correct answer: 335.851 km. Explanation: The range R is given by v02 sin(2 θ0) g (2150 m/s)2 sin 45.4◦ = 9.8 m/s2 = 335.851 km . R= 007 (part 2 of 2) 10.0 points How long was it in the air? 1. 319.039 2. 142.012 3. 96.466 4. 169.326 5. 351.341 6. 220.215 7. 264.121 8. 104.188 9. 232.938 10. 493.495 Correct answer: 169.326 s. Explanation: The time in the air is R t= v0x R = v0 cos θ0 3 3.35851 × 105 m (2150 m/s) cos 22.7◦ = 169.326 s . = 008 (part 1 of 4) 10.0 points Horizontal Projection 40319 A ball is thrown horizontally from the top of a building 47.7 m high. The ball strikes the ground at a point 108 m from the base of the building. The acceleration of gravity is 9.8 m/s2 . Find the time the ball is in motion. 1. 2.39046 2. 1.96396 3. 3.17516 4. 3.12005 5. 2.79212 6. 2.28125 7. 2.82482 8. 2.10926 9. 1.98463 10. 2.13331 Correct answer: 3.12005 s. Explanation: In this solution, we take the origin of the x and y axes at the base of the building. The positive y direction is taken to point upward, so that the vertical acceleration is −g. The time the ball is in flight is found from y = y0 + vy0 t + 1 a y t2 2 1 0 = y 0 + 0 − g t2 2 r 2 y0 t= g s 2 (47.7 m) = 9.8 m/s2 = 3.12005 s . 009 (part 2 of 4) 10.0 points Find the initial velocity of the ball. 1. 38.0337 Version 001 – Quest 2 Review Vectors &amp; 2d kinematics – tubman – (Phys1A2122p3) 2. 34.6149 3. 39.5346 4. 61.3605 5. 10.9751 6. 31.0983 7. 29.6678 8. 15.041 9. 11.2234 10. 42.8439 4 1. -31.085 2. -25.5859 3. -30.5765 4. -28.4169 5. -33.0708 6. -29.1657 7. -17.542 8. -25.8905 9. -16.6829 10. -18.783 Correct answer: 34.6149 m/s. Explanation: There is no acceleration in the horizontal direction. Therefore, we have x = x0 + vx0 t x = 0 + vx0 t x vx0 = t 108 m = 3.12005 s = 34.6149 m/s . 010 (part 3 of 4) 10.0 points Find the x component of its velocity just before it strikes the ground. 1. 35.9132 2. 29.6678 3. 10.9751 4. 21.3483 5. 34.6149 6. 38.515 7. 19.5451 8. 56.0463 9. 42.8439 10. 58.5836 Correct answer: −30.5765 m/s. Explanation: The y component of the velocity just before the ball strikes the ground is vy = vy0 + ay t = 0−gt = −(9.8 m/s2 ) (3.12005 s) = −30.5765 m/s . 012 (part 1 of 3) 10.0 points AP B 1993 MC 64 66 84565 A ball is thrown and follows the parabolic path shown. Point Q is the highest point on the path and points P and R are the same height above the ground. Q P R Correct answer: 34.6149 m/s. Explanation: Since there is no acceleration in the horizontal direction, the x component of the velocity is constant. So vx = vx0 = 34.6149 m/s . 011 (part 4 of 4) 10.0 points Find the y component of its velocity just before it strikes the ground. How do the speeds of the ball at the three points compare? Air friction is negligible. 1. k~vQ k < k~vR k < k~vP k 2. k~vQ k < k~vP k = k~vR k correct Version 001 – Quest 2 Review Vectors &amp; 2d kinematics – tubman – (Phys1A2122p3) 3. k~vP k = k~vR k < k~vQ k 5 9. 4. k~vR k < k~vQ k < k~vP k 5. k~vP k < k~vQ k < k~vR k Explanation: The speed of the ball in the x-direction is constant. Because of gravitational acceleration, the speed in the y-direction is zero at point Q. Since points P and R are located at the same point above ground, by symmetry they have the same vertical speed component (though they do not have the same velocity), so vQ < vP = vR . Explanation: Since air friction is negligible, the only acceleration on the ball after being thrown is that due to gravity, which acts straight down. 014 (part 3 of 3) 10.0 points Which diagram best indicates the direction of the net force, if any, on the ball at point Q? 1. 2. 013 (part 2 of 3) 10.0 points Which diagram best indicates the direction of the acceleration, if any, on the ball at point R? correct 3. 4. 1. 5. 2. 6. 3. 4. correct 5. 6. 7. The ball is in free fall and there is no acceleration at any point on its path. 8. 7. 8. 9. The ball is in free-fall and there is no acceleration at any point on its path. Explanation: By Newton’s second law, the force is in the direction of the acceleration (downward). 015 (part 1 of 3) 10.0 points AP B 1994 FR 1 short 95850 Version 001 – Quest 2 Review Vectors &amp; 2d kinematics – tubman – (Phys1A2122p3) A ball of mass 0.6 kg, initially at rest, is kicked directly toward a fence from a point 20 m away, as shown below. The velocity of the ball as it leaves the kicker’s foot is 17 m/s at angle of 57 ◦ above the horizontal. The top of the fence is 7 m high. The ball hits nothing while in flight and air resistance is negligible. The acceleration due to gravity is 9.8 m/s2 . m/ s b b b b b b b b b b b b b 17 b b b b 7m ◦ b 57 b b 20 m Determine the time it takes for the ball to reach the plane of the fence. 1. 1.49296 2. 1.86943 3. 1.70916 4. 1.37341 5. 1.55884 6. 1.51383 7. 1.79945 8. 1.35642 9. 2.16009 10. 2.05111 Correct answer: 2.16009 s. Explanation: Let : θ = 57 ◦ and d = 20 m . The horizontal component of the velocity is constant, so vhoriz = v0 cos θ = (17 m/s) cos 57 ◦ = 9.25886 m/s . The horizontal motion defines the time of flight: vhoriz t = d t= 6 d vhoriz 20 m = 9.25886 m/s = 2.16009 s . b 016 (part 2 of 3) 10.0 points How far above the top of fence will the ball pass? Consider the diameter of the ball to be negligible. 1. 0.207175 2. 1.42074 3. 1.01814 4. 0.962449 5. 0.437954 6. 0.305372 7. 0.403716 8. 0.933905 9. 0.81545 10. 0.706199 Correct answer: 0.933905 m. Explanation: The vertical component of the initial velocity is vvert = v0 sin θ = (17 m/s) sin 57 ◦ = 14.2574 m/s . The height of the ball during its flight is given by 1 g t2 2 = (14.2574 m/s) (2.16009 s) 1 − (9.8 m/s2 ) (2.16009 s)2 2 = 7.93391 m . y = vvert t − Therefore, the distance that the ball passes above the fence is ∆y = (7.93391 m) − (7 m) = 0.933905 m . 017 (part 3 of 3) 10.0 points Version 001 – Quest 2 Review Vectors &amp; 2d kinematics – tubman – (Phys1A2122p3) What is the vertical component of the velocity when the ball reaches the plane of the fence? 1. -6.455 2. -6.71698 3. -6.59675 4. -7.18939 5. -2.87927 6. -4.20553 7. -6.9115 8. -4.59697 9. -6.01041 10. -7.03111 Correct answer: −6.9115 m/s. Explanation: Thus the the vertical component of the velocity when the ball reaches the plane of the fence is vvert = v0 sin θ − g t = (17 m/s) sin 57 ◦ − (9.8 m/s2 ) (2.16009 s) = −6.9115 m/s . Vertical Component of Velocity (m/s) This is verified by analyzing the graph below Vertical Velocity vs Time 25 20 15 10 5 0 −5 −10 −15 −20 −25 0 0.5 1.0 1.5 Time (s) 2.0 2.5 7