md1 - This print-out should have 18 questions...

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Version 072/ABACA – midterm 01 – Turner – (58120) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The speed of an arrow fired from a compound bow is about 64 m / s. An archer sits astride his horse and launches an arrow into the air, elevating the bow at an angle of 53 above the horizontal and 1 . 4 m above the ground. The acceleration of gravity is 9 . 81 m / s 2 . 64 m / s 53 1 . 4 m range What is the arrow’s range? Assume: The ground is level. Ignore air resistance. 1. 261.907 2. 24.4111 3. 323.976 4. 389.927 5. 53.9349 6. 537.785 7. 402.411 8. 160.421 9. 490.125 10. 484.748 Correct answer: 402 . 411 m. Explanation: Let : v o = 64 m / s , θ = 53 , h = 1 . 4 m , and g = 9 . 81 m / s 2 . v o θ h range In the projectile motion, we have x = ( v o cos θ ) t t = x v o cos θ horizontally, and y = h + ( v o sin θ ) t 1 2 g t 2 vertically. Thus y = h + x v o sin θ v o cos θ 1 2 g x 2 v 2 o cos 2 θ = 0 when the arrow lands. Thus 2 v 2 o h cos 2 θ + 2 x v 2 o sin θ cos θ g x 2 = 0 2 v 2 o h cos 2 θ x v 2 o sin 2 θ + g x 2 = 0 x = v 2 o sin 2 θ ± radicalBig v 4 o sin 2 2 θ + 8 g v 2 o h cos 2 θ 2 g . Since v 4 o sin 2 2 θ + 8 g v 2 o h cos 2 θ = (64 m / s) 4 (sin 2 106 ) + (8) (9 . 81 m / s 2 ) × (64 m / s) 2 (1 . 4 m) (cos 2 53 ) = 1 . 56655 × 10 7 m 2 / s 2 , then x = (64 m / s) 2 (sin 106 ) 2 (9 . 81 m / s 2 ) + radicalbig 1 . 56655 × 10 7 m 2 / s 2 2 (9 . 81 m / s 2 ) = 402 . 411 m . 002 10.0 points Consider the plot describing motion along a straight line with an initial position of 10 m.
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Version 072/ABACA – midterm 01 – Turner – (58120) 2 2 1 0 1 2 3 4 1 2 3 4 5 6 7 8 9 time (s) velocity (m/s) What is the position at 9 seconds? 1. 29.0 2. 23.5 3. 31.5 4. 30.0 5. 21.0 6. 19.5 7. 28.0 8. 41.5 9. 44.0 10. 24.0 Correct answer: 24 m. Explanation: 1 2 3 4 5 6 7 8 9 0 1 2 3 4 1 2 time (s) velocity (m/s) The acceleration during the first 2 seconds is a = Δ v Δ t = 3 m / s 0 m / s 2 s 0 s = 1 . 5 m / s 2 . The position at 2 seconds is 10 m plus the area of the triangle (shaded above) x = 10 m + 1 2 (2 s 0 s)(3 m / s 0 m / s) = 13 m ; however, it can also be calculated: x = x i + v i ( t f t i ) + 1 2 a ( t f t i ) 2 = 10 m + (0 m / s) (2 s 0 s) + 1 2 (1 . 5 m / s 2 )(2 s 0 s) 2 = 13 m . The acceleration during the second time interval is a = Δ v Δ t = 4 m / s 3 m / s 6 s 2 s = 0 . 25 m / s 2 . The position at 6 seconds is 13 m plus the area of the trapezoid x = 13 m + 1 2 (6 s 2 s)(4 m / s + 3 m / s) = 27 m ; however it can also be calculated: x = x i + v i ( t f t i ) + 1 2 a ( t f t i ) 2 = 13 m + (3 m / s) (6 s 2 s) + 1 2 (0 . 25 m / s 2 )(6 s 2 s) 2 = 27 m . The acceleration during the third time in- terval is a = Δ v Δ t = 2 m / s 0 m / s 9 s 6 s = 0 . 666667 m / s 2 . Finally the position at 9 seconds is 27 m plus the area of the triangle x = 27 m + 1 2 (9 s 6 s)( 2 m / s 0 m / s) = 24 m ; however it can also be calculated: x = x i + v i ( t f t i ) + 1 2 a ( t f t i ) 2 = 27 m + (0 m / s) (9 s 6 s) + 1 2 ( 0 . 666667 m / s 2 )(9 s 6 s) 2 = 24 m .
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Version 072/ABACA – midterm 01 – Turner – (58120) 3 003 10.0 points In the figure below the left-hand cable has a tension T 1 and makes an angle of 46 with the horizontal. The right-hand cable has a tension T 3 and makes an angle of 53 with the horizontal. A W 1 weight is on the left and a W 2 weight is on the right.
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