Midterm 1 Solutions (1)

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1 © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. BINGHAMTON UNIVERSITY Department of Physics, Applied Physics and Astronomy S O L U T I O N S PHYS 121: Midterm Exam I Wednesday, Sept. 20, 2023 Fall 2023 INSTRUCTIONS: 1. Carefully read all problems and solve all of them. Box your answers. Observe the point value for each problem, as they may differ. All 4 long answer problems must be done in pen ; failure to do so will result in forfeiting your access to a re-grade. All final answers for multiple choice questions must be placed in your Gradescope bubble sheet using a pen or a pencil . You will not receive credit for the multiple choice questions if you do not write your name on the top of the Gradescope bubble sheet. 2. It is your responsibility to print your name on ALL pages of this exam. For grading, the exam will be separated, and pages could get mixed up if your name is not clearly printed at the top of all pages, thereby resulting in a lower grade. 3. Answers to each problem should be written on the specific sheet that contains the question. Use both sides of the same sheet for your answers. Anything placed on a different page will not be graded. Any scrap paper given to you will be discarded; do not use this for your answers. 4. Show all your work and reasoning to receive full credit . Be sure your work is clearly shown, and your answers are circled with appropriate units. Don’t make the graders have to guess what you are trying to convey. 5. Cross out all unwanted work, as incorrect statements may cost you points. 6. This examination is to be taken without the aid of any notes, textbooks, electronic devices (except non-alphanumeric calculators), or other individuals. Any violations of this condition will be referred to the Academic Honesty Committee, and this particular exam will earn you no credit while also negating your opportunity for a grade replacement. 7. If a problem has multiple parts, and a latter part requires an answer from an earlier part for which you could not solve, state an assumed answer for the earlier part , and then use it in the latter part. 8. You may use the formula sheet attached to the back of this exam. You may also remove the formula sheet for easier viewing.
2 © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. Part I: Multiple Choice Section (2.5 Points Each) Answer in pen or pencil on your Gradescope bubble sheet. Write your name and B# on the top of the Gradescope bubble sheet as failing to do so will result in a loss of credit from this section. You may write on the multiple choice questions, but only the Gradescope bubble sheet will be graded and retained. Each question is worth 2.5 points. Choose the closest answer. Some questions may have 4 choices, while other questions have 5 choices. 1. Using significant figures, how many sig figs would the sum of 0.9090 + 0.009 have? a. Two b. Three c. Four d. Five 2. For the measurement shown, which is the correct statement of length with the associated uncertainty? 3. Rufus swims straight across a river when there is no current, taking him 7 minutes to reach the other side. The next day he again aims straight across and swims at the same speed relative to still water as the previous day. However, a current of 1 m/s is flowing perpendicular to the direction he initially aims to swim. Which of the following is a true statement? a. Rufus swims straight across the river b. Rufus’s magnitude of velocity relative to earth is slower c. Rufus still crosses the river in 7 minutes. d. Rufus crosses to a spot upstream (against the current) 4. A wind blows horizontal to the ground, during which an aerodynamic mass is dropped from a height. Which description best depicts what happens? a. The mass lands at the same time as if there was no wind. b. The mass lands at a later time than if there was no wind. c. The mass is in freefall. d. The mass is moving in one dimension. 5. What condition must be true for two objects moving in one dimension that collide? a. They have the same displacement b. They move the same distance c. They have the same position d. They have the same acceleration 6. The net force on an object in freefall is best described as a. Zero b. 1 Newton c. 9.8 Newtons d. equal to its weight e. none of these choices 7. A buoy floats in still water; that is, the buoy is supported by the water. The reaction force is: a. The force on the water from the buoy b. The force on the earth from the water c. The force of gravity on the buoy d. The gravitational pull of the buoy upon the earth 8. An object is suspended from a massless rope (shown). But, if the rope did have mass, then how would the tension at spot a compare to the tension at spot b ? a. T a > T b b. T a < T b c. T a = T b d. T a = T b = 0 Due to the weight of the rope 9. A box sits on an inclined plane. Jorge begins putting extra mass in the box, to get it to slide down the inclined plan. Which statement is true? a. As mass is added, the pull of gravity increases, and the box will slide down the incline. b. As mass is added, the pull of gravity decreases, and the box will remain stationary on the incline. c. As mass is added, the pull of gravity increases, and the friction decreases, so the box will slide downward. d. As mass is added, the pull of gravity increases, and the friction increases, so the box will remain stationary. 10. A car is parked on the road. To change the coefficient of friction between the car and the road, what must occur? a. Increase the mass of the tires b. Decrease the mass of the tires c. Change what the tires are made of d. Increase the weight of the car Last Name First Name a. (3.9±1) cm b. (3.9±1) mm c. 3.9 cm ± .1 mm d. 3.9 cm ± 1 mm a b
3 Answer in PEN © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. PHYS 121 Midterm I Wednesday, Sept. 20, 2023 1. Rufus is on top of a 20.0 m high building and accidentally drops his prized banana (whoops!). Jorge, directly below Rufus, aims his slingshot to intercept the banana with a stone before the banana can hit the ground. Jorge fires his stone straight up from the ground at a speed of 19.8 m/s just as the banana is dropped. a) If the banana were not in the way, calculate the maximum height the stone could go. b ) If the stone were not in the way, calculate the impact velocity of the banana upon contacting the ground. c) Calculate the time of collision between the banana and stone. d) Calculate the position above the ground at which the banana and stone collide. ?) ??? ?ℎ? ?????: 𝑉 2 = 𝑉 0 2 − 2?ℎ → 0 2 = 19.8 2 − 2(9.80)ℎ → ℎ = 20.0? ?) ??? ?ℎ? ??????: 𝑉 2 = 𝑉 0 2 − 2?ℎ → 𝑉 = √0 2 − 2(9.80)(−20. ) = 19.7989 ≈ 19.8 ?/? ?) ? 𝑏 = 20 − 4.9? 2 ??? ? 𝑠 = 19.8? − 4.9? 2 𝐴? ????𝑖?𝑖??, ? 𝑏 = ? 𝑠 𝐴???? ????𝑖?𝑖?? ????𝑖?𝑖??: 20 − 4.9? 2 = 19.8? − 4.9? 2 → ? = 1.01? ?) ? 𝑏 = 20 − 4.9(1.01) 2 = 15.0? 𝑂? ? 𝑠 = 19.8(1.01) − 4.9(1.01) 2 = 15.0? Last Name First Name For Graders Only [24 Pts ] 6pt 6pt 8pt 4pt
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4 Answer in PEN © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproducti on of this material is prohibited without the author’s consent. PHYS 121 Midterm I Wednesday, Sept. 20, 2023 2. PART A: A penalty kick in soccer is made 11.0 m from the goal. The top of the goal post is 2.44 m above the ground. Assuming the goalie is not a factor: if the soccer ball is kicked straight toward the goal (starting at ground level) with a velocity of 58.1 m/s, (directed at) 15.0 degrees above the ground, then calculate: a) the time at which the ball reaches the goal. b) whether the ball goes under the goal post or goes over the goal post. ?) ? ? = 11.0 = 58.1 cos(15.0) ? → ? = 0.1960 ≈ 0.196? ?) ? ? = 58.1 sin(15.0) (0.196) − 4.9(0.196) 2 = 2.76? ?? ?𝑂?? 𝑂𝑉?? ?ℎ? ???? ????. PART B: Rufus has decided to compete in a dogsled race which requires each racer to use only three dogs to pull their sleds with. After Rufus has attached Dog A , Dog B , and Dog C to his sled (by using a massless rope to do so for each dog), Jorge accidently fires the starting gun, scaring the three dogs enough to cause each of them to run in a different direction away from the sled until all their ropes become tight. Amazingly, the sled does not move as the three dogs continue to pull on it simultaneously, each of them doing so with a separate (but constant) force of their own: Dog A pulls on the sled with a force of F A = 18.4 N, (directed at) 30.0° South of East , while Dog B pulls on the sled with a force of F B = 16.0 N, (directed at) due North . c) What condition must be true for the sled to behave as described in the situation above? (Write an expression and/or words to explain your answer for this part.) d) Based on this condition, calculate the force F C (magnitude and direction) that Dog C is pulling on the sled with. c) ∑ ? on sled = 0 → For the sled not to move, the sum of all forces acting on it must be equal to zero . d) In general: ∑ ? on sled = ? A + ? B + ? C = 0 → ? C = −? A − ? B → ? C = −(? A + ? B ) so: ? A,x + ? B,x = ? A cos 𝜃 A + ? B cos 𝜃 B = (18.4 N) cos(330°) + (16.0 N) cos(90°) = 15.935 N ? A,y + ? B,y = ? A sin 𝜃 A + ? B sin 𝜃 B = (18.4 N) sin(330°) + (16.0 N) sin(90°) = 6.80 N |? A + ? B | = (? A,x + ? B,x ) 2 + (? A,y + ? B,y ) 2 = √(15.935 N) 2 + (6.80 N) 2 = 17.325 N ≈ 17.3 N 𝜃 𝐹 A +𝐹 B = tan −1 ( 𝐹 A,y +𝐹 B,y 𝐹 A,x +𝐹 B,x ) = tan −1 ( 6.80 N 15.935 N ) = tan −1 (0.42673) = 23.109° ≈ 23.1° ? A + ? B = 17.3 N @ 23.1° North of East ∴ ? C = −(? A + ? B ) = 17.3 N @ 23.1° South of West For Graders Only Last Name First Name [ 24 Pts ] 4pt 6pt 4pt 10pt
5 Answer in PEN © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. PHYS 121 Midterm I Wednesday, Sept. 20, 2023 3. Two blocks, Block M and Block m , are connected to each other by a massless string that passes over a massless , frictionless pulley as shown in the diagram. Block M , with a mass of M = 3.00 kg , is suspended in the air while Block m , with a mass of m = 2.00 kg , is sitting on an inclined plane which makes an angle of θ = 30.0° with respect to the horizontal. The coefficient of kinetic friction between Block m and the surface of the incline is known to be μ k = 0.250 , and the suspended block is observed to immediately move downward once this system is released from rest . a) Using the figures below, draw and label all forces acting on each block: b) Using Newton’s 2nd Law, write the x -direction and y -direction equations of motion (i.e., sum the forces) for Block m . (Note: these two expressions should be written using symbols only .) c) Using Newton’s 2nd Law, write the equation of motion (i.e., sum the forces) for Block M . (Note: this expression should be written using symbols only .) d) Calculate the magnitude of the kinetic frictional force F f,k acting on Block m . e) Calculate the magnitude of the acceleration a for the (two-block) system. f) Calculate the tension T of the string connecting the two blocks to each other. a) See the figures above. b) In the x -direction: ∑ ? x = ? − ? f,k − ?? sin 𝜃 = ?? Eq. 1 In the y -direction: ∑ ? y = ? N − ?? cos 𝜃 = 0 Eq. 2 c) Taking the direction of “ a to be positive: ∑ ? = 𝑀? − ? = 𝑀? Eq. 3 d) Rearrange Eq. 2 to find the normal force F N : ? N = ?? cos 𝜃 ∴ ? f,k = 𝜇 k ? N = 𝜇 k ?? cos 𝜃 ? f,k = (0.250)(2.00 kg)(9.80 m/s 2 ) cos(30.0°) = 4.2435 N ≈ 4.24 N e) Add Eq. 1 and Eq. 3 to eliminate the tension T and then solve for the acceleration a : 𝑀? − ? f,k − ?? sin 𝜃 = ?? + 𝑀? 𝑀? − ? f,k − ?? sin 𝜃 = (? + 𝑀)? ? = 𝑀? − ? f,k − ?? sin 𝜃 ? + 𝑀 ? = (3.00 kg)(9.80 m/s 2 ) − 4.2435 N − (2.00 kg)(9.80 m/s 2 ) sin(30.0°) 2.00 kg + 3.00 kg = 3.0713 m/s 2 ≈ 3.07 m/s 2 f) Rearrange Eq. 3 to find the tension T : ? = 𝑀(? − ?) = (3.00 kg)(9.80 m/s 2 − 3.0713 m/s 2 ) = 20.186 N ≈ 20.2 N Last Name First Name For Graders Only [ 27 Pts ] 6pt mg F f,k T Mg T x y F N 7pt 3pt 4pt 5pt 2pt
6 © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. © 2023; J. Nale Reproduction of this material is prohibited without the author’s consent. Binghamton University Department of Physics, Applied Physics and Astronomy PHYS 121: Midterm Exam #1 Reference Sheet Wednesday, Sept. 20, 2023 Possible Useful Information: For ?? 2 + ?? + ? = 0, ? = −? ± √? 2 − 4?? 2? If ∅ + 𝜃 = 90°, sin(∅) = cos(𝜃) tan(𝜃) = sin(𝜃) cos(𝜃) sin(45°) = 1 √2 cos 2 (𝜃) − sin 2 (𝜃) = cos(2𝜃) cos 2 (𝜃) + sin 2 (𝜃) = 1 sin(𝜃) cos(𝜃) = 1 2 sin(2𝜃) ? = 9.80 m/s 2 , toward Earth′s center Mass of Earth = 5.97 × 10 24 kg Along the Surface of Earth: Possible Useful Equations: ∆? ⃗ = ? ⃗ − ? 0 𝑣 avg = ∆? ∆? ? avg = ∆𝑣 ∆? 𝑣 avg = 𝑣 ⃗ + 𝑣 0 2 ? 1,2 = −? 2,1 𝑣 ⃗ = 𝑣 0 + ? ⃗? 𝑣 2 = 𝑣 0 2 + 2? ⃗∆? ⃗ ∆? ⃗ = 𝑣 0 ? + 1 2 ? ⃗? 2 ? net = ?? ⃗ ? fr,k = 𝜇 k ? N Vertically upward (toward the sky): out of the page (+ z ) Vertically downward (toward the ground): into the page ( z ) South ( y ) East (+ x ) West ( x ) North (+ y )
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