Midterm Exam 1 October 2, 2012

Transcription

1 Midterm Exam 1 October 2, 2012 Name: Instructions 1. This examination is closed book and closed notes. All your belongings except a pen or pencil and a calculator should be put away and your bookbag should be placed on the floor. 2. You will find one page of useful formulae on the last page of the exam. 3. Please show all your work in the space provided on each page. If you need more space, feel free to use the back side of each page. 4. Academic dishonesty (i.e., copying or cheating in any way) will result in a zero for the exam, and may cause you to fail the class. In order to receive maximum credit, each solution should have: 1. A labeled picture or diagram, if appropriate. 2. A list of given variables. 3. A list of the unknown quantities (i.e., what you are solving for). 4. Graphical representations of the motion (e.g., position, velocity, acceleration vs. time), if appropriate. 5. A labeled 1D or 2D coordinate axis system, if appropriate. 6. The full equation or equations needed to solve the problem (e.g., appropriate equations of motion). 7. An algebraic solution of the unknown variables in terms of the known variables. 8. A final numerical solution, including units, with a box around it. 9. An answer to additional questions posed in the problem, if any. 1

2 1. One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its highest point, comes to within 10 cm of the end of the ramp without going off. You give the puck a push, releasing it with a speed of 6.0 m{s when it is 9 m from the end of the ramp. The puck s speed after traveling 2.5 m is 5.0 m{s. Are you a winner? Solution: This is a 1D kinematics question. We are given the puck s initial velocity v Di, and it s velocity after it has traveled 2.5 m. We are also given the length of the ramp, x f, and we are asked to find whether the puck stops within 10 cm of the end of the ramp. To solve this problem we need the 1D kinematic equation relating velocities, distance, and acceleration. We need to find the acceleration from the two velocities and the two speeds at the start of the puck s trip up the slide: v 2 f v 2 i 2apx f x i q (1) 2apx f x i q v 2 i v2 f (2) a v 2 i v2 f 2px f x i q (3) a p6 m{sq2 p5 m{sq 2 (4) 2p2.5q a 2.2 m{s 2 (5) We can use the same equation again, using the starting and ending velocity and the acceleration that we just found, where v f 0: (6) v 2 f v 2 i 2apx f x i q (7) 2

3 0 v 2 i 2apx f x i q (8) 2apx f x i q v 2 i (9) px f x i q v2 i 2a (10) p6 m{sq 2 2p2.2 m{s 2 q (11) 8.2 m (12) Unfortunately, you do not win! The puck stops well before the winning distance. To win, the puck has to stop at a distance between 8.9 and 9 m. (13) 3

4 2. As a science project, you drop a watermelon off the top of the Empire State Building, 350 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a constant speed of 40 m{s. How fast is the watermelon going when it passes Superman? Solution: This is a 1D kinematics question. We are given the watermelon and Superman s starting position at 350 m and Superman s constant velocity of 40 m{s. We also know that the acceleration of gravity is 9.8m{s 2. To solve this problem we need use the 1D kinematic equations for the watermelon and Superman: x Sf x Si v Si t x W f x W i v W i t 1 2 a Sp tq 2 x Si v Si t, (14) 1 2 a W p tq 2 1 x W i 2 a W p tq 2. (15) We re trying to find the speed of the watermelon just as it passes Superman, or when the x Sf x W f. Setting these two equations equal to each other, we have: x Si v Si t x W i 1 2 a W p tq 2 (16) (17) Since the initial positions are the same, x Si x W i, and this equation becomes: v Si t 1 2 a W p tq 2 (18) (19) 4

5 We can factor t, and substitute a W g, and the equation simplifies to: v Si 1 g t (20) 2 (21) Solving for the time, t: t 2v Si g 2p 40 mq sec (22) 9.8 m{s (23) To find the velocity of the watermelon: v f v i a t (24) v f g t (25) v f 9.8 m{s 2 p8.2 secq 80.4 m{s (26) (27) 5

6 3. Let A 9î 3ĵ, B 7î 5ĵ, and D 3A 2B. (a) Write vector D in component form. (b) Draw a coordinate system and on it show vectors A, B, and D. (c) What are the magnitude and direction of vector D? Solution: (a) To write D in component form we first multiply A by 3 and B by 2. 3A 3 9î 3 3ĵ 27î 9ĵ (28) 2B 2 p 7qî 2 5ĵ 14î 10ĵ (29) (30) Next we add the x- and y-components: D 3A 2B (31) p27î 9ĵq p 14î 10ĵq (32) p27 14qî p 9 10qĵ (33) 41î 19ĵ. (34) (b) (c) The magnitude of the vector D is given by D a 41 2 p 19q 2 (35) 45.2, (36) 6

7 and the direction θ (measured positive counter-clockwise from the x-axis) is equal to θ tan 1 Dy D x tan or (37) 7

8 4. The figure shows three ropes tied together in a knot. One of your friends pulls on a rope with 3.0 units of force and another pulls on a second rope with 5.0 units of force. How hard and in what direction must you pull on the third rope to keep the knot from moving? Solution: This is a vector problem that uses x- and y- components of vectors. The three vectors must be balanced. This means that the x- and y - components of the three vectors must be balanced. Fx 0 (38) Fy 0 (39) The 3 unit force vector is only in the x-direction. The 5 unit force vector has x- and y- components. The x-component of the 5 unit force vector is found from SOH- CAH-TOA: (40) cos θ F x 5 (41) F x 5 cos θ 5 cosp60q 2.5 units (42) sin θ F y 5 (43) F y 5 sin θ 5 sinp60q 4.33 units (44) (45) 8

9 Now we add the x-components: Fx (46) For the unknown vector to balance the other forces on the knot, To find the magnitude of the force, (47) F x 0.5 (48) F y 4.33 (49) (50) D a p.5q 2 p4.33q 2 (51) 4.36 units (52) and the direction θ (measured positive counter-clockwise from the x-axis) is equal to θ tan 1 Dy D x tan S of W or θ (53) 9

10 5. A stunt-man drives a car at a speed of 35 m{s off a 25 m high cliff. The road leading to the cliff is inclined upward at an angle of 15. (a) How far from the base of the cliff does the car land? (b) What is the car s impact speed? Solution: This is a 2D projectile motion problem. We are given the initial height of the car above the ground, y i 25 m, its initial velocity, v i 35.0 m{s, and the angle above the horizontal with which the car leaves the cliff, θ 15. We are asked about the final distance from the base of the cliff and the car s impact speed. a) To solve the problem we first determine the initial velocity of the ball in the horizontal (x) and vertical (y) dimensions: v xi v i cos θ p35 m{sq cos p15 q 33.8 m{s, (54) v yi v i sin θ p35 m{sq sin p15 q 9.1 m{s. (55) Next, we use the vertical kinematic equation of projectile motion to find the time it takes the car to hit the ground, resulting in y f 0: y f 0 y i v yi t 1 2 gp tq2, (56) 1 2 gp tq2 v yi t y i 0, (57) 4.9 t t 25 0 (58) (59) 10

11 Using the quadratic equation and substituting in values, t b?b 2 4ac (60) 2a t 9.1 ap 9.1q 2 4p4.9qp 25q (61) 2p4.9q t 3.4 sec (62) Next, we use the horizontal kinematic equation of projectile motion to find the distance the car lands from the bottom of the cliff: (63) x f x i v xi t (64) v xi t (65) p33.8 m{sqp3.4 secq (66) 114 m (67) b)to find the impact speed, we need to find the final x- and y- components of velocity. The x-component of the final velocity is: The y-component of the final velocity is: (68) v fx v ix 33.8 m{s (69) (70) v fy v iy g t 9.1 m{s p9.8 m{s 2 qp3.4 secq (71) To find the magnitude of the velocity or the impact speed, 24.2 m{s (72) (73) v a p33.8q 2 p 24.2q 2 (74) 41.6 m{s (75) (76) 11

12 6. You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 50 m away, making a 2 angle with the ground. How fast was the arrow shot? Solution: This is a 2D projectile motion problem. We are told that the arrow is pointed in the horizontal direction initially, that the range, or horizontal distance the arrow travels is x f 50 m, and that the angle at which the arrow hits the ground with respect to the horizontal is θ 15. We are asked to find the initial speed of the arrow. a) To start the problem, let s use the kinematic equation for x where the initial position, x i 0: x f x i v ix t v ix t (77) Then, let s use the kinematic equation for y where the initial velocity v iy 0: (78) v yf v yi g t g t (79) Since the velocity in the x-direction is a constant, let s set v ix v fx v x. Now, use the angle at which the arrow hits the ground to relate v x and v y. (80) tan θ v y v x (81) v x v y tan θ (82) (83) 12

13 From the equation above, we know that v y g t, which gives v x v y tan θ g t tan θ (84) (85) Plugging this into the equation for x f, we get x f v x t g t t (86) tan θ g t2 tan θ Find the time by rearranging the equation to get (87) (88) t d To find the initial velocity, t 2 x f tan θ (89) g d x f tan θ p50 mq tanp2q t g 9.8 m{s 2 t 0.4 sec (90) v x v x v y tan θ g t tan θ (91) v y tan θ p9.8 m{s2 qp0.4 secq tanp2q (92) v x 112 m{s (93) Note: We drop the negative sign on g because we re keeping track of the positive and negative values, and we know that v x is positive. The negative in the equation would result from the fact that the velocity is in the negative y-direction and the angle is in the 4th quadrant. (94) 13

14 7. As the earth rotates, what is the speed of (a) a physics student in Miami, Florida, at latitude 25, and (b) a physics student in Fairbanks, Alaska, at latitude 66? Ignore the revolution of the earth around the sun. The radius of the earth is 6390 km. Solution: This is a uniform angular acceleration problem in which we are given the radius of the earth, r e 6390 km, and the latitude (angle θ with respect to the equator) of each physics student. (a) From the geometry of the problem, the linear distance of the student from the rotational axis of the earth is r r e sin p90 θq (95) p6390 kmq sinp65 q (96) 5791 km. (97) The speed of the student in Miami is just equal to the circumference divided by the period, T, which is just one day: v 2πr (98) T 2π 5791 km 1 day (99) 1 day sec km s 1 (100) 421 m{s. (101) (b) This solution proceeds just like above but with θ 66. In this case the radius is 2600 km, and v m{s 189 km s 1. 14

15 8. A rock stuck in the tread of a 56.0 cm diameter bicycle wheel has a tangential speed of 2.80 m{s. When the brakes are applied, the rock s tangential deceleration is 1.30 m{s 2. (a) What are the magnitudes of the rock s angular velocity and angular acceleration at t 1.70 s? (b) At what time is the magnitude of the rock s acceleration equal to g? Solution: This is a non-uniform angular acceleration problem. We are given the size of the wheel, r 28 cm 0.28 m, the initial tangential speed of the rock, v i 2.80 m{s, and the tangential deceleration of the rock when the brakes are applied, a T 1.30 m{s 2. (a) In order to determine the final angular velocity, ω f of the rock after t 1.70 sec, we first have to determine the initial angular velocity: w i v i r 2.8 m{s 0.28 m 10 rad{sec. (102) The angular acceleration is a constant and given by: α a T r 1.30 m{s m 4.64 rad{sec2. (103) Finally, using the appropriate rotational kinematic equation and substituting, we obtain ω f ω i α t (104) 10 rad{sec p4.64 rad{sec 2 q p1.7 secq (105) 2.11 rad{sec. (106) 15

16 (b) In order to find the time at which the magnitude of the angular acceleration of the rock equals g, we first need to find the linear velocity of the rock at that time. We want a g a a 2 T a 2 r, where a T is given in the problem and a r v 2 {r. Substituting and solving for v we get g d a 2 T v 2 r 2 (107) g 2 a 2 v 4 T (108) r 2 ñ v 4 r 2 pg 2 a 2 T q (109) v r 1{2 pg 2 a 2 T q1{4 (110) p0.28 mq 1{2 rp9.8 m{s 2 q 2 p1.30 m{s 2 q 2 s 1{4 (111) 1.65 m{s. (112) To find the time at which the rock reaches this velocity, we use the 1D kinematic equation for velocity and solve for the time: v f v i a T t (113) ñ t v i v f (114) a T 2.80 m{s 1.65 m{s 1.30 m{s 2 (115) 0.88 sec. (116) 16

17 Kinematics and Mechanics x f x i v xf v xi v 2 xf v2 xi y f y i v yf v yi v 2 yf v2 yi θ f θ i ω f ω i ω f 2 ω i 2 s rθ c 2πr v 2πr T v t ωr v xi t v yi t a x t 1 2 a xp tq 2 2a xpx f x i q a y t 1 2 a yt 2 2a ypy f y i q ω i t α t a r v2 r ω2 r a t rα 2α θ 1 2 α t2 17