Seat: PHYS 1500 (Fall 2008) Exam #1, V1 Name: 5 pts 1. Roughly, what is the mass of F. Robicheaux, your instructor? (a) 7 g (b) 70 g (c) 700 g (d) 7 kg (e) 70 kg (f) 700 kg 5 pts 2. You are holding a picture against a wall by applying a force of 6.3 N in the x-direction and 12.7 N in the y-direction. What is the force that the picture exerts on you? Explain. The force that the picture exerts on you is -6.3 N in the x-direction and -12.7 N in the y-direction. The force that the picture exerts on you is equal in magnitude and opposite in direction from the force that you apply to the picture. Newton s 3rd Law. 5 pts 3. An archer shoots an arrow horizontally at a target; the arrow leaves the bow with a speed of 75 m/s. The arrow is aimed directly at the center of the target, but it hits 52 cm lower. Ignore air resistance. What is the distance of the target from the arrow? This problem is equivalent to that asked of you in the lab where you launched a ball to hit a target. You know v 0x, v 0y = 0, a x = 0, a y = 9.8 m/s 2, and y. There are many ways to solve this problem. I solved it by using the y information to obtain t. I then used x = v 0x t to find the distance of the target. y = v 0y t + (1/2)a y t 2 = 4.9 m s 2 t2 t 2 = 0.52 m 4.9 m/s = 0.106 2 s2 t = 0.33 s x = 75 m 0.33s = 24.4 m (V 1) s t 2 = 0.35 m 4.9 m/s = 0.0714 2 s2 t = 0.27 s x = 65 m 0.27s = 17.4 m (V 2) s t 2 = 0.71 m 4.9 m/s = 0.145 2 s2 t = 0.38 s x = 85 m 0.38s = 32.4 m (V 3) s
10 pts 4. You are driving in a car and want to speed up to beat a red light. 1.0 s after you start accelerating, you have a displacement of 15.0 m; 4.0 s after you start accelerating, you have a displacement of 80.0 m from your 0 s position. Assume you are moving in a straight line and your acceleration is constant. (a) What is your acceleration? (b) What was your velocity at 0 s? This problem is equivalent to that asked of you in the lab where you rolled a cart down a slight incline. You know two displacements at two different times. You do not know the initial velocity. There are many ways to solve this problem. I will solve it by having two equations with two unknowns. 15 m = v 0 1 s + 0.5 a 1 s 2 15 m s = v 0 + 0.5 s a 80 m = v 0 4 s + 0.5 a 16 s 2 20 m s = v 0 + 2.0 s a Use the 1st equation to find an expression for v 0 and substitute that into the 2nd equation. v 0 = 15 m s 0.5 s a 20 m s = 15 m s 0.5 s a+2.0 s a 5 m s = 1.5 s a a = 3.33 m/s 2 v 0 = 15 m s 0.5 s 3.33 m/s2 = 13.3 m/s (V 1) 15 m = v 0 1 s + 0.5 a 1 s 2 15 m s = v 0 + 0.5 s a 100 m = v 0 4 s + 0.5 a 16 s 2 25 m s = v 0 + 2.0 s a Use the 1st equation to find an expression for v 0 and substitute that into the 2nd equation. v 0 = 15 m s 0.5 s a 25 m s = 15 m s 0.5 s a+2.0 s a 10 m s = 1.5 s a a = 6.66 m/s 2 v 0 = 15 m s 0.5 s 6.66 m/s2 = 11.7 m/s (V 2) 12 m = v 0 1 s + 0.5 a 1 s 2 12 m s = v 0 + 0.5 s a 80 m = v 0 4 s + 0.5 a 16 s 2 20 m s = v 0 + 2.0 s a Use the 1st equation to find an expression for v 0 and substitute that into the 2nd equation. v 0 = 12 m s 0.5 s a 20 m s = 12 m s 0.5 s a+2.0 s a 8 m s = 1.5 s a a = 5.33 m/s 2 v 0 = 12 m s 0.5 s 5.33 m/s2 = 9.33 m/s (V 3)
5 pts 5. The x-component of an object s position is positive and getting larger with time. Check the 1 correct answer. (a) The velocity is positive and the acceleration is positive. (b) The velocity is positive and the acceleration is negative. (c) The velocity is positive and the acceleration may be positive, negative or zero. (d) The velocity is negative and the acceleration is positive. (e) The velocity is negative and the acceleration is negative. (f) The velocity is negative and the acceleration may be positive, negative or zero. 5 pts 6. After winning a baseball game, one player drops a glove while another tosses a glove into the air. Just after the gloves are released, how do the accelerations of the two gloves compare? Explain. The accelerations are the same because the acceleration from gravity doesn t depend on the velocity or mass of the glove. 5 pts 7. An 8.0 kg crate is sitting on a frictionless incline. It is held in place by a rope that is clamped at the top. What is the tension in the rope? This is the same kind of problem as you needed to solve in the lab with the two masses with one on an incline. The tension in the string has to cancel out the fraction of the force from gravity that is along the incline. The component of the force of gravity along the incline is mg sin θ where θ is the angle between the incline and horizontal. T = 8 kg 9.8 m s 2 sin 40 = 50.4 N (V 1) T = 7 kg 9.8 m s 2 sin 40 = 44.1 N (V 2) T = 6 kg 9.8 m s 2 sin 40 = 37.8 N (V 3)
10 pts 8. You kick a ball (1.2 kg) so that its initial speed is 11.2 m/s with an initial direction that is 35 from the vertical. It hits a wall that is 8.0 m away. How high above the ground does the ball hit the wall? Ignore the effects of air. This is a variation of the homework problems from Secs 3.3 and 3.4. You know x, v 0x, v 0y, a x = 0, and a y = 9.8 m/s 2 and you are being asked for y. There are many ways to solve this problem. Since a x = 0 you can use x = v 0x t to find t. Once you have t, you can use y = v 0y t + (1/2)a y t 2 to find y. v 0x = 11.2 m s sin 35 = 6.42 m s t = 8.0 m 6.42 m/s = 1.25 s v 0y = 11.2 m s cos 35 = 9.17 m s y = 9.17 m s 1.25 s + 1 2 ( 9.8 m/s2 ) (1.25 s) 2 = 3.77 m (V 1) v 0x = 12.3 m s sin 35 = 7.05 m s t = 9.5 m 7.05 m/s = 1.35 s v 0y = 12.3 m s cos 35 = 10.1 m s y = 10.1 m s 1.35 s + 1 2 ( 9.8 m/s2 ) (1.35 s) 2 = 4.68 m (V 2) v 0x = 10.1 m s sin 35 = 5.79 m s t = 8.4 m 5.79 m/s = 1.45 s v 0y = 10.1 m s cos 35 = 8.27 m s y = 8.27 m s 1.45 s + 1 2 ( 9.8 m/s2 ) (1.45 s) 2 = 1.70 m (V 3)
5 pts 9. A person is standing in an elevator that is moving upward at constant speed. The upward normal force exerted by the elevator floor on the person is (a) larger than the person s weight measured at home. (b) equal to the person s weight measured at home. (c) smaller than the person s weight measured at home. 5 pts 10. A rifle, at a height H above ground, fires a bullet parallel to the ground. At the same instant and at the same height, a second bullet is dropped from rest. In the absence of air resistance, which bullet strikes the ground first? Explain. This is like the demo done in class where the one ball was dropped and the other was kicked to the side. They strike the ground at the same time, because the y-component of the motion does not depend on the x- component if you neglect the effects of air. 5 pts 11. You throw a rock straight up. It rises to height of 22.0 m before falling back to the ground. (a) What was the speed that you launched the rock? (b) How long did it take to reach the highest point? This is the same as the homework problem 2.45 in Ed. 8 (2.43 in Ed. 7). You know x, v = 0, and a = 9.8 m/s 2. With this information you can find v 0 and t. There are many ways to solve this problem. I used v 2 = v 2 0 + 2a x to find v 0 and then v = v 0 + at to find t. 0 = v 2 0 + 2( 9.8 m/s 2 )22m v 2 0 = 431.2 m 2 /s 2 v 0 = 20.8 m/s 0 = 20.8 m/s + ( 9.8 m/s 2 )t t = 20.8 m/s = 2.12 s (V 1) 9.8 m/s2 0 = v 2 0 + 2( 9.8 m/s 2 )18m v 2 0 = 352.8 m 2 /s 2 v 0 = 18.8 m/s 0 = 18.8 m/s + ( 9.8 m/s 2 )t t = 18.8 m/s = 1.92 s (V 2) 9.8 m/s2 0 = v 2 0 + 2( 9.8 m/s 2 )29m v 2 0 = 568.4 m 2 /s 2 v 0 = 23.8 m/s 0 = 23.8 m/s + ( 9.8 m/s 2 )t t = 23.8 m/s = 2.43 s (V 3) 9.8 m/s2
10 pts 12. A string is stretched between two poles. The distance between the poles is 14 m. A 6.0 kg mass is attached to the middle of the string. The string sags 0.8 m. What is the tension in each string? Ignore the mass of the string. This is the same as the homework problem 4.23 in Ed. 8 (4.21 in Ed. 7). You need to account for all of the forces acting on the mass. There is a force from each half of the string and a force down from gravity. Because the mass is in the middle, the figure is symmetrical. You need to find the x- and y-components of all the forces. The x-component of the force of gravity is 0 and the x-components from each half of the string cancels out. The y-component of the forces have to add to 0; if they don t, the mass is accelerating. To find the angle, you need to use trigonometry. y component : F y = ma y gives 2T sin θ mg = 0 T = mg 2 sin θ tan θ = 0.8 m = 0.114 θ = 6.52 7 m 6.0 kg 9.8 m/s2 T = = 259 N (V 1) 2 sin 6.52 tan θ = 0.8 m = 0.133 θ = 7.59 6 m 5.0 kg 9.8 m/s2 T = = 185 N (V 2) 2 sin 7.59 tan θ = 0.8 m = 0.0941 θ = 5.38 8.5 m 7.0 kg 9.8 m/s2 T = = 366 N (V 3) 2 sin 5.38
Equations Basic Mathematic Formulas sin θ = h o /h cos θ = h a /h tan θ = h o /h a h 2 = h 2 o + h 2 a A circ = πr 2 Circum of circ = 2πr V sph = 4π 3 r3 A sur of sph = 4πr 2 x = b ± b 2 4ac 2a Chapter 2 x = x f x i v = x/ t ā = (v f v i )/(t f t i ) = v/ t v = v 0 + at x = 1 2 (v 0 + v)t x = v 0 t + 1 2 at2 v 2 = v 2 0 + 2a x Chapter 3 A x = A cos θ A y = A sin θ A = A 2 x + A 2 y tan θ = A y /A x r = r f r i v av = r/ t a av = ( v)/ t v x = v 0x + a x t x = 1 2 (v 0x + v x )t x = v 0x t + 1 2 a xt 2 v 2 x = v 2 0x + 2a x x v y = v 0y + a y t y = 1 2 (v 0y + v y )t y = v 0y t + 1 2 a yt 2 v 2 y = v 2 0y + 2a y y Chapter 4 F = m a Fg = G m 1m 2 r 2 11 N m2 G = 6.67259 10 kg 2 F 12 = F 21 f s µ s n f k = µ k n w = mg