# TIME OF COMPLETION DEPARTMENT OF NATURAL SCIENCES. PHYS 1111, Exam 2 Section 1 Version 1 October 30, 2002 Total Weight: 100 points

Save this PDF as:

Size: px
Start display at page:

Download "TIME OF COMPLETION DEPARTMENT OF NATURAL SCIENCES. PHYS 1111, Exam 2 Section 1 Version 1 October 30, 2002 Total Weight: 100 points"

## Transcription

1 TIME OF COMPLETION NAME DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Section 1 Version 1 October 30, 2002 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on eight (8) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 80 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:20 p.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN

2 MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. A a. 27,491 furlongs/fortnight. (5.00 yards/second)(1.00 furlong/220 yards)(14 x 24 x 3600 seconds/1.00 fortnight) = (5) b. 13,674 furlongs/fortnight. c. 6,221 furlongs/fortnight. d. 2,749 furlongs/fortnight. 27,491furlong/fortnight 2. On the Moon, the acceleration due to gravity is only about 1/6 of that on Earth. An astronaut whose weight on Earth is 600 N travels to the lunar surface. His mass as measured on the Moon will be: a. 600 kg. (5) b. 100 kg. W = m g m = W/g m = (600 N)/(9.80 m/s 2 ) = 61.2 kg c kg. d kg. 3. When we subtract a displacement vector from a velocity vector, the result is: a. A velocity. (5) b. An acceleration. c. Another displacement. d. Result doesn t have physical meaning. 4. Which of the following is an example of the type of force which acts at a distance? a. Gravitational.

3 (5) b. Magnetic. c. Electrical. d. All of the above. 5. Imagine a book sitting on a horizontal table. One of the forces acting on a book is a normal force exerted by the table (let s call it action). What force does form the Newton s 3 rd law pair (reaction)? e. Weight of the book. (5) f. Weight of the table. g. Normal force exerted on the table by the book. h. Friction. 6. Take a look at the graph representing one-dimensional motion of a body. Is the instantaneous velocity of the body at point A: (5) a. Positive. b. Negative. c. Zero. d. Cannot be determined from the graph. 7. Coasting due west on your bicycle at 8.00 m/s, you encounter a sandy patch of road 7.20 m across. When you leave the sandy patch your speed has been reduced to 6.50 m/s. (15) a. Assuming the bicycle slows with constant acceleration, what was its acceleration in the sandy patch? Give both magnitude and direction.

4 V 2 = V a ( x- x 0 ) a = (V 2 - V 2 0 )/ 2( x- x 0 ) V 0 = 8.00 m/s; V = 6.50 m/s; x 0 = 0 m; x = 7.20 m a = {(6.50 m/s) 2 - (8.00 m/s) 2 }/{2 x (7.50 m)} = m/s 2 west or 1.51 m/s 2 east b. What time did it take you to cross the patch? V = V 0 + a t t = (V - V 0 )/a t = (6.50 m/s 8.00 m/s) / (-1.51 m/s 2 ) = s c. What was your average velocity on the sand? V average = (x - x 0 )/(t - t 0 ) = (7.20 m)/(0.993 s) = 7.25 m/s 8. Deep inside an ancient physics text you discover two vectors: A: 45.0 o B: o Not content with these hoary relics, you are asked to find a new vector R = A - B. Find the magnitude and direction of vector R (15) A: o B: o Ax = A cos( A) = 45.0m cos(150 o ) = m Ay = A sin( A) = 45.0m sin(150 o ) = 22.5 m Bx = B cos( B) = 30.0m cos(-15.0 o ) = 29.0 m By = B sin( B) = 30.0m sin(-15.0 o ) = m Rx = Ax - Bx = m = m Ry = Ay - By = 22.5 m m = 30.3 m 2 R = (Rx 2 + Ry ) 1/2 = ( (-68.0 m) 2 + (30.3 m) 2 ) 1/2 = 74.4 m

5 R = tan -1 (Ry /Rx) = tan -1 (30.3 m /-68.0 m) = o o = 156 o (Rx < 0) 9. A soccer ball is kicked with a speed of 9.50 m/s at an angle of 40.0 o above the horizontal direction. Some time later the ball lands at the same level from which it was kicked. a. Find the components of the ball s initial velocity. Vi: o Vx0 = V0 cos( 0) = 9.50m/s cos(40.0 o ) = 7.28 m/s Vy0 = V0 sin( 0) = 9.50m/s sin(40.0 o ) = 6.11 m/s (20)

6 b. What are the components of the ball s velocity at the top of the trajectory? Vx = 7.28 m/s Vy = 0 m/s c. How long does it take the ball to reach the top of the trajectory? Vy = Vy0 - gt (Vy0 - Vy) /g = t t = (6.11 m/s 0 m/s)/9.80 m/s 2 = s d. How far does the ball go in the horizontal direction? x = x0 + Vx0 ( t) = 0 m + (7.28 m/s) (1.25 s) = 9.07 m 10. During your winter break, you enter a dogsled race in which students replace the dogs. Wearing cleats for traction, you begin the race by pulling on a rope attached to the sled with a force of 150 N at 25 o with the horizontal. The mass of the sled is 80.0 kg and there is negligible friction between the sled and ice. a. Draw a free-body diagram. (20) b. Find the acceleration of the sled.

7 Ax = A cos( ) = (150 N) cos( ) = 136 N Ay = A sin( ) = (150 N) sin( ) = 63.4 N nx = n cos(9 ) = 0 ny = n sin(9 ) = n wx = w cos(270 ) = 0 wy = w sin(270 ) = (80.0 kg) (9.80 m/s 2 ) = -784 N Fx = m ax Ax = m ax ax = Ax /m = (136.0 N)/(80.0 kg) = 1.70 m/s 2 c. Find the normal force exerted by the surface on the sled. Fy = m ay ay = 0 ny + wy + Ay = 0 n 784 N 63.4 N = 0 n = 721 N Problem 5

8

9 TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 2 October 4, 2005 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 10:30 a.m. Stop: 11:45 a.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE

10 CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Which of the following operations will not change a vector? a. Translate it parallel to itself. (5) b. Rotate it. c. Multiply it by a constant factor. d. Add a constant vector to it. 2. Suppose that an object travels from one point in space to another. Make a comparison between the displacement and the distance traveled. a. The displacement is either greater than or equal to the distance traveled. (5) b. The displacement is always equal to the distance traveled. c. The displacement is either less than or equal to the distance traveled. d. The displacement can be either greater than, smaller than, or equal to the distance traveled. 3. The position, x, of an object is given by the equation x = A + Bt +Ct 2, where t refers to time. What are the dimensions of A, B, and C? a. Distance, distance, distance. (5) b. Distance, time, time 2. c. Distance, distance/time, distance/ time 2. d. Distance/time, distance/ time 2, distance/ time 3.

11 4. In the diagram shown, the unknown vector is a. (5) b. c. d. Cannot be determined from the diagram. 5. Which of Newton's laws best explains why motorists should buckle-up? a. The first law. (5) b. The second law. c. The third law. d. The law of gravitation. 6. A net force F accelerates a mass m with an acceleration a. If the same net force is applied to mass 3m, then the acceleration will be a. 3a. (5) b. 2a. c. a/2. d. a/3.

12 7. A projectile is fired with an initial speed of 40.2 m/s at an angle of 35.0 o above the horizontal on a long flat firing range. Determine a. The components of the projectile s initial velocity. V i : o V x0 = V 0 cos( 0 ) = (40.2 m/s) cos(35.0 o ) = 32.9 m/s V y0 = V 0 sin( 0 ) = (40.2 m/s) sin(35.0 o ) = 23.1 m/s b. The maximum height reached by the projectile. V 2 y = V 2 y0 - g (y y 0 ) (y y 0 ) = (V 2 y - V 2 y0 )/(-g) (y 0) = (0 2 (23.1 m/s) 2 )/(-9.81 m/s 2 ) y = 54.4 m c. The time it takes to reach the maximum height. V y = V y0 - gt (V y0 - V y ) /g = t t = (23.1 m/s 0 m/s)/(9.81 m/s 2 ) = 2.35 s d. The projectile s range. x = x 0 + V x0 (2t) = 0 m + (32.9 m/s) (4.70 s) = 155 m Bonus (5 extra points): Find velocity of the projectile 1.00 s after firing. V x = V x0 = 32.9 m/s V y = V y0 gt = (23.1 m/s) (9.81 m/s 2 )(1.00 s) = 13.3 m/s V = (V 2 x + V 2 y ) 1/2 = ( (32.9 m/s) 2 + (13.3 m/s) 2 ) 1/2 = 35.5 m/s = tan -1 (V y /V x ) = tan -1 ((13.3 m/s) /(32.9 m/s)) = 22.0 o

13 8. A stone is thrown vertically upward with a speed of 18.0 m/s. a. How fast is it moving when it reaches a height of 7.00 m? V 2 y = V 2 y0 - g (y y 0 ) V 2 y = (18.0 m/s) 2 - (9.81 m/s 2 )((7.00 m) 0) V y = +/ m/s b. How long is required to reach this height? V y = V y0 - gt (V y0 - V y ) /g = t t 1 = (18.0 m/s 16.0 m/s)/(9.81 m/s 2 ) = s t 2 = (18.0 m/s m/s)/(9.81 m/s 2 ) = 3.47 s c. Why are there two answers to (b)? Because the stone reaches this point twice: on its way up and on its way down. 9. A student decides to move a box of books into her dormitory room by pushing on the box. She pushes with a force of 100 N at an angle of 15.0 o with the horizontal. The box has a mass of 25.0 kg, and we are going to neglect friction between the box and the floor for simplicity. a. Draw a free body diagram.

14 b. Find the acceleration of the box. A x = A cos(-15.0 o ) = (100 N) cos(-15.0 o ) = 96.6 N A y = A sin(-15.0 o ) = (100 N) sin(-15.0 o ) = N n x = n cos(90.0 o ) = 0 n y = n sin(90.0 o ) = n w x = w cos(-90.0 o ) = 0 w y = w sin(-90.0 o ) = (25.0 kg) (9.81 m/s 2 ) = -245 N F x = m a x A x = m a x a x = A x /m a x = (96.6 N)/(25.0 kg) = 3.86 m/s 2 c. Find the normal force the box exerts on the floor. F y = m a y a y = 0 n y + w y + A y = 0

15 n 245 N 25.9 N = 0 n = 271 N Note: this is the normal force that the FLOOR exerts on the BOX, but the third Newton s Law states that this force is equal in magnitude and opposite in direction to the force that the BOX exerts on the FLOOR. 10. Vector A has a magnitude of m/s and points in a direction 20.0 o below the x axis. A second vector, B, has a magnitude of 80.0 m/s and points in a direction 95.0 o above the x axis. Find a new vector R = A + B. Find the magnitude and direction of vector R. A: o B: o A x = A cos( A ) = (127 m/s) cos(-20.0 o ) = 119 m/s A y = A sin( A ) = (127 m/s) sin(-20.0 o ) = m/s B x = B cos( B ) = (80.0 m/s) cos(95.0 o ) = m/s B y = B sin( B ) = (80.0 m/s) sin(95.0 o ) = 79.7 m/s R x = A x + B x = 119 m/s + (-6.97 m/s) = 112 m/s R y = A y + B y = m/s m/s = 36.3 m/s R = (R x 2 + R y 2 ) 1/2 = ( (112 m/s) 2 + (36.3 m/s) 2 ) 1/2 = 118 m/s q R = tan -1 (R y /R x ) = tan -1 ((36.3 m/s) /(112 m/s)) = 18.0 o TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 1 October 2, 2006 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10)

16 problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. The metric prefix for one one-hundredth is a. Milli. b. Centi.

17 (4) c. Kilo. d. Mega. 2. An object moving in the +x axis experiences an acceleration of 2.0 m/s 2. This means the object is a. Traveling at 2.0 m in every second. (4) b. Traveling at 2.0 m/s in every second. c. Changing its velocity by 2.0 m/s. d. Increasing its velocity by 2.0 m/s in every second. 3. When an object is released from rest and falls in the absence of friction, which of the following is true concerning its motion? a. The speed of the falling object is proportional to its mass. (4) b. The speed of the falling object is proportional to its weight. c. The speed of the falling object is inversely proportional to its surface area. d. None of the above is true. 4. Ignoring air resistance, the horizontal component of a projectile's velocity a. Is zero. (4) b. Remains constant. c. Continuously increases. d. Continuously decreases.

18 5. When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed? a. It is zero. (4) b. It is less than its initial speed. c. It is equal to its initial speed. d. It is greater than its initial speed. 6. If you exert a force F on an object, the force which the object exerts on you will a. Depend on whether or not the object is moving. (4) b. Depend on whether or not you are moving. c. Depend on the relative masses of you and the object. d. Always be F. 7. Vector V 1 is 6.6 units long and points along the negative x axis. Vector V 2 points at 45º to the positive x axis. a. What are the x and y components of each vector? V 1x = V 1 cos( 1 ) = (6.60) cos(180 o ) = V 1y = V 1 sin( 1 ) = (6.60) sin(180 o ) = 0 V 2x = V 2 cos( 2 ) = (8.50) cos(45.0 o ) = 6.01 is 8.5 units long and

19 V 2y = V 2 sin( 2 ) = (8.50) sin( 45.0 o ) = 6.01 b. Determine the sum V 1 V 2 (magnitude and angle). R x = V 1x + V 2x = (- 6.60) + (6.01) = R y = V 1y + V 2y = (0) + (6.01) = R = ( R x + R 2 y ) 1/2 = ( (0.59) 2 + (6.01) 2 ) 1/2 = 6.04 R = tan -1 (R y /R x ) = tan -1 ((6.01)/(-0.59)) = o o = 95.6 o 8. A projectile is shot from the edge of a cliff with an initial speed of 65.0 m s at an angle of 37.0º with the horizontal. a. Determine the components of the initial velocity of the projectile. V 0x = V 0 cos( 0 ) = (65.0 m/s) cos(37.0 o ) = 51.9 m/s V 0y = V 0 sin( 0 ) = (65.0) sin( 37.0 o ) = 39.1 m/s b. If it takes 10.5 s for the projectile to reach the base of the cliff, how high is the cliff? y = y 0 + V y0 t ½ g t 2 0 = y 0 + V y0 t ½ g t 2 y 0 = - V y0 t + ½ g t 2 = - (39.1 m/s) (10.5 s) + ½ (9.81 m/s 2 ) (10.5 s) 2 y 0 = 130 m c. Determine the range X of the projectile as measured from the base of the cliff. x = x 0 + V x0 t x = (0 m) + (51.9 m/s) (10.5 s) =545 m Bonus (5 points): Find the maximum height above the cliff top reached by the projectile.

20 2 V y = V 2 y0-2 g (y-y 0 ) 0 = V 2 y0-2 g (y-y 0 ) (y-y 0 ) = V 2 y0 /(2 g) = 77.9 m y = 208 m 9.The driver of a truck slams on the brakes when he sees tree blocking the road. The truck slows down uniformly with an acceleration of m/s 2 for 4.20 s. If the truck was moving at 27.0 m/s the moment the driver sees the tree, a. With what speed does the truck hit the tree? V = V 0 + a t V = (27.0 m/s) + (-5.60 m/s 2 )(4.20 s) = 3.48 m/s b. How long are the skid marks? x = x 0 + V 0 t + ½ a t 2 x = 0 + (27.0 m) (4.20 s) + ½ (-5.60 m/s 2 ) (4.20 s) 2 x = 64.0 m c. What is the average velocity of the truck over these 4.20 s? V av = (x - x 0 )/(t - t 0 ) = (64.0 m)/(4.20 s) = 15.2 m/s 10. At the instant a race began, a 65-kg sprinter exerted a force of 720 N on the starting block at a 22º angle with respect to the ground. a. Draw a free body diagram for the sprinter.

21 b. What was the horizontal acceleration of the sprinter? A x = A cos(22.0 o ) = (720 N) cos(22.0 o ) = 668 N A y = A sin(22.0 o ) = (720 N) sin(22.0 o ) = 270 N n x = n cos(90.0 o ) = 0 n y = n sin(90.0 o ) = n w x = w cos(-90.0 o ) = 0 w y = w sin(-90.0 o ) = (65.0 kg) (9.81 m/s 2 ) = -638 N F x = m a x A x = m a x a x = A x /m a x = (668 N)/(65.0 kg) = 10.3 m/s 2

22 c. If the force was exerted for 0.32 s, with what speed did the sprinter leave the starting block? V = V 0 + a t V = (0 m/s) + (10.3 m/s 2 )(0.32 s) = 3.30 m/s TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 1 October 1, 2007 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination.

23 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. The velocity, V, of an object is given by the equation V = A + Bt, where t refers to time. What are the dimensions of A, and B? A) distance/time, distance/time 2. (4) B) distance/time, distance 2 /time 2. C) distance, distance/ time 2 D) distance/time 2, distance/ time 3

24 2. Can an object's velocity change direction when its acceleration is constant? Support your answer with an example. A) No, this is not possible because it is always speeding up. (4) B) No, this is not possible because it is always speeding up or always slowing down, but it can never turn around. C) Yes, this is possible, and a rock thrown straight up is an example. D) Yes, this is possible, and a car that starts from rest, speeds up, slows to a stop, and then backs up is an example. 3. A ball is thrown straight up, reaches a maximum height, then falls to its initial height. Make a statement about the direction of the velocity and acceleration as the ball is coming down. A) Both its velocity and its acceleration point upward. (4) B) Its velocity points upward and its acceleration points downward. C) Its velocity points downward and its acceleration points upward. D) Both its velocity and its acceleration point downward. 4. An object is moving with constant non-zero acceleration in the +x axis. The position versus time graph of this object is A) a horizontal straight line. (4) B) a vertical straight line. C) a straight line making an angle with the time axis. D) a parabolic curve. 5. Ignoring air resistance, the horizontal component of a projectile's acceleration A) is zero. B) remains a non-zero constant.

25 (4) C) continuously increases. D) continuously decreases. 6. A child's toy is suspended from the ceiling by means of a string. The Earth pulls downward on the toy with its weight force of 8.0 N. If this is the "action force," what is the "reaction force"? A) The string pulling upward on the toy with an 8.0-N force. (4) B) The ceiling pulling upward on the string with an 8.0-N force. C) The string pulling downward on the ceiling with an 8.0-N force. D) The toy pulling upward on the Earth with an 8.0-N force. 7. Vector V 1 is 20.7 m long and points at 55.0 o to the positive x axis. Vector V 2 points at o to the positive x axis. a. What are the x and y components of each vector? V 1x = V 1 cos( 1 ) = (20.7 m) cos(55.0 o ) = 11.9 m V 1y = V 1 sin( 1 ) = (20.7 m) sin(55.0 o ) = 17.0 m V 2x = V 2 cos( 2 ) = (15.0 m) cos(-15.0 o ) = 14.5 m V 2y = V 2 sin( 2 ) = (15.0 m) sin( o ) = m b. Determine the vector V 1 V 2 (magnitude and angle). is 15.0 m long and R x = V 1x - V 2x = (11.9 m) - (14.5 m) = m

26 R y = V 1y - V 2y = (17.0 m) - (-3.88 m) = 20.9 m R = ( R x 2 + R y 2 ) 1/2 = ( (-2.60 m) 2 + (20.9 m) 2 ) 1/2 = 21.0 m R = tan -1 (R y /R x ) = tan -1 ((20.9 m)/(-2.60 m)) = o o = 97.1 o 8. A diver running 1.8 m s dives out from the edge of a vertical cliff at an angle of 25.0 o with the horizontal. He hits the water 5.00 m away from the base of the cliff. a. What are the components of the initial velocity? V x0 = V 0 cos( 0 ) = (1.80 m/s) cos(25.0 o ) = 1.63 m/s V y0 = V 0 sin( 0 ) = (1.80 m/s) sin( 25.0 o ) = m/s x = x 0 + V x0 t b. How long was the diver in the air? t = (x - x 0 )/V x0 t = (5.00 m)/(1.63 m/s) = 3.07 s c. How high was the cliff? y = y 0 + V y0 t ½ gt 2 y 0 = y - V y0 t + ½ gt 2 y 0 = 0 (0.761 m/s) (3.07 s) + ½ (9.81 m/s 2 )(3.07 s) 2 = 43.9 m Bonus (5 points): How high above the edge of the cliff did the diver rise? 2 V y = V 2 y0-2 g (y-y 0 ) 0 = V 2 y0-2 g (y-y 0 ) (y-y 0 ) = V 2 y0 /(2 g)

27 (y m) = (0.761 m/s) 2 /(2 g) y = m (from the ground) or m from the edge of the cliff 9. A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m s ) in the first 15.0 m of the race. V 2 = V a (x-x 0 ) a. What is the average acceleration of this sprinter? a = (V 2 - V 0 2 )/(2(x-x 0 )) a = ((11.5 m/s) 2 0)/(2(15.0 m)) = 4.41 m/s 2 V = V 0 + at b. How long does it take her to reach that speed? t =(V - V 0 )/a = ((11.5 m/s) 0)/( 4.41 m/s 2 ) = 2.61 s c. What is her average velocity during this time? V av = x/ t = (15.0 m)/(2.61 s) = 5.75 m/s 10. A box sits at rest on a rough 30º inclined plane. Mass of the box is 10.0 kg. a. Draw a free body diagram including all the forces acting on the box.

28 b. Find the magnitude of the normal force exerted on the box by the incline. f x = f cos(0 o ) = f f y = f sin(0 o ) = 0 n x = n cos(90.0 o ) = 0 n y = n sin(90.0 o ) = n w x = w cos(-120 o ) = (10.0 kg) (9.81 m/s 2 ) cos(-120 o ) = N w y = w sin(-120 o ) = (10.0 kg) (9.81 m/s 2 ) sin(-120 o ) = N F y = m a y f y + n y + w y = 0 n N = 0 n = 85.0 N c. Find the magnitude of the friction force acting on the box. F x = m a x f x + n x + w x = 0 f N = 0

29 f = 13.8 N TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 1 February 16, 2005 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on eight (8) pages.

30 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 2. A furlong is a distance of 220 yards. A fortnight is a time period of 2 weeks. A race horse is running at a speed of 5 yards per second. What is its speed in furlongs per fortnight? e. 13,674 furlongs/fortnight. f. 6,221 furlongs/fortnight.

31 (5) g. 2,749 furlongs/fortnight. h. 27,491 furlongs/fortnight 2. On the Moon, the acceleration due to gravity is only about 1/6 of that on Earth. An astronaut whose weight on Earth is 600 N travels to the lunar surface. His mass as measured on the Moon will be: e. 600 kg. (5) f. 100 kg. g kg. h kg. 3. When we subtract a displacement vector from a velocity vector, the result is: a. A velocity. (5) b. An acceleration. c. Another displacement. d. Result doesn t have physical meaning. 4. Which of the following is an example of the type of force which acts at a distance? a. Gravitational. (5) b. Magnetic. c. Electrical. d. All of the above. 5. Imagine a book sitting on a horizontal table. One of the forces acting on a book is a normal force exerted by the table (let s call it action). What force does form the Newton s 3 rd law pair (reaction)? e. Normal force exerted on the Earth by the table.

32 (5) f. Normal force exerted on the table by the Earth. g. Normal force exerted on the table by the book. h. Normal force exerted on the Earth by the book. 6. Take a look at the graph representing one-dimensional motion of a body. The instantaneous velocity of the body at point A is e. Positive. (5) f. Negative. g. Zero. h. Cannot be determined from the graph. 7. Coasting due west on your bicycle at 8.00 m/s, you encounter a sandy patch of road 7.20 m across. When you leave the sandy patch your speed has been reduced to 6.50 m/s. a. Assuming the bicycle slows with constant acceleration, what was its acceleration

33 in the sandy patch? Give both magnitude and direction. V 0 = 8.00 m/s V= 6.50 m/s x x 0 = 7.20 m V 2 = V a (x x 0 ) a = (V 2 - V 0 2 )/[2(x x 0 )] = m/s 2 (15) b. What time did it take you to cross the patch? V = V 0 + a t t = (V - V 0 )/a = s c. What was your average velocity as you were crossing the patch? V av = (x x 0 )/t = (7.20 m) / (0.993 s) = 7.25 m/s 8. Deep inside an ancient physics text you discover two vectors: A: 45.0 o B: o Not content with these hoary relics, you are asked to find a new vector R = A - B. Find the magnitude and direction of vector R A x = A cos( A ) = (45.0 m) cos(150.0 o ) = m A y = A sin( A ) = (45.0 m) sin(150.0 o ) = 22.5 m B x = B cos( B ) = (30.0 m/s) cos(-15.0 o ) = 30.0 m B y = B sin( B ) = (30.0 m/s) sin(-15.0 o ) = m R x = A x - B x = (-40.0 m) - (30.0 m) = m R y = A y - B y = (22.5 m) - (-7.76 m) = 30.3 m R = ( R x 2 + R y 2 ) 1/2 = ( (-70.0 m) 2 + (30.3 m) 2 ) 1/2 = 76.3 m

34 R = tan -1 (R y /R x ) = = tan -1 ((30.3 m)/(-70.0 m)) = o o = 157 o (15) 9. A soccer ball is kicked with a speed of 9.50 m/s at an angle of 40.0 o above the horizontal direction. Some time later the ball lands at the same level from which it was kicked. V 0 : o a. Find the components of the ball s initial velocity. V x0 = V 0 cos( 0 ) = (9.50 m/s) cos(40 o ) = 7.28 m/s V y0 = V 0 sin( 0 ) = (9.50 m/s) sin(40 o ) = 6.11 m/s (20) b. What are the components of the ball s velocity at the top of the trajectory? V x = V x0 = 7.28 m/s V y = 0 m/s c. How long does it take the ball to reach the top of the trajectory? V y = V y0 - gt t = - (V y - V y0 )/g = s x = x 0 + V x0 t d. How far does the ball go in the horizontal direction? x = (0 m) + (7.28 m/s) (2 x s) = 9.08 m 10. During your winter break, you enter a dogsled race in which students replace the dogs. Wearing cleats for traction, you begin the race by pulling on a rope attached to the sled with a force of 150 N at 25 o with the horizontal. The mass of the sled is 80.0 kg and there is negligible friction between the sled and ice. a. Draw a free-body diagram.

35 b. Find the acceleration of the sled. A x = A cos( A ) = (150 N) cos(25.0 o ) = 136 N A y = A sin( A ) = (150 N) sin(25.0 o ) = 63.4 N w x = w cos( w ) = (80.0 kg)(9.80 m/s 2 ) cos(-90.0 o ) = 0 N w y = w sin( w ) = (80.0 kg)(9.80 m/s 2 ) sin(-90.0 o ) = -784 N n x = n cos( n ) = n cos(90.0 o ) = 0 N n y = n sin( n ) = n sin(90.0 o ) = n F X = m a X 136 N = (80.0 kg) a X a X = 1.70 m/s 2 c. Find the normal force exerted by the surface on the sled.

36 F Y = m a Y = N -784 N + n = 0 n = 721 N TIME OF COMPLETION NAME_SOLUTION SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 1 February 20, 2006 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are five (5) multiple choice and four (4) calculation problems. Work all multiple choice problems and 4 calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT

37 TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. If 1 inch = 2.54 cm, and 1 yd = 36 in., how many meters are in 7.00 yd? (4) a m. b m. c. 640 m. d m. (7.00 yd)(36 in/1 yd)( m/1 in) = 6.40 m 2. Suppose that an object travels from one point in space to another. Make a comparison between the displacement and the distance traveled. a. The displacement is either greater than or equal to the distance traveled. b. The displacement is always equal to the distance traveled. (4) c. The displacement is either less than or equal to the distance traveled. d. The displacement can be either greater than, smaller than, or equal to the distance traveled. 3. Ignoring air resistance, the horizontal component of a projectile's velocity

38 a. Is zero. (4) b. Remains constant. c. Continuously increasing. d. Continuously decreasing. 4. Ignore this problem. 5. Mass and weight a. Both measure the same thing. (4) b. Are exactly equal. c. Are two different quantities. d. Are both measured in kilograms. 6. A person pushes a 14.0-kg lawn mower at constant speed with a force of F 88.0 N directed along the handle, which is at an angle of 45.0º to the horizontal. Making a highly unrealistic assumption that there is no friction between the mower and the ground,. a. Draw the free-body diagram showing all forces acting on the mower.

39 b. Calculate the acceleration of the mower. A x = A cos(-45.0 o ) = (88.0 N) cos(-45.0 o ) = 62.2 N A y = A sin(-45.0 o ) = (88.0 N) sin(-45.0 o ) = N n x = n cos(90.0 o ) = 0 n y = n sin(90.0 o ) = n w x = w cos(-90.0 o ) = 0 w y = w sin(-90.0 o ) = (14.0 kg) (9.81 m/s 2 ) = -137 N F x = m a x A x = m a x a x = A x /m a x = (62.2 N)/(14.0 kg) = 4.44 m/s 2 c. Calculate the magnitude of the normal force that the ground exerts on the mower. F y = m a y

40 a y = 0 n y + w y + A y = 0 n + (-137 N) + (-62.2 N) = 0 n = 199 N 7. A projectile is shot from the edge of a cliff with an initial speed of 65.0 m s at an angle of 37.0º with the horizontal. The projectile lands on the ground 10.0 s later. a. What are the components of the initial velocity? V 0 : o V x0 = V 0 cos( 0 ) = (65.0 m/s) cos(37.0 o ) = 51.9 m/s V y0 = V 0 sin( 0 ) = (65.0 m/s) sin(37.0 o ) = 39.1 m/s b. How high is the cliff? y = y 0 + V y0 t ½ g t 2 0 = y 0 + V y0 t ½ g t 2 y 0 = ½ g t 2 - V y0 t y 0 = ½ (9.81 m/s 2 ) (10.0 s) 2 - (39.1 m/s) (10.0 s) = 99.5 m c. Determine the range of the projectile as measured from the base of the cliff. x = x 0 + V x0 (t) = 0 m + (51.9 m/s) (10.0 s) = 519 m d. At the instant just before the projectile hits the ground, find the magnitude of the velocity. V x = V x0 = 51.9 m/s V y = V y0 g t = (39.1 m/s) - (9.81 m/s 2 ) (10.0 s) = m/s V = (V x 2 + V y 2 ) 1/2 = 78.6 m/s Bonus (5 points): Find the maximum height above the cliff top reached by the projectile.

41 2 2 V y = V y0 2 g (y y 0 ) (y y 0 ) = (V 2 y - V 2 y0 )/(-2 g) (y 0) = (0 2 (39.1 m/s) 2 )/(-2 x 9.81 m/s 2 ) y = 77.9 m 8. A fugitive tries to hop on a freight train. The fugitive starts from rest and accelerates at 2 a 4.0 m s to his maximum speed of 8.0 m s. a. How long does it take him to reach the maximum speed? V = V 0 + a t V = 0 + a t t = V/a = (8.00 m/s)/(4.00 m/s 2 ) = 2.00 s b. What is the distance traveled to reach the maximum speed? x = x 0 + V 0 t + ½ a t 2 = ½ a t 2 x = ½ (4.00 m/s 2 )(2.00 s) 2 = 8.00 m c. What is the average velocity of the fugitive during this time interval? V av = x/ t = (8.00 m)/(2.00 s) = 4.00 m/s 9. A car is driven 215 km 25º north of west and then 85 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram. A: o B: o A x = A cos( A ) = (215 km) cos(155 o ) = -195 km A y = A sin( A ) = (215 km) sin(155 o ) = 90.9 km

42 B x = B cos( B ) = (85.0 km) cos(225 o ) = km B y = B sin( B ) = (85.0 km) sin(225 o ) = km R x = A x + B x = -195 km + (-60.1 km) = -255 km R y = A y + B y = 90.9 km + (-60.1 km) = 30.8 km R = (R x 2 + R y 2 ) 1/2 = ( (-255 km) 2 + (30.8 km) 2 ) 1/2 = 257 km q R = tan -1 (R y /R x ) = tan -1 ((30.8 km) /(-255 km)) = o +180 o = 173 o TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 1 February 19, 2007 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points.

43 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. If you are 5'10'' tall, what is your height in meters? (1 in = 2.54 cm.) a. 1.5 m. (4) b. 1.6 m. c. 1.7 m. d. 1.8 m. 2. Suppose that a car traveling to the West (-x direction) begins to slow down as it approaches a

44 traffic light. Make a statement concerning its acceleration. a. The car s acceleration is positive. (4) b. The car s acceleration is negative. c. The acceleration is zero. d. A statement cannot be made using the information given. 3. A ball is thrown with a velocity of 20.0 m/s at an angle of 60.0 above the horizontal. What is the horizontal component of its instantaneous velocity at the exact top of its trajectory? a m/s. (4) b m/s. c m/s. d. Zero. 4. The acceleration of an object is inversely proportional to a. The net force acting on it. (4) b. Its position. c. Its velocity. d. Its mass. 5. Action-reaction forces a. Sometimes act on the same object. (4) b. Always act on the same object. c. May be at right angles.

45 d. Always act on different objects. 6. The acceleration due to gravity is lower on the Moon than on Earth. Which of the following is true about the mass and weight of an astronaut on the Moon's surface, compared to Earth? a. Mass is less, weight is same. (4) b. Mass is same, weight is less. c. Both mass and weight are less. d. Both mass and weight are the same. 7. Vector V 1 is 9.60 units long and points along the positive x axis. Vector V 2 points at 75.0 o to the positive x axis. a. What are the x and y components of each vector? V 1 : o V 2 : o V 1x = V 1 cos( 1 ) = (9.60 units) cos(0 o ) = 9.60 units V 1y = V 1 sin( 1 ) = (9.60 units) sin(0 o ) = 0 V 2x = V 2 cos( 2 ) = (7.55 units) cos(75.0 o ) = 1.95 units V 2y = V 2 sin( 2 ) = (7.55 units) sin(75.0 o ) = 7.29 units is 7.55 units long and b. Determine the sum V 1 V 2 (magnitude and angle). R x = V 1x + V 2x = 9.60 units units = 11.6 units R y = V 1y + V 2y = units = 7.29 units R = (R x 2 + R y 2 ) 1/2 = ( (11.6 units) 2 + (7.29 units) 2 ) 1/2 = 13.7 units R = tan -1 (R y /R x ) = tan -1 ((7.29) /(11.6)) = 32.1 o 8. A football is kicked at ground level with a speed of 18.0 m s at an angle of 35.0º to the

46 horizontal. a. What are the components of the initial velocity? V 0 : o V x0 = V 0 cos( 0 ) = (18.0 m/s) cos(35.0 o ) = 14.7 m/s V y0 = V 0 sin( 0 ) = (18.0 m/s) sin(35.0 o ) = 10.3 m/s b. How much later does it hit the ground? y = y 0 + V y0 t ½ g t 2 0 = 0 + V y0 t ½ g t 2 0 = V y0 ½ g t 2 V y0 /g = t t = 2.10 s c. How far does it travel in the horizontal direction? x = x 0 + V x0 (t) = 0 m + (14.7 m/s) (2.10 s) = 30.9 m 9. A helicopter is ascending vertically with a speed of 5.20 m s. At a height of 125 m above the Earth, a package is dropped from a window. a. How much time does it take for the package to reach the ground? [Hint: The package s initial speed equals the helicopter s.] y = y 0 + V y0 t ½ g t 2 0 = (125 m) + (5.20 m/s) t ½ g t 2 t = 5.61 s b. How fast does the package move just before it hits the ground? V = V 0 g t = (5.20 m/s) - (9.81 m/s 2 ) (5.61 s) = m/s

47 c. What is the acceleration of the package just before it heats the ground? Acceleration due to gravity, 9.81 m/s 2 downward 10. During your winter break, you enter a dogsled race in which students replace the dogs. Wearing cleats for traction, you begin the race by pulling on a rope attached to the sled with a force of 100 N at 15 o with the horizontal. The mass of the sled is 60.0 kg and there is negligible friction between the sled and ice. a. Draw a free body diagram for the sled. (20) b.find the acceleration of the sled.

48 c. Find the normal force exerted by the surface on the sled. TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 1 February 20, 2008 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT

49 TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Which of the following is a scalar quantity? A) Velocity. (4) B) Force. C) Speed. D) Acceleration. 2. The density of a solid object is defined as the ratio of the mass of the object to its volume. The dimension of density is A) [M]/[L]. (4) B) [L] 3 /[M]. C) [M]/[L] 3. D) [M][L] A ball is thrown straight up. When it reaches its highest point, A) Its velocity is zero. (4) B) Its acceleration is zero. C) Both its velocity and acceleration are zero. D) Neither is zero.

50 4. Vector A is along the x-axis and vector B is along the y-axis. Which one of the following statements is correct? A) The x-component of vector A is equal to the x-component of vector B. (4) B) The y-component of vector A is equal to the y-component of vector B. C) The x-component of vector A is equal and opposite to the x-component of vector B. D) Vector A has a zero y-component and vector B has a zero x-component. 5. Under which of the following conditions would a car have a westward acceleration? A) The car going eastward and its speed is increasing. (4) B) The car going eastward and its speed is decreasing. C) The car going westward and its speed is constant. D) The car going eastward and its speed is constant. 6. When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal velocity. This means that A) The parachutist s acceleration is equal to g. (4) B) The force of air resistance is equal to zero. C) The effect of gravity has died down. D) The force of air resistance is equal to the weight of the parachutist.

51 7. You are a player on a reality TV show. You and two other contestants are brought to the center of a large, flat field. Each is given a meter stick, a compass, a calculator, a shovel, and following two displacements: 72.4 m 32.0 o east of north, 57.3 m 36.0 o south of west. These two displacements lead to the point where the keys to a new Porsche are buried. Two players start measuring immediately, but you (knowledge is power!) first calculate where to go and reach the point first. How far do you go and in what direction? A x = A cos( A ) = (72.4 m) cos(58.0 o ) = 38.4 m A y = A sin( A ) = (72.4 m) sin(58.0 o ) = 61.4 m B x = B cos( B ) = (57.3 m) cos(216 o ) = m B y = B sin( B ) = (57.3 m) sin(216 o ) = m R x = A x + B x = (38.4 m) + (-46.4 m) = m R y = A y + B y = (61.4 m) + (-33.7 m) = 27.7 m 2 R = ( R x + R 2 y ) 1/2 = ( (-8.00 m) 2 + (27.7 m) 2 ) 1/2 = 28.8 m R = tan -1 (R y /R x ) = tan -1 ((27.7 m)/(-8.00 m)) = o +180 =106 o 11. A rock is tossed off a cliff at a speed of 20.0 m/s at an angle of 60.0 o to the horizontal. The rock lands at the base of the cliff 4.03 s later. a. What are the components of the initial velocity? V 0 : o V x0 = V 0 cos( 0 ) = (20.0 m/s) cos(60.0 o ) = 10.0 m/s

52 V y0 = V 0 sin( 0 ) = (20.0 m/s) sin(60.0 o ) = 17.3 m/s b. How high is the cliff? y = y 0 + V y0 t ½ gt 2 0 = y 0 + V y0 t ½ gt 2 y 0 = - V y0 t + ½ gt 2 y 0 = - (17.3 m/s)(4.03 s) ½ (9.80 m/s 2 )(4.03 s) 2 = 9.94 m V x = V x0 = 10.0 m V y = V y0 - gt c. What is the angle of impact? V y = (17.3 m/s) (9.80 m/s 2 )(4.03 s) = s = tan -1 (V y /V x ) = tan -1 ((-22.2 m/s)/(10.0 m/s)) = o Bonus (5 points): How high above the base of the cliff does the rock rise? V 2 y = V 2 y0-2 g (y y 0 ) (y y 0 ) = (V 2 y - V 2 y0 )/(- 2 g) V y = 0 m/s (y y 0 ) = (V 2 y0 )/(2 g) = 15.3 m 12. A student throws a water balloon vertically downward from the top of the building. The ballooh leaves the thrower s hand with a speed of 6.00 m/s. Air resistance may be ignored, so the water balloon is in free fall after it leaves the trower s hand.

53 a.what is its speed after falling for 2.00 s? (20) b. How far does it fall in 2.00 s? c. What is its velocity after falling 10.0 m? 13. A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.50 m/s. The worker s force makes an angle of 30.0 o with the horizontal. There is a constant friction force acting on the box with a magnitude of 22.0 N.

54 d. Draw a free body diagram including all the forces acting on the box. (20) e. What is the magnitude of the force the worker must apply to maintain the motion?. f. Find the magnitude of the normal force the floor exerts on the box. TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 1 June 10, 2004 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages.

55 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 80 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems multiple choice problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 10:15 a.m. Stop: 11:35 a.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. If L stands for length, T for time, and M for mass, the dimensions of acceleration are? (5) i. ML 2. j. ML/T. k. L/T 2.

56 l. L/M. 2. Which one of the following statements is NOT true in case of 2-dimentional motion? a. If the acceleration is zero, the speed must be constant. (5) b. If the speed is constant, the acceleration must be zero. c. If the acceleration is zero, the direction of motion must be constant. d. If the speed and direction of the motion are constant, the acceleration must be zero. 3. Figure below represents the parabolic trajectory of a projectile going from point A to point C. What is the direction of the projectile s acceleration at point B? (5) i. Up and to the right. j. Down and to the left. k. Straight up. l. Straight down. 4. You have two vectors, A (6.00; -3.00) and B (-3.00; 4.00). Which one of the following vectors is A + B? a. (9.00; -7.00). (5) b. (3.00; 1.00). c. (-18.0; -12.0). d. (3.00; -7.00). 5.Imagine a physics textbook laying on a horizontal table. One of the forces acting on a textbook is a

57 normal force exerted by the table (let s call it action). What force does form the Newton s 3 rd law pair (reaction)? a. Gravitational force exerted on the textbook by the Earth. (5) b. Gravitational force exerted on the table by the textbook. c. Normal force exerted on the table by the textbook. d. Normal force exerted by the Earth on the table. 6. Take a look at the graph representing one-dimensional motion of a body. Is the instantaneous velocity of the body at point A: (5) i. Positive. j. Negative. k. Zero. l. Cannot be determined from the graph. 7. A joyful physics student throws his cap into the air with an initial velocity of 24.5 m/s at 36.9 o from the horizontal. Another physics student catches the cap at the same height it was thrown. a. What are the components of the cap s initial velocity? V0: o V0x = V0 cos( 0) = 24.5 m/s cos(36.9 o ) = 19.6 m/s V0y = V0 sin( 0) = 24.5 m/s sin(36.9 o ) = 14.7 m/s b. How long is the cap in the air?

58 y = y 0 + V 0y t ½ g t 2 0 = 0 + V 0y t ½ g t 2 0 = (V 0y ½ g t) t (V 0y ½ g t) = 0 t = 2 V 0y /g = 2 (14.7 m/s)/ (9.80 m/s 2 ) = 3.00 s c. What is the maximum height of the cap s trajectory? y = y 0 + V 0y t ½ g t 2 y = (0 m) + (14.7 m/s)(1.50 s) ½ (9.80 m/s 2 ) (1.50) 2 = 11.0 m d. How far does the cap travel in horizontal direction? x = x0 + V0x (t) = 0 m + (19.6 m/s) (3.00 s) = 58.8 m 8. A bullet traveling at 350 m/s strikes a telephone pole and penetrates a distance of 12.0 cm before stopping. a. Assuming that acceleration is constant, find the acceleration of the bullet inside the pole. V 2 = V a ( x- x 0 ) V = 0 m/s; V 0 = 350 m/s; x- x 0 = m a = (V 2 - V 2 0 )/ (2 ( x- x 0 )) a = -510,417 m/s 2 b. How long does it take for the bullet to stop? V = V 0 + a t t = (V - V 0 )/a t = (0 m/s 350 m/s) / (-510,417 m/s 2 ) = 6.84 x 10-4 s c. What is the average velocity of the bullet inside the pole?

59 V average = (x - x 0 )/(t - t 0 ) = (0.120 m)/(6.84 x 10-4 s) = 175 m/s 9. Two cats pull horizontally on the ropes attached to a toy mouse. Cat A exerts a force of 130 N at 20 O north of east and cat B exerts a force of 90 N due west. Find the magnitude of the resultant force on the toy and its direction. A: o B: o Ax = A cos( A) = (130 N) cos(20 o ) = 122 N Ay = A sin( A) = (130 N) sin(20 o ) = 44.5 N Bx = B cos( B) = (90.0 N) cos(180 o ) = N By = B sin( B) = (90.0 N) sin(180 o ) = 0 N Rx = Ax - Bx = 122 N 90.0 N = 32.0 N Ry = Ay - By = 44.5 N 0 N = 44.5 N 2 2 R = (Rx + Ry ) 1/2 = ( (32.0 N) 2 + (44.5 N) 2 ) 1/2 = 54.8 N R = tan -1 (Ry /Rx) = tan -1 (44.5 N /32.0 N) = 54.3 o 10. A 30.0 kg sled is coasting with constant velocity of 5.00 m/s over perfectly smooth, level ice. It enters a rough stretch of ice 20.0 m long in which the force of friction exerted by the ice on the sled is 12.0 N. a.draw a free body diagram indicating all forces acting on the sled while on rough stretch. b. Calculate the magnitude of the normal force the ice exerts on the sled. nx = n cos(9 ) = 0 ny = n sin(9 ) = n fx = f cos(180 ) = -f = N fy = f sin(180 ) = 0 wx = w cos(270 ) = 0 wy = w sin(270 ) = (30.0 kg) (9.80 m/s 2 ) = -294 N Fy = m ay = N + n = 0 n = 294 N

60 c. Calculate the acceleration of the sled while on rough stretch. Fx = m ax N = (30.0 kg) a x a x = m/s 2 Bonus (Worth 5 points): What is the coefficient of kinetic friction between ice and the sled? f = n = f/n = (12.0 N)/ (294 N) = TIME OF COMPLETION NAME SOLUTION

61 DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 1 June 20, 2005 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 11:25 a.m. Stop: 12:50 p.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR

62 PARTIAL CREDIT. 3. If a is acceleration, v is velocity, x is position, and t is time, then which equation is NOT dimensionally correct? m. t = x/v. (5) n. a = v 2 /x. o. v = a/t. p. t 2 = 2x /a 2. Which of the following expresses a principle initially stated by Galileo and later incorporated into Newton s laws of motion? i. An object s acceleration is inversely proportional to its mass. (5) j. For every action is an equal but opposite reaction. k. The natural condition for a moving object is to remain in motion. l. The natural condition for a moving object is to come to rest.. 3. A 7.00 kg bowling ball experiences a net force of 5.00 N. What is its acceleration? m m/s 2. n m/s 2. (5) a = F/m = (5.00 N)/(7.00 kg) o m/s 2. p m/s Which of the following is an example of the type of force which acts at a distance? a. Strong nuclear. b. Weak nuclear.

63 (5) c. Electromagnetic. d. All of the above. 5. Imagine a box sitting on a horizontal table. One of the forces acting on a box is a normal force exerted by the table (let s call it action). What force does form the Newton s 3 rd law pair (reaction)? q. Normal force exerted on the Earth by the table. (5) r. Normal force exerted on the table by the Earth. s. Normal force exerted on the table by the box. t. Normal force exerted on the Earth by the box. 6. Vector A points north and vector B points east. If C = B A, then vector C points: m. North of east. (5) n. South of east. o. North of west. p. South of west. 7. A projectile is fired with an initial speed of 51.2 m/s at an angle of 44.5 o above the horizontal on a long flat firing range. Determine a. The components of the projectile s initial velocity. V 0 : o V x0 = V 0 cos( 0 ) = (51.2 m/s) cos(44.5 o ) = 36.5 m/s

64 V y0 = V 0 sin( 0 ) = (51.2 m/s) sin(44.5 o ) = 35.9 m/s b. The maximum height reached by the projectile. V 2 y = V 2 y0-2 g (y y 0 ) (y y 0 ) = (V 2 y - V 2 y0 )/(- 2 g) V y = 0 m/s (y y 0 ) = (V 2 y0 )/(2 g) = 65.8 m Time to reach the top: c. The total time in the air. V y = V y0 gt 1/2 t 1/2 = V y0 /g = 3.66 s Total time: t = 2 t ½ t = 7.33 s d. The projectile s range. x = x 0 + V x0 t x = (0 m) + (36.5 m/s) (7.33 s) = 267 m Bonus (5 extra points): Find velocity of the projectile 1.50 s after firing. V x = V x0 = 36.5 m/s V y = V y0 gt = (35.9 m/s)- (9.80 m/s 2 )(1.50 s) = 21.2 m/s V =((36.5 m/s) 2 + (21.2 m/s) 2 ) 1/2 = 42.2 m/s = tan -1 (V y /V x ) = = tan -1 ((21.2 m/s)/(36.5 m/s)) = 30.0 o 9. A stone is thrown vertically upward with a speed of 10.0 m/s from the edge of the cliff

65 5.0 m high. a. How much later does it reach the bottom of the cliff? y = y 0 + V y0 t ½ gt 2 y = 5.00 m y 0 = 0 m V y0 = m/s 0 = t 4.9 t t 2-10 t - 5 = 0 t = 2.46 s V y = V y0 - gt b. What is its speed just before hitting? V y = (10.0 m/s) (9.80 m/s 2 )(2.46 s) = s V y = 14.1 s To the top: c. What total distance did it travel? V y 2 = V y0 2-2 g (y y 0 ) (y y 0 ) = (V y 2 - V y0 2 )/(- 2 g) V y = 0 m/s y - y 0 = (V y0 2 )/(2 g) = 5.10 m Total distance traveled: 5.10 m m m = 15.2 m 10. Jets at JFK International airport accelerate from rest at one end of a runaway, and must attain takeoff speed before reaching the other end of the runaway. A plane has acceleration of 11.0 m/s 2 and takeoff velocity of 90.0 m/s.

66 V 0 = 0 m/s V= 90.0 m/s a = 11.0 m/s2 a. What is the minimum runaway required for this airplane? V 2 = V a (x x 0 ) (x x 0 ) = (V 2 - V 0 2 )/(2 a) (x x 0 ) = 368 m V = V 0 + a t b. How long time does it take the plane to attain the takeoff speed? t = (V - V 0 )/a = 8.18 s c. What is the average velocity of the plane in the process? V av = (x x 0 )/t = (368 m) / (8.18 s) = 45.0 m/s 10. Vector A has a magnitude of 50.0 m and points in a direction 20.0 o below the x axis. A second vector, B, has a magnitude of 70.0 m and points in a direction 50.0 o above the x axis. Find a new vector R = A + B. Find the magnitude and direction of vector R. A x = A cos( A ) = (50.0 m) cos(-20.0 o ) = 47.0 m A y = A sin( A ) = (50.0 m) sin(-20.0 o ) = m B x = B cos( B ) = (70.0 m) cos(50.0 o ) = 45.0 m B y = B sin( B ) = (70.0 m) sin( 50.0 o ) = 53.6 m R x = A x + B x = (47.0 m) + (45.0 m) = 92.0 m R y = A y + B y = (-17.1 m) + (53.6 m) = 36.5 m R = ( R x 2 + R y 2 ) 1/2 = ( (92.0 m) 2 + (36.5 m) 2 ) 1/2 = 99.0 m

67 R = tan -1 (R y /R x ) = tan -1 ((36.5 m)/(92.0 m)) = 21.6 o TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 1 Version 1 June 20, 2007 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination.

68 4. The total weight of the examination is 150 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 1:30 p.m. Stop: 2:45 p.m PROBLEM POINTS CREDIT TOTAL 150 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Which of the following is a vector quantity? a. Time (6) b. Mass. c. Displacement. d. Speed.

69 2. Suppose that a car traveling to the East (+ x direction) begins to slow down as it approaches a traffic light. Make a statement concerning its acceleration. a. The car s acceleration is positive. (6) b. The car s acceleration is negative. c. The acceleration is zero. d. A statement cannot be made using the information given. 3. If the position versus time graph of an object is a horizontal line, the object is a. Moving with constant non-zero velocity. (6) b. Moving with constant non-zero acceleration. c. At rest. d. Moving with infinite velocity. 4. The acceleration of an object is linearly proportional to a. The net force acting on it. (6) b. Its position. c. Its velocity. d. Its mass. 5. A ball is thrown straight up. When it reaches its highest point, PHYS 1111 Exam 1, Version 1 Fall

70 a. Both its velocity and its acceleration are zero. (6) b. Its velocity is zero and its acceleration is not zero. c. Its velocity is not zero and its acceleration is zero. d. Neither its velocity nor acceleration is zero. 6. A ball is thrown straight into the air. Ignore air resistance. At the highest point of its trajectory, the net force acting on it is a. Equal to its weight. (6) b. Greater than its weight. c. Less than its weight. d. Zero. 7. Vector V 1 is 12.5 units long and points along the negative x axis. Vector V 2 (30) points at 25.0 o to the positive x axis. a. What are the x and y components of each vector? is 4.35 units long and V 1x = V 1 cos( 1 ) = (12.5) cos(180 o ) = V 1y = V 1 sin( 1 ) = (12.5) sin(180 o ) = 0 V 2x = V 2 cos( 2 ) = (4.35) cos(25.0 o ) = 3.94 V 2y = V 2 sin( 2 ) = (4.35) sin( 25.0 o ) = 1.84 PHYS 1111 Exam 1, Version 1 Fall

71 b. Determine the sum V 1 V 2 (magnitude and angle). R x = V 1x + V 2x = (- 12.5) + (3.94) = R y = V 1y + V 2y = (0) + (1.84) = R = ( R x + R 2 y ) 1/2 = ( (-8.56) 2 + (1.84) 2 ) 1/2 = 8.76 R = tan -1 (R y /R x ) = tan -1 ((1.84)/(-8.56)) = o o = 168 o 14. A ball is thrown from the top of a 100-m building with an initial speed of 15.0 m/s at an angle of 40.0º to the horizontal. The ball hits the ground 5.60 s later. a. What are the components of the initial velocity? V x0 = V 0 cos( 0 ) = (15.0 m/s) cos(40.0 o ) = 11.5 m/s V y0 = V 0 sin( 0 ) = (15.0 m/s) sin( 40.0 o ) = 9.64 m/s x = x 0 + V x0 t b. How far from the base of the building does the ball land? x = (0 m) + (11.5 m/s) (5.60 s) =64.4 m c. What is the maximum height that the ball reaches? 2 V y = V 2 y0-2 g (y-y 0 ) 0 = V 2 y0-2 g (y-y 0 ) (y-y 0 ) = V 2 y0 /(2 g) (y- 100 m) = (9.64 m/s) 2 /(2 g) y = 105 m (from the ground) or 4.74 m from the top of the building PHYS 1111 Exam 1, Version 1 Fall

72 Bonus (8 points): Find the angle of the impact. V x = V x0 = 11.5 m/s V y = V y0 - gt = (9.64 m/s) (9.81 m/s 2 )(5.60 s) = m/s = tan -1 (V y /V x ) = tan -1 ((-45.3 m/s)/(11.5 m/s)) = o 15. A bullet traveling at 350 m/s strikes a telephone pole and penetrates a distance of 12.0 cm before stopping. a. What is the acceleration of the bullet assuming it is constant? V 2 = V a (x-x 0 ) a = (V 2 - V 2 0 )/(2(x-x 0 )) a = (0 2 (350 m/s) 2 )/(2(0.120 m)) = m/s 2 = x 10 5 m/s 2 V = V 0 + at b. How long did it take for the bullet to stop? t =(V - V 0 )/a = (0 (350 m/s))/( x 10 5 m/s 2 ) = 6.86 x 10-4 s 10. Dr. K is trying to move a box full of physics textbooks (total mass of 30 kg) up the incline by exerting a constant force of 200 N parallel to the surface of the incline. Luckily for her, the surface of the incline is frictionless. The surface is inclined to the horizontal at an angle of 35 o. g. Draw a free body diagram including all the forces acting on the box. PHYS 1111 Exam 1, Version 1 Fall

73 h. Find the magnitude of the normal force exerted on the box by the incline. A x = A cos(0 o ) = (200 N) cos(0 o ) = 200 N A y = A sin(0 o ) = (200 N) sin(0 o ) = 0 N n x = n cos(90.0 o ) = 0 n y = n sin(90.0 o ) = n w x = w cos( o ) = (30.0 kg) (9.81 m/s 2 ) cos( o ) = -169 N w y = w sin( o ) = (30.0 kg) (9.81 m/s 2 ) sin( o ) = -241 N F y = m a y A y + n y + w y = 0 n -241 N = 0 n = 241 N i. Find the acceleration of the box. PHYS 1111 Exam 1, Version 1 Fall

74 F x = m a x A x w x = m a x a x = (A x + w x )/m a x = (200 N 169 N)/(30.0 kg) = 1.03 m/s 2 TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 1 Section 2 PHYS 1111 Exam 1, Version 1 Fall

75 Version 2 February 22, 2010 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on six (6) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and three (3) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 5:00 p.m. Stop: 6:15 p.m. PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. A bird flies 3.00 km due west and then 2.00 km due north. What is the magnitude of the PHYS 1111 Exam 1, Version 1 Fall

76 bird s displacement? u km. (5) v km. w km. x km. 2. The gas pedal in a car is sometimes referred to as the accelerator. Which other controls on the vehicle can be used to produce acceleration? m. The breaks. (5) n. The steering wheel. o. The gear shift. p. All of the above. 3.Two vectors appear as in figure below. Which combination points directly to the left? (5) a. P Q b. P Q PHYS 1111 Exam 1, Version 1 Fall

77 c. Q P d. Q P 4. A ball fired from a canon at point 1 follows a trajectory shown below. Air resistance may be neglected. Four possible vectors are also shown in the figure. Which vector best represents the ball s velocity at point 2? a. A. (5) b. B. c. C. d. D. 5. A ball fired from a canon at point 1 follows a trajectory shown above. Air resistance may be neglected. Four possible vectors are also shown in the figure. Which vector best represents the ball s acceleration at point 4? a. A. (5) b. B. c. C. d. D. PHYS 1111 Exam 1, Version 1 Fall

### Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion

Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion Conceptual Questions 1) Which of Newton's laws best explains why motorists should buckle-up? A) the first law

### B) 286 m C) 325 m D) 367 m Answer: B

Practice Midterm 1 1) When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal velocity. This means that A) the acceleration is equal to g. B) the force of

### Chapter 7: Momentum and Impulse

Chapter 7: Momentum and Impulse 1. When a baseball bat hits the ball, the impulse delivered to the ball is increased by A. follow through on the swing. B. rapidly stopping the bat after impact. C. letting

### Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

### Speed A B C. Time. Chapter 3: Falling Objects and Projectile Motion

Chapter 3: Falling Objects and Projectile Motion 1. Neglecting friction, if a Cadillac and Volkswagen start rolling down a hill together, the heavier Cadillac will get to the bottom A. before the Volkswagen.

### Exam 1 Review Questions PHY 2425 - Exam 1

Exam 1 Review Questions PHY 2425 - Exam 1 Exam 1H Rev Ques.doc - 1 - Section: 1 7 Topic: General Properties of Vectors Type: Conceptual 1 Given vector A, the vector 3 A A) has a magnitude 3 times that

### Physics: Principles and Applications, 6e Giancoli Chapter 2 Describing Motion: Kinematics in One Dimension

Physics: Principles and Applications, 6e Giancoli Chapter 2 Describing Motion: Kinematics in One Dimension Conceptual Questions 1) Suppose that an object travels from one point in space to another. Make

### PHY121 #8 Midterm I 3.06.2013

PHY11 #8 Midterm I 3.06.013 AP Physics- Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension

### Chapter 3 Practice Test

Chapter 3 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which of the following is a physical quantity that has both magnitude and direction?

### 2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration.

2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration. Dynamics looks at the cause of acceleration: an unbalanced force. Isaac Newton was

### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Vector A has length 4 units and directed to the north. Vector B has length 9 units and is directed

### Newton s Laws of Motion

Section 3.2 Newton s Laws of Motion Objectives Analyze relationships between forces and motion Calculate the effects of forces on objects Identify force pairs between objects New Vocabulary Newton s first

### Projectile Motion 1:Horizontally Launched Projectiles

A cannon shoots a clown directly upward with a speed of 20 m/s. What height will the clown reach? How much time will the clown spend in the air? Projectile Motion 1:Horizontally Launched Projectiles Two

### 1. Newton s Laws of Motion and their Applications Tutorial 1

1. Newton s Laws of Motion and their Applications Tutorial 1 1.1 On a planet far, far away, an astronaut picks up a rock. The rock has a mass of 5.00 kg, and on this particular planet its weight is 40.0

### 8. As a cart travels around a horizontal circular track, the cart must undergo a change in (1) velocity (3) speed (2) inertia (4) weight

1. What is the average speed of an object that travels 6.00 meters north in 2.00 seconds and then travels 3.00 meters east in 1.00 second? 9.00 m/s 3.00 m/s 0.333 m/s 4.24 m/s 2. What is the distance traveled

### Physics Kinematics Model

Physics Kinematics Model I. Overview Active Physics introduces the concept of average velocity and average acceleration. This unit supplements Active Physics by addressing the concept of instantaneous

### Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4 Forces and Newton s Laws of Motion continued Clicker Question 4.3 A mass at rest on a ramp. How does the friction between the mass and the table know how much force will EXACTLY balance the gravity

### AP Physics C Fall Final Web Review

Name: Class: _ Date: _ AP Physics C Fall Final Web Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. On a position versus time graph, the slope of

### v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

### Physics 11 Assignment KEY Dynamics Chapters 4 & 5

Physics Assignment KEY Dynamics Chapters 4 & 5 ote: for all dynamics problem-solving questions, draw appropriate free body diagrams and use the aforementioned problem-solving method.. Define the following

### B) 40.8 m C) 19.6 m D) None of the other choices is correct. Answer: B

Practice Test 1 1) Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.60 s and reaches the level of the top of the pole after a total elapsed time

### Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting

### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) The following four forces act on a 4.00 kg object: 1) F 1 = 300 N east F 2 = 700 N north

### 5.1 Vector and Scalar Quantities. A vector quantity includes both magnitude and direction, but a scalar quantity includes only magnitude.

Projectile motion can be described by the horizontal ontal and vertical components of motion. In the previous chapter we studied simple straight-line motion linear motion. Now we extend these ideas to

### Physics 2101, First Exam, Fall 2007

Physics 2101, First Exam, Fall 2007 September 4, 2007 Please turn OFF your cell phone and MP3 player! Write down your name and section number in the scantron form. Make sure to mark your answers in the

### Ch 6 Forces. Question: 9 Problems: 3, 5, 13, 23, 29, 31, 37, 41, 45, 47, 55, 79

Ch 6 Forces Question: 9 Problems: 3, 5, 13, 23, 29, 31, 37, 41, 45, 47, 55, 79 Friction When is friction present in ordinary life? - car brakes - driving around a turn - walking - rubbing your hands together

### VELOCITY, ACCELERATION, FORCE

VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how

### 1 of 7 9/5/2009 6:12 PM

1 of 7 9/5/2009 6:12 PM Chapter 2 Homework Due: 9:00am on Tuesday, September 8, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]

### Web review - Ch 3 motion in two dimensions practice test

Name: Class: _ Date: _ Web review - Ch 3 motion in two dimensions practice test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which type of quantity

### 2008 FXA DERIVING THE EQUATIONS OF MOTION 1. Candidates should be able to :

Candidates should be able to : Derive the equations of motion for constant acceleration in a straight line from a velocity-time graph. Select and use the equations of motion for constant acceleration in

### Chapter 4 Dynamics: Newton s Laws of Motion. Copyright 2009 Pearson Education, Inc.

Chapter 4 Dynamics: Newton s Laws of Motion Force Units of Chapter 4 Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal

### Newton s Third Law. object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1

Newton s Third Law! If two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1!! Note on notation: is

### Review Chapters 2, 3, 4, 5

Review Chapters 2, 3, 4, 5 4) The gain in speed each second for a freely-falling object is about A) 0. B) 5 m/s. C) 10 m/s. D) 20 m/s. E) depends on the initial speed 9) Whirl a rock at the end of a string

### Work, Energy & Momentum Homework Packet Worksheet 1: This is a lot of work!

Work, Energy & Momentum Homework Packet Worksheet 1: This is a lot of work! 1. A student holds her 1.5-kg psychology textbook out of a second floor classroom window until her arm is tired; then she releases

### 5. Forces and Motion-I. Force is an interaction that causes the acceleration of a body. A vector quantity.

5. Forces and Motion-I 1 Force is an interaction that causes the acceleration of a body. A vector quantity. Newton's First Law: Consider a body on which no net force acts. If the body is at rest, it will

### TEACHER ANSWER KEY November 12, 2003. Phys - Vectors 11-13-2003

Phys - Vectors 11-13-2003 TEACHER ANSWER KEY November 12, 2003 5 1. A 1.5-kilogram lab cart is accelerated uniformly from rest to a speed of 2.0 meters per second in 0.50 second. What is the magnitude

### Chapter 4 Newton s Laws: Explaining Motion

Chapter 4 Newton s s Laws: Explaining Motion Newton s Laws of Motion The concepts of force, mass, and weight play critical roles. A Brief History! Where do our ideas and theories about motion come from?!

### Tennessee State University

Tennessee State University Dept. of Physics & Mathematics PHYS 2010 CF SU 2009 Name 30% Time is 2 hours. Cheating will give you an F-grade. Other instructions will be given in the Hall. MULTIPLE CHOICE.

### PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?

1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always

### SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Exam Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s.

### PHYSICS 149: Lecture 4

PHYSICS 149: Lecture 4 Chapter 2 2.3 Inertia and Equilibrium: Newton s First Law of Motion 2.4 Vector Addition Using Components 2.5 Newton s Third Law 1 Net Force The net force is the vector sum of all

### 2After completing this chapter you should be able to

After completing this chapter you should be able to solve problems involving motion in a straight line with constant acceleration model an object moving vertically under gravity understand distance time

### Physics I Honors: Chapter 4 Practice Exam

Physics I Honors: Chapter 4 Practice Exam Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Which of the following statements does not describe

### Worksheet #1 Free Body or Force diagrams

Worksheet #1 Free Body or Force diagrams Drawing Free-Body Diagrams Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.

### 9. The kinetic energy of the moving object is (1) 5 J (3) 15 J (2) 10 J (4) 50 J

1. If the kinetic energy of an object is 16 joules when its speed is 4.0 meters per second, then the mass of the objects is (1) 0.5 kg (3) 8.0 kg (2) 2.0 kg (4) 19.6 kg Base your answers to questions 9

### AP Physics B Practice Workbook Book 1 Mechanics, Fluid Mechanics and Thermodynamics

AP Physics B Practice Workbook Book 1 Mechanics, Fluid Mechanics and Thermodynamics. The following( is applicable to this entire document copies for student distribution for exam preparation explicitly

### 1) The gure below shows the position of a particle (moving along a straight line) as a function of time. Which of the following statements is true?

Physics 2A, Sec C00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to ll your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry

### PHYS 117- Exam I. Multiple Choice Identify the letter of the choice that best completes the statement or answers the question.

PHYS 117- Exam I Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Car A travels from milepost 343 to milepost 349 in 5 minutes. Car B travels

### Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel

Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel Name: Lab Day: 1. A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the

### Units DEMO spring scales masses

Dynamics the study of the causes and changes of motion Force Force Categories ContactField 4 fundamental Force Types 1 Gravity 2 Weak Nuclear Force 3 Electromagnetic 4 Strong Nuclear Force Units DEMO spring

### Chapter 4 Dynamics: Newton s Laws of Motion

Chapter 4 Dynamics: Newton s Laws of Motion Units of Chapter 4 Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal

### Physics Notes Class 11 CHAPTER 3 MOTION IN A STRAIGHT LINE

1 P a g e Motion Physics Notes Class 11 CHAPTER 3 MOTION IN A STRAIGHT LINE If an object changes its position with respect to its surroundings with time, then it is called in motion. Rest If an object

### Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces Units of Chapter 5 Applications of Newton s Laws Involving Friction Uniform Circular Motion Kinematics Dynamics of Uniform Circular

### DISPLACEMENT AND FORCE IN TWO DIMENSIONS

DISPLACEMENT AND FORCE IN TWO DIMENSIONS Vocabulary Review Write the term that correctly completes the statement. Use each term once. coefficient of kinetic friction equilibrant static friction coefficient

### Review Assessment: Lec 02 Quiz

COURSES > PHYSICS GUEST SITE > CONTROL PANEL > 1ST SEM. QUIZZES > REVIEW ASSESSMENT: LEC 02 QUIZ Review Assessment: Lec 02 Quiz Name: Status : Score: Instructions: Lec 02 Quiz Completed 20 out of 100 points

### Q1. (a) State the difference between vector and scalar quantities (1)

Q1. (a) State the difference between vector and scalar quantities....... (1) (b) State one example of a vector quantity (other than force) and one example of a scalar quantity. vector quantity... scalar

### Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION

1 P a g e Inertia Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION The property of an object by virtue of which it cannot change its state of rest or of uniform motion along a straight line its own, is

### Lecture 07: Work and Kinetic Energy. Physics 2210 Fall Semester 2014

Lecture 07: Work and Kinetic Energy Physics 2210 Fall Semester 2014 Announcements Schedule next few weeks: 9/08 Unit 3 9/10 Unit 4 9/15 Unit 5 (guest lecturer) 9/17 Unit 6 (guest lecturer) 9/22 Unit 7,

### C B A T 3 T 2 T 1. 1. What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N

Three boxes are connected by massless strings and are resting on a frictionless table. Each box has a mass of 15 kg, and the tension T 1 in the right string is accelerating the boxes to the right at a

### Vectors; 2-D Motion. Part I. Multiple Choice. 1. v

This test covers vectors using both polar coordinates and i-j notation, radial and tangential acceleration, and two-dimensional motion including projectiles. Part I. Multiple Choice 1. v h x In a lab experiment,

### Physics 590 Homework, Week 6 Week 6, Homework 1

Physics 590 Homework, Week 6 Week 6, Homework 1 Prob. 6.1.1 A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 35 m/s. At the same time it has a horizontal

### F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26

Physics 23 Exam 2 Spring 2010 Dr. Alward Page 1 1. A 250-N force is directed horizontally as shown to push a 29-kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force,

### Chapter 4: Newton s Laws: Explaining Motion

Chapter 4: Newton s Laws: Explaining Motion 1. All except one of the following require the application of a net force. Which one is the exception? A. to change an object from a state of rest to a state

### F13--HPhys--Q5 Practice

Name: Class: Date: ID: A F13--HPhys--Q5 Practice Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A vector is a quantity that has a. time and direction.

### B Answer: neither of these. Mass A is accelerating, so the net force on A must be non-zero Likewise for mass B.

CTA-1. An Atwood's machine is a pulley with two masses connected by a string as shown. The mass of object A, m A, is twice the mass of object B, m B. The tension T in the string on the left, above mass

### PHY231 Section 1, Form B March 22, 2012

1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate

### 356 CHAPTER 12 Bob Daemmrich

Standard 7.3.17: Investigate that an unbalanced force, acting on an object, changes its speed or path of motion or both, and know that if the force always acts toward the same center as the object moves,

### 10.1 Quantitative. Answer: A Var: 50+

Chapter 10 Energy and Work 10.1 Quantitative 1) A child does 350 J of work while pulling a box from the ground up to his tree house with a rope. The tree house is 4.8 m above the ground. What is the mass

### Curso2012-2013 Física Básica Experimental I Cuestiones Tema IV. Trabajo y energía.

1. A body of mass m slides a distance d along a horizontal surface. How much work is done by gravity? A) mgd B) zero C) mgd D) One cannot tell from the given information. E) None of these is correct. 2.

### Projectile motion simulator. http://www.walter-fendt.de/ph11e/projectile.htm

More Chapter 3 Projectile motion simulator http://www.walter-fendt.de/ph11e/projectile.htm The equations of motion for constant acceleration from chapter 2 are valid separately for both motion in the x

### Catapult Engineering Pilot Workshop. LA Tech STEP 2007-2008

Catapult Engineering Pilot Workshop LA Tech STEP 2007-2008 Some Background Info Galileo Galilei (1564-1642) did experiments regarding Acceleration. He realized that the change in velocity of balls rolling

### 3. KINEMATICS IN TWO DIMENSIONS; VECTORS.

3. KINEMATICS IN TWO DIMENSIONS; VECTORS. Key words: Motion in Two Dimensions, Scalars, Vectors, Addition of Vectors by Graphical Methods, Tail to Tip Method, Parallelogram Method, Negative Vector, Vector

### ACTIVITY 1: Gravitational Force and Acceleration

CHAPTER 3 ACTIVITY 1: Gravitational Force and Acceleration LEARNING TARGET: You will determine the relationship between mass, acceleration, and gravitational force. PURPOSE: So far in the course, you ve

### Newton s Laws of Motion

Physics Newton s Laws of Motion Newton s Laws of Motion 4.1 Objectives Explain Newton s first law of motion. Explain Newton s second law of motion. Explain Newton s third law of motion. Solve problems

### Problem Set #8 Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.01L: Physics I November 7, 2015 Prof. Alan Guth Problem Set #8 Solutions Due by 11:00 am on Friday, November 6 in the bins at the intersection

### Supplemental Questions

Supplemental Questions The fastest of all fishes is the sailfish. If a sailfish accelerates at a rate of 14 (km/hr)/sec [fwd] for 4.7 s from its initial velocity of 42 km/h [fwd], what is its final velocity?

### Lecture Presentation Chapter 4 Forces and Newton s Laws of Motion

Lecture Presentation Chapter 4 Forces and Newton s Laws of Motion Suggested Videos for Chapter 4 Prelecture Videos Newton s Laws Forces Video Tutor Solutions Force and Newton s Laws of Motion Class Videos

### Chapter 3 Falling Objects and Projectile Motion

Chapter 3 Falling Objects and Projectile Motion Gravity influences motion in a particular way. How does a dropped object behave?!does the object accelerate, or is the speed constant?!do two objects behave

### PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7. February 13, 2013

PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7 February 13, 2013 0.1 A 2.00-kg object undergoes an acceleration given by a = (6.00î + 4.00ĵ)m/s 2 a) Find the resultatnt force acting on the object

### Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of

### III. Applications of Force and Motion Concepts. Concept Review. Conflicting Contentions. 1. Airplane Drop 2. Moving Ball Toss 3. Galileo s Argument

III. Applications of Force and Motion Concepts Concept Review Conflicting Contentions 1. Airplane Drop 2. Moving Ball Toss 3. Galileo s Argument Qualitative Reasoning 1. Dropping Balls 2. Spinning Bug

### AP Physics Newton's Laws Practice Test

AP Physics Newton's Laws Practice Test Answers: A,D,C,D,C,E,D,B,A,B,C,C,A,A 15. (b) both are 2.8 m/s 2 (c) 22.4 N (d) 1 s, 2.8 m/s 16. (a) 12.5 N, 3.54 m/s 2 (b) 5.3 kg 1. Two blocks are pushed along a

### Understanding the motion of the Universe. Motion, Force, and Gravity

Understanding the motion of the Universe Motion, Force, and Gravity Laws of Motion Stationary objects do not begin moving on their own. In the same way, moving objects don t change their movement spontaneously.

### Mass, energy, power and time are scalar quantities which do not have direction.

Dynamics Worksheet Answers (a) Answers: A vector quantity has direction while a scalar quantity does not have direction. Answers: (D) Velocity, weight and friction are vector quantities. Note: weight and

### UNIT 2D. Laws of Motion

Name: Regents Physics Date: Mr. Morgante UNIT 2D Laws of Motion Laws of Motion Science of Describing Motion is Kinematics. Dynamics- the study of forces that act on bodies in motion. First Law of Motion

### ANSWER KEY. Reviewing Physics: The Physical Setting THIRD EDITION. Amsco School Publications, Inc. 315 Hudson Street / New York, N.Y.

NSWER KEY Reviewing Physics: The Physical Setting THIRD EDITION msco School Publications, Inc. 315 Hudson Street / New York, N.Y. 10013 N 7310 CD Manufactured in the United States of merica 1345678910

### Newton s Law of Motion

chapter 5 Newton s Law of Motion Static system 1. Hanging two identical masses Context in the textbook: Section 5.3, combination of forces, Example 4. Vertical motion without friction 2. Elevator: Decelerating

### Chapter 11 Equilibrium

11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium of

### 2) When you look at the speedometer in a moving car, you can see the car's.

Practice Kinematics Questions Answers are at the end Choose the best answer to each question and write the appropriate letter in the space provided. 1) One possible unit of speed is. A) light years per

### Chapter 6. Work and Energy

Chapter 6 Work and Energy ENERGY IS THE ABILITY TO DO WORK = TO APPLY A FORCE OVER A DISTANCE= Example: push over a distance, pull over a distance. Mechanical energy comes into 2 forms: Kinetic energy

### http://www.webassign.net/v4cgikchowdary@evergreen/assignments/prev... 1 of 10 7/29/2014 7:28 AM 2 of 10 7/29/2014 7:28 AM

HW1 due 6 pm Day 3 (Wed. Jul. 30) 2. Question Details OSColPhys1 2.P.042.Tutorial.WA. [2707433] Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 (a) The graph below plots the position versus time

### Newton s Laws. Physics 1425 lecture 6. Michael Fowler, UVa.

Newton s Laws Physics 1425 lecture 6 Michael Fowler, UVa. Newton Extended Galileo s Picture of Galileo said: Motion to Include Forces Natural horizontal motion is at constant velocity unless a force acts:

### Chapter 5 Newton s Laws of Motion

Chapter 5 Newton s Laws of Motion Sir Isaac Newton (1642 1727) Developed a picture of the universe as a subtle, elaborate clockwork slowly unwinding according to well-defined rules. The book Philosophiae

### Physics Section 3.2 Free Fall

Physics Section 3.2 Free Fall Aristotle Aristotle taught that the substances making up the Earth were different from the substance making up the heavens. He also taught that dynamics (the branch of physics

### 4 Gravity: A Force of Attraction

CHAPTER 1 SECTION Matter in Motion 4 Gravity: A Force of Attraction BEFORE YOU READ After you read this section, you should be able to answer these questions: What is gravity? How are weight and mass different?

### charge is detonated, causing the smaller glider with mass M, to move off to the right at 5 m/s. What is the

This test covers momentum, impulse, conservation of momentum, elastic collisions, inelastic collisions, perfectly inelastic collisions, 2-D collisions, and center-of-mass, with some problems requiring

### PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be