Surname Given Names Student Number University of Guelph Department of Physics PHYS*1020DE Introductory Physics Instructor: R.L. Brooks Midterm Examination 26 February 2003 90 Minutes INSTRUCTIONS: This is a closed book exam. You may use a calculator and the enclosed formula sheet. For the 15 multiple choice questions, circle the letter preceding the correct answer AND mark your answer on the computer answer sheet provided. Do all work on this paper and hand in BOTH the answer sheet and this question package. Give solutions to the three problems on the question sheets themselves. Use the reverse side if needed. Show all work; partial marks are possible for the three problems.
1. Which of the quantities given below is NOT a vector? A. displacement D. velocity B. force E. power C. acceleration 2. A ball is thrown straight up from a height of 5 m with a velocity of 5 m/s. Ignoring air friction how fast is the ball moving when it hits the ground? A. 45 m/s D. 11.1 m/s B. 18.7 m/s E. 8.5 m/s C. 13.3 m/s 3. A train accelerates as it enters a valley from the top of a hill. It accelerates uniformly at 0.04 m/s 2 for 4 minutes and during this time travels 3.0 km. What was the magnitude of the velocity at the start of the acceleration? A. 2.9 m/s D. 168 m/s B. 7.7 m/s E. 670 m/s C. 13 m/s 4. A ball is tossed straight up into the air and returns to your hand. Ignoring friction, you know that A. On the way up, the ball had a downward acceleration. B. On the way down, the ball had an upward acceleration. C. At the top of its flight, the ball had no acceleration. D. The ball's initial acceleration was zero. E. The ball never had any acceleration. 5. On a free-body diagram you want to draw all of the forces acting on a body to help you to determine if there is a net force or not. The net force should not be drawn. On a free-body diagram of a skier moving straight down an incline there should be: A. 1 force; the force of gravity acting on the skier B. 2 forces; the force of gravity and the normal force. C. 3 forces; the force of gravity, the normal force and any friction force slowing down the skier. D. 4 forces; the force of gravity, the normal force, the friction force, and the force accelerating the skier down the incline. E. 5 forces; all of the above plus the component of the normal force opposing gravity. 6. If vector A is 8 m west and vector B is 5 m north then vector A-B is: A. 9.4 m south-west D. 13 m 58E south of east B. 9.4 m 32E north of west E. 13 m 58E west of north C. 9.4 m 32E south of west
Below is the velocity curve for an automobile which has smoothly accelerated over a time interval of 10 seconds. The next two questions test your ability to interpret this curve. 7. After 6 seconds the automobile is travelling at: A. 50 m/s D. 12 m/s B. 22 m/s E. 17 m/s C. 5 m/s 8. The automobile s acceleration at time t=0 is zero and is also zero at time t=10s. The largest acceleration occurs at time t=5s. About what is the value of this acceleration? A. 5 m/s 2 D. 6.5 m/s B. 25 m/s 2 E. 0 m/s C. 12.5 m/s 2 9. A person is standing in an elevator stopped at the 14th floor. By Newton s third law, the force that is equal and opposite to the person pushing down on the elevator is: A. The tension in the cable pulling up on the elevator. B. The normal force of the elevator pushing up on the person. C. The elevator pulling up on the earth. D. The person pulling up on the earth. E. The earth pulling down on the person. 10. An experiment is used to determine the speed of light in a transparent material. The measured value is 1.96 10 8 m/s and the accepted value is 2.06 10 8 m/s. The percent error in the measurement is: A. 1% D. 10% B. 2% E. 15% C. 5%
11. A small ice-cube is released from rest at the top edge of a hemispherical bowl. When it reaches the bottom of the bowl, its speed is 1.8 m/s. Using conservation of energy the radius of the bowl is: A. 5 m D. 11 cm B. 11 m E. 17 cm C. 5 cm 12. A coin is placed flat on a record at a distance of 10 cm from the centre of the record. Calculate the magnitude of the centripetal acceleration of the coin if the record is playing at 33 1/3 rpm (revolutions per minute). A. 1.2 m/s 2 D. 1.9 10 2 m/s 2 B. 1.6 m/s 2 E. 4.4 10 3 m/s 2 C. 2.2 m/s 2 13. A refrigerator can remove 792 kilo-joules of heat every hour. If it ran at 100% efficiency it would be tated at A. 792 kw D. 13.2 kw B. 792 W E. 13.2 W C. 220 W 14. A box with a mass of 20 kg is dragged 3.0 m across the floor. The coefficient of kinetic friction is 0.45. About how much frictional heat is generated? A. 27 J D. 265 J B. 44 J E. 590 J C. 130 J 15. A 20 m long ramp connects the ground to an outdoor stage 5 m above ground level. A 5 kg crate, starting from rest, slides down the ramp and reaches the ground moving at 4 m/s. Using energy considerations you know that: A. A small fraction of the crate s initial energy was lost to friction on the way down. B. Over 80% of the crate s initial energy was lost to friction on the way down. C. There is no friction on the ramp. D. All of the crate s initial energy was lost to friction on the way down. E. One needs the coefficient of friction to answer this question. 3 PROBLEMS: Do all work on these sheets. If you cannot get some intermediate result that is needed subsequently, assume some value, show your work, and you will get full marks for the parts answered correctly.
1. An cannon ball is shot into the air at an angle of 43 to the horizontal. It achieves a maximum height of 200 m before striking the ground. (a) What is the cannon ball's initial velocity? (3 pts) (b) How long is the ball in the air? (5 pts) (c) How far is the landing position from the launch position? (2 pts) Solution: First solve the up/down part of the problem. You know the ball starts from rest, rises to 200m where it stops briefly, and falls to the earth. Consider first the motion from start to maximum height. Then v 2 - v 0 2 = 2 a (y - y 0 ) y 0 = 0. y = 200 m a = -9.8 m/s 2 v 0 =? v = 0 m/s at the top So -v 0 2 = -19.6 m/s 2 X 200 m v 0 2 = 3920 m 2 /s 2 v 0 = 62.6 m/s UP so this is v 0y The initial angle is 43 to the horizontal so that means v 0y = v 0 sin 43 v 0x = v 0 cos 43 We know v 0y, we want v 0 so the first equation can be solved to obtain (a) v 0 = 91.8 m/s at 43 to the horizontal For the time in the air, again use the vertical motion to write: y = y 0 + v 0y t + 1/2 a t 2 The ball starts at y=0 but it ends at y=0 if we consider the up and down motion which lasts the entire time that we are interested in. We also know v 0y from the previous section and a=-g so we know that as well. We get 0 = 0 + 62.6 m/s t + 1/2 (-9.8)m/s 2 t 2 62.6 m/s = 4.9 m/s 2 t and solve for t (b) t = 12.8 s The horizontal distance is just v 0x multiplied by this time as no acceleration acts horizontally. That is, x = x 0 + v 0x t + 1/2 a t 2 x 0 = 0 a = 0 (c) x = 91.8 m/s cos 43 12.8 s = 859 m
2. Two sleds are connected by a horizontal rope, and the front sled is pulled by another horizontal rope with a tension of magnitude 30 N. The sled masses are: front 7.0 kg; rear, 5.0 kg. The coefficient of kinetic friction for both sleds is 0.15. (a) Determine the frictional force on each of the sleds (3 pts) (b) Determine the acceleration of the sleds (3 pts) (c) Determine the tension in the connecting rope (4 pts.) The frictional force is given by F k = µ k F N The normal force is just the weight of the sleds and the coefficient of friction has been given. Hence for each of the sleds we have: F k = 0.15 9.8 m/s 2 7.0 kg (front sled) (a) = 10.3 N (front sled) = 0.15 9.8 m/s 2 5.0 kg (rear sled) (a) = 7.4 N (rear sled) The net force on the sleds horizontally is just 30 N to the right and 10.3 + 7.4 N to the left or 12.3 N to the right net force. This net force accelerates the sleds to the right. The total mass is 12 kg, so using F = m a we get an acceleration of (b) 1.0 m/s 2 to the right. Now the net force that accelerates the sleds is just the difference between the tension and the frictional force. The tension on the lead rope was given as 30 N. To find the tension in the connecting rope we must realize that it provides the pull for the rear sled. That tension must satisfy F net = T - F k the same equation used previously. We know that on the rear sled F k = 7.4 N and we know that the net force is accelerating the rear sled. So the net force on the rear sled must be F net = m a = 5.0 kg 1.0 m/s 2 = 5 N So 5 N = T - 7.4 N and solving for T we get: (c) T = 12.4 N
3. A pen with initial speed of 1.0 m/s slides across a horizontal desk and loses 0.02 J of energy to friction after sliding some distance and coming to a stop. The frictional force is 0.10 N in magnitude. (a) How far did the pen slide across the desk? (3 pts) (b) What was the pen's mass? (3 pts) (c) What was the coefficient of kinetic friction? (4 pts) Use energy for this one. The initial energy of the pen is all kinetic, given by 1/2 m v 2 where v is the given velocity. This energy was lost to friction through the work that the frictional force did on the pen. That work is W and is given by F k x, or force times distance is work. W = F x x 0.02 J = 0.10 N x (a) x = 0.2 m or 20 cm Since the pen lost ALL of its energy, 0.02 J must be its initial kinetic energy so W = E k and E k = 1/2 m v 2 (b) 0.02 J = 1/2 m (1.0 m/s) 2 m = 2 0.02 J /(1.0 m/s) 2 m = 0.04 kg or 40g The frictional force was 0.10N and this must be µ k F N where F N is the pen's weight. 0.10 N = µ k 0.04 kg 9.8 m/s 2 µ k = 0.10 N / (0.04 kg 9.8 m/s 2 ) (c) µ k = 0.26 coefficient of kinetic friction