UIC Physics 105 1 st Midterm Practice Exam Fall 2014 Best if used by September 30 PROBLEM POINTS SCORE Multiple Choice Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 48 11 9 10 10 12 Total 100 Page 1 of 8
MULTIPLE CHOICE QUESTIONS (2-3 points each) Clearly circle the letter of the best answer MCQ01 [3 points]: Later this semester, you ll be given the following equation describing pressure (P): P = ρgh where ρ is mass density (mass per volume), g is the acceleration due to gravity, and h is depth (a length). What are the dimensions of P? (A) (B) (C) (D) Answer (D): MCQ02 [3 points]: Find the length of the side x shown in the diagram to the right (A) 5 cm Answer (B): (B) 9 cm ACB 90 CAB 60 and CBA 30 (C) 12 cm 12 tan 60 tan 30 or (D) 14 cm = 3 So, x = 9 cm MCQ03 [2 points]: For a vector of magnitude 6.0 m and a y-component of 4.0, which of the following are possible angles that the vector makes with the positive x-axis: (A) 38 (B) 62 Answer (C): sin sin = 42 or 180 42 = 138 (C) 138 42 as an answer is not listed = 138 (D) 145 MCQ04 [2 points]: Figure to the right shows the position-time graph of an object moving along x-axis. The total distance traveled by the object during 0.5 t 7 s is (A) 20 m (A) 0 m Answer (D): d = 15 ( 5) + 5 ( 15) = 20 m (B) 5 m (C) 10 m (D) 20 m MCQ05 [2 points]: The motion of a particle is described in the velocity vs. time graph shown to the right. In the time interval from 0 to 6 s, the speed of the particle (A) increases (B) decreases (C) increases and then decreases (D) decreases and then increases Answer (D): The first 4 s the particle s velocity and acceleration are in opposite directions, i.e. it undergoes deceleration until the speed of the particle becomes zero (at t = 4 s). During the last 4 seconds the particle's velocity and acceleration are in the same direction, i.e. it speeds up. Page 2 of 8
MCQ06 [3 points]: The velocity-versus-time graph for a particle moving along x-axis is shown in the figure to the right. At t = 0 s the particle has position x0 = 2 m. The average velocity of the particle over time interval 0 t 8 s is (A) 0 m/s (B) 2 m/s (C) 4 m/s (D) 8 m/s Answer (A):. The final position x we can find as where x is the area under curve. = 0 over 0 t 8 s = 0 m/s MCQ07 [2 points]: In MCQ06, the instantaneous acceleration of the particle at t = 6 s is (A) 0 m/s 2 (B) 1 m/s 2 Answer (B): lim is a slope of a tangent line at t. So, = 1 (C) 2 m/s 2 m/s2 (D) 4 m/s 2 MCQ08 [2 points]: A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, giving the car a deceleration of 3.50 m/s2. How far does the car travel after the driver put on the brakes before it comes to a stop? (A) 134 m (B) 146 m (C) 152 m (D) 168 m Answer (B): v = 0 m/s, v0 = 32.0 m/s and a = 3.5 m/s 2. From 2 = 32 2 /2/ (3.5) = 146 m MCQ09 [2 points]: A freely falling body is found to be moving downwards at 27 m/s at one instant. If it continues to fall, one second later the object would be moving with downward velocity closest to (A) 270 m/s (B) 37 m/s (C) 27 m/s (D) 10 m/s Answer (B):, where 27 m/s and 10 m/s 2 27 10 1 37 m/s or 37 m/s, downward MCQ10 [3 points]: Two particles start from the origin of the horizontal x-y plane. Particle A moves along y axis with speed v. Particle B moves along +x axis with the same speed v. The magnitude of the velocity of the particle B with respect to A is Answer (E): (A) 2 m/s 45 below x-axis Velocity of A along x-axis is 0 and along y-axis is m/s (B) m/s 45 below x-axis Velocity of B along x-axis is m/s and along y-axis is 0 (C) 0 m/s velocity of B with respect to A along x-axis is 0 m/s, and (D) m/s 45 above x-axis velocity of B with respect to A along y-axis is 0 m/s (E) 2 m/s 45 above x-axis So, 2 at 45 above x-axis Page 3 of 8
MCQ11 [3 points]: A stone is thrown horizontally at speed of 5.0 m/s from the top of a cliff 78.4 m high. How long does it take the stone to reach the bottom of the cliff? (A) 2.5 s (B) 4.0 s (C) 5.0 s (D) 6.0 s Answer (B):,, where = 0 m, = 78.4 m and, = 0. = 4.0 s. MCQ12 [2 points]: A player kicks a football from ground level at 27 m/s at an angle of 35 above the horizontal. Calculate the distance the ball travels (horizontally) before it hits the ground. (A) 34.9 m Answer (B): (B) 69.8 m sin2 sin2 35 = 69.8 m (C) 72.4 m. (D) 82.9 m MCQ13 [2 points]: Two perpendicular forces, one of 45.0 N directed upward and the second of 60.0 N directed to the right, act simultaneously on an object with a mass of 25.0 kg. What is the magnitude of the resultant acceleration of the object? (A) 1 m/s 2 (B) 1.41 m/s 2 (C) 2 m/s 2 (D) 3 m/s 2 Answer (D): the magnitude of the net force is = 75 N = 75/25 = 3 m/s2 MCQ14 [2 points]: With action-reaction forces, (A) the action force is created first Answer (C): In Newton s 3 rd law, equal means: (B) the reaction force is created first a) The action and reaction forces are exactly the same in magnitude (C) the forces are created at the same time b) The action and reaction forces occur at exactly the same time (D) both forces already existed MCQ15 [2 points]: A 1-lb pendulum bob is held at an angle = 60 from the vertical by a horizontal Force as shown in the figure to the right. The tension in the string supporting the pendulum bob (in pounds) is (A) 0.5 lb (B) 1.0 lb (C) 1.2 lb (D) 2.0 lb Answer (D): From translational equilibrium along vertical line T cos = mg T = mg/ cos = 1/cos(60) = 2 lb Page 4 of 8
MCQ16 [3 points]: A 110 kg object is supported by two ropes attached to the ceiling. What is the tension T in the right-hand rope? (A) 465 N (B) 542 N (C) 937 N (D) 1326 N MCQ17 [3 points]: A person who normally weighs 200 pounds is standing on a scale inside elevator. The elevator is moving upwards with a speed of 7 m/s, and then begins to slow down at a rate of 5 m/s 2. Before the elevator begins to slow down, the reading of the scale is R1 and while the elevator is slowing down, the reading of the scale is R2. Use g = 10 m/s 2. The scale readings are (A) R1 = 200 lb and R2 = 100 lb (B) R1 < 200 lb and R2 = 0 lb (C) R1 < 200 lb and R2 = 100 lb (D) R1 > 200 lb and R2 = 100 lb MCQ18 [2 points]: A block with mass m and contact area A slides down an inclined plane with friction, covering a distance L in time T. How much time does it take another block with the same mass and composition, but contact area 2A, to slide down the same length? (A) (B) T (C) T 2 (D) T 3 MCQ19 [2 points]: A mass with m = 120 kg is attached to the bottom of a block-and-tackle pulley system as depicted in the figure to the right. How much force F is required to just keep the mass at its current position? Use g = 10 m/s 2. (A) 300 N (B) 400 N (C) 600 N (D) 1200 N Answer (B): The frictional force f = N is independent of area of contact. If the mass and composition of these two blocks are the same then the frictional force is also the same. So, for both blocks it will take the same time to slide down the distance L Answer (B): Answer (A):, a = 0 = 200 lb, a = g/2 400 N = 100 lb MCQ20 [3 points]: A homeowner pushes a lawn mower across a horizontal patch of grass with a constant speed by applying a force F. The arrows in the diagram shown to the right correctly indicate the directions but not necessarily the magnitudes of the various forces on the lawn mower. Which of the following relations among the various force magnitudes, W(weight), f(friction), N(normal force), F(applied force) is CORRECT? (A) F > f and N > W (B) F < f and N = W (C) F > f and N < W (D) F = f and N > W Answer (A): Since F is at a downward angle, the normal force is increased. Since a component of F balances the frictional force f, F itself must be larger than f. Page 5 of 8
SHORT PROBLEMS You must show your work and write your answers clearly and legibly SP1: In a 400 m race, a runner accelerates uniformly from rest during the first 24.0 s and then runs at a constant speed for another 20 s when he reaches the finish line. (a) [4 points] Sketch a velocity versus time graph for the runner. (b) [7 points] Find the runner s top (maximum) speed. From the sketch in part (a), during the first 24 s, (area under curve for time interval 0 t 24 s) Next 20 seconds, So, / ), where = 400 m, = 24 s and = 20 s = 12.5 m/s SP2: Once upon a time, a famous zoologist was hunting antelope for a zoo with a dart gun that fired darts with a range of 100 m at a speed of 25 m/s. If he shot a dart at an antelope at rest 50 m away, and the antelope instantly began to accelerate at a constant rate away from the dart, running out of range of the dart just in time: (a) [4 points] Find the maximum time the antelope can remain in range of the dart. The maximum time will be the time it takes the dart to reach its range limit, i.e., where = 0. So, / = 100/25 = 4 s (b) [5 points] Find the minimum speed of the antelope as it runs out of range of the dart. First, let s find the antelope s acceleration:, where 50 m, = 0 m/s = 6.25 m/s 2. The minimum speed of the antelope we can find as = 6.24 4 = 25 m/s Page 6 of 8
SP3: A golfer hits the ball with a club from an elevated tee that lands on the green close to the cup a distance of 165 m from where he hit it as shown in the figure to the right. The spot where it lands is 5 m below the spot where the ball is hit. The ball leaves the club with a horizontal component of velocity equal to 50.0 m/s. (Ignore air resistance) (a) [4 points] How long does the ball take to get to the green?, /, = 3.3 s (b) [6 points] What is the vertical component of the ball's initial velocity?, where = 0 m, = 5 m and 9.8 m/s 2 So,, = 14.7 m/s SP4: Three boxes are stacked in an elevator as shown in the figure to the right. The elevator is traveling downwards with a constant speed of vy = 5.0 m/s. The acceleration due to gravity has magnitude g = 9.81 m/s 2 and points downward, as usual. (a) [6 points] Calculate the magnitude of the contact force of the 1 kg block on the 2 kg block. [Hint: sketch a free body diagram for the 2 kg block]. 0 i.e. 0, where is the weight of 3 kg block So, = (3 + 2) 9.81 = 49 N (b) [4 points] Calculate the net force acting on the 1 kg block. Elevator is traveling at constant speed (a = 0), so all forces are canceled, i.e. the net force acting on any one of the blocks is equal to zero (Fnet = 0). Page 7 of 8
SP5: Block A with weight 3W, slides down an inclined plane S of slope angle = 36.9 at a constant speed while plank B, with weight W, rests on top of A. The plank is attached by cord to the wall as shown in the figure below. (a) [6 points] Find the normal force exerted by the plane S on the block A in terms of and W (weight). Block A, y axis:, 3cos cos 0 3cos 4W cos y N A f BA x A f AS 3Wsin 3W (b) [6 points] If the coefficient of kinetic friction is the same between A and B and between S and A, determine its value Block A, x axis:, sin 0, where 3 cos cos 3 sin 3 cos cos0 tan tan 36.9 0.45 Page 8 of 8