PHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009

PHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively. In this exam, assume that the magnitude of the acceleration due to earth s gravity at the surface of the earth is g = 9.80 m/s 2. Problems 1 through 15 are worth 2 points each 1. Which one of the following four choices is a right-handed Cartesian coordinate system? Using the right-hand rule, one finds that answer B is the only example of a right-handed Cartesian coordinate system. 2. Consider two vectors A and B, with nonzero magnitudes of A and B, respectively. The quantity A ( B A) is equal to. A. A 2 B B. A 2 B C. AB D. AB 2 E. 0 By definition the angle between either A or B and the vector B A is 90. Therefore, from the definition of the scalar product, we have A ( B A) = A B A cos(90 ) = 0 1

3. A particle is moving along the x-axis. The x-component x of the position of the particle is plotted versus time t in the figure. The x-component v x of the velocity of the particle at time t = 2.0 s is m/s. A. 0 B. 2 C. 2 D. 4 E. 4 The velocity v x at time t is the slope of x(t) at that time. Thus, v x (t = 2.0 s) = x t = 4.0 m = 2.0 m/s. 2.0 s 4. A person throws a ball straight upwards. At the instant that the ball reaches its maximum height, which one of the following five statements is true? A. The acceleration of the ball is zero and the velocity of the ball is zero. B. The acceleration of the ball is upwards and the velocity of the ball is upwards. C. The acceleration of the ball is zero and the velocity of the ball is downwards. D. The acceleration of the ball is downwards and the velocity of the ball is zero. E. The acceleration of the ball is downwards and the velocity of the ball is downwards. 5. In separate experiments, a ball is launched from horizontal ground at x = 0 and y = 0 with four different trajectories as shown in the figure, where the positive y-axis points upwards and the x-axis is horizontal. As seen in the figure, the height that the ball reaches in each trajectory is the same. For which trajectory does the ball take the longest time after launch to hit the ground? (Ignore effects of air friction) A. The time that the ball is in the air is the same for all four trajectories B. Trajectory A C. Trajectory B D. Trajectory C E. Trajectory D y (m) 50 A B C D 40 30 20 10 0 0 20 40 60 80 100 120 x (m) The time the ball takes to hit the ground is governed by the vertical component of the motion only. The vertical height is the same for each trajectory, and therefore so is the time of flight. 2

6. A ball is going counterclockwise in a circle at a speed that is decreasing as time passes. Of the choices A, B, C, D, and E in the figure at the right, the best representation of the direction of the (total) acceleration of the ball is. (Ignore gravity) The radial acceleration is in the direction of C, and since the speed is decreasing the tangential acceleration is in the direction E opposite to that of the velocity. Adding these two components together gives the total acceleration in a direction between C and E, which is qualitatively in the direction of D. 7. A particle is going around in a circle with fixed radius R in the x-y plane as shown in the figure. The angle between the position vector of the particle and the positive x-axis is given by = 0 At 2, where 0 and A are positive constants and t > 0 is the time. As usual, positive values of are measured counterclockwise. Which one of the following five statements is false? A. The z-component of the angular velocity of the particle is negative. B. The z-component of the angular acceleration of the particle is negative. C. The tangential acceleration of the particle and the velocity of the particle are in opposite directions. D. The angular acceleration of the particle is in the same direction as the angular velocity of the particle. E. The radial acceleration of the particle is perpendicular to the velocity of the particle. From the quoted expression for, we get z = d = 2At and dt = d z z = 2A. Thus both dt z and z are negative and have the same sign (they are in the same direction). Therefore answers A, B and D are all true. Since z and z are both negative and the +z direction is pointed out of the page (towards you), from the circular right-hand rule the particle is going around clockwise and the speed of the particle is increasing with time. The tangential acceleration of the particle is therefore in the same direction as the velocity of the particle and hence answer C is false. The radial acceleration of a particle going around in a circle is always perpendicular to the velocity of the particle, so answer E is true. 3

8. Which one of the following five statements is true? A. An object tends to come to a stop if there is no net force on it to keep it moving. B. If a heavy truck collides head-on with a light car, during each instant of the collision the magnitude of the force of the truck on the car is the same as the magnitude of the force of the car on the truck. C. If a net force applied to a 1 kg mass produces an acceleration of the mass of magnitude 2 m/s 2, the same net force applied to a 10 kg mass produces an acceleration of the 10 kg mass of magnitude 20 m/s 2. D. If a car is accelerating forwards, there exists a force on the driver that pushes the driver backwards into the seat. E. The magnitude of the gravitational force of the earth on a person on the surface of the earth is much larger than the magnitude of the gravitational force of the person on the earth. Answer B is true because this is just a statement of Newton s 3 rd law. Answer E is false because it violates Newton s 3 rd law. Answers A, C and D are false because they violate Newton s 2 nd law. 9. A block of mass m slides down a frictionless inclined plane that is at an angle to the horizontal as shown in the figure at the right. The magnitude of the acceleration of the block is. A. g B. g / cos C. g cos D. g tan E. g sin The component of the gravitational force pointing down the sloping surface is F = mg sin. Therefore from Newton s 2 nd law the magnitude of the acceleration of the block is a = F/m = g sin. 10. An Atwood Machine consists of two masses m 1 and m 2 hanging by an ideal massless string that passes over an ideal massless frictionless pulley as shown in the figure. Here, m 2 > m 1. The tension in the string is T. Which one of the following five statements is true? A. T < m 2 g B. T < m 1 g C. T = m 1 g D. T = m 2 g E. The tension T in the string changes with time as the masses accelerate. Since m 2 > m 1, m 2 will accelerate downwards, which means that T < m 2 g, and m 1 will accelerate upwards, which means that T > m 1 g. 4

11. A ball of mass m is attached to a taut ideal string and goes around a vertical circle of radius R in a counterclockwise direction as shown in the figure at the right. Earth s gravity acts on the ball. As the ball goes around the circle, the tension T in the string changes. When the ball is at the bottom of the circle, the speed of the ball is the value v and the tension in the string is then T =. A. mg B. mg + mv 2 /R C. mg mv 2 /R D. mv 2 /R E. 0 When the ball is at the bottom of the circle, the net upward force on the ball is T mg. Using Newton s 2 nd law, we set this equal to the mass times the centripetal acceleration upwards: mv 2 /R = T mg. Solving for T gives T = mg + mv 2 /R. 12. A wooden block of mass m is sitting at rest on a horizontal table. The coefficient of static friction and the coefficient of kinetic friction between the block and the table are both nonzero. A horizontal force is applied to the block starting at time t = 0 with a magnitude F that increases with t according to F = At, where A is a positive constant. At time t 1, the block begins to slide. The magnitude f of the frictional force of the table on the block versus t is best represented by plot. The static friction force has the same magnitude as the magnitude of the applied force until the object starts sliding, at which point the kinetic friction force is slightly less that the maximum static friction force and is independent of the block s speed. Thus the answer is E. 5

13. A person sequentially moves a block of mass m counterclockwise along four sides 1, 2, 3, and 4 of a vertical square, each of length L, in the directions as shown. The upward direction on the page is the upward direction above the earth s surface and the horizontal direction on the page is the horizontal direction along the earth s surface. The initial and final speeds of the block at the ends of each of the four paths are zero. The net work W net done on the block by the gravitational force of the earth as the block is moved completely around the square once is W net =. A. 0 B. mgl C. 2mgL D. 4mgL E. 2mgL For each leg of the trip, we have W grav = mg(y 2 y 1 ) where the positive y-axis points upwards. Thus we get W 1 = 0, W 2 = mgl, W 3 = 0, W 4 = mgl, W net = W 1 + W 2 + W 3 + W 4 = 0. 14. Which one of the following five statements is false? A. According to the Work-Energy Theorem, the net work done on an object during a time interval t is equal to the change in the object s kinetic energy during that time interval. B. The work done on an object by a Hooke s law spring during a time interval t can be positive, negative or zero. C. The work done on an object by the gravitational force of the earth during a given time interval only depends on the initial and final heights of the object above the earth s surface, and not on the path between the initial and final positions of the object. D. If an object is sliding along a stationary surface, the work done on the object by the kinetic friction force between the object and the surface over a time interval is positive. E. The kinetic energy of an object can never be negative. The kinetic friction force of an object sliding on a stationary surface is always antiparallel to the velocity. Therefore the power of the force P = F v is negative, and the work W = P t done over any time interval is also negative. Therefore answer D is false. 15. At some instant of time, a person exerts a constant horizontal force of magnitude 50 N in the î direction on a crate that is moving in a straight line in the î direction at a speed of 0.50 m/s on a horizontal floor. The power delivered to the crate by the force exerted by the person at that instant of time is W. A. 1 B. 12 C. 25 D. 37 E. 50 P = F v = Fv = (50 N)(0.50 m/s) = 25 W. 6

Problems 16 through 30 are worth 4 points each 16. A vector B is given by B = (3.00 m)î (4.00 m)ĵ. The magnitude of B is. A. (0.600)î (0.800)ĵ B. (0.600)î + (0.800)ĵ C. 1.00 m D. 3.50 m E. 5.00 m B = B 2 x + B 2 y = (3.00 m) 2 + (4.00 m) 2 = 5.00 m. 17. Two vectors A and B satisfy the relationships A B = 1.73 m 2 and A B = 1.00 m 2. The angle between the two vectors is degrees. (The angle must be between zero and 180 degrees) A. 30 B. 60 C. 90 D. 150 E. 175 A B = ABcos and A B = ABsin. Dividing the 2 nd equation by the first gives A B = arctan A B 1.00 m2 = arctan 1.73 m 2 = 30 or 150. Since the angle must be between zero and 180 degrees, we take = 150. 18. A particle is moving along the x-axis. The x-component of the position of the particle is given by x = 4.00 m (2.00 m/s 2 )t 2, where t is the time in seconds. The x-component of the average velocity of the particle between the times t = 0.00 s and t = 2.00 s is m/s. A. 4 B. 2 C. 0 D. 2 E. 4 x(t = 0.00 s) = 4.00 m. x(t = 2.00 s) = 4.00 m. v av x = x t = (4.00 m) (4.00 m) 2.00 s = 4.00 m/s. 19. A baseball is launched vertically upwards with an initial speed of 29.4 m/s. The time that it takes for the ball to return to the same height at which it was launched is s. (Ignore effects of air friction on the motion of the ball) A. 2 B. 4 C. 6 D. 8 E. 10 Use the free-fall formula y y 0 = v 0 y t 1 2 gt 2. Setting y = y 0 gives t = 2v 0 y g 7 = 2(29.4 m/s) 9.80 m/s 2 = 6.00 s.

20. A projectile is launched near the surface of the earth at time t = 0. The direction pointing straight upwards is the positive y-direction and the horizontal direction is the x-direction. The (x, y) coordinates of the projectile versus time t are given by x = 3.00 m + (12.0 m/s)t y = 2.00 m + (16.0 m/s)t (4.90 m/s 2 )t 2 The initial launch speed of the projectile at time t = 0 is m/s. A. 15 B. 20 C. 25 D. 30 E. 35 We have v x = dx = 12.0 m/s dt v v y = dy dt = 16.0 m/s (9.80 m/s2 )t = v 0 x x (t = 0) = 12.0 m/s v 0 y = v y (t = 0) = 16.0 m/s Therefore, v 0 = 2 v 0 x 2 + v 0 y = (12.0 m/s) 2 + (16.0 m/s) 2 = 20.0 m/s. 21. A ball is launched horizontally from a cliff at a height h = 44.1 m above the ground as shown in the figure. The initial speed of the ball at launch is 20.0 m/s. The horizontal distance L that the ball goes before hitting the ground is m. (Ignore effects of air friction) A. 20 B. 40 C. 60 D. 80 E. 100 From the vertical motion, calculate the time for the ball to hit the ground. Then use that value in the horizontal motion to get L. First use the free-fall formula for the vertical direction y y 0 = v 0 y t 1 2 gt 2 = 0 1 2 gt 2. Thus, t = 2(y 0 y) g = 2(44.1 m) = 3.00 s. 2 9.80 m/s Then L = x x 0 = v 0 x t = (20.0 m/s)(3.00 s) = 60.0 m. 8

22. A boat crosses a river in a straight line perpendicular to the river from point A to point B as shown in the figure. The width of the river is w = 100 m. The river is flowing towards the right at a speed of v river = 3.00 m/s. The speed of the boat with respect to the water is 5.00 m/s. The time that it takes the boat to cross the river is seconds. A. 25 B. 50 C. 75 D. 100 E. 125 The relative velocities are related by the relative velocity expression v BG = v BW + v WG where B refers to the boat, W to the water and G to the ground, as shown qualitatively in the figure at the right. We want to know the speed v B G in order to calculate the time for the boat to cross the river. From the figure and using the Pythagorean theorem we see that v B G = 2 v B W 2 v W G = (5.00 m/s) 2 (3.00 m/s) 2 = 4.00 m/s. Thus the time that the boat takes to cross the river is t = w v C G = 100 m 4.00 m/s = 25.0 s. 23. Two different forces are exerted on a particle of mass m = 2.50 kg. The two forces are: F 1 = (15.0 N)î + (10.0 N)ĵ (20.0 N)ˆk F 2 = (15.0 N)î (13.0 N)ĵ + (24.0 N)ˆk The magnitude of the acceleration of the particle is m/s 2. A. 1 B. 2 C. 3 D. 4 E. 5 The net force on the particle is F = F 1 + F 2 = (3.0 N)ĵ + (4.0 N)ˆk. The magnitude of the force is F = F y 2 + F z 2 = (3.0 N) 2 + (4.0 N) 2 = 5.0 N. Thus from Newton's 2nd law the magnitude of the acceleration is a = F m = 5.0 N 2.50 kg = 2.0 m/s2. 9

24. Three blocks with masses m 1 = 5.00 kg, m 2 = 3.00 kg and m 3 = 2.00 kg are sliding on a frictionless surface and are accelerating together towards the right due to a force of magnitude 50.0 N that pushes on mass m 1 towards the right as shown in the figure. The magnitude of the force that block m 1 exerts on m 2 is N. A. 5 B. 10 C. 15 D. 20 E. 25 All three blocks move together and are accelerated by a single net force. Thus the magnitude of the acceleration of all three blocks, and of each block, is a = F m 1 + m 2 + m 3 = 50.0 N 5.00 kg + 3.00 kg + 2.00 kg = 5.00 m/s2. The force that is exerted on m 2 by m 1 has to accelerate both m 2 and m 3 at the above value a, giving from Newton s 2 nd law F 1 on 2 = (m 2 + m 3 )a = [(3.00 kg) + (2.00 kg)](5.00 m/s 2 ) = 25.0 N. 25. A block with mass m = 10.0 kg slides down an inclined plane that is at an angle = 30 to the horizontal, as shown in the figure. The coefficient of kinetic friction between the inclined plane and the block is 0.29. The magnitude of the acceleration of the block is m/s 2. A. 0.6 B. 1.2 C. 1.8 D. 2.4 E. 3.0 Let the x-direction point along the downward sloping inclined plane. Then the x-component of the gravitational force directed down the incline is F grav x = mg sin. The x-component of the kinetic friction force is f kx = μ k n = μ k mg cos. The net force down the incline is thus F net x = mg(sin μ k cos). From Newton s 2 nd law, the magnitude of the acceleration is then a = F net x m = g(sin μ k cos) = (9.80 m/s 2 )[sin(30 ) (0.29)cos(30 )] = 2.4 m/s 2. 10

26. A 50.0 kg person is standing on a bathroom scale inside an elevator. The elevator is moving downwards with a speed given by v elevator = 1.50 m/s + (2.80 m/s 2 )t, where t is the time in seconds. The bathroom scale reads N. A. 200 B. 350 C. 500 D. 650 E. 800 The y-axis points upwards. The y-component of the velocity of the person is given as v y = 1.50 m/s (2.80 m/s 2 )t. Thus the acceleration of the person is a y = dv y dt = 2.80 m/s 2. The net force on the person in the y direction is F net y = n mg where n is the upward normal force exerted by the scale and mg is the downward gravitational force. Setting F net y = ma y from Newton s 2 nd law gives n = m(g + a y ) = (50.0 kg)[(9.80 m/s 2 ) (2.80 m/s 2 )] = 350 N, which is what the scale reads. 27. A car of mass 1500 kg is going around a curve of radius 40.8 m on a road whose surface is horizontal (not banked). The coefficient of static friction between the tires and the road is 1.00. The maximum speed at which the car can go around the curve without sliding off the road is m/s. A. 10 B. 15 C. 20 D. 25 E. 30 The magnitude of the radial (centripetal) force needed to keep the car on the road is F rad = mv 2 /R. Setting this equal to the maximum static friction force μ s mg gives v max = μ s gr = (1.00)(9.80 m/s 2 )(40.9 m) = 20.0 m/s. 28. A 0.204 kg block slides at constant speed around a circular path of radius 0.955 m on a horizontal table. The coefficient of kinetic friction between the block and the table is 0.500. The work done on the block by the kinetic friction force during one complete revolution of the block around the circle is J. A. 6 B. 3 C. 0 D. 3 E. 6 The component of the constant kinetic friction force on the block in the instantaneous direction of motion of the block is f k = μ k mg where μ k is the coefficient of kinetic friction. Thus using the fact that the circumference of the circle is 2R, the work per revolution done by kinetic friction is W = f k (2R) = (μ k mg)(2r) = (0.500)(0.204 kg)(9.80 m/s 2 )2(0.955 m) = 6.00 J. 11

29. A 1.00 kg cart is moving up a curving frictionless surface as shown in the figure at the right. The initial speed of the cart is 10.0 m/s. After the height of the cart above the ground has increased by h = 4.29 m as shown, the speed of the cart is now m/s. Hint: use the Work-Energy Theorem. A. 1 B. 2 C. 3 D. 4 E. 5 The work done by gravity on the cart is W = mg(y 2 y 1 ). According to the Work-Energy theorem, this is equal to the change in kinetic energy: W = mg(y 2 y 1 ) = 1 2 m(v 2 2 v 2 1 ). Thus, v 2 = v 1 2 2g(y 2 y 1 ) = (10.0 m/s) 2 2(9.80 m/s 2 )(4.29 m) = 3.99 m/s. 30. A block of mass m on a frictionless horizontal surface is attached to the right-hand end of a spring with spring constant k = 100 N/m as shown in the figure at the right. The left-hand end of the spring is attached to a stationary wall. The equilibrium position of the right-hand end of the spring is at x = 0. When the right-hand end of the spring moves from x 1 = 0.300 m to x 2 = +0.100 m, the net work done by the spring on the block is J. A. 4 B. 2 C. 0 D. 2 E. 4 W = 1 2 k(x 2 2 x 2 1 ) = 1 2 (100 N/m)[(0.100 m)2 (0.300 m) 2 ] = 4.00 J. 12