PHYSICS 221, FALL 2010 EXAM #1 Solutions WEDNESDAY, SEPTEMBER 29, 2010

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1 PHYSICS 1, FALL 010 EXAM 1 Solutions WEDNESDAY, SEPTEMBER 9, 010 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. Problems 1 through 15 are worth points each 1. The speed of a car is 7.0 km/hour, which can also be expressed as m/s. A. 4 B. 8 C. 1 D. 16 E km/h = 7.0 km h 1000 m km $ 1 h $ % & 3600 s% & = 0.0 m / s. Two vectors F and G are shown in the figure. Which one of the other five vectors shown is equal to G F? A. B. C. D. E. A B C D E Using the rule for vector addition, one sees from the figure that G F = G + ( F) = B 3. For the vectors in the figure in problem, which magnitude of the vector product is largest? A. D E B. F G C. D C D. A D E. B B The magnitude of a vector product is the magnitude of the first vector times the component of the second vector that is perpendicular to the first vector. Therefore one sees that D E = 0, F G = FG = ( ) =, B B = 0. The largest value is D C. D C = DC = = 4, A D = AD = 1() =, and 1

2 4. A particle is moving along the x-axis. The x- component v x of its instantaneous velocity is plotted versus time t in the figure at the right. The instantaneous acceleration of the particle at time t = 4.0 s is a x = m/s. A. B. 1 C. 0 D. 1 E. At t = 4.0 s, a x = dv x dt = v x t =.0 m/s.0 s = 1.0 m / s. 5. For the particle with the velocity versus time plotted in the figure in problem 4, during what time interval(s) is the particle moving in the +x direction? A. 0 s to 1 s and 3 s to 6 s B. 0 s to s and 5 s to 6 s C. 0 s to 1 s only D. 0 s to s only E. s to 5 s only The particle is moving in the +x direction when v x is positive. These time intervals are given in B. 6. A cannon ball is launched from ground level at an angle of 60.0 above the horizontal with an initial speed of 50.0 m/s. The speed of the cannon ball at the top of its trajectory is m/s. Neglect any influence of air friction. A. 0 B. 5 C. 30 D. 35 E. 40 When the ball is at the top of its trajectory, the ball is moving horizontally since the vertical component of the velocity is zero there. Then the speed is just the magnitude of the horizontal component of the velocity, which does not change with time so it is the initial x-component of the velocity: v = v x = v 0 cos 0 = (50.0 m/s)cos( 60 ) = 5.0 m / s.

3 7. A circular disk is spinning in a clockwise direction with a particle attached to its circumference as shown. The angular velocity of the particle is pointed. A. toward the right on the page B. toward the left on the page C. out of the page D. into the page E. upwards on the page Using the circular right-hand rule, one sees that the angular velocity of the particle is pointed into the page. 8. Two balls are attached to a rigid disk that is rotating counterclockwise at angular speed ω. Ball 1 is at a radius R from the center of the disk and Ball is at a radius of R as shown in the figure. The speed of Ball 1 is v 1. The speed of Ball is v =. A. v 1 / B. v 1 C. v 1 D. 4 v 1 E. 8 v 1 Both balls have the same angular speed ω since they are both rigidly attached to the spinning disk. The speed of a ball is v = rω, where r is the radius of the respective circular motion, so one gets v v 1 = r r 1 = R R =. Therefore v = v 1. 3

4 9. Two blocks slide together toward the right on a horizontal frictionless surface as shown. An external contact force F 1 is applied to Block 1 that has mass m 1. As the blocks accelerate toward the right, Block 1 pushes on Block that has a mass m = m 1 /. The contact force applied by Block 1 on Block is F =, A. F 1 / 3 B. F 1 / C. F 1 D. F 1 E. 3 F 1 Treating both blocks together as a unit, the net external force is F 1. Then according to Newton s nd law, the acceleration of both blocks as a unit is therefore a F = 1 M = F 1 3m 1 / = F 1, where 3 m 1 M = (3 / )m 1 is the total mass of both blocks. Then again using Newton s nd law, the net force on Block is F = m a = m 1 3 F 1 m 1 $ % & = F If a particle is moving with constant velocity, then. A. there are no forces applied to the particle. B. a nonzero net force is applied to the particle. C. the net force applied to the particle is zero. D. a friction force on the particle must be present. E. exactly one nonzero force is applied to the particle. Newton s 1 st law states that since the acceleration of the particle is zero, this requires that the net force applied to the particle is zero. The net force is the sum of all forces on the particle. There can be two or more forces applied to the particle that vectorially add to zero. 11. An Atwood Machine consists of two blocks hanging by an ideal massless string that passes over an ideal massless frictionless pulley as shown in the figure. If the masses of the two blocks are the same (m 1 = m = m), then the tension in the string is T =. A. 4 mg B. mg C. 3 mg / D. mg E. mg / Let the positive y-axis be pointed upwards. The free body diagram for either mass has only the downward gravitational force mg ĵ on the mass and the upward tension force T ĵ from the string. Since the accelerations of both masses are zero (the Atwood Machine is balanced ), from Newton s 1 st law, these two forces add to zero giving the force magnitudes as T = mg. 4

5 1. A roller coaster car is going over the top of a curve of radius R at speed v as shown. The total mass of the car and the passenger is M. The horizontal arrow shows the instantaneous velocity of the car. The magnitude of the normal force exerted upwards by the track on the car is. A. M g v % $ R & ' B. M g + v $ R % & C. M v $ R g % & ' D. Mg E. Mv / R Let the positive y-direction be upwards. From a free body diagram for the roller coaster car, the net vertical force on the car is F net y = n Mg, where n is the normal force from the track. Then using Newton s nd law F net y = Ma y we set this equal to the mass M times the centripetal acceleration a rad y = v / R, which is negative because it is directed towards the center of the circle in the y direction: n Mg = Mv / R. Solving for n gives n = M $ g v R % & '. 13. Two constant horizontal forces F 1 and F, with F 1 = F, act on a block sliding on a horizontal table as shown. When the block moves to the right from initial position x = x i to final position x = x f as shown, which one of the following answers is true regarding the work W 1 done by F 1 and the work W done by F? A. W = W 1 B. W = W 1 C. W = W 1 / D. W is positive E. W 1 is negative Use W = F r = F x x. Since F 1x and Δx are positive and F x is negative, and F 1x = F x, one sees that W is negative, W 1 is positive and W = W 1. 5

6 14. A force with a magnitude of 3.00 N is exerted on one end of a spring (the other end is fixed in position) in a direction parallel to the axis of the spring, and the spring becomes compressed by 1.50 cm. The spring constant of the spring is N/m. A. 100 B. 00 C. 300 D. 400 E. 500 k = F x x = 3.00 N m = 00 N / m. 15. A wood block of mass M slides in a straight line on a horizontal wood table with some initial speed. The coefficient of kinetic friction of the block with the table is µ k. After sliding a distance D, the block comes to a stop. The initial kinetic energy of the block is. A. MgD B. µ k MgD C. µ k MgD / D. MgD / E. MgD The work done by kinetic friction is W = F x x = F x D = µ k nd = µ k MgD, where n = Mg is the magnitude of the normal force of the table on the block and the negative sign comes from the fact that the friction force is directed opposite to the direction of motion of the block. By the Work-Energy Theorem, W is equal to the change in kinetic energy of the block: W = K K 1 = 0 K 1. Substituting our expression for W gives the initial kinetic energy K 1 = µ k MgD. Problems 16 through 30 are worth 4 points each 16. Two vectors A and B are given by A = (3.00 m)î + (4.00 m)ĵ B = (3.00 m/s)î (4.00 m/s)ĵ The angle between A and B is degrees. A. 74 B. 86 C. 98 D. 10 E. 106 A B = ABcos, where is the angle between the two vectors. A B = A x B x + A y B y = (3.00 m)(3.00 m/s) + (4.00 m)(4.00 m/s) = 7.00 m /s. A = A x + A y = (3.00 m) + (4.00 m) = 5.00 m. B = B x + B y = (3.00 m/s) + (4.00 m/s) = 5.00 m/s. Therefore, A B & = arccos $ % AB ' ( = arccos * )7.00 m /s -, + (5.00) m /s /. =

7 17. For the two vectors A and B in problem 16, their vector product is A B =. A. (1.0 m /s)î + (1.0 m /s)ĵ B. (1.0 m /s)î (1.0 m /s)ĵ C. (4 m /s)ˆk D. (4 m /s)ˆk E. 4 Since A and B are in the x-y plane, from the right-hand rule their vector product is along the z- axis: A B = (A x B y A y B x )ˆk = [(3.00 m)(4.00 m/s) (4.00 m)(3.00 m/s)]ˆk = (4.0 m /s)ˆk. 18. A particle is moving along the x-axis. The x- component v x of its instantaneous velocity is plotted versus time t in the figure at the right. The change Δx = x x 1 of the x-coordinate of the particle between times t 1 = 1.0 s and t = 5.0 s is m. (The change in a quantity is the final value minus the initial value) A. 4 B. C. 0 D. E. 4 In general, the change in position between times t 1 and t is x = t t 1 v x (t)dt. This is just the area under the v x (t) plot between times t 1 and t. From the figure, the area under the plot between times of 1.0 s and 5.0 s is x = 1 (.0 m/s)(1.0 s) + 1 (.0 m/s)(1.0 s) + 1 (.0 m/s)(.0 s) =.0 m. 7

8 19. A car is moving in a straight line in the positive x-direction with an initial speed of 0.0 m/s. The brakes are applied and the acceleration of the car is then constant in time with a value a x =.00 m/s until the car stops. The time it takes the car to come to a stop after the brakes are applied is s. Ignore any influence of air resistance. A. 4 B. 6 C. 8 D. 10 E. 1 For constant acceleration, one has v x = v 0 x + a x t. Setting v x = 0 gives t = v 0 x a x = 0.0 m/s.00 m/s = 10.0 s. 0. A ball is thrown straight up into the air from a height of.00 m above the ground with an initial speed of 5.0 m/s. The height of the ball above the ground when the ball momentarily comes to a stop is m. Ignore any influence of air resistance. A. 6 B. 30 C. 34 D. 38 E. 4 For free-fall (constant acceleration) motion, one has v y = v 0 y g(y y 0 ). Solving for y gives y = y 0 + v 0 y v y g = (.00 m) + (5.0 m/s) 0 (9.80 m/s ) = 33.9 m. 1. A ball is launched horizontally at a speed v 0 = 5.0 m/s from a height h = 31.9 m above the ground as shown in the figure. The speed of the ball just before it hits the ground is m/s. Ignore any effects due to air friction. A. 5 B. 35 C. 45 D. 55 E. 65 The component of the velocity in the x-direction is constant: v x = v 0 x = 5.0 m/s. The y- component of the velocity when the ball hits the ground, using the formula for the solution of Problem 0, is v y = v 0 y g(y y 0 ) = 0 (9.80 m/s )( m) = 5.0 m/s. The speed of the ball when it hits the ground is v = v x + v y = (5.0 m/s) + (5.0 m/s) = (5.0 m/s) = 35.4 m / s. 8

9 . A particle goes around in a circle of radius 3.00 m in the x-y plane. The angular position of the particle with respect to the positive x-axis is given by = 5.60 rad (.50 rad/s )t, where t is the time in seconds. The magnitude of the (total) acceleration of the particle at time t = s is m/s. A. 3 B. 1 C. 1 D. 30 E. 39 The angular acceleration is constant in time with the value z = d dt = 5.00 rad/s. The time-dependent angular velocity is z = d dt = (5.00 rad/s )t. At time t = s, z (0.447 s) = (5.00 rad/s )(0.447 s) =.4 rad/s. Then at t = s the magnitude of the acceleration of the particle is a = a tan + a rad = R + 4 = (3.00 m) (5.00 rad/s ) + (.4 rad/s) 4 = 1.3 m / s. 3. A bird flies 100 km directly north from Ames with respect to the ground. There is a wind present with a speed of 10.0 m/s that is blowing directly east with respect to the ground. If the speed of the bird with respect to the air is 15.0 m/s, it takes the bird minutes to fly the 100 km. A. 75 B. 100 C. 15 D. 150 E. 175 Let B stand for bird, A stand for air, and G stand for ground. Then the rule for adding relative velocities becomes v B G = v B A + v A G, as shown in the figure. Let north = y and east = x. Then using the Pythagorean Theorem, we see from the figure that v B G = v B A v A G = (15.0 m/s) (10.0 m/s) = 11. m/s. Thus the time it takes to fly 100 km directly north is t = m 11. m/s 1 min $ 60 s % & = 149 min. 9

10 4. Two forces F 1 and F act on a 1.00 kg block with F 1 = (3.00 N)î (4.00 N)ĵ (5.00 N)ˆk. If the acceleration of the block is a = (4.00 m/s )î (8.00 m/s )ˆk, the second force is F =. A. (7 N)î (4 N)ĵ (13 N)ˆk B. (4 N)î (8 N)ˆk C. (1 N)î + (4 N)ĵ (3 N)ˆk D. (1 N)î (8 N)ĵ (9 N)ˆk E. (8 N)î + (8 N)ĵ + (10 N)ˆk From Newton s nd law, we have F 1 + F = m a. Thus, F = m a F 1 = (1.00 kg)[(4.00 m/s )î (8.00 m/s )ˆk] [(3.00 N)î (4.00 N)ĵ (5.00 N)ˆk] = (1.00 N)î + (4.00 N)ĵ (3.00 N)ˆk 5. Two blocks are sitting at rest on a table, one on top of the other as shown. Block 1 has mass m 1 = 10. kg and Block has mass m = 5.1 kg. The magnitude of the contact force of the table on the bottom Block 1 is N. (gravity is present) A. 50 B. 75 C. 100 D. 15 E. 150 With respect to the table, one can consider the two blocks to be a single system with mass M = m 1 + m = 15.3 kg. Gravity pulls this mass downwards with a force F grav = Mg = (15.3 kg)(9.80 m/s ) = 150 N. From Newton s 1 st law, the normal force of the table on Block 1 has to have the same magnitude but point upwards. 10

11 6. An Atwood Machine consists of two blocks hanging by an ideal massless string that passes over an ideal massless frictionless pulley as shown in the figure. The mass of Block 1 is m 1 = 5.00 kg and the mass of Block is m = kg. The magnitude of the acceleration of either block is m/s. A. 9.8 B. 7.7 C. 6.5 D. 4.9 E. 3.3 Let the positive x-axis point upwards along the string on the left side of the pulley and downwards along the string on the right side. The net force on both masses as a system is F x = (T m 1 g) + (m g T ) = g(m m 1 ). The total mass of the system is M = m + m 1. Using Newton s nd law, one obtains the acceleration of either block as a x = F x M = g m m kg 5.00 kg = (9.80 m/s ) m + m kg kg = 4.90 m / s. 7. A block of mass M = 5.00 kg hangs at rest from a massless ring by two massless ropes 1 and as shown. Rope 1 is horizontal and rope makes an angle θ = 35 with the vertical. The tension in rope is T = N. A. 40 B. 45 C. 50 D. 55 E. 60 One can determine the tension in rope just by considering the vertical component of the net force on M and setting it equal to zero according to Newton s 1 st law: F net y = T cos Mg = 0, giving T = Mg cos = (5.00 kg)(9.80 m/s ) = 59.8 N. cos(35 ) 11

12 8. A 1500 kg car is going around a curve of radius 63.8 m on a horizontal flat (not banked) road. If the coefficient of static friction of the tires with 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. 5 B. C. 19 D. 16 E. 13 The maximum static friction force that the road can exert on the car is f s max = µ s n = µ s mg. Setting this equal to the centripetal force mv / R gives the maximum speed to be v max = µ s gr = (1.00)(9.80 m/s )(63.8 m) = 5.0 m/s. 9. A toy gun has an ideal massless spring in it with a spring constant of 400 N/m. When the gun is cocked, the spring is compressed by.00 cm. Then when the gun is fired, the spring pushes on and accelerates a kg rubber bullet, and the bullet leaves the gun at a speed of m/s. A. 4 B. 6 C. 8 D. 10 E. 1 The work that the spring does on the bullet when the gun is fired is W = 1 kx, where x is the compression of the spring and k is the spring constant. According to the Work-Energy Theorem, this work is converted to kinetic energy K = 1 mv of the bullet when the gun is fired. Equating K to W gives v = x k m = (0.000 m) 400 N/m kg = 4.00 m / s. 30. A 000 kg car accelerates in a straight line on a horizontal surface at a constant magnitude of 5.00 m/s starting from rest. When the car reaches a speed of 30.0 m/s, the instantaneous power required to continue to accelerate the car at the same value is kw. A. 100 B. 150 C. 00 D. 50 E. 300 The power needed to accelerate the car is P = F v = F x v x. But from Newton s nd law, F x = ma x. Then we have P = F x v x = ma x v x = (000 kg)(5.00 m/s )(30.0 m/s) = W = 300 kw. 1