PHY121 #8 Midterm I


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1 PHY11 #8 Midterm I 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 in the string between the blocks is: A) F B) F C) F 3 D) 1 F E) 1 F 3 Applying Newton s second law in the absence of friction, F = (3 kg)a. Therefore, a = F/3. When you isolate the first block, F#9 net = FT = ma =(1 kg)(f/3). Therefore, FT also is equal to F/3. #. When an object of weight W is suspended from the center of a massless string as shown above, the tension at any point in the string is: A) W cosθ B) W cosθ C) W cosθ D) W cosθ E) W cosθ The downward weight force is equal to the sum of the upward components of the two tension forces. Each tension has an upward component of FT (cos θ). Since the system is in static equilibrium, W = #3 #5 (FT cos θ). 3. A satellite of mass M moves in a circular orbit of radius R with constant speed v. True statements include which of the following? I. Its angular velocity is v/r. II. Its tangential acceleration is zero. III. The magnitude of its centripetal acceleration is constant. A) I only B) II only C) I and III only D) II and III only E) I, II, and III The first statement is the definition of angular velocity. The second statement is true because the object is moving at constant speed. The third statement is true since ac = v/r, speed is constant, and the statement specifically refers to magnitude. 1
2 4. A projectile is fired from a cannon horizontally from the edge of a 15 m cliff at 75 m/s. How far from the bottom of the cliff does the projectile land? A) 15 m B) 75 m C) 550. m D) 1,390 m E) 6,880 m The time in the air can be determined from the vertical displacement: 15 = (1/)(9.8)t, or t = 5 seconds. The range is the horizontal velocity times the time in the air: x = v x t = (75)(5) = 1390 m. 5. A rope is tied to the handle of a bucket, which is then whirled in a vertical circle of radius 60.0 cm. The mass of the bucket is 3.00 kg. At the lowest point in its path, the tension in the rope is 50.0 N. What is the speed of the bucket? A) 1.0 m/s B).03 m/s C) 3.00 m/s D) 4.06 m/s E) 5.00 m/s At the bottom of the vertical circle, the centripetal force equals the tension force minus the weight of the bucket. Therefore, we can state that: mv /r = F T F g. Substituting our values, we get: (3.00)v /(0.60) = 50.0 (3.00)(9.8), and v =.03 m/s. 6. For the bucket and rope in problem #6, what is the critical speed below which the rope would become slack when the bucket reaches the highest point in the circle? When the rope is about to become slack, the tension force approaches zero. Therefore, at the top of the circle, F c = F g, and (3.00)v /r = (3.00)(9.8). The speed is now.4 m/s. A) 0.6 m/s B) 1.8 m/s C).4 m/s D) 3. m/s E) 4.8 m/s #10 7. To weigh a fish, a person hangs a tackle box of mass 3.5 kg and a cooler of mass 5.0 kg from the ends of a uniform rigid pole that is suspended by a rope attached to its center, as shown above. The system balances when the fish hangs at a point ¼ of the rod s length from the tackle box. What is the mass of the fish? A) 1.5 kg B).0 kg C) 3.0 kg D) 6.0 kg E) 6.5 kg The sum of the torques is zero since the system is in static equilibrium. Therefore, (5.0)(9.8)(l/) + (3.5)(9.8)(l/) + m(9.8)(l/4) = 0. The mass of the fish is 3.0 kg. #11
3 8. The velocity of a projectile at launch has a horizontal component v x and a vertical component v y. When the projectile is at the highest point of its trajectory, which of the following show the vertical and horizontal components of its velocity and the vertical component of its acceleration? Vertical Velocity Horizontal Velocity Vertical Acceleration A) v y v x 0 B) v y 0 0 C) 0 v x 0 D) 0 0 g E) 0 v x g The acceleration of a projectile is constant g, or 9.8 m/s. The vertical velocity is zero at the top of AP its path Physics while the Straight horizontal Line velocity Motion remains AP Exam constant Multiple throughout Choice Questions the object s  Kinematics trajectory. AP Physics Straight Line Motion AP Exam Multiple Choice Questions  Kinematics #1 #1 9. The graph above shows the velocity as a function of time for an object moving in a straight line. Which of the following graphs shows the corresponding displacement as a function of time for the same time interval? The correct answer is D. The first segment of the graph represents uniform acceleration, which is parabolic on a displacementtime graph. The second segment is constant positive velocity, which is a straight line on a displacementtime graph. The third segment is also parabolic on the displacement time graph, with the line s slope approaching zero as the object s velocity approaches zero. # # #3 #3 10. A ball is thrown straight downward with an initial speed of 5 m/s. It strikes the ground after.0 seconds. How high is the building? A) 0 m B) 30 m C) 50 m D) 70 m E) 80 m x = v o t + 1/at = (5)() + ½(9.8)() = 70 m. The initial velocity is negative since the ball was thrown downward. 3
4 11. It takes the planet Jupiter 1 years to orbit the Sun once. What is the average distance from Jupiter to the Sun? The mass of the Sun is 1.99 x kg. A).4 x m B) 3.9 x m C) 5. x m D) 7.8 x m E) 9.7 x m It is necessary to derive the expression for radius using the mass of the Sun and Jupiter s period. mv /r =GmM/r, or 4π r 3 =GMT. #19 #1 1. A block of mass 5.0 kilograms lies on an inclined plane, as shown above. The horizontal and vertical supports for the plane have lengths of 4.0 meters and 3.0 meters, respectively. The coefficient of friction between the plane and the block is The magnitude of the force F necessary to pull the block up the plane with a constant speed is most nearly: A) 30 N B) 41 N C) 49 N D) 50 N E) 58 N The angle of the incline is 37, since tan θ = (3/4). If the block moves up the incline at constant speed, the sum of all forces is zero. Therefore, F = F f +mg(sinθ) = µmg(cosθ) + mg(sinθ) = 41 N. 13. Assuming a frictionless, massless pulley (shown above), determine the acceleration of the blocks once they are released from rest. A) + m g B) m + m g C) m g D) m + g # E) m m m + g The acceleration of the from a systems perspective can be found from the expression: ( + m )a = m g g. Therefore, the acceleration is a = (m )g divided by the sum of the masses. #0 4
5 #10 #7 14. A nearly massless rigid rod with masses attached to its ends is pivoted about a horizontal axis as shown above. When released from rest in a horizontal orientation, the rod begins to rotate with an angular acceleration of magnitude: A) 1g 7l B) 1g 5l C) 1g 4l D) 5g 7l E) g l We first need to find the moment of inertia and net torque. Since the rod is massless, the moment of inertia is I = Σmr = (3Mo)l +(Mo)(l) =7Mol. The sum of the torques = (Mog)(l) +(3Mog)(l) = Mogl. The angular acceleration is the net torque divided by the moment of inertia, or Mogl/(7Mol) = g/7l. 15. A wheel starts from rest and accelerates at a uniform rate to reach an angular speed of 6.0 rad/s while #8 What is the constant angular acceleration of the wheel? turning through.0 revolutions. A) 0.4 rad/s B) 1.4 rad/s C).9 rad/s D) 4. rad/s E) 9.0 rad/s Applying the angular kinematics equation, vf = vo + αθ, we see that 6.0 = α()(π). 16. If all of the forces acting on an object balance so that the net force is zero, then: A) the object must be at rest. B) the object s speed will decrease. C) the object will follow a parabolic trajectory. D) the object s direction of motion can change, but not its speed. #11 E) none of the above. If the forces are balanced, the object may be at rest or it may be moving at a constant velocity. Therefore, its speed and direction cannot change. It is also incorrect to say that it must be at rest. 5
6 17. Assuming the lower arm has a mass of.8 kg and its center of gravity is 1 cm from the elbowjoint pivot, how much force must the extensor muscle in the upper arm (F M ) exert on the lower arm for the hand to hold a 7.5 kg shotput. A) 1.0 x 10 N B) 5.0 x 10 N C) 1.0 x 10 3 N D) 1.5 x 10 3 N E).0 x 10 3 N The system is in static equilibrium, so it is easiest to approach the problem by finding an expression for the sum of the torques. The clockwise torques are exerted by the weight of the arm (.8)(9.8)(0.1m) and the force of the shotput on the arm (7.5)(9.8)(0.30m). The counterclockwise torque is exerted by the extensor muscle (+F M )(0.05m). Solving for F M, we find the downward force is equal to approximately 1000 N. 18. The components of three vectors are: Vector xcomponent ycomponent S T U What is the angle of the resultant vector S + T + U as measured from the positive x axis? The positive x axis corresponds to the east direction and the positive y axis corresponds to north. A) 4 NE B) 4 SE C) 66 NE D) 66 SE E) 33 SE The sum of the xcomponents is +.0, and the sum of the ycomponents is Therefore, the resultant points in the southeast direction, and tan 1 (4.5/.0) = 66 SE. 6
7 19. At the surface of the Earth, an object of mass m has a weight of w = mg. If the object is transported to a height above the surface that is twice the radius of the Earth, then, at the new location: A) its mass is m and its weight is w. B) its mass is m and its weight is w. C) its mass is m and its weight is w 4. D) its mass is m and its weight is w 4. E) its mass is m and its weight is w 9. The mass of the object is the same regardless of its altitude. If it is elevated to a height at r above the surface of the Earth, its new distance from the Earth s center of mass is 3r. Since g = GM/r, if we triple r, the weight is 1/9 of its original value. 0. A 0 kg block is suspended vertically by a.0m long, 5.0mm diameter rope. Young s modulus for this rope is 1.5 x 10 9 N/m. The system is in static equilibrium. By how much does the rope stretch? A) 3.0 mm B) 6.3 mm C) 9.3 mm D) 13 mm E) 17 mm We apply the elasticity equation with Young s modulus since the rope is a rigid object: F = (YA/L)ΔL. The force is provided by the weight of the block. Substituting values, we get: (0)(9.8) = (1.5 x 10 9 )(π)(0.005) (ΔL)/(.0), and the change in length is equal to 13 mm. 7
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