BROCK UNIVERSITY. PHYS 1P21/1P91 Solutions to Mid-term test 26 October 2013 Instructor: S. D Agostino

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1 BROCK UNIVERSITY PHYS 1P21/1P91 Solutions to Mid-term test 26 October 2013 Instructor: S. D Agostino 1. [10 marks] Clearly indicate whether each statement is TRUE or FALSE. Then provide a clear, brief, detailed explanation for why the statement is true or false. If the statement is false, make sure to include a correction. You are welcome to include diagrams and/or formulas in your explanation if you wish. (a) When a car brakes suddenly, a passenger in the car is thrown forward by the force of the car s motion. (b) A car moving along a circular track at a constant speed does not accelerate, because its speed is constant. (c) The net force acting on an object is always in the same direction as the object s acceleration, even when the object moves along a curved path. (d) The net force acting on a car moving along a circular track at a constant speed is zero, by Newton s first law, which states that a body in motion tends to stay in motion. (e) When you are at rest sitting on a chair, the gravitational force acting on you and the normal force from the chair acting on you are action-reaction forces according to Newton s third law, and are therefore equal in magnitude but opposite in direction. Solution: (a) FALSE. There is no force that throws the passenger forward; furthermore the motion of the car doesn t carry along a force with it. The passenger has a tendency to continue moving forward at a constant speed, a phenomenon we call inertia, and will do so unless a force causes acceleration. If the passenger has a seat belt on, then the seat belt will cause the passenger to slow down along with the car. Thus, the force on the passenger is towards the back of the car, not in the forward direction. (b) FALSE. Even though the speed is constant, the car s direction continually changes, and therefore the car s velocity continually changes too. Therefore, the car does accelerate. Because the speed of the car is constant, the acceleration is directed towards the centre of the circle. (c) TRUE. Newton s second law is a vector equation: F = m a. This means that the net force acting on an object is in the same direction as the object s acceleration, no matter what the path of the object is. (d) FALSE. We concluded in Part (b) that a car moving in a circle has non-zero acceleration. Therefore, by Newton s second law, the net force acting on the car is also non-zero.

2 Furthermore, it s NOT true that Newton s first law states that a body in motion tends to stay in motion. Newton s first law is more specific, and states that a body in motion tends to continue moving in a straight line at a constant speed, unless it is forced to move in another way. (e) FALSE. Just because two forces have equal magnitude and opposite direction doesn t make them action-reaction pairs of forces. The gravitational force acting on you and the normal force acting on you can t possibly be action-reaction forces because they act on the same object: You. Remember that Newton s third law says that if A exerts a force on B, then B exerts an equal and opposite force on B. Thus, in an action-reaction pair of forces, each force acts on a different object. 2. [4 marks] Alice walks along a straight sidewalk. Her position-time graph is given below. x t v t (a) Draw Alice s velocity-time graph in the space provided above. (b) Briefly describe Alice s motion in words for each of the four segments of her motion.

3 Solution: The velocity-time graph is drawn above. Alice moves in the negative direction for the first 2 time units at a constant speed of 1 speed unit. Alice suddenly stops after 2 time units, and remains stopped for an additional 2 time units. Then Alice suddenly begins moving in the positive direction for the next 2 time units, at a constant speed of 1.5 speed units. Then Alice suddenly switches direction and moves in the negative direction for 2 time units, at a constant speed of 1 speed unit. During each of the four segments of the motion, Alice s acceleration is zero, because her velocity is constant. Of course Alice must accelerate between each of the four described motion segments, but for simplicity we ll ignore these accelerations. (For all you physics majors, the accelerations between the four segments are not really describable, because the motion between the four segments is unphysical. But first-year problems typically use such unphysical position-time graphs because they lack curves and are therefore considered easier to deal with.) 3. [4 marks] An airplane is headed north at a constant speed of 200 km/h with respect to the air. The prevailing wind blows from west to east at a constant speed of 40 km/h with respect to the ground. (a) Determine the magnitude and direction of the airplane s velocity with respect to the ground. (b) For the same wind velocity, determine the direction that the airplane must head in (with a constant speed of 200 km/h with respect to the air) so that its velocity with respect to the ground is in the north direction. Solution: (a) v AG v PA θ v PG Let v PA represent the velocity of the plane with respect to the air (pilots call the magnitude of this vector the plane s air speed) and let v PG represent the velocity of the plane with respect to the ground (pilots call the magnitude of this vector the plane s ground speed). The velocity of the wind with respect to the ground is denoted by v AG. Therefore, v PG = v PA + v AG which is represented by the vectors in the diagram.

4 The given information is v PA = 200 km/h v AG = 40 km/h Applying Pythagoras s theorem to the vectors in the diagram, v PG 2 = v PA 2 + v AG 2 v PG 2 = v PG = v PG = 204 km/h Using trigonometry, we can determine the angle in the diagram: tan θ = θ = 11.3 East of North Thus, the velocity of the plane with respect to the ground is 204 km/h at an angle of 11.3 East of North. (b) To move straight North, the plane must head somewhat West of North, as in the following diagram: v AG v PA θ v PG sin θ = θ = 11.5 West of North Thus, the plane must head 11.5 West of North for the plane s path to be straight North. 4. [4 marks] A golf ball is struck from the ground with an initial speed of 140 km/h at an angle of 40 above the horizontal. Determine (a) the time at which the ball reaches maximum height. (b) the maximum height reached by the ball. (c) the time at which the ball reaches the ground again. (Use a symmetry argument and the result of Part (a).) (d) the horizontal distance travelled.

5 Solution: First convert the initial speed from km/h to m/s: 140 km h = 140 km h 1000 m 1 km 1 h = 38.9 m/s 3600 s (a) The ball reaches maximum height when v y = 0. v y = a y t 1 0 v i sin 40 = 9.8 t 1 (38.9) sin 40 t 1 = 9.8 t 1 = 2.55 s (b) Using the time from Part (a), the relevant displacement in the vertical direction is y = v iy t a y ( t 1 ) 2 = 38.9 sin 40 (2.55) (2.55)2 y = 31.9 m (c) By symmetry, the time needed for the ball to hit the ground is twice the time needed to reach the maximum height: t 2 = = 5.10 s (d) Using the time from Part (c), the horizontal displacement for the entire journey is x = v ix t 2 = v i cos 40 t 2 = 38.9 (cos 40 ) (5.1) x = 152 m 5. [3 marks] In the diagram, the mass of block A is 5 kg and the mass of block B is 10 kg. The coefficient of kinetic friction between block A and the table it slides on is 0.4. The string is massless and does not stretch, and the pulley is massless and frictionless. Determine the acceleration of the blocks and the tension in the string. (Assume that the blocks will move, so that you can ignore static friction.)

6 A B Solution: First draw free-body diagrams for each block: T n f T 5g 10g Figure 1: A free-body diagram for block A. For the horizontal direction, take right to be the positive direction. Figure 2: A free-body diagram for block B. For the vertical direction, take down to be the positive direction. This choice is compatible with the choice of positive direction in the free-body diagram for block A. Next make sure you choose positive directions for each block that are compatible; I ve chosen up and to the right as the positive directions for block A, and down as the positive direction for block B. Using the free-body diagram for block A, we can apply Newton s second law of motion for each of the vertical and horizontal directions to obtain

7 T f = 5a (1) n 5g = 0 (2) Applying Newton s second law to block B, using the free-body diagram for block B for guidance, we obtain 10g T = 10a (3) Additionally, recall that there is a connection between the frictional force on block A and the normal force acting on block A: f = µ k n (4) Now we are faced with four equations in four unknowns, which you can solve in a number of ways. Here s my way: From equations (2) and (4), f = µ k (5g) = (0.4)(5g) = 2g Substituting 2g for f into equation (1), we get T 2g = 5a T = 2g + 5a (5) Substituting the expression for T from equation (5) into equation (3), we get 10g (2g + 5a) = 10a 8g = 15a a = 8 15 g a = 5.2 m/s 2 Thus, from equation (5) we get, T = 2g + 5a T = 45.7 N Do these results seem reasonable? Is there any way to check them?

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