Exam 1 Practice SOLUTIONS Physics 111Q.B

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1 Exam 1 Practice SOLUTIONS Physics 111Q.B Instructions This is a collection of practice problems for the first exam. The first exam will consist of 7-10 multiple choice questions followed by 1-3 problems (depending on length). Multiple Choice 1. A stone is thrown at an angle of 45 above the horizontal x-axis in the +x-direction. If air resistance is ignored, which of the velocity versus time graphs shown above best represents v x versus t and v y versus t, respectively? v x vs. t v y vs. t (a) I IV (b) II I (c) II III (d) II V (e) IV V 1

2 2. Two projectiles are launched from level ground and both spend the same amount of time in the air. Which of the following is NOT true? (a) The two projectiles reach the same maximum height. (b) The initial velocities of the two projectiles have identical vertical components. (c) The two projectiles travel the same distance horizontally. (d) The accelerations of the two projectiles after the launch are identical. 3. On the d=95.6 mile trip from Philadelphia to New York, a student has an average speed of v avg =64.7 mph. Which of the following is/are true? (a) I only. (b) II only. (c) I and II. (d) II and III. (e) I an III. I The time for the trip is t = d v avg II The slope of the displacement vs. time graph is always v avg. III The average acceleration is vavg t. 4. For an object moving in a circle at constant speed, how must the centripetal force be changed to double that speed (while keeping the radius fixed)? (a) Decrease by a factor of 4. (b) Decrease by a factor of 2. (c) Increase by a factor of 2. (d) Increase by a factor of A rock is thrown vertically upward with intial speed v 1. Assume that there is a force of air resistance proportional to v, where v is the velocity of the rock. Which of the following is correct? (a) The acceleration of the rock is always equal to g. (b) The acceleration of the rock is equal to g only at the top of the flight. (c) The acceleration of the rock is always less than g. (d) The speed of the rock upon returning to its starting point is v In which of the following is the net force on a wooden block equal to zero? (a) Sitting on a desk. (b) Falling off of the desk. (c) Sliding down a ramp at constant velocity. (d) (a) and (b). (e) (a)and(c). (f) (b) and (c). 2

3 Problems 1. The first graph below shows the position vs. time for an object moving in one dimension. Sketch, as accurately as you can, the velocity vs. time and acceleration vs. time graphs on the coordinate axes below. Also, describe in words what is happening at points A and B in terms of position, velocity, and acceleration. At A, the object momentarily has zero velocity and it has reached its maximum positive x-displacement. The acceleration is constant and negative. The object is turning around and coming back to x=0. At B, the particle stops for some time at a negative x-location. The velocity and acceleration are both zero, though the velocity had been constant and negative just before arriving at B. 3

4 2. You play a game at the carnival in which you shoot a toy cannon at a target. If you hit the target, you win a prize (perhaps that giant monkey you ve always wanted!). The target hangs from the ceiling so that it is 1 m above the ground and 3 m away from the cannon. The cannon is also 1 m off the ground so that the launch height is the same as the target height. Analyze this problem assuming that air resistance is neglible. (a) The cannon is set to launch at an angle of 30 with respect to the horizontal, as shown in the figure above. Draw a free body diagram for the cannon ball immediately after it has left the cannon. Label all of the forces on the cannon ball. Also include the initial velocity, v. Draw and label the components of v (in terms of the angle θ). There is only one force on the cannon ball after it has left the cannon: the force of gravity, pulling straight down. (b) Find the initial speed v such that the cannon ball will hit the target. (There is some additional workspace on the next page.) Let s analyze the motion to the highest point, halfway through the motion. At this point, there is no velocity in the y-direction and the cannon ball has traveled half of the distance to the target. What we know is summarized in the table below. x motion y motion x = 1.5 m y =? v 1x = v cos30 v y1 = v sin 30 v x2 = v cos30 v y2 = 0 a x = 0 a y = g t =? t =? 4

5 Obtain an equation for v and t from the y motion: v 2y = v sin 30 = v 1y g t 0 = v sin 30 g t t = Obtain an equation for v and t from the x motion: x = x = g t (.5)v 9.8 v 1x t 1 2 g t2 v 1x t 1.5 = v cos30 t t = 1.5 (.866)v Now we have two equations with two unknowns and we can solve for v:.5v 9.8 = v (.866)(.5)v 2 = (1.5)(9.8) v = 5.83 m/s (1) (c) Given the initial speed that you just calculated, show that an initial launch angle of 60 will also hit the target. Explain why there are two angles that will hit the target for the same speed. If we use a launch angle of 60 instead, the x and y components of the velocity will change: v 1y = v sin60 = (.866)v v 1x = v cos60 = (.5)v Let s assume that x is unknown and subsitute our new velocity values into our equations from (b) and then solve for x. If we do this, we should find that x is 1.5 m. Substituting this new value for the velocity in the x direction into our equations from (b), we obtain: t = x (.5)v Substituing into our equations from (b) for the y motion, we obtain: t = (.866)v 9.8 Now we have two equations with two unknowns and we can solve for x:.5v 9.8 = x.866v (.866)(.5)(5.83) 2 = ( x)(9.8) x = 1.5 m This is exactly the same as for the 30 launch angle. As it turns out, there are always two possible launch angles that will hit a target for a give initial speed. The steeper angle corresponds to a smaller x velocity and a larger y velocity. Since v x is smaller, the projectile won t travel as far unless it has more time which is exactly what the larger v y gives it. The smaller angle corresponds to the opposite case when v x is large but due to the small v y, there is not much time for the projectile to move. 5

6 (d) Now imagine that at the moment that you fire the cannon the cord holding the target is cut. What angle should you use to hit the target? Hint: This should require NO calculation. Since the cannon ball and the target will fall exactly the same distance while the cannon ball is on its way to the target, the launch angle should be 0. 6

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