D.W. Poppy Secondary School Physics 12. KinematicsMidterm2. Name:

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1 W. Poppy Secondary School Physics 12 KinematicsMidterm2 Name: Directions: Fill in the scantron form with the following information: 1. ID number (student number) 2. Name at top of form 3. Name bubbled into the columns labelled "C1C2C3C4" 4. Test Version Number (shown below); bubble in spaced labelled "version"on the form Version # 0

2 1. The graph below shows the distance travelled by an object plotted against time. What is the object's average speed during the time interval t =10 s to t =20 s? 0.40 m/s 0.60 m/s 0.80 m/s 1.7 m/s v av = total distance t v av = 6/10 = 0.6 m/s 2. An object travels in a circle of radius 'r'. After one complete revolution, what is its displacement from the starting point? 0 2r 2π 2πr Since the object is back at the starting position, P = 0. Displacement = P = If a body has an instantaneous acceleration of +5 m/s 2, it MUST be moving. True False Acceleration is the rate of change of velocity. An object can be stationary for an instant and still have an acceleration. Think of a ball thrown vertically upward, at the top the velocity will be zero, but still have an acceleration of 9.81 m/s 2 downward. 4. An athlete completes two laps of a circular track of radius 50.0 m. The total displacement of the athlete at the end of the run is: 0 m 50.0 m m 314 m E. 728 m Displacement is a vector quantity. Displacement = P f - P i Since the athlete is back at the starting point P = 0 KinematicsMidterm2 Page 1 Version 0 Mr. Furse

3 5. The graph below shows the displacement of a cart as a function of time. It is a parabola with a constantly increasing slope. Which is the acceleration graph for the cart? From graphical analysis, D varies directly with t 2 From analyzing the equation d = v i t at 2, this too gives us a parabola if a = constant So the graph with constant acceleration represents the acceleration for the object. 6. The graph shows the motion of an automobile during a seven hour period. Which of the following statements is FALSE? The car returned to its starting place after 7 h. The total distance travelled in the 7 h was 100 km. The car's motion was uniform for the first 2 h. The car was moving at all times during the 7 h period. Horizontal line on a D-t graph indicates no motion. So the car was not moving at all times. KinematicsMidterm2 Page 2 Version 0 Mr. Furse

4 7. The graph shown below displays velocity v versus time t for a moving object. The slope of this graph represents the object s mass. momentum. acceleration. displacement. slope of graph = rise/run rise/run = change in velocity / change in time change in velocity / change in time = acceleration 8. The area under a velocity-time represents: acceleration change in acceleraton speed displacement Displacement = area under a velocity-time graph. d = v x t 9. A car accelerates uniformly from 3.5 m/s to 12.5 m/s in 4.5 s. The magnitude of the acceleration is: 0.28 m/s m/s m/s m/s 2 a = v t a = ( ) 4.5 a = 2.0 m/s A car travelling at 15 m/s accelerates at 3.0 m/s 2 for 5.0 s. What is its final speed? 30 m/s 23 m/s 20 m/s 45 m/s = v i + at = x 5 = 30 m/s KinematicsMidterm2 Page 3 Version 0 Mr. Furse

5 11. A car travelled up a hill at a constant speed of 10.0 m/s and returned down the hill at 20.0 m/s. If the time to turn around is ignored, what was the average speed for the total trip? m/s 15 m/s 16.7 m/s v av = total D total t Since t up = 2 s, then d up = d down = 20 m, and t down = 0.5 x t up total D = 40 m, total t = 3 s v av = 40 3 v av = 13.3 m/s 12. How far would a car travel in 6.0 s if its initial speed is 2.0 m/s and its acceleration is 2.0 m/s 2? 12 m 14 m 24 m 36 m E. 48 m d = v i t at 2 d = 2.0 x x 2.0 x d = 48 m 13. A ball moving with an initial velocity of 100 m/s north is given an acceleration of 10 m/s 2 south. What will its velocity be after 6.0 s? 40 m/s north 40 m/s south 60 m/s south 60 m/s north Assum North = +ve = v i + at = x 6 = 40 m/s north 14. In landing, a jet plane decelerates uniformly and comes to a stop in 38 s, covering a distance of m along the runway. What was the jet s landing speed when it first touched the runway? 2.1 m/s 39 m/s 79 m/s 170 m/s KinematicsMidterm2 Page 4 Version 0 Mr. Furse

6 known quantities: = 0 acceleration is negative vi v f = vi 2ad vi = 2ad a = 2d vi v f = vi at vi = at a = t combining the two equations for acceleration we get: 2 vi vi = 2d t 2d vi = t v i = 3000/38 v i = 78.9 m/s 15. A cannon fires a shell at 500 m/s at an angle of 30 above the horizontal. The horizontal component of the velocity is: 250 m/s 433 m/s 577 m/s 612 m/s E. 866 m/s v x = v cosø v x = 500 x cos30 v x = 433 m/s 16. A 1.5 kg ballistic cart, moving at 2.8 m/s, ejects a 25 g ball vertically with an initial velocity of 6.5 m/s. What is the magnitude of the initial velocity of the ball relative to the ground? 7.1 m/s 6.5 m/s 2.8 m/s 9.3 m/s KinematicsMidterm2 Page 5 Version 0 Mr. Furse

7 R = (2.8) 2 + (6.5) 2 R = 7.08 m/s 17. A dart is thrown horizontally toward X with a speed of 20 m/s. It hits a point Y 0.1 s later. The distance XY will be approximately: 2m 1 m 0.5 m 0.1 m E m The initial vertical velocity = 0 if it is thrown horizontally. We are concerned with the vertical displacement XY. t at 2 where = displacement (XY), v iy = 0, a = 9.81 m/s 2 = 4.9 (0.1) 2 = m Use the following information to answer the next 2 question(s). A missle is fired horizontally with an initial velocity of 45 m/s from the top of a building 75 m high. 18. How long is the missle in the air? 2.77 s 3.91s 4.59 s 15.3 s Examine the motion in the vertical direction. v iy = 0, a = 9.81 m/s 2 down, = -75 m -75 = t 2 t = (75 4.9) t = 3.91 s KinematicsMidterm2 Page 6 Version 0 Mr. Furse

8 19. What is the horizontal range of the missle? 125 m 176 m 207 m 689 m time for missle to reach ground is found from the vertical values. t t 2 where v iy = 0 = -75 m. -75 = -4.9t 2 t = (75 4.9) = 3.91 s Horizontal range will be = v x t = 45 x 3.91 = 176 m 20. A projectile is fired with an initial velocity of 120 m/s at an angle of 30 above the horizontal. If air resistance is negligible, how much time elapses before the projectile strikes the ground at the same elevation from which it was fired? 6.1 s 11 s 12 s 21 s The vertical displacement will be 0 m since the projectile lands at the same elevation. Since the initial velocity is upward, we will define up as positive and downward as negative. So this gives us: = 0 m and v iy = v sinø a = m/s 2 t t 2 0 = 60 t t t 2 =60 t t = 0 s or t = s t = 12.2 s Use the following information to answer the next 2 questions (21-22). A cat on the floor jumps up with an initial velocity of 5.00 m/s at an angle of 75.0 to the horizontal. Air resistance can be ignored. 21. What is the maximum height reached by the cat? 1.19 m 1.23 m 1.28 m 2.38 m KinematicsMidterm2 Page 7 Version 0 Mr. Furse

9 At the maximum height, the cat will have zero vertical velocity. y = 0, v iy = 5 x sin 75 y 2 2-2a = (v iy 2 - y 2 ) 2a = ( ) (2 x 9.81) = 1.19 m 22. What horizontal distance has been travelled by the cat when it lands on the floor? m 1.28 m 4.93 m 5.10 m When the cat reaches the floor, the vertical displacement will be zero. t -0.5 at 2 where v iy = v sin ø v iy = 5 x sin75 0 = 4.83t t 2 t (-4.9 t +4.83) = 0 either t = 0 or the bracket is zero. t = t = 0.99 s = v ix t where v ix = v cosø = 5 x cos75 = 5 x cos75 x 0.99 = 1.28 m 23. An object is projected as shown in the diagram. Calculate the time elapsed before it strikes the ground. (4.00 marks) KinematicsMidterm2 Page 8 Version 0 Mr. Furse

10 (4.00 marks) We are only concerned with the vertical displacement, which will be 50 m below the starting point. Since the initial velocity is upward, we will define up as positive and downward as negative. So this gives us: = -50 m and v iy = v sinø a = m/s 2 t t 2-50 = 125 t t t t - 50 = 0 use the quadratic equation to solve for t. t = 25.9 s KinematicsMidterm2 Page 9 Version 0 Mr. Furse