1 / 63

Physics 1201Q Review

Physics 1201Q Review. 1.7 The Components of a Vector. Example A displacement vector has a magnitude of 175 m and points at an angle of 50.0 degrees relative to the x axis. Find the x and y components of this vector.

terrel
Download Presentation

Physics 1201Q Review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Physics 1201Q Review

  2. 1.7 The Components of a Vector Example A displacement vector has a magnitude of 175 m and points at an angle of 50.0 degrees relative to the x axis. Find the x and y components of this vector.

  3. A hiker walks 100m north, 130m northeast, and 120m south. Find the displacement vector and the angle measured from the positive x axis. N W E S D θ

  4. 3.2 Equations of Kinematics in Two Dimensions

  5. 3.2 Equations of Kinematics in Two Dimensions

  6. 2.4 Equations of Kinematics for Constant Acceleration Example 6 Catapulting a Jet Find its displacement.

  7. 2.4 Equations of Kinematics for Constant Acceleration

  8. 2.6 Freely Falling Bodies Example 12 How High Does it Go? The referee tosses the coin up with an initial speed of 5.00m/s. In the absence if air resistance, how high does the coin go above its point of release?

  9. 2.6 Freely Falling Bodies

  10. Free Fall Example • What is the maximum height the ball reaches? • How long does it take to reach the maximum height? • How long is the ball in the air total? • What is the velocity of the ball just before it hits the ground?

  11. 3.3 Projectile Motion Example 3 A Falling Care Package The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. Determine the time required for the care package to hit the ground.

  12. 3.3 Projectile Motion

  13. 3.3 Projectile Motion

  14. 3.3 Projectile Motion Example 4 The Velocity of the Care Package What are the magnitude and direction of the final velocity of the care package?

  15. 3.3 Projectile Motion

  16. 3.3 Projectile Motion

  17. Projectile Motion Example • A canon is fired with a muzzle velocity of 1000 m/s at an • angle of 30°. The projectile fired from the canon lands in the water • 40 m below the canon. • What is the range of the projectile. • What id the velocity of the projectile in x and y when it lands • What is the landing angle of the projectile vo Known vo = 1000m/s θ = 30° Δy = -40m Unknown X, vx,vy,θ, vf voy 30° vox

  18. Projectile Motion Example Use y info to find t -40m = (500m/s)t + ½(-9.8m/s2)t2 4.9t2 – 500t – 40 = 0 a b c Solve for t using quadratic equation t = -(-500) ± (-500)2 – 4(4.9)(-40) 2(-40) Take the positive root: t = 102.1s

  19. Projectile Motion Example Calculate x: Calculate vy

  20. Projectile Motion Example Find landing angle and velocity vx = 866m/s θ vy = -500.8m/s vf vf = (866m/s)2 + (-500.8m/s)2 = 1000.4m/s tan θ = -500.8m/s = -0.578 866m/s θ = tan-1(-0.578) θ = -30.04°

  21. Inclined Plane Problem • M1 = 8kg • M2 = 22kg • W = mg • W1 = (8kg)(9.8m/s2) = 78.4N • W2 = (22kg)(9.8m/s2)= 215.6N Take the direction of motion as positive and use Newton’s Second Law to write equations For m1 (taking up the plane as positive) ΣFx = T - W1sin30 = m1a For m2 (taking down as positive) ΣFy = W2 – T = m2a

  22. Inclined Plane Problem T - W1sin30 = m1a eq. 1 W2 – T = m2a eq. 2 W2 - W1sin30 = m1a + m2a W2 - W1sin30 = a m1 + m2 215.6N – (78.4N)(sin 30) = 5.88m/s2 8kg + 22kg • M1 = 8kg • M2 = 22kg • W = mg • W1 = (8kg)(9.8m/s2) = 78.4N • W2 = (22kg)(9.8m/s2)= 215.6N Solve for T using eq. 2 215.6N – T = (22kg)(5.88m/s2) T = 86.2N

  23. 4.9 Static and Kinetic Frictional Forces The sled comes to a halt because the kinetic frictional force opposes its motion and causes the sled to slow down.

  24. 4.9 Static and Kinetic Frictional Forces • Suppose the coefficient of kinetic friction is 0.05 and the total • mass is 40kg. • What is the kinetic frictional force? • How far does the child slide before coming to a stop?

  25. Child on a sled continued How far does the child slide before coming to a stop? Calculate the deceleration of the child: ΣFx = max ax = ΣFx = -20N = -0.5m/s2 m 40kg v = v0 + 2aΔx Δx = v - v0 0m/s – 4m/s = = 4m 2a2(-0.5m/s2)

  26. A block having a mass of 5kg is pulled with a force of 20N acting at 30° above the horizontal. The coefficient of friction between the block and the table is 0.2. Find the acceleration of the block.

  27. 5.4 Banked Curves • On an unbanked curve, the static frictional force • provides the centripetal force. • A car rounds a curve having a 100m radius • Travelling at 20m/s. What is the minimum • Coefficient of friction between the tires and • the road required? • Fc = f =Fn • mv2 = mg • r •  = v2 = (20m/s)2 • gr (9.8m/s2)(100m) • = 0.41 Fn f W = mg

  28. 5.5 Satellites in Circular Orbits Fg = Fc

  29. 5.5 Satellites in Circular Orbits Example 9: Orbital Speed of the Hubble Space Telescope Determine the speed of the Hubble Space Telescope orbiting at a height of 598 km above the earth’s surface.

  30. 5.5 Satellites in Circular Orbits = 42r3 GM

  31. 5.5Satellites in Circular Orbits Global Positioning System = 42r3 GM T = (24 hours)(3600s/hour) = 86400s T2GMe 42 r = 3 r = (86400s)2(6.67 x 10-11 Nm2/kg2)(5.98 x 1024kg) 42 3 r = 42250474m = distance from center of the earth to GPS r = Re + h  h = r – Re = 42250474m – 6380000m = 35870474m = 22,300 mi

  32. 6.2 The Work-Energy Theorem and Kinetic Energy Example 4 Deep Space 1 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If the 56.0-mN force acts on the probe through a displacement of 2.42×109m, what is its final speed?

  33. 6.2 The Work-Energy Theorem and Kinetic Energy

  34. 6.2 The Work-Energy Theorem and Kinetic Energy

  35. 6.5 The Conservation of Mechanical Energy Conceptual Example 9 The Favorite Swimming Hole The person starts from rest, with the rope held in the horizontal position, swings downward, and then lets go of the rope. Three forces act on him: his weight, the tension in the rope, and the force of air resistance. Calculate the person’s final speed if he starts from A height of 8m?

  36. What is the swimmer’s velocity at its lowest point if the rope is 8m long? vf = 2gho = (2)(9.8m/s2)(8m) = 12.5m/s

  37. Example: A marble having a mass of 0.15 kg rolls along the path shown below. A) Calculate the Potential Energy of the marble at A (v=0)B) Calculate the velocity of the marble at B.C) Calculate the velocity of the marble at C A 10m C B 3m

  38. Example 7: Two marbles collide in an elastic head-on collision. The first marble has a mass m1 = 0.25kg and a velocity of 5m/s. The second has mass m2 = 0.8kg and is initially at rest. Find the velocities of the marbles after the collision. v1i – v2i = v2f – v1f Eq 3 5m/s – 0 = v2f – v1f v1f = v2f – 5m/s (substitute below) Eq 2 (0.25kg)(5m/s) + 0 = (0.25)v1f + (0.8kg)v2f 1.25kgm/s = (0.25kg)(v2f – 5m/s) + (0.8kg)v2f 2.5kgm/s = (0.25kg)(v2f) + (0.8kg)v2f v2f = 2.38m/s v1f = 2.38m/s – 5m/s = -2.62m/s

  39. 7.3 Collisions in One Dimension Example 8 A Ballistic Pendulim The mass of the block of wood is 2.50-kg and the mass of the bullet is 0.0100-kg. The block swings to a maximum height of 0.650 m above the initial position. Find the initial speed of the bullet.

  40. 7.3 Collisions in One Dimension Apply conservation of momentum to the collision:

  41. 7.3 Collisions in One Dimension Applying conservation of energy to the swinging motion:

  42. 7.3 Collisions in One Dimension

  43. 8.3 The Equations of Rotational Kinematics

  44. 8.3 The Equations of Rotational Kinematics Example 5 Blending with a Blender The blades are whirling with an angular velocity of +375 rad/s when the “puree” button is pushed in. When the “blend” button is pushed, the blades accelerate and reach a greater angular velocity after the blades have rotated through an angular displacement of +44.0 rad. The angular acceleration has a constant value of +1740 rad/s2. Find the final angular velocity of the blades.

  45. 8.3 The Equations of Rotational Kinematics

  46. 8.5 Centripetal Acceleration and Tangential Acceleration Example 7 A Discus Thrower Starting from rest, the thrower accelerates the discus to a final angular speed of +15.0 rad/s in a time of 0.270 s before releasing it. During the acceleration, the discus moves in a circular arc of radius 0.810 m. Find the magnitude of the total acceleration.

  47. 8.5 Centripetal Acceleration and Tangential Acceleration

  48. 8.6 Rolling Motion The tangential speed of a point on the outer edge of the tire is equal to the speed of the car over the ground.

  49. 8.6 Rolling Motion Example 8 An Accelerating Car Starting from rest, the car accelerates for 20.0 s with a constant linear acceleration of 0.800 m/s2. The radius of the tires is 0.330 m. What is the angle through which each wheel has rotated?

  50. 8.6 Rolling Motion

More Related