* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Topic 2 Mechanics Part 3 and 4 projectile, friction,10
Coriolis force wikipedia , lookup
Specific impulse wikipedia , lookup
Derivations of the Lorentz transformations wikipedia , lookup
Fictitious force wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Relativistic mechanics wikipedia , lookup
N-body problem wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Classical mechanics wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Relativistic angular momentum wikipedia , lookup
Hunting oscillation wikipedia , lookup
Equations of motion wikipedia , lookup
Four-vector wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Seismometer wikipedia , lookup
Velocity-addition formula wikipedia , lookup
NOTES AND SOLUTIONS TOPIC 2 PART 3 AND 4 Topic 2 New- Projectile Motion, Friction SL and HL Topic 10 ( NEW) Gravitational Field HL ONLY ONLY Text Tsokos pp. 132-157 1 Topic 2 New- Projectile Motion 1. Components of Motion Objects launched at an angle are called projectiles and have two components: vx and vy . vx is always constant in magnitude and direction throughout the flight. vy decreases going up due to gravity pushing down. vy increases going down due to gravity. 2 2. Projectiles at an Angle Objects launched at an angle are called projectiles and have two components: vx and vy . Vx is always constant in magnitude and direction throughout the flight. V y decreases going up due to gravity pushing down. Vy increases going down due to gravity. vy at the top = 0. vx is constant throughout the flight and NOT zero at the top. vy0 = initial velocityy Do not confuse vy0 with vy at the top = 0. Time in the air make sure you understand how to calculate: Time in the air (t) = tup + t down . Remember from free fall tup = tdown. So total time = 2 x tup From free fall formula ( see below) v = u + gt : tup = u = initial vy g 9.8 = vy0 9.8 Formulas Projectiles vy = v at the top = 0 so: height of object: y = vy tup + ½ g (tup )2 Data Booklet Formulas – Free Fall v = u + gt ( g = - 9.8 ms-2 ) s = ut + ½ g t2 height s= (v+ u )t 2 v2 = u2 + 2gs x = vx t total = Range Don’t confuse total time ( ttotal) with tup tup = vy0 g Range (x) = horizontal displacement Range x uses total time: ttotal = 2 x tup x = vx t total *height of object: y = vy tup + ½ g (tup )2 Height of object ( y) only depends on time up: g = -9.8ms-2 y = vy tup + ½ g (tup )2 3 Fig. 3.21 – Range (x) = horizontal displacement from point of origin = Vx t Range is longest at 45 degrees. Example Range : Two identical balls are shot with an initial velocity of 20.0 ms-1. One ball is shot from an angle of 15.00 and another is shot from an angle of 60.00. Find the range ( x) and height (y) of each. See solution next page. Ball shot at 15.00 4 Ball shot at 60.00 5 6 3. Horizontal Projectile Motion Formulas Vx doesn´t change. Vy changes due to gravity. 7 Example . 3.8 – determining how long the ball is in the air See solution next page…… 5 Ex.3 - Solution Thinking It Through. In looking at the components of motion, we find that part (a) involves the time it takes the ball to fall vertically, analogous to a ball dropped from that height. Thls time is also the time the ball travels in the horizontal direction .The horizontal speed is constant ,so we can find the horizontal distance, requested in part (b). Solution. Writing the data with the origin chosen as the point from which the ball is thrown and downward taken as the negative direction, we have: y = -25.0m vx = 8.25 ms-1 a) b) 6 General Questions and Problems : Solutions next page …..--> 1. On Cartesian axes, the x component of a vector is generally associated with : a) cosine b) sine c) tangent d) none of the above 2. Can an x component of a vector be greater than the magnitude of the vector? Explain. 3. A golf ball is hit with an initial velocity of 35ms-1 at an angle less than 450 above the horizontal. The horizontal component of the velocity is : a) greater than b) equal to or c) less than the vertical component Explain the answer above: 4. If the initial velocity of the ball is 35 ms-1 and the ball is hit at an angle of 370 , what are the horizontal and vertical components of the velocity? 5. If the magnitude of a velocity vector is 7.0 ms-1 and the x component is 3.0 ms-1, what is the y component? 6. The x component of the velocity vector has an angle of 370 to the x axis and a magnitude of 4.8 ms-1 . a) what is the magnitude of the velocity vector? b) what is the magnitude of the y component ? 7 7. 7 General Questions and Problems : Solutions 1. On Cartesian axes, the x component of a vector is generally associated with : a) cosine b) sine c) tangent d) none of the above Answer - a 2. Can an x component of a vector be greater than the magnitude of the vector? Explain. Answer – No. Both components , X and Y, add together to get the resultant. 3. A goofball is hit with an initial velocity of 35ms-1 at an angle less than 450 above the horizontal. The horizontal component of the velocity : a) greater than b) equal to or c) less than the vertical component Answer – a) Explain the answer above: Range (x) = horizontal displacement from point of origin = Vx t Conceptual Note : Projectiles shot at angles less than 45o will have a larger Vx than Vy . Basically the horizontal component of velocity increases as the angle gets smaller or “ closer to the ground” . This can be proven mathematically because cos of angles ( θ) less than 45 are numerically greater than sin of angles (θ) less than 45 Fig. 3.21 – range is longest at 45 degrees 8 8 4. If the initial velocity of the ball is 35 ms-1 and the ball is hit at an angle of 370 , what are the horizontal and vertical components of the velocity? Answer : Vx = V cos θ = 28ms-1 Vy = V sin θ = 21ms-1 5. If the magnitude of a velocity vector is 7.0 ms-1 and the x component is 3.0 ms-1, what is the y component? Answer: V2 = Vx2 + Vy2 Vy = √V2 – Vy2 Vy = 6.3ms-1 6. The x component of the velocity vector has an angle of 37 0 to the + x axis and a magnitude of 4.8 ms-1 . a) what is the magnitude of the velocity vector? Vx = V cos θ V = Vx = cos θ 6.0 ms-1 b) what is the magnitude of the y component ? Vy = V sin θ = 3.6ms-1 9 9 7. 1010 More Example Questions: see solutions next page …..--> 1. 2. 1111 More examples - Solutions 1. D 2. 1212 Topic10 New (HL ONLY) - Gravitational Field Tsokos pp. 142-146 This is a complicated topic. The book goes into too much detail so focus on the following . concepts , formulas and example problems . These formulas are given in the data booklet Work being done to move masses is stored as gravitational potential energy. 1313 NOTE : Do not confuse gravitational potential energy formula ( Ep ) with the gravitational force formula ( Fg ) . The gravitational potential energy formula is derived form the gravitational force formula ( Fg ) as follows: Fg = GMm R2 given in data booklet Energy = Work done = Force x distance Distance = R Work = Ep = Fg x R Work = Ep = GMm X R R2 Work = Ep = GMm R in data booklet 1414 Gravitational Potential (V) V=W m V= GMm r . m = GM r Example: The gravitational field close to the surface of a planet is uniform. The gravitational potential difference between the surface of the planet and a point 30.0m above is 15 Jkg -1. a) What is the work done to raise a mass of 1.0 kg from the surface of the planet to 30.0 m above the surface? b) What is the work done to raise a mass of 4.0 kg from the surface of the planet to 30.0 m above the surface? c) What is the work done to raise a mass of 4.0 kg from the surface of the planet to 10.0 m above the surface? 1515 M is the planet The escape velocity is derived by the following way: Change in gravitational potential energy is equal to the change in kinetic energy (½ m v 2) . The mass of the object does not matter and velocity is solved in the following way: Ep = GMm = ½ m v2 R 1616 Examples - see solutions next page Example 1 – Escape Velocity Example 2 - Gravitational Potential Energy A satellite in orbit about Earth moves to another orbit that is closer to the surface of Earth. When the satellite moves into the orbit closer to earth, which of the following correctly describes the change in speed of the satellite and the change in ts gravitational potential energy ? Choice Speed A B C D decreases decreases increases increases Gravitational potential energy decreases increases increases decreases 17 17 Example 3 Example 4 Example 5 1818 Equipotential surfaces and fields 1919 Example 6 2020 Gravity vs Electricity 2121 SOLUTIONS Example 2 - Gravitational Potential Energy 2222 2323 Example 5 Example 6 2424 TOPIC 2 MECHANICS PART 4 NEW - FRICTION Basic Friction Basically 4 forces acting on the box: W = weight of the box = N N = normal force of table pushing up on the box = W Fa = applied force to the right fs = force of static friction between the box and the surface of the table. This opposes the applied force. f s = Fa Text Tsokos p. 74 : 12 - Worked Example Answer : F = 1343 N Frictional force = fs = horizontal component Fx = Fa = 1163 F = pulling force ( resultant force of Fx and Fy ) Since Fx = F cos 300 F= Fx = 1343 N Cos 30 2525 Friction in Detail : Static and Kinetic Formulas fs = µs N fk = µk N N=w=mg F=ma Types of Friction: Static friction – when the frictional force is sufficient to prevent relative motion between surfaces. Sliding/ kinetic friction - static friction is overcome and there is relative motion between surfaces in contact. Rolling friction – rotating motion with no sliding or slipping. Use fs = µs N µs = coefficient of static friction and depends on type of surfaces involved, for example glass on glass is .94 and dry rubber on dry concrete is 1.2 . Once there is motion use fk = µk N . µk is less than µs due to object in motion and varies with the slow speed of surfaces involved. At fast speeds ( 10 m / s or more ) µk is fairly constant. when there is no motion 2626 Examples - check answers next page Example 1 A box of mass m = 20.0 kg and the coefficients of static and kinetic friction between the box and the table are 0.800 and 0.500 respectively. A string is attached to the box and used to pull the box directly horizontal at no angle Draw a free body diagram. Label the normal force ( N) the weight ( W) the force due to static friction ( fs) and the applied force in the string ( Fa): A) what is the minimum tension in the string to start the box in motion. B) after the box is moving and the tension is kept constant, what is the acceleration of the box? 2727 Example 2 A box of mass m = 40.0 kg and the coefficients of static and kinetic friction between the box and the table are 0.650 and 0.500 respectively. A string is attached to the box and used to pull the box directly horizontal at no angle. A) what is the minimum tension in the string to start the box in motion. B) after the box is moving and the tension is kept constant, what is the acceleration of the box? 28 Answers to examples 1 and 2 – Friction Example 1 a) Fa = fs = s N N = mg = (40.0)(9.8) = 392 N Fa = (392) (0.8) = 314 N b) Fnet = Fa - fk From a) above need Fa of 314 N to get crate moving fk = k N = (0.5)(314) = 157 N Fnet = Fa - fk = 314 N – 157 N = 157 N a = Fnet m = = 3.93 ms-2 157 N 40.0 Example 2 a) Fa = fs = s N N = mg = (40.0)(9.8) = 392 N Fa = (392) (0.65) = 255 N b) Fnet = Fa - fk From a) above need Fa of 255 N to get crate moving fk = k N = (0.5)(392) = 196 N Fnet = Fa - fk = 255 N – 196 N = 59 N a = Fnet m = 59 N 40.0 = 1.48 ms-2 29 MISCELLANEOUS PROBLEMS Wilson Buffa pp. 137 – 139 : 74, 75, 78, 82 – 85 , 96 Solutions at end of packet 78. Is it easier to push a lawn mower at an angle ? Draw a free body diagram and explain. 30 31 32 SOLUTIONS MISCELLANEOUS PROBLEMS Wilson Buffa pp. 137 – 139 : 74, 75, 78, 82 – 85 , 96 33 34 35 36 37 38