1206 - Concepts in Physics Friday, October 16th
Notes Assignment #4 due Wednesday, October 21 st in class (no later than noon) There are still assignments #1 and #2 in my office to be picked up... If you would like to get a copy of the solutions for assignment #1,#2,#3 (after Monday s tutorial), #4 (after class next Friday), you can come by my office and make a copy - they are in my handwriting I will offer an extra tutorial during reading week, Monday October 26 th at 10:30 am in F055
Rules for midterm Calculators are allowed, but without programmable options Provided formula sheet (on webpage) can be printed out and used The only constant you have to know is the acceleration due to gravity on earth (g = 9.80 m/s 2 ). Other s will be provided. Time to complete exam is 90 min Graphs and sketches can be made in pencil, otherwise pens should be used. In case of sickness a doctors note must be provided
Basic math These are assumed to be known - some where discussed in class - other s are given in your textbook Re-arranging equations Pythagorean theorem Area of basics geometric forms (square, triangular, rectangular, trapeze, circle Vectors Trigonometry (sin, cos, tan)
Format of midterm There will be a number of multiple choice questions (or short questions) testing your understanding of the concepts. These should not take a lot of time (1-2 min each) and don t require complicated calculations. The rest will be very similar to the assignments (especially #3 and #4) As in the assignment, you will find the possible points for each single one given
What can you do to prepare? Look at the lecture notes (examples) Look at the corresponding chapters in the textbook (1-10) Make sure you understand the assignments Do some of the problems in the textbook (see selection on lecture note webpage) Talk to each other to see if you open questions and the same understanding of the concepts If you have questions come by my office or email
Sample of multiple choice 1.) Power is defined as a) Time multiplied by work b) Time divided by work c) Work divided by time d) Work times Work 2.) Potential gravitational energy has a dependence on the mass of an object that is a) Quadratic b) Linear c) A sinus function d) no dependence 3.) When an object moves with constant velocity on the x axis its acceleration is a) constantly changing b) also constant c) twice a big as its velocity d) non existent 4.) The area of an triangle is given by a) half the base times the height b) two time the base times the height c) the base times the base d) the height times the height 2.) The unit for work is a) Watt b) Joule c) Newton d) Workunit 3.) When an object is in equilibrium the sum of the externally applied forces is a) bigger than the internal forces b) smaller then the internal forces c) zero d) not of interest
Sample of multiple choice 1.) Power is defined as a) Time multiplied by work b) Time divided by work c) Work divided by time d) Work times Work 2.) Potential gravitational energy has a dependence on the mass of an object that is a) Quadratic b) Linear c) A sinus function d) no dependence 3.) When an object moves with constant velocity on the x axis its acceleration is a) constantly changing b) also constant c) twice a big as its velocity d) non existent 4.) The area of an triangle is given by a) half the base times the height b) two time the base times the height c) the base times the base d) the height times the height 2.) The unit for work is a) Watt b) Joule c) Newton d) Work unit 3.) When an object is in equilibrium the sum of the externally applied forces is a) bigger than the internal forces b) smaller then the internal forces c) zero d) not of interest
Sample of short questions 1.) What does Newton s third law say? 2.) What is the difference between a scalar and a vector quantity? 3.) Show that the unit Newton is kg * m/s 2! 4.) Draw the path of a stone that is thrown horizontally from a 25m high building. 5.) What is the displacement of a speedboat that accelerates for 3s with 2m/s 2?
1.) What does Newton s third law say? Action = Reaction (Whenever a body exerts a force on a second body, the second body exerts and oppositely directed force of equal magnitude on the first body) 2.) What is the difference between a scalar and a vector quantity? A scalar quantity has only a magnitude, while a vector quantity has both direction magnitude. 3.) Show that the unit Newton is kg * m/s 2! Newton is unit for force. Force is given as mass times acceleration. The unit for mass is kg and the unit for acceleration is m/s 2, therefore Newton is defined as kg m/s 2. 4.) Draw the path of a stone that is thrown horizontally from a 25m high building. The path is described by a parabola 5.) What is the displacement d of a speedboat that accelerates for 3s with 2m/s 2? v [m/s] 6 4 2 t [s] 1 2 3 d = 1/2 * 3s * 6m/s = 9 m
Note on units Units are very important - they have to be correct and are a useful tool for checking your calculation The unit calculations are part of problem solving and you will loose points, if they are missing
Topics for midterm Kinematics (one and two dimensions) Dynamics (one and two dimensions) Energy, Work, Power, Momentum Conservation of energy and momentum We will go through every one today and look at some example problem step-by-step
Kinematics o Kinematics describes the motion of objects o Important vector quantities are: - displacement d - velocity v - acceleration a o For constant acceleration we have a special set of equation o Motion in perpendicular directions can be calculated independently o circular motion can be described similar to linear motion by replacing velocity v with angular velocity ω, acceleration a by angular acceleration α and displacement d by angular displacement Θ o Projectile motion o Freely falling bodies
How to tackle a kinematics problem Ask yourself a few questions Is there acceleration and is it constant? No acceleration means, you only have to look at velocity, time and displacement according to v = d/t For constant acceleration we had a special set of equations (see next slide) Is the motion linear or circular? How many dimensions are involved?
Reminder: for details see lecture Wednesday October 7th Rotational Motion (α constant) ω = ω0 + αt Θ = 1/2 (ω0 + ω) t Linear Motion (a constant) v = v0 + at d = 1/2(v0 + v)t Θ = ω0t + 1/2αt 2 d = v0t + 1/2at 2 ω 2 = ω0 2 + 2αθ v 2 = v0 2 + 2ad
Example A baseball player hits a home run, and the ball lands in the left field seats, 7.5 m above the point at which it was hit. It lands with a velocity of 36 m/s at an angle of 28 below the horizontal. Ignoring air resistance, find the initial velocity with which the ball leaves the bat. This has two dimensions - the ball changes height during the movement as well as covers ground There is no rotation, therefore we have only translation (linear motion) There is acceleration due to gravity, so we can use our special set of equations Let s define the vertical as y direction and the flight direction between the bat and the seats (where the ball lands) as x direction. Let s make a drawing: +y v0y =? vx +x Θ y = 7.5m vy 28 v = 36 m/s v0x To find the initial velocity, we must determine its magnitude v0y and direction θ. These components are related to the horizontal and vertical components of the initial velocity: v0 = sqrt [v0x 2 + v0y 2 ] and θ = tan -1 (v0y/v0x) Since air resistance is being ignored the horizontal component of the velocity vx remains constant. Thus v0x = vx = v cos28 = +32 m/s
The value for v0y can be obtained from: vy 2 = v0y 2 + 2ayy (see table) v0y = sqrt[vy 2-2ayy] y = 7.5 m, ay = -9.80 m/s 2, vy = -36 sin28 m/s Using these numbers, we obtain: v0y = 21 m/s We choose the positive sign because the initial velocity points upwards. We can now determine the magnitude of the initial velocity by using our results for the x and y components: v0 = sqrt [v0x 2 + v0y 2 ] = 38 m/s and the direction: θ = tan -1 (v0y/v0x) = 33
How to tackle a dynamics problem Dynamic problems involve forces! Draw a free body diagram Are all forces acting along one axis? If we need more than one dimension, what are the components? Remember that an acceleration must be caused by a force Can we make use of Newton s law?
Example A truck is hauling a trailer along a level road. The mass of the truck is m1 = 8500 kg an that of the trailer is m2 = 27000 kg. The two move along the x axis with an acceleration of ax = 0.78 m/s 2. Ignoring the retarding forces of friction and air resistance, determine (a) the tension T in the horizontal drawbar between the trailer and the truck and (b) the force D that propels the truck forward. Trailer Truck ax = 0.78 m/s 2 T +x T D +x Only components along the x axis are sufficient to solve the problem. Newton s second law tells us that the sum of all forces in x direction is equals the mass times the acceleration in x direction: ΣFx = m ax The trailer: T = m2 ax = 27000 kg * 0.78 m/s 2 = 21000 N (a) The truck: D - T = m1 ax According to Newtons third law T must be equal to T in magnitude. D = m1 ax + T = (8500 kg)(0.78 m/s 2 ) + 21000 N = 28000 N (b)
Work, Energy, Power Work, Energy and Power are scalar quantities. Let s review the definitions: Work done on an object by a constant force is W = (F cosθ)s, where F is the magnitude of the force, s the magnitude of the displacement and Θ the angle between the force and the displacement. The unit of work is Nm = J (Joule) Mechanical energy is the sum of potential energy (gravitational) and kinematic energy. Potential energy (gravitational) PE is stored in an object just because of its location above the ground - it is linear dependent of mass and height of the object and can be written as PE = mgh. Kinematic energy KE is energy stored in an object due to its movement, it is linear dependent on mass and quadratically dependent on the velocity of the object KE = 1/2mv 2. Work and Energy are related: The work done by the net external nonconservative forces is given by the difference between total mechanical energy in the end and in the beginning: W = Ef - E0 (Work-Energy-theorem) Power is the average rate at which work is done, and it is obtained by dividing work by time. Since work is given by force times distance, we can also find that power is force times velocity.
Example A 0.20 kg rocket in a fireworks display is launched from rest and follows an erratic flight path to reach the point P. Point P is 29 m above the starting point. In the process, 425 J of work is done on the rocket by the nonconservative force generated by the burning propellant. Ignoring air resistance and the mass lost due to the burning propellant, find the speed vf of the rocket at the point P. The only nonconservative force acting on the rocket is the force generated by the burning propellant, and the work done by this force is W = 425 J. We can use the work-energy theorem to find the final speed. W = Ef - E0 = (1/2mvf 2 + mghf) - (1/2mv0 2 + mgh0) The rocket starts from rest, therefore v0 = 0. It starts at h0 - we define that as our zero. As a result we only have to take the final energy into account an can re-arrange for vf: W = 1/2mvf 2 + mghf --> vf 2 = 2(W - mghf)/m vf = sqrt {2(W - mghf)/m} = sqrt {2(425 J - 0.20kg * 9.80 m/s 2 * 29m)/ 0.20 kg} = 61 m/s
Conservation of energy and momentum Example: A ballistic pendulum can be used to measure the speed of a projectile, such as a bullet. The ballistic pendulum consists of a block of wood (mass m2 = 2.50 kg) suspended by a wire of negligible mass. A bullet (mass m1 = 0.0100 kg) is fired with a speed v01. Just after the bullet collides with it, the block (with the bullet in it) has a speed vf and then swings to a maximum height of 0.650 m above the initial position. Find the speed v01 of the bullet, assuming that air resistance is negligible. m1 = 0.0100 kg m2 = 2.50 kg hf = 0.650 m m2 m1 + m2 hf v01 vf There are two parts to this event. First we have a completely inelastic collision between the bullet and the block. The second is the resulting motion of the block and bullet as they swing upward. The total momentum is conserved during the collision. For the swinging upwards we also can apply the principle of energy conservation. Momentum conservation: (m1+ m2) vf = m1v01 --> v01 = (m1+ m2)/m1 * vf Use energy conservation to determine vf : (m1+ m2)ghf = 1/2(m1+ m2)vf 2 vf = sqrt(2ghf) Now insert this in equation above: v01 = (m1+ m2)/m1 * sqrt(2ghf) = (0.0100 kg + 2.50 kg)/0.0100 kg * sqrt(2*9.80m/s 2 *0.650m) = 896 m/s
Formula sheet Not quite ready - it will be posted on Monday