STRAIGHT-LINE MOTION UNDER CONSTANT ACCELERATION

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1 STRAIGHT-LINE MOTION UNDER CONSTANT ACCELERATION Problems involving a body moving in a straight line under constant acceleration have five relevant variables: u = Initial velocity in m/s v = Final velocity in m/s a = Acceleration in m/s/s r = Displacement t = Time in s If any three of these variables are known, substitution into an appropriate formula can solve for any one of the remaining quantities. Five useful formulae are: v = u + at r = ut + 1 at r u + v = t v = u + ar r = vt 1 at When attempting these types of questions, write down the variables provided in the question (there should be at least three variables supplied) and the variable to be determined. Then select the formula with the relevant components. In all quantitative questions assume zero air resistance! EXAMPLE 5 A cannon ball is dropped from the top of a building and it hits the ground after three seconds. How high is the building? Solution There are three known values but they may not all be obvious. Since the ball is dropped it can be inferred that initial velocity (u) is zero. It is also assumed knowledge that acceleration near the Earth s surface is 9.8 ms -. The time for the descent is 3.0 s. Relevant variables are u, a, t & r. The appropriate formula is: r = ut + 1 at 1 r = r = m The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 1

2 EXAMPLE 6 A sandbag is dropped from a hot air balloon that is rising directly upwards at 6 ms -1. The point of release is 33 m above the ground. Assume the acceleration due to gravity is 9.8 ms -. What is: (b) The velocity of the sandbag when it hits the ground? The time the sandbag is in the air? Solution As always, first identify the known and required quantities. In this example you will also need to distinguish up from down. i.e.: u = 6 m/s a = -9.8ms - r = -33 m v =? t =? (The negative symbol indicates that acceleration is downwards.) (Again the negative sign tells us that the displacement is downwards.) (We will be expecting a negative value as it will be travelling downwards when it hits the ground.) The relevant formula is: v = u + ar v = 6 + ( x -9.8 x 33) v = 68 v = 6.1 ms -1 Of course we all know that the complete answer is 6.1 ms -1 (or 6.1 ms -1 down) as it will be travelling downwards at the point of impact. Direction needs to be described in the answer because velocity is a vector. (b) Now to determine the time the object is in flight. Let s use: v = u + at (however there are other formulae that could be used as we now know four of the quantities) -6.1 = x t t = (-6.1 6) / -9.8 t = 3.8 seconds In many questions on motion if you don t correctly assign the positive and negative signs to designate up and down it is very likely that the final answer will be wrong. Don t forget that up is usually positive. You must state this to ensure it is clear what you are doing. The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page

3 QUESTION 1 A car accelerates away from the traffic lights uniformly 5 ms -. How fast is it travelling at t = 4 s? (b) What is the car s displacement at t = 4 s? (c) How far does the car travel in the 1st second of motion? (d) How far does the car travel in the nd second of motion? QUESTION 13 A small furry object is dropped from the top of a 150 m high building. Calculate: The time taken for the small furry object to reach the ground. (b) The velocity with which the small furry object hits the ground. The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 3

4 QUESTION 14 A stunt woman is shot from a cannon at the top of a 0 m high platform. She has an initial vertical velocity of 0 ms -1 and she lands on a safety net at ground level sometime after launch. Calculate: Hint: As in many projectile questions there are two parts to the stunt woman s motion, the upwards part and the downwards part. It is often helpful to treat these two parts separately. The time taken to reach maximum height. (b) The maximum height reached (above the ground). (c) The total time of flight. The velocity with which the stunt woman hits the safety net located on the ground below the tower. The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 4

5 NEWTON S LAWS OF MOTION Newton s Laws of Motion should be well understood by now as they were covered in the Preliminary course. Here is a brief summary of these laws. NEWTON S FIRST LAW The velocity of an object will remain constant unless a net force acts on the object. If at rest, an object will continue in this same state of rest. If an object is in motion with an eastward velocity of 8 m/s, it will continue in this same state of motion (8 m/s, East). If in motion with a leftward velocity of 0 m/s, it will continue in this same state of motion (0 m/s, left). The state of motion of an object is maintained, and no force needs to be applied in order for an object to keep moving with a constant velocity. In fact, a force must be applied in order to slow it down (or speed it up or change its direction). The term inertia is often used to describe this phenomenon. Some examples of this include: The sensation of being thrown sideways when the bus you are travelling in turns a bend. Being jolted backwards when the train departs the station. The old magician s trick of pulling the tablecloth from the table set with cutlery and crockery. Can you think of some more examples? NEWTON S SECOND LAW If a net force acts on an object, the acceleration that is produced is directly proportional to the force. F net = ma The acceleration of a body is proportional to the applied force. The acceleration is also in the direction of the applied force (remember, force and acceleration are both vectors). The unit of force is the Newton (N). A force is usually drawn as an arrow where the size of the arrow is reflective of the magnitude of the force and the direction of the arrow reveals the direction in which the force is acting. Since forces are vectors, the effect of an individual force upon an object is often cancelled by the effect of another force. The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 5

6 NEWTON S THIRD LAW This is the most difficult of the laws to understand: If Object A exerts a force on Object B, then Object B will exert a force of the same size but in the opposite direction on Object A. It can also be expressed as: For every action, there is an equal and opposite reaction. The two forces (action and reaction): Are equal in magnitude; Act in opposite directions; and Act on separate objects (i.e. one force acts on Object A, the other force acts on Object B). The last point is very important it is the one that students most often forget. Some common misunderstandings associated with this law include: 1. If the forces are equal and opposite then they cancel each other out and nothing happens. This is wrong because it fails to recognise that the equal and opposite forces act on separate objects so the forces on each object are not necessarily balanced.. Inanimate objects can t push (or pull). An object does not need to be alive or moving to apply a force to another object. Your chair is pushing you up against your weight. If you are heavier than the person next to you then your chair is pushing you harder. The chair does not need to be moving to push you, nor does it need to be able to calculate your weight in order to push back with the right force. It just applies an equal and opposite force to the one being applied to it. Some examples include: Forces of repulsion between two north magnetic poles. Force of gravity between you and Earth. When the garden hose is turned on too hard it thrashes backwards. What action/reaction pairs of forces are involved in walking? Walking requires a degree of friction between your foot and the ground. When walking in a northerly direction, in what direction does the friction of ground act on your feet? The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 6

7 QUESTION 15 A sports car of mass 900 kg and initial velocity of 50 ms -1 North brakes heavily and comes to rest in a time of 3 s. Calculate: (b) The magnitude and direction of the force that was exerted on the car by the ground. The magnitude and direction of the force that was exerted on the ground by the car. Solution QUESTION 16 A 1 kg trolley, which is initially at rest, is acted on by a force of 0 N. Calculate the distance that it will move in a time of s. Solution The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 7

8 PROJECTILE MOTION A projectile is any object that is moving under the influence of gravity alone (ignoring any air resistance). Examples of projectiles include: Objects thrown or dropped through the air. A satellite in orbit above the atmosphere of a planet or moon. A force is often involved in initially setting the object in motion, while it is moving through the air, no force other than gravity, acts on it (remember, we are ignoring air resistance). EXAMPLE 7 Explain whether a rock and an aeroplane are both projectiles or not. Solution A rock can be a projectile when thrown into the air but an aeroplane cannot because planes have engines providing thrust and wings to provide a lift force. QUESTION 17 Which of the following is not a projectile? A B C D A golf ball in flight A high diver after leaving the platform A rocket during launch A pen falling off a table The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 8

9 Galileo was the first person to successfully analyse and describe the motion of projectiles, as well as formulate a law of inertia. He published his work in the mid 1600 s. He was able to show that the path, or trajectory, of a projectile is curved (it is, in fact, parabolic). Galileo s work was later built upon by such scientists and mathematicians as Kepler and Newton. Galileo realised that: The horizontal velocity of a projectile is independent of the vertical velocity. The horizontal velocity of a projectile does not change during its motion. Only the vertical component of the velocity changes and it increases exponentially, with time, towards the Earth. v trajectory projectile mg At any given point in the motion, the velocity vector is always a tangent to the path. Note: The vector mg = Weight force = The only force acting on the object = Net force This net force does not change at all throughout the motion of the projectile. The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 9

10 Here is something to remember about net forces: If the net force is in the same direction as the velocity of the object, the object speeds up: v F net object speeds up If the net force is in the opposite direction to the velocity of the object, the object slows down: v F net object slows down But, if the net force acts at right angles to the velocity vector, then the speed of the object in the direction of that vector does not change. v speed in the direction of v remains constant F net The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 30

11 Now consider the following diagram: v y v m v x mg = F net Notice that the velocity vector, v, has two components: A horizontal component, v x. A vertical component v y. The net force, mg, has: An effect on v y. No effect on v x. Thus, there is no acceleration horizontally (which is to say that v x remains constant throughout the motion) but there is indeed a vertical acceleration. The next diagram shows the path taken by a projectile that has been thrown horizontally. The position of the projectile is shown at equal time intervals. Notice that it travels at constant velocity horizontally (for it covers equal distances in equal time intervals) but it is accelerating vertically (it covers greater distances in successive equal time intervals). As one would expect, it is moving at constant speed horizontally, but it is speeding up vertically. The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 31

12 The diagram below shows how the velocity vector changes as the projectile moves along its trajectory. Also shown are the horizontal and vertical components of the velocity vector. Of course, the horizontal component stays constant but the vertical component changes. The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 3

13 We usually handle projectile motion problems by breaking up the motion into horizontal and vertical components. The vertical component of the motion is independent of the horizontal component. The formulae in below appear in the data sheet that will be given in your HSC Physics exam. For the horizontal component, we use: Δx = u x t v = Δx t For the vertical component, we use the constant acceleration formulae: v = u + at v y = u y + a y Δy Δy = u y t + ½ a y t *Useful but not supplied on data sheet: u y + v Δy = y t The meanings of the symbols above and their units are as follows: Note: The following list of definitions is assumed knowledge and will not be given on your data sheet in exams. v x = final velocity in the horizontal direction (m/s) u x = initial velocity in the horizontal direction (m/s) Δx = horizontal displacement, sometimes called the range (m) t = time (s) v y = final velocity in the vertical direction (m/s) u y = initial velocity in the vertical direction (m/s) Δy = vertical displacement (m) a y = acceleration in the vertical direction = acceleration due to gravity *On the data sheet, the acceleration due to gravity is given as g = 9.8 ms - The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 33

14 EXAMPLE 8 A dare devil motorcyclist recklessly rides off the top of a 180 m high building with an initial horizontal speed of u = 3 m/s. 0 m/s u 180 m Calculate: (b) (c) The time of flight. The horizontal distance from the base of the building to the point where the dare devil hits the ground. The velocity with which the dare devil hits the ground. Solution Here, we are only interested in the vertical component: u x = 0 a y = -9.8 m/s t =? Δy = -180 m (i.e. if an object, initially at rest, accelerates at -9.8 m/s, at what time is its displacement -180 m?) Δy = u y t + ½ a y t as u y = 0 m/s, this becomes ( 9. ) 180 = t 1 8 t = -180 x -9.8 t = 6.06 s (b) Now we are only interested in the horizontal component: Δx v = t 3 = x/6.06 x = m The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 34

15 (d) At the point of impact, the bike and rider are moving thus: path V The vector, V, represents the final velocity. It is tangential to the path and has both horizontal and vertical components: V V V V H The horizontal component, V x, is obviously 3 m/s. To find the vertical component, V y, proceed thus: Vertical component: U y = 0 a y = -9.8 m/s v y =? Δy = -180 m v = u + aδy v = v y = 59.4 m/s ( )( ) Or (this is better), knowing that t = 6.06 s, find v by using v = u + at. The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 35

16 We now know that V y = 59.4 m/s. We can therefore draw the appropriate diagram, thus: 3 V 59.4 and use some simple trigonometry to find: v = (3 ) + (59.4) v = 63.7 m/s θ = tan-1(3 / 59.4) θ = 1. ο Therefore v = 63.7 m/s at 1. o to the vertical (or 68.8 o to the horizontal) QUESTION 18 A golf ball is whacked at an initial velocity of 100 m/s at an angle of elevation of 30 from the top of a 10 m high tower. Calculate: The time taken to reach maximum height. (b) The maximum height reached (above ground level). The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 36

17 (c) The time of flight. (e) The horizontal distance from the base of the tower to the point where the ball hits the ground. (f) The velocity with which the ball hits the ground. The School For Excellence 016 Trial Exam Revision Lectures Physics Book 1 Page 37