Combined Notes - Unit 1 Edexcel Physics A-level PDF

            April 2001 Number 13 Motion I – Uniform Acceleration This...

            April 2001 Number 13 Motion I – Uniform Acceleration This Factsheet will cover :
the basic definitions of speed, velocity and acceleration
the use of the equations of motion for uniform acceleration
the application of the equations of motion to projectile motion 1. Basic Concepts It is important to note when using these equations that u, v, a and s can be negative as well as positive. Displacement is the distance and direction of a body relative to a  Positive and negative values of displacement (s) refer to positions given reference point (usually the point at which it starts). As it has each side of the starting point – for example, if a positive both a length and a direction, it is a vector. Its SI unit is the metre displacement refers to positions above the starting point, then (m). (eg “200 metres east of my house” is a displacement) negative ones will refer to positions below it.  Positive and negative values of velocity (u or v) refer to its Velocity is the rate of change of displacement. Accordingly, direction. For example, if you throw a ball up in the air, then if its velocity is also a vector. Its SI unit is metres per second (ms-1). initial velocity – when going up – is positive, then its final velocity – (eg 3 ms-1 north is a velocity) when it is coming down – will be negative.  Any acceleration that has the opposite sign to the velocity will act total change in displaceme nt as a retardation – in other words it will slow the body down. So, if Average velocity = the velocity is positive, a retardation will be negative. time taken Approach to problems using equations of motion Acceleration is the rate of change of velocity – so, again, is a 1. Check that the body is moving with constant acceleration! vector. Its SI unit is metres per second2 (ms-2). 2. Write down any of u, v, a, s, t that you know 3. Note down which of u, v, a, s and t that you want (e.g. write a = ?) total change in velocity 4. Decide which equation to use by looking at which of the variables Average acceleration = time taken you have got written down in steps 2 and 3. For example, if you have got values for u, t and s, and you want a value for v, then you look for the equation with u, t, s and v in it. 5. Substitute the values you know in, then rearrange. 6. Check that the answer makes sense. Speed and velocity. Speed is the rate of change of distance. Example 1. A particle is moving in a straight line with constant acceleration. It passes point P with speed 2 ms-1. Ten seconds later, it The speed of a body at any instant is equal to the magnitude of its passes point Q. The distance between P and Q is 40 metres. Find the velocity at that instant. However, this is not generally the case for speed of the particle as it passes point Q. average speed and average velocity. We know: u = 2ms-1 s = 40 m t = 10 s To see why this is, imagine walking 5 m North in 2 s, then 5 m South in We want: v = ? 2 s. Since you end up at the point where you started, your overall displacement is zero – so your average velocity is zero. However, you Since we have u, s, t and v involved, use s = ½ (u + v)t have travelled a distance of 10 m in 4 s, so your average speed is: Substituting in: 10
4 = 2.5 ms-1. 40 = ½ (2 + v)10 40 = 5(2 + v) The average speed will only be the magnitude of the average velocity if 40 = 10 + 5v the body concerned is moving in a straight line, without reversing its 30 = 5v direction – since then, the distance it moves will always be equal to the v = 6 ms-1 magnitude of its displacement. Example 2. A particle is moving in a straight line with constant acceleration 0.2 ms-2. After it has moved a total of 20m, its speed is 8ms-1. Find its initial speed. 2. Equations of motion for uniform acceleration a = 0.2 ms-2 s = 20m v = 8 ms-1 u = ? v = u + at u = initial velocity So use v2 = u2 + 2as v2 = u2 + 2as v = final velocity s = ut + ½at2 a = acceleration 82 = u2 + 2(0.2)(20) s = ½ (u + v)t s = displacement from the starting point 64 = u2 + 8 t = time 56 = u2 u = 7.48 ms-1 (3SF) 1 Motion I – Uniform Acceleration
    Tips: c) u = 20 ms-1 s = 0 (since it has returned to ground) a = -10ms-2 1. It is usually easier to put the values in before rearranging the t=? equations. 2. Take particular care with negative values. On many So we use calculators, if you type in –22, you will get the answer –4, s = ut + ½ at2 rather than the correct value of 4. This is because the 0 = 20t + ½ (-10)t2 calculator squares before “noticing” the minus sign. 0 = 20t – 5t2 0 = 5t(4 – t) So t = 0 or 4 Example 3. A particle starts from rest and moves with constant We want t = 4, since t=0 is when particle was thrown upwards. acceleration 0.5 ms-2 in a straight line. Find the time it takes to travel a distance of 8 metres. NB: If you are unhappy about the factorising of 20t – 5t 2 used above, then consult Factsheet 15 Maths for Physics: Algebraic Manipulation. u = 0 (as it starts from rest) a = 0.5 ms-2 s=8m t=? Use s = ut + ½at2 Example 5. A child throws a stone vertically upwards from the top of a 8 = 0t + ½ (0.5)t2 cliff with speed 15 ms-1. Five seconds later, it hits the sea below the 8 = 0.25t2 cliff. Taking g = 9.8 ms-2, calculate 32 = t2 a) the velocity of the stone when it hits the sea t =
32 = 5.66 s (3 SF) b) the height of the cliff Vertical motion under gravity We will take upwards as positive If we assume that air resistance can be ignored, then any body moving a) u = 15 ms-1 a = -9.8ms-2 t = 5 s v=? under gravity has acceleration g downwards, where g = 9.81 ms-2. So use v = u + at v = 15 + (-9.8)(5) Since g is constant (for bodies moving close to the earth’s surface, this v = 15 – 49 = -34ms-1 is a good approximation), the equations of motion can be used. The So its velocity is 34 ms-1 downwards (because of the minus sign) same strategy as before should be used, but in addition, the following should be borne in mind: b) u = 15ms-1 a = -9.8ms-2 t = 5 s s=?  Direction is important. You should always decide which direction So use s = ut + ½ at2 you are taking as positive at the start of the problem. You may find s = 15(5) + ½(-9.8)(5)2 it helpful to always take upwards as positive s = 75 – 122.5 = -47.5 m  The acceleration will always be g downwards – or –g, if you are So the sea is 47.5 m below the cliff (minus sign – and we’d expect this!) taking upwards as positive So the height of the cliff is 47.5 metres  The body will carry on going upwards until v = 0  When the body returns to the ground, s = 0 Derivation of the Equations of Motion Some exam boards require you to derive the equations of motion. Exam Hint:- Here are some common mistakes: If yours does, you must learn the following:  Thinking that v = 0 when the body returns to the ground – it isn’t!  Thinking that a = 0 when the body is at its highest point – a change in velocity v  u 1. We know a = = doesn’t change! time t  Ignoring directions – always ask yourself whether a displacement So, multiplying up: at = v – u or velocity is up or down, and so whether it should be put as So, rearranging: u + at = v positive or negative.  Not using the value of g given in the question – you may be told to displacement take it as 9.8, 9.81 or 10ms-2 – and you must use the value you are 2. We know average velocity = time given. But average velocity = ½ (u + v) s So ½ (u + v) = Example 4. A ball is thrown vertically upwards with speed 20 ms-1. t Taking g = 10 ms-2, calculate So ½ (u + v)t = s a) Its velocity after 1.5 seconds. b) The height to which it rises 3. From 1, we know v = u + at. c) The time taken for it to return to the ground. Substituting this into s = ½ (u + v)t, we get: s = ½ (u + u + at)t We will take upwards as positive s = ½ (2u + at)t a) u = 20 ms-1 a = -10ms-2 t =1.5s v=? s = (u + ½ at)t So we use v = u + at s = ut + ½ at2 v = 20 + (-10)(1.5) v = 5ms-1 vu 4. From 1, by rearranging we know t = a b) u = 20 ms-1 v = 0 (as we want the highest point) a = -10ms-2 Substituting this into s = ½ (u+v)t, we get: s=? (v  u) So we use v2 = u2 + 2as s = ½ (u + v) a 0 = 202 + 2(-10)(s) 0 = 400 – 20s (u  v) ( v  u ) (u  v)( v  u ) v 2  u 2 20s = 400 s=   2 a 2a 2a s = 20 m. 2as = v2 – u2 2 Motion I – Uniform Acceleration
    Typical Exam Question: Experimental Determination of g A rocket accelerates from rest for 20s with a constant upward This experiment determines g by finding the time a ball-bearing acceleration of 10ms-2. At the end of 20s the fuel is used up and it takes to fall from rest through a measured distance. The completes its flight under gravity alone. Assuming that air resistance can apparatus is shown below. be neglected and taking g = 9.8ms -2, calculate the: (a) speed reached after 20s. Electromagnet Starting the timer (b) height after 20s. cuts off current to (c) maximum height reached. electromagnet (d) speed just before the rocket hits the ground. Steel ball-bearing Taking upwards as positive: Electronic timer (a) u = 0, a = 10ms-2, t = 20 s, v = ? h Small permanent v = u + at magnet holds the v = 0 + 10(20)
= 200ms-1
Hinge plate against the contact at A (b) u = 0, a = 10ms-2, t = 20 s, s = ? Timer stops when s = ut + ½at2 connection at A s = 0(20) + ½ (10)(202)
A is broken s = 2000 m
Metal plate (c) Need to find height reached while moving under gravity. So start from point where fuel runs out. 2h u = 200ms-1 (from (a)) a = -9.8 ms-2 v = 0 s = ? Since u = 0, we have h = ½ gt2, so g = t2 v2 = u2 + 2as 0 = 2002 + 2(-9.8)(s)
To ensure reasonable accuracy,  the timer must be accurate to 0.01 seconds. 200 2 s= = 2040m (3 SF)
 h is measured from the bottom of the ball-bearing, so the size 2  9.8 of the ball-bearing does not introduce errors. So total height = 2000 + 2040 = 4040m (3 SF)
 the experiment should be repeated a number of times and the (d) Starting from highest point: average found. s = -4040m a = -9.8 ms-2 u = 0 v = ?  the current in the electromagnet should be reduced to the v2 = u2 + 2as minimum that will hold the ball-bearing. This reduces the v2 = 2(-4040)(-9.8) = 79184
chance of a delay in the ball-bearing being released. v = 281 ms-1 (3 SF)
3. Projectiles – motion in two dimensions under gravity Example 1. A vase is thrown out of a first floor window, which is 5 m If you throw an object, it follows a parabolic path (shown below). above the ground, with a horizontal velocity of 4ms-1. Taking g = 10ms-2, find a) The time taken for the vase to hit the ground b) The horizontal distance it travels. c) Its speed as it hits the ground. We will take upwards as positive a) Since this involves the vertical position of the vase, we must consider the vertical motion Vertically: u = 0 (since thrown horizontally) a = -10ms-2 To deal with this situation, we consider the horizontal and vertical s = -5m (since it is going downwards) t = ? components of the motion seperately. Again, we assume that air Using s = ut + ½ at2: resistance can be neglected -5 = 0(t) + ½ (-10)t2  horizontally, there is no resultant force on the object, so its -5 = -5t2 velocity is constant (and there is no acceleration) t = 1 second.  vertically, gravity is the only resultant force. So its acceleration is g downwards b) We must consider horizontal motion u = 4 ms-1 a = 0 t = 1 s (from a)) s = ? The general approach to problems is very similar to that used in the previous section, but the following should be borne in mind: Using s = ut + ½ at2:  In each part of the question, you must decide whether you need to s = 4(1) = 4m. use the horizontal or the vertical motion  Again, take care with directions and signs c) This is the magnitude of its velocity.  At the highest point, v = 0 vertically We need both horizontal and vertical components.  If it returns to the same level at which it started, s = 0 vertically. Vertically: u = 0 a = -10ms-2 t = 1s (from a))  To find its velocity, you need to find the resultant of its vertical and v = u + at horizontal speeds (see Factsheet 02 Vectors & Forces ) v = 0 + (-10)(1) = -10ms-1  If it is thrown at a speed U and angle  to the horizontal, then
the horizontal component of velocity is Ucos Horizontally, velocity is constant, so v = 4ms-1
the vertical component of velocity is Usin (NB: some exam boards only consider bodies projected horizontally Resultant speed = ( 10)2  4 2 = 10.8 ms-1 or vertically, rather than at an angle). 3  Motion I – Uniform Acceleration
    Example 2.A ball is thrown from ground level with a speed of 20 ms-1 Exam Workshop at an angle of 30o to the horizontal. Taking g = 9.8ms-2, find: This is a typical poor student's answer to an exam question. The a) The greatest height it reaches comments explain what is wrong with the answers and how they b) The time taken for the ball to return to ground level. can be improved. The examiner's answer is given below. c) The horizontal distance the ball travels in this time. A dart player throws a dart horizontally. By the time it reaches Take upwards as positive: the dartboard, 3.00m away, it has fallen a height of 0.200m. a) This is vertical motion. Taking g as 9.81ms -2 , find: u = 20sin30o ms-1 a = -9.8 ms-2 v=0 s=? (a) The time of flight s = ½ gt2 2as = v2 – u2 0.2 = 5t2
2(-9.8)s = 0 – (20sin30o)2 t = 0.2 s -19.6s = -100 Examiner’s comment: The candidate has used a correct method, s = 100  19.6 = 5.10 m but has used g = 10ms-2. Read the question! b) Again, vertical motion as its level is referred to. (b) The initial velocity u = 20sin30o ms-1 a = -9.8ms-1 s = 0 t = ? 3 = u0.2
u= 15ms-1 s = ut + ½ at2 Examiner’s comment: Full marks would have been awarded here 0 = 20sin30ot + ½ (-9.8)t2 as a “follow through” from the error in part a), but for the fact the 0 = 10t – 4.9t2 candidate has not given the direction of the velocity, just its 0 = t(10 – 4.9t) magnitude. t = 0 (not applicable) or 104.9 So t = 2.04 s (3 SF) (c) The magnitude and direction of the velocity as it is just about to hit the dartboard. c) Horizontal motion: v = gt = 2 ms-1

u = 20cos30oms-1 a = 0 t = 2.04 s (from b)) s = ? 15 + 2 = 17ms-1 s = ut + ½ at2 Examiner’s comment: The candidate started correctly, by s = 20cos30o (2.04) = 35 m (2 SF) considering the horizontal and vertical components of the velocity. However, the candidate does not seem to appreciate Questions that the two components must be combined as vectors. No attempt has been made to find the direction – the fact that this 1. Explain the difference between velocity and speed was asked for should have alerted the candidate to the inadequacy of his/her method. 2. A particle is moving in a straight line with constant acceleration. It passes through point A with speed 4ms-1, and 2 seconds later, Examiner’s Answers through point B with speed 3ms-1. Find a) Vertical: s = -0.2 m a = - 9.81ms-2 u = 0 t = ? a) Its acceleration s = ut + ½ at2 b) The distance between A and B -0.2 = -4.905t2
t = 0.202 s
(3 SF) 3. A ball is thrown vertically upwards from ground level with speed b) Horizontally: s = 3m a = 0 t = 0.202s u = ? 25ms-1. Taking g = 10ms-2, find s = ut + ½ at2 a) Its speed when it is 2m above ground level 3 = u(0.202)
b) Its greatest height u = 14.9 ms-1 (3 SF)
4. A ball is thrown horizontally from the top of a cliff with speed c) Vertically: u = 0 t =0. 202 s a = -9.81 ms-2 10ms-1, and later falls into the sea. The cliff is 50m high. v = u + at Taking g = 10ms-2, find v = (-9.81)(0.202)
= -1.98 ms-1
a) The time taken for the ball to reach the sea b) The distance from the bottom of the cliff that it lands 14.9 c) The magnitude and direction of its velocity as it reaches the sea 1.98 5. A ball is thrown from ground level with a speed of 28.2 ms-1 at an angle of 45o to the horizontal. Taking g = 9.81ms-2, find a) The time it takes to reach its greatest height b) The time taken to travel a horizontal distance of 30m magnitude =
(14.92 + 1.982)
= 15 ms-1
(2 SF) at angle tan-1(1.98/14.9)
= 7.6o (2SF) below horizontal
There is a tree 30 m from the point from which the ball was thrown. The ball just passes over the top of it. c) Find the height of the tree Answers 1. See page 1 2. a) a= -0.5 ms-2 b) 7 m Acknowledgements: This Factsheet was researched and written by Cath Brown Curriculum Press, Unit 305B The Big Peg, 120 Vyse Street, Birmingham B18 6NF. 3. a) 24 ms-1 (2 SF) b) 31 m (2 SF) Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a 4. a) 3.2 s (2 SF) b) 32 m (2 SF) registered subscriber. They may be networked for use within the school. No part of these Factsheets may be reproduced, stored in a retrieval system or transmitted in any other form or by any other means without the c) 34ms-1 73o below horizontal prior permission of the publisher. ISSN 1351-5136 5. a) 2.03 s (3 SF) b) 1.50 s (3 SF) c) 18.9 m 4  Physics Factsheet April 2003 Number 52 Displacement-time and Velocity-time Graphs This Factsheet explains how motion can be described using graphs, in The size of the sprinter’s maximum velocity can be obtained from the graph particular how displacement-time graphs and velocity-time graphs can be by calculating the gradient of the second section of the graph, beyond the used. two second point. Displacement-time graphs The gradient is best calculated by drawing a right angled triangle as shown Displacement, plotted on the vertical axis, represents the straight line in the diagram below. The height or ‘rise’ and length or ‘run’ of the triangle distance away from a start point. Time, plotted on the horizontal axis, is are then easily read from the graph and used to calculate the velocity. the time taken after the start. Since velocity = displacement/time, the gradient of a displacement- displacement (m) 100 time graph also represents velocity. The steeper the gradient the larger the velocity. 80 A straight line with a constant gradient will represent an object travelling with constant velocity. 60 rise = 100-5 A curved line with a gradient that changes will represent an object = 95 m travelling with a varying velocity. 40 20 displacement displacement run = 10-2 = 8 s 0 time (s) 0 2 4 6 8 10 constant gradient = constant velocity steeper gradient displacement rise 95 Velocity = = gradient = = = 11.86 ms-1 = larger constant time run 8 velocity Calculating velocity from a displacement-time graph The gradient of a displacement-time graph is equal to velocity. rise time time Velocity = gradient = run displacement Calculating instantaneous velocities from displacement-time graphs increasing gradient Calculating the gradient of a graph that does not have a convenient straight = increasing velocity line portion requires a tangent to be drawn to the curve. (so it is accelerating) Consider the displacement-time graph below, which shows a constantly changing gradient indicating that the velocity of the moving object is constantly changing. time displacement The graph below is a displacement-time for a 100 metre sprinter. The tangent sprinter is slower at the beginning as it takes some time to reach full speed. This is shown by the shallow gradient during the first two seconds of the race, at the start of the graph. As the race progresses the sprinter reaches top speed and is able to maintain this maximum velocity for the rest of the P rise race. This is shown by the gradient of the graph being constant after the first two seconds. run displacement (m) time 100 80 The instantaneous velocity of the moving object at point P will be given by shallow, but the gradient of the curve at this point. increasing gradient Calculating the gradient of the curve at this point is done by drawing a 60 as sprinter constant gradient = tangent to the curve. The tangent is the straight line that just touches the 40 accerates constant, maximum, curve of the graph and has the same gradient as the graph at this point. The velocity gradient of the tangent can then be calculated in exactly the same way as 20 described previously, by forming a large right angled triangle and reading the ‘rise’ and ‘run’ of the triangle. 0 time (s) 0 2 4 6 8 10 1 Displacement-time and Velocity-time Graphs Physics Factsheet Exam Hint: It is a good idea to make the sides of your gradient triangle Velocity-time graphs as long as possible. The reason for this is that a small mistake in a large Velocity, plotted on the vertical axis, represents the velocity moving away number is not significant but a small mistake in a small number could from the start point. The time taken, plotted on the horizontal axis, easily be. Make sure you draw as long a tangent as you can – in order represents the time taken since the start. to make your gradient calculation as accurate as possible. change in velocity Since acceleration = time Look along your tangent, by holding you graph paper up to your eye. the gradient of a velocity-time graph also represents acceleration. You can see how good it is and whether or not it just touches the curve The steeper the gradient the larger the acceleration at one point. A straight line with a constant gradient will represent an object travelling with constant acceleration. A curved line with a gradient that changes will represent an object Calculating instantaneous velocities from displacement- travelling with a varying acceleration. time graphs - The instantaneous velocity can be calculated from a curved displacement-time graph by drawing a tangent to the curve at the velocity velocity steeper gradient place where the velocity is required. The gradient of the tangent to the = greater acceleration curve will be equal to the instantaneous velocity at that point. constant gradient = constant acceleration Typical Exam Question The table of results below were taken for an object being dropped and falling under gravity. Distance fallen (m) 0.0 2.0 4.0 6.0 8.0 10.0 time time Time taken (s) 0.00 0.63 0.99 1.08 1.25 1.40 velocity velocity (i) Plot a graph of distance fallen (on the vertical axis) against increasing increasing = time taken (on the horizontal axis). zero gradient gradient acceleration (ii) Explain why your graph is not a straight line = zero acceleration (iii) Calculate the velocity of the object after 1.00 second. (constant velocity) Answer (i) Graph paper would be supplied with a question like this one. Choose an axis scale that allows the plotted points to fill as much as the graph time time paper as possible. A mark may be deducted if your points don’t fill more than half of the graph paper. The graph below represents the velocity-time graph for a freefalling skydiver. 10 9 velocity constant velocity (terminal velocity) distance fallen (m) 8 tangent as air resistance = weight 7 6 5 4 3 2 rise 1 large acceleration time (s) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 run time (ii) The increasing gradient of the graph shows an increasing velocity, The gradient is initially large as the skydiver is accelerating with the in other words acceleration. ! acceleration due to gravity. The acceleration is caused by the gravitational force acting on the The gradient of the graph gradually decreases showing the acceleration of object. ! the skydiver to be decreasing as the air resistance on the skydiver increases. Eventually the air resistance on the skydiver is equal to his weight; there (iii) The graph shown as the answer to part (i) has a tangent drawn at is no resultant force so there is no acceleration. The skydiver falls at a time of 1.00 second. constant velocity, shown by the horizontal line on the graph; zero gradient Velocity = gradient of tangent ! implies zero acceleration and constant velocity. rise 9.2 -0 = = = 9.5 m/s ! The size of the initial acceleration of the skydiver can be determined by run 1.4 -0.5 ! calculating the initial gradient of the graph. This is done in exactly the same way as for any other graph, by taking a rise and run from the graph, as we looked at with displacement – time graphs, using a right angled triangle drawn on the graph. Calculating acceleration from a velocity-time graph The gradient of a velocity-time graph is equal to acceleration. The height or ‘rise’ and length or ‘run’ of a part of the graph is measured. change in velocity rise Acceleration = = gradient = time run 2 Displacement-time and Velocity-time Graphs Physics Factsheet The total displacement during a journey can also be calculated from a velocity-time graph. The area beneath the line on a velocity-time graph Velocity is a vector gives the total displacement. Velocity is a vector. A vector is a measured quantity that is described by a magnitude, (or size), and a direction. The graph below is a velocity-time graph for an accelerating car. The graph For motion along a straight line this means that moving in one direction is a straight line showing that the car has constant acceleration. The along the line will be a positive velocity and moving in the opposite displacement of the accelerating car after this 5 second journey can be direction will be called a negative velocity. determined by calculating the area beneath the graph. The line of the graph forms a triangle with the horizontal axis so the area Displacement-time graph and velocity-time graph for a bouncing ball of the triangle can be calculated. The displacement-time and velocity-time graphs for a bouncing ball are specifically mentioned in several A-level specifications. The two graphs velocity (m/s) below are for a ball that is initially dropped from someone’s hand and allowed to bounce on the floor. 30 displacement time (s) 5 time Total displacement = area beneath graph = area of triangle shape = ½ × base × height = ½ × 5 × 30 = 75m velocity first bounce second bounce Calculating the total displacement from a velocity-time graph The total displacement is equal to the area beneath the line on a velocity- time graph for the time considered. Exam Hint: The majority of velocity-time graphs that will be used in an time exam will consist of sections of constant acceleration or constant velocity. This means that the graph can be split into a series of triangles and rectangles when calculating the area beneath the graph. Velocity as a vector Velocity is a vector quantity. This means that velocities are described by Displacement-time graph two things; the size or magnitude of the velocity and the direction of the Zero displacement is defined as the floor. velocity. The gradient of the displacement-time graph is velocity. The gradient The size of the velocity is simply described by a number with a unit of the graph is negative and becomes increasingly large as the ball falls in the usual way, e.g a cyclist moving at 5 ms-1. and speeds up. The direction of the velocity in journeys that can only go back and forth When the ball hits the ground, it bounces back up and the gradient in a straight line is described by adding a sign to the size of the velocity. becomes positive. A positive sign would mean travelling in one direction and a negative The gradient then decreases until the ball is at the top of its path. sign would mean travelling along the same line but in the opposite The ball then drops downwards once more. direction. Therefore, a swimmer who is swimming lengths of the pool, there and back, would have a positive velocity while swimming to the Velocity-time graph far end of the pool but a negative velocity while swimming back to the The ball is dropped from rest and so the initial velocity is zero. start. Velocity downwards has been given a negative sign and so the velocity Velocity-time graphs can also show negative velocities by having then becomes a bigger negative number as the ball accelerates negative values plotted on the vertical axis. downwards. The swimmer, swimming at a constant velocity of 0.50 ms-1, would have The gradient of the graph is acceleration and this is constant a velocity-time graph as shown below. The velocity switches from at –9.81 ms-2 as this is acceleration due to gravity. positive to negative as the swimmer turns around and starts to swim in the When the ball bounces it rapidly comes to a stop before bouncing back, opposite direction along the pool. upwards, with a positive velocity. The ball will then slow down until, at the top of its path, it will velocity (m/s) instantaneously have zero velocity before heading back towards the ground. 0.5 time (s) -0.5 3 Displacement-time and Velocity-time Graphs Physics Factsheet Exam Workshop This is a typical poor student’s answer to an exam question. The (c) Without any calculation state what the area beneath your comments explain what is wrong with the answers and how they velocity-time graph represents and what the value should be. can be improved. The examiner’s answer is given below. Area beneath the graph represents length of race and it should be 500m The graph below represents the displacement of a drag racing car along a straight track. The candidate would be awarded both marks for this part of the displacement (m) question but more detail could have been given for the first part of the 500 500 answer by mentioning the total displacement of the car from the start position. 400 400 Examiner’s Answer ! rise 400-100 300 300 (a) (i) velocity = gradient of graph = = = 37.5 ms-1! run 16-8 200 200 (ii) Instantaneous velocity = gradient of tangent rise 115-0! = = = 14.4 ms-1 ! 100 100 run 10-2 Please note that actual numbers for rise and run will vary 0 0 time (s) depending on the size of the line used to calculate the gradient but 00 22 44 66 88 10 10 12 12 14 14 16 16 18 18 20 20 the final answers should all be very similar. (a) (i) Calculate the instantaneous velocity of the car 12 seconds (b) velocity (m/s) after the start. displacement 250 −1 velocity = = = 21 ms time 12 40 The student has simply substituted values of displacement and time 35 from the point at 12 seconds on the graph. No attempt has been made 30 to determine the gradient of this straight line portion of the graph. 25 (ii) Calculate the velocity of the car 6.0 seconds after the start. displacement 50 20 velocity = = = 8.3 ms−1 time 6 15 Again, the values from the point at 6 seconds have been substituted. 10 The graph is a curve at this point and a tangent should be drawn on the graph in order to calculate an instantaneous gradient. 5 (b) On the axes below sketch a velocity-time graph for the car over 0 time (s) the same period of time. 0 2 4 6 8 10 12 14 16 18 20 (c) The area beneath the graph represents the total displacement of the car, which is the distance the car has travelled along the straight track.! The area beneath the graph should be the total displacement given on the displacement-time graph = 500m.! Even though the question says ‘sketch’ values should be placed on the vertical axis as we have just calculated 2 points from the first part of the question. Typical Exam Question (b) Over what distance did the sprinter race? The graph below is an idealised velocity-time graph for a sprinter. (c) What was the average velocity of the sprinter for the entire race? velocity (m/s) 10 Answer 9 change in velocity ! rise 10 ! 8 (a) acceleration = = gradient = = = 5 ms-2.! time run 2 7 (b) The displacement of the sprinter will give the length of the race. 6 5 The area beneath the graph gives the displacement. ! 4 The graph can be split up into a triangle for the first 2 seconds and a 3 rectangle for the final 9 seconds. 2 The total displacement will be given by the sum of the two areas. 1 0 time (s) Displacement = total area beneath the graph ! = (½ × 2 × 10) +! (9 × 10) = 100 m ! 0 1 2 3 4 5 6 7 8 9 10 11 total displacement 100 (c) average velocity = = = 9.1ms-1 ! (a) What is the initial acceleration of the sprinter? total time taken 11 ! 4 Displacement-time and Velocity-time Graphs Physics Factsheet Qualitative (Concept) Test Quantitative Test Answers 1. What does the gradient of a displacement-time graph represent? 1. (a) In the final minute of the dive. 2. How would the gradient of a curved graph be calculated? (b) 4 minutes 3. What does the gradient of a velocity-time graph represent? rise 18 (c) velocity = gradient = = = 0.15 ms -1 4. What does the area beneath a velocity-time graph represent? run 2 × 60 5. What is a vector and how is the vector nature of velocity in a straight change in velocity line shown? 2. (b) acceleration = time taken 6. Sketch the displacement-time and velocity-time graph of a bouncing (15-0) ball and label the important features of both. = = 0.50 ms -2 (30-0) Quantitative (Calculation) Test change in velocity rise (c) acceleration = = gradient = 1. The graph below represents the depth of a scuba diver during a 15 time taken run minute dive. (0-15) = = -0.30 ms -2 (90-40) Depth (m) deceleration = 0.30 ms -2 18 (d) Total distance = area beneath graph 16 = (½ × 30 × 15) + (10 × 15) + (½ × 50 × 15) = 750m 14 total distance 750 (e) average velocity = = = 8.3 ms -1 12 time 90 10 8 6 4 2 time (mins) 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (a) During which period of the dive was the diver ascending the quickest? (b) How long did the diver stay at the bottom of the sea, a depth of 18m? (c) What was the vertical velocity of the diver during his descent? 2. The table of results below were taken for a cyclist travelling along a straight road. Velocity (ms -1) 0 5 10 15 15 12 9 6 3 0 Time taken (s) 0 10 20 30 40 50 60 70 80 90 (a) Draw a graph of velocity on the vertical axis against time on the horizontal axis for the journey. (b) What is the initial acceleration of the cyclist? (c) Calculate the deceleration of the cyclist in the final 50 seconds of the journey. (d) Calculate the total distance that the cyclist travelled along the straight road. (e) Calculate the average velocity of the cyclist for the entire journey. Acknowledgements: This Physics Factsheet was researched and written by Jason Slack. The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF. Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136 5  Physics Factsheet www.curriculumpress.co.uk Number 63 Solving problems on projectiles This Factsheet looks at the motion of objects moving under gravity. They are not being driven forward but may have been given a start velocity in any direction. This applies to many sports like golf, tennis, discus, ski jump, target shooting, long jump and stunt riding. The same approach also applies to charged particles moving in a uniform electric field. Any air resistance is ignored in these questions. You need to know the standard equations of motion (Factsheet 13); the ones used most commonly here will be v = u + at, 2as = v2 - u2 and s = ut + ½at2 The up direction is taken as positive for displacement, velocity and acceleration - hence acceleration due to gravity, g, is negative. You should also be familiar with vector addition and resolving into components (Factsheet 2) General Points The strategy is to consider the horizontal and vertical components of the motion separately, and use the equations of motion in each direction. The vertical acceleration is always -9.8 ms-2 (due to gravity) The horizontal acceleration is always zero The projectile will continue to move upwards until the vertical The horizontal velocity is constant component of its velocity is zero Strategies There are various quantities that you are commonly asked to find. These are the strategies to find them: Time to the highest point use v = u + at vertically, with v = 0 Greatest vertical height use 2as = v2 - u2 vertically, with v = 0, or use s = ut + ½at2 if the time is known Time taken for it to reach a particular height use s = ut + ½at2 vertically (if it is returning to the same height at which it started - eg returning to the ground - take s = 0) Total horizontal distance travelled find the time taken for it to finish its journey (as above), then use s = ut horizontally Types of question There are three common situations, when an object is: 1. Thrown vertically upwards or downwards 2. Projected with a horizontal velocity 3. Projected at an angle to the horizontal The first one is not conventionally called a projectile, but can be examined, and understanding it helps with the last two. 1. Object thrown vertically If the particle is thrown downwards, the graphs are as follows: No horizontal component - consider vertical motion only. Start velocity: u has +ve value if thrown upwards, −ve value if downwards velocity-time graph displacement-time graph There is vertical acceleration g = - 9.8 ms-2. time time If the object is thrown upwards, it continues to rise until v = 0 u straight line, parabola gradient -9.8 For an object thrown vertically v = u − 9.8t u = initial vertical velocity s = ut −4.9t2 v = velocity vertically upward vertical vertical s = displacement vertically upward velocity displacement If the particle is thrown upwards, the graphs are as follows: Worked example: A ball is thrown straight up with a speed of 23 ms-1. Find the time taken to reach the top of flight and the height gained. velocity-time graph displacement-time graph vertical vertical Here we have v= 23 – 9.8 t velocity displacement So at the top of the flight, t = 23/9.8 = 2.35 seconds u straight line, parabola Since we have found the time, use it to find the height: gradient -9.8 s = 23t − ½ (9.8)t2 = 23(23/9.8) − ½ (9.8)(23/9.8)2 time = 27. 0 ms-1 projectile reaches Note that if you were unsure about the accuracy of the value calculated for time the time, it would be possible to calculate the height without it, using the highest point Note: This assumes the projectile equation 2as = v2 - u2 returns to the same vertical level from which it started. Exam Hint Note how the original, rather than rounded, time has been used in the second part of the calculation to avoid rounding errors. 1 63 - Solving problems on projectiles Physics Factsheet 2. Object projected horizontally Start velocity has no vertical component, only a horizontal one Typical Exam Question Acceleration is g = - 9.8 ms-2 vertically; no horizontal acceleration A dartboard is 3.0 m away. A dart is thrown horizontally from Object continues to move until it falls to the ground. height 1.90 m and hits the board at 1.60 m. Calculate the (a) time of flight The horizontal velocity is not altered by acceleration downwards. (b) initial speed of the dart A B (c) speed of the dart when it hits the board. (a) whatever the horizontal speed, it takes the same time to fall 0.30 m. 2h 2×0.30 from h = ½ g t2 (1mark), = = 0.247s (1 mark) g 9.8 (b) During this time it travels the 3.0 m (1 mark). distance 3.0 so the speed = = = 12.1 ms-1 (1 mark) time 0.247 (c) Note the question asks for speed so you don’t need the angle but you Neither is the acceleration downwards changed by the horizontal motion. need the magnitude of the resultant velocity. They are independent. the vertical velocity v = at = -9.8 × 0.247 = - 2.42 ms-1 downwards (1 mark) combine this with the horizontal velocity using Pythagoras’ theorem All objects launched horizontally or dropped from the same height will hit the ground at the same time. Their vertical component of 12.1 ms-1 velocity is the same at all times. -2.42 ms-1 magnitude of resultant velocity = 12.12 + (-2.42 ) = 12.34 ms-1 Exam Hint: If the question asks for a velocity at a certain time or point 2 with no direction mentioned then it will mean resultant velocity. You (1 mark) should give both magnitude and direction (angle) in your answer. Vh (the minus doesn’t affect the answer because it is squared but it is consistent α to put it in) Vv resultant velocity Typical Exam Question For an object projected horizontally A boy throws a stone horizontally off a cliff. It hits the sea 2 seconds vv = − 9.8t vh = uh uh = start horizontal velocity later, at a distance of 40 m from the foot of the cliff. sv = −4.9t2 sh = uht vv ,vh = velocity components Calculate the 2(-9.8)sv = vv2 sv ,sh = displacement components (a) height of the cliff (b) initial speed of the stone (c) direction in which the stone is moving when it strikes the water. Vertical Motion (a) Consider vertical motion: sv = −4.9t2 s = -4.9(22) = -19.6m velocity-time graph displacement-time graph Cliff is 19.6m high time time (b) Consider horizontal motion: s = ut straight line, parabola 40 = u2 gradient -9.8 u = 20ms-1 (c) Direction it is moving in is direction of velocity vh = 20ms-1 vv = -9.8(2) = -19.6ms-1 vertical vertical velocity displacement 20.0 ms-1 α Angle = tan-1 (19.6/20) -19.6 ms-1 = 44o Horizontal Motion So direction of motion is 44o below the horizontal velocity-time graph displacement-time graph Typical Exam Question horizontal horizontal straight line, A bullet is fired horizontally at 850ms-1 from a height of 1.8 m above velocity displacement gradient uh the ground. Calculate the (a) time it takes to hit the ground uh horizontal (b) horizontal distance it travels straight line, (a) Consider vertical motion: s = - 4.9t2 -1.8 = -4.9t2 time time t = 0.61 seconds (b) Consider horizontal motion: s = 850t s = 950(0.60609) = 580m (2SF) 2 63 - Solving problems on projectiles Physics Factsheet 3. Launched at an angle to the horizontal Exam Hint You may see very complicated equations in the textbooks Start velocity has both horizontal and vertical components but there is no point trying to remember them, just work through step Acceleration is g = - 9.8 ms-2 vertically; no horizontal acceleration by step using your basic equations. Object continues to move until it falls to the ground. In many cases, the object returns to the same height form which it Typical Exam Question started, but this is not always the case (eg object thrown upwards from A boy throws a ball with a speed of 40ms-1 at an angle of 30o to the a cliff) horizontal. Assuming the ball is projected from ground level, find top of flight (a) the time taken for the ball to reach its greatest height (b) the horizontal distance travelled by the ball Start vertical speed = 40sin30o Start horizontal speed = 40cos30o (a) Consider vertical motion: Use v = u + at 0 = 40sin30o - 9.8t start 20 = 9.8t t = 2.0 seconds (2 SF) The components of the start velocity are given by (b) Here we must find the time taken for the ball to return to the ground first, u uh = ucosα then use this value to find the horizontal distance travelled. α uv = usinα Because the motion is symmetrical (the ball returns to the same height from For a general projectile which it started), this is double the time to the highest point. vv =uv− 9.8t vh = uh uv, uh = start velocity components So t = 2(20/9.8) sv = uvt− 4.9t2 sh = uht vv ,vh = velocity components s = 40cos30o (20/9.8) = 71m (2 SF) 2(-9.8)sv = vv2− uv2 sv ,sh = displacement components Typical Exam Question Vertical Motion A girl stands at the top of a cliff, and throws a stone with a speed of 20ms-1 at an angle of 10o above the vertical. It hits the sea 3 seconds velocity-time graph displacement-time graph later. Find vertical vertical (a) the height of the cliff velocity displacement (b) the highest distance above the cliff reached by the stone uv straight line, (c) the time at which the velocity is at an angle of 45o below the gradient -9.8 parabola horizontal time Start vertical speed = 20sin10o Start horizontal speed = 20cos10o projectile reaches (a) Using s = ut + ½at2 vertically: highest point time s = 20sin10t − 4.9t2 s = 20sin10 (3) − 4.9(32) = -34 Horizontal Motion Height is 34m (2 SF) velocity-time graph displacement-time graph NB: the negative sign just means the projectile has finished below its starting point. horizontal horizontal velocity straight line, displacement gradient uh (b) Using 2as = v2 - u2 vertically 2(-9.8)(s) = 0 - (20sin10o)2 uh horizontal s = (20sin10o)2/ 19.6 straight line, = 0.62m NB: we cannot assume it reaches the highest point halfway through the time motion, is the motion is not symmetrical - the stone ends up lower than it time starts. Example (long jump) If the maximum running speed is about 11 ms-1 and you could take off (c) vv = 20sin10o - 9.8t vh = 20cos10o at 45o with this speed, find the theoretical maximum distance which 20cos10o could be jumped. NB: we must include the negative 45o Start vertical speed = 11sin45o Start horizontal speed = 11cos45o o sign on the vertical velocity, -(20sin10 - 9.8t) Need to find time for jumper to return to ground. because we are considering its Using s = ut + ½at2 vertically: component downwards rather 0 = 11sin45ot − 4.9t2 than upwards 0 = t(11sin45o - 4.9t) t = 11sin45o/4.9 = 1.6 seconds (2 SF) -(20sin10o - 9.8t) So tan45o = 20cos10o Distance jumped: use horizontal motion tan45o = 1 so -(20sin10o - 9.8t) = 20cos10o s = ut 9.8t = 20cos10o + 20sin10o s = 11cos45o (1.58738) = 12 m (2 SF) t = 2.4 seconds (2SF) 3 63 - Solving problems on projectiles Physics Factsheet Qualitative test Answers 1. State the vertical velocity of a projectile at its highest point. 1. v = 23 - 9.8(2) = 3.4 ms-1 2. Sketch the displacement-time and velocity-time graphs for (a) an object thrown vertically upwards 2. 2as = v2 - u2 (b) an object thrown vertically downwards 2(-9.8)(15) = 0 - u2 294 = u2 3. Sketch horizontal and vertical displacement-time graphs for an object u = 17 ms-1 (2 SF) projected horizontally from a window 3. s = ut + ½at2 s = - 4.9t2 = -4.9(2.52) Quantitative (calculation) test = -31 1. Calculate the vertical velocity of an object thrown downwards at Depth = 31m (2SF) 23 ms-1 after 2.0 s. 4. (a) Horizontal motion: 2. A ball is thrown vertically upwards with speed u. It reaches a height s = 8(3.6) of 15 metres. Calculate u. s = 28.8 s = 29m (2 SF) 3. A stone is dropped down a well. It hits the water 2.5 seconds later. Find the depth of the well. (b) vv = -9.8(3.6) = -35.28 vh = 8 Angle = tan-1 (35.28/8) 4. An object is in the air for 3.6 s. It was launched from a tower = 77o below horizontal horizontally with velocity 8 ms-1 (a) how far away from the tower does it land? 5. (a) 302 + 402 = 50ms-1 (b) what angle with the horizontal does the final velocity have? (b) vh = 30 ms-1 vv = 40 − 9.8(3) = 10.6 ms-1 5. A stone is catapulted from ground level. Its initial velocity has v = 302 +10.62 horizontal component 30 ms-1 and vertical component 40ms-1 v = 32 ms-1 (2SF) Find: (a) Its initial speed (c) Using 2as = v2 - u2 vertically (b) Its speed 3 seconds after it is projected 2(-9.8)s = 0 − 402 (c) The maximum height it reaches s = 1600/19.6 = 82 m ( 2SF) (d) Its horizontal range (d) Using s = ut + ½at2 vertically 6. A stone is thrown with a speed of 20ms-1 at an angle of 40o to the 0 = 40t − 4.9t2 horizontal. Five metres from the point from which it is projected there 0 = t(40 − 4.9t) is a wall of height 4 metres. t = 40/4.9 = 8.1633 seconds Determine whether the stone will hit the wall. So range = 30 (8.1633) = 240 m (2 SF) 7. A girl is attempting to throw a ball across her bedroom onto a shelf. The shelf is 2m above floor level and the horizontal distance from the 6. uh = 20cos40o uv = 20sin40o girl to the shelf is 5m Need to find the height of stone after it has travelled 5m horizontally The girl throws the ball with speed u ms-1 at an angle of 30o to the Horizontal motion: horizontal. It lands on the nearest edge of the shelf. 5 = 20cos40ot 5 t = 5/(20cos40o) (a) Show the time of flight of the ball is ucos30o At this time, vertical height is given by: (b) Find the value of u s = 20sin40ot − 4.9t2 s = 20sin40o (5/20cos40o) − 4.9(5/20cos40o)2 s = 3.67 m So stone will hit wall, since its height is less than 4 metres here 7. uh = ucos30o uv = usin30o (a) Using s = ut horizontally: Acknowledgements: This Physics Factsheet was researched and written 5 = ucos30ot by Celia Bloor. t = 5/ucos30o The Curriculum Press,Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU. (b) When t = 5/ucos30o, sv = 2 Physics Factsheets may be copied free of charge by teaching staff or 2 = usin30o(5/ucos30o) - 4.9(5/ucos30o)2 students, provided that their school is a registered subscriber. 2 = 5sin30o/cos30o - 24.5/u2cos230o No part of these Factsheets may be reproduced, stored in a retrieval 2 = 2.8868 - 32.6667/u2 system, or transmitted, in any other form or by any other means, without 32.6667 = 0.8868u2 the prior permission of the publisher. u = 6.1 ms-1 ISSN 1351-5136