In Chapter 6 we will see that the value of g varies with position and with height above the Earth’s surface and also in the absence of air resistance, the acceleration of free falling objects is independent of their mass. This was first noted by Galileo who is reputed to have timed the duration of fall for different objects dropped from the top of the Leaning Tower of Pisa. The fact that the acceleration of free fall is independent of an object’s mass has far reaching significance in Physics and is discussed in more detail in “Topic H. (General relativity)”
If you carry out an experiment to measure g and obtain a value say of 9.4 m s–2 then make sure that you calculate
your error using the correct method. For example do not
assume that the value of g is 9.8 m s–2 and hence compute
your error in measurement as ± 4%. You do not know what the value of g is at your location and that is why you are measuring it. One correct way to calculate the error is along the lines indicated in the preceeding exercise, i.e., using error bars.
2.1.5 SOLVEPROBLEMSINVOLVING
THE
EQUATIONS
OFUNIFORMLY
ACCELERATEDMOTION
Exercise 2.1 (b)
1. A stone is dropped down a well and a splash is heard 2.4 s later. Determine the distance from the top of the well to the surface of the water?
2. A girl stands on the edge of a vertical cliff and she throws a stone vertically upwards. The stone eventually lands in the sea below her. The stone leaves her hand with a speed of 15 m s-1 and the
height of the cliff is 25 m.
Calculate
i. the maximum height reached by the stone. ii. the time to reach the maximum height. iii. the speed with which the stone hits the sea. iv. the time from leaving the girl’s hand that it
takes the stone to hit the sea.
3. A sprinter starts off down a track at a speed of 10 m s-1. At the same time a cyclist also starts
off down the track. The cyclist accelerates to a top speed of 20 m s-1 in 4.0 s. Ignoring the
acceleration of the sprinter, determine the distance from the start that the cyclist will pass the sprinter.
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2.1.6 THE
EFFECTOFAIRRESISTANCE
So far in this discussion of motion we have ignored the effects of air resistance. When an object moves though the air, it is subjected to a retarding force and this force can be very complicated particularly if the object moves at high speed. The force will also depend on the shape of the object and its mass. Think of the way in which a feather floats to Earth compared to the way in which a stone falls. However, for spherical objects moving at relatively low speeds, experiment shows that the retarding force due to air resistance, sometimes referred to as the drag force, is directly proportional to the speed of the object (provided that the density of the air stays constant). Effectively this means that as the object moves faster and faster, the drag force gets greater and greater until in fact it reaches a value equal to the value of the force accelerating the object. When this occurs the object will no longer accelerate and will move with constant velocity. (We will discuss the effect of forces on the motion of objects in much more detail in the next section). The constant velocity that the object attains is called the terminal velocity. If a stone is dropped from a balloon that is at a height of 5000 m, then, if we ignore air resistance, the velocity with which it strikes the ground is about 320 m s-1 (about the speed of
sound). Because of air resistance the actual speed is much less than this. For example if you fell out of the balloon, your terminal velocity would be about 60 m s-1. Still not a
very comfortable speed with which to strike the ground.
Exercise 2 on page 39 will help you become familiar with the idea of terminal velocity.
2.1.7 GRAPHICAL
REPRESENTATION
OFMOTION
A very useful way to describe the motion of an object is to sketch a graph of its kinematics properties. This is best illustrated by means of an example. Consider the motion of a hard rubber ball that is dropped onto a hard surface. The ball accelerates uniformly under gravity. On each impact with the surface it will lose some energy such that after several bounces it will come to rest. After each successive bounce it will leave the ground with reduced speed and reach a reduced maximum height. On the way down to the surface it speeds up and on leaving the surface it slows down.
The graph in Figure 210 shows how we can best represent how the velocity changes with time i.e., the velocity–time graph. ve lo ci ty time O A B C D E F G H
Figure 210 Changes of velocity with time
The ball leaves the hand at point O and accelerates uniformly until it hits the ground at A. At A it undergoes a large acceleration during which its velocity changes from positive to negative (being zero at B). The change in velocity between B and C is less than the change in velocity between A and B since the rebound velocity is lower than the impact velocity. The ball accelerates from C to D at which point it is at its maximum height and its velocity is zero. Notice that even though its velocity is zero its acceleration is not. The ball now falls back to the surface and hits the surface at point E. Neglecting air resistance the speed of the ball at points C, and E will be the same. The process now repeats.
The lines OA, CE and FG are parallel and the gradient of these lines is the acceleration of free fall.g.
The lines AC, EF and GH are also parallel and the gradient of these lines is equal to the acceleration of the ball whilst it is in contact with the surface. The lines should not be vertical as this would mean that the acceleration would be infinite.
The acceleration of the ball at points such as A, C and E is not equal to g.
Sketching an acceleration–time graph taking into account the acceleration of the ball whilst it is in contact with the surface is a little tricky however, you might like to try it as an exercise. For all other parts of the ball’s motion it has a constant acceleration, g. The speed-time graph is interesting and this is left as a question at the end of the chapter.
The displacement–time graph is also interesting and this is also left as a question at the end of the chapter.
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Exercise 2.1 (c)
1. A ‘glider’ bounces backwards and forwards between the ‘buffers’ of a linear air-track.
Neglecting friction, which one of the graphs below best represents how the
(a) velocity, (b) acceleration, (c) displacement and (d) speed
of the glider varies with time?
time time
time time
A. B.
C. D.
2. For the example of the stone dropped from a balloon and striking the ground and taking into account air resistance sketch
i. a distance–time graph. ii. a velocity–time graph. iii. an acceleration–time graph.
2.1.8 ANALYSISOFMOTIONGRAPHS
A distance-time graph will help us to understand the concept of instantaneous speed (or velocity).
Consider the distance-time graph of an object moving with constant speed as shown in Figure 212
In this situation equal distances are covered in equal times and clearly the speed is equal to the gradient of the graph – in this case, 10 m s-1. The average speed is equal to the
instantaneous speed at all points.
Figure 212 Distance-time graph
If the velocity is not constant we can still find average speeds and instantaneous speeds from displacement–time graphs. To demonstrate this, let us look at the distance– time graph for a falling ball this is shown in Figure 213.
/ m ec na tsi d time / s 0 0.2 0.4 0. 6 0.8 1.0 1.2 1.4 0 1 2 3 4 5 6 1.8 0.4
Figure 213 Distance- time for a falling ball
From the graph we see that the time it takes the ball to fall 1.0 m is 0.4 s. The average speed over this distance is therefore 2.5 m s-1. (Remember that speed is the magnitude
of velocity.)
To find say the instantaneous speed at 1.0 m we find the gradient of the curve at this point.
To do this we draw the tangent to the point as shown. From the tangent that is drawn, we see that the slope of the line is 1.8 (= Δs) divided by 0.4 (= Δt) = 4.5 m s-1
d/ ec na tsim time / s 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120
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This is actually what we mean when we write an instantaneous speed as s a vP Δs t Δ --- = Δt→0
vP is the instantaneous speed at the point P and
as
s
Δ
t
Δ
--- Δt→0 is the gradient of the displacement–time
graph at the point P.
(Those students who are familiar with calculus will recognise this process as differentiation. The equation for the above graph
is s = kt2such that the derivative ds dt
--- = 2kt is the
gradient at any time t and hence the instantaneous speed).
When sketching or plotting a displacement-time graph we have to bear in mind that displacement is a vector quantity. Consider for example, the situation of an object that leaves point A, travels with uniform speed in a straight line to point B, returns to point A at the same constant speed and passes through point A to a point C. If we ignore the accelerations at A and B and regard the point A as the zero reference point, then a sketch of the displacement-time graph will look like that shown in Figure 214.
time di sp la ce m en t A B C t1 2t1
Figure 214 Displacement-time Graph
VELOCITY(SPEED) - TIMEGRAPHS
The graph in Figure 215 shows how the velocity of the falling ball varies with time.
Figure 215 Velocity-time graph for a falling ball
The acceleration is the change of velocity____________ time and in thisand in this case this is equal to the gradient of the straight line and is equal to 10 m s -2. This is a situation of constant
acceleration but even when the acceleration is not constant the acceleration at any instant is equal to the gradient of the velocity–time graph at that instant.
In section 2.1.2 we saw that we defined instantaneous acceleration as
a = Δv ___
Δt as Δt → 0
Students familiar with calculus will recognise that acceleration is the derivative
a = dv ___ dt which can also be written as:
a = d __ dt
(
ds __ dt)
= d 2 s ___ dt2We can also determine distances from speed–time graphs. This is easily demonstrated with the graph of Figure 216 which shows the speed time graph for constant speed.
time / s 0 2 4 6 8 10 12 14 0 10 20 30 40 sp eed / m s –1
Figure 216 Speed-time graph to determine distance
time / / s 0 0.2 0.4 0.6 0. 8 1.0 1.2 1.4 0 1 2 3 4 5 6 7 8 9 10 ve lo ci ty / ms –1
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The distance travelled is just velocity × time. So at a constant speed of 20 m s-1 after 10 s the object will have
travelled 200 m. This is of course equal to the area under the line between t = 0 and t = 10 s.
If the velocity is not constant then the area under a velocity–time graph will also be equal to the displacement. So, for the falling ball, we see from the velocity time graph, Figure 215, that the distance travelled after 1.0 s is equal to the area of the triangle of base 1.0 s and height 10 m s-1
equals = ½ × 1.0 s × 10 m s-1 = 5.0 m
Example
A train accelerates uniformly from rest to reach a speed of 45 m s-1 in a time of 3.0 min. It then travels at this speed for a
further 4.0 min at which time the brakes are applied. It comes to rest with constant acceleration in a further 2.0 min.
Draw the speed-time graph for the journey and from the graph calculate
(a) the magnitude of the acceleration between 0 and 3.0 min
(b) the magnitude of the acceleration after the brakes are applied
(c) the total distance travelled by the train
Solution
The graph is shown plotted below
(a) the acceleration of the train in the first 3 minutes is the gradient of the line AB.
Therefore, we have, a =
45
180
= + 0.25 m s -2(b) the acceleration of the train after the brakes are applied is the gradient of the line
CD = -
45
120
= - 0.38 m s -2(c) the distance travelled by the train is the total area under the graph.
Total area
= area of triangle ABE + area BCFE + area of triangle CDF
= 17550 = 18000 m
= 18 km.
ACCELERATION-TIMEGRAPHS
If the acceleration of an object varies it is quite tricky to calculate the velocity of the object after a given time. However, we can use an acceleration-time graph to solve the problem. Just as the area under a velocity-time graph is the distance travelled then the area under an acceleration- time graph is the chane in speed achieved. For the falling ball the acceleration is constant with a value of 10 m s–2.
A plot of acceleration against time will yield a straight line parallel to the time axis, Figure 218. If we wish for example to find the speed 3.0 s after the ball is dropped, then this is just the area under the graph:
0 2 4 6 8 10 12 0 1 2 3 4 5 time / s accelera tio n / m s –2
Figure 218 Acceleration - time graph
We could of course have found the speed directly from the definition of uniform acceleration i.e.
speed = acceleration × timeExample
speed / m s –1 time / s 0 100 200 300 400 500 600 0 10 20 30 40 50
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Example
The acceleration of an object increases uniformly at a rate of 3.0 m s-2 every second. If the object starts from rest,
calculate its speed after 10 s.
Solution
A plot of the acceleration-time graph for this situation is shown in the graph below.
30 2 4 6 8 10 12 acceleration / m s–2 time /s 10 20
The speed attained by the object after 10 s is the area under the line.
Therefore, speed = area enclosed by acceleration–time graph
= 1-2 × 10 × 30 = 150 m s-1
To summarise :
• the gradient of a displacement-time graph is equal to the velocity
• the gradient of a velocity-time graph is equal to the acceleration
• the area under a velocity -time graph is equal to the distance.
• the area under an acceleration-time graph is the change in velocity.
2.1.9 RELATIVE
MOTION
Suppose that you are standing on a railway station platform and a train passes you travelling at 30 m s-1. A passenger
on the train is walking along the corridor in the direction of travel of the train at a speed of 1 m s-1. From your point of view the speed of the person is 31 m s-1. In another situation a car A is travelling at a speed of 25 m s-1 and
overtakes a car B that is travelling in the same direction at a speed of 20 m s-1. The speed of car A relative to car B
is 5 m s . Clearly the determination of speed (and therefore velocity and acceleration) depends on what it is measured relative to.
Generally speaking, if the velocity of a particle A relative to an assigned point or reference frame O is VA and the velocity of a particle B relative to the same point is VB, then the velocity of A relative to B is the vector difference VA – VB (See Topic 1.3)
The example that follows illustrates how relative velocity is very important in situations such as plotting a correct course for an aircraft or for a boat or ship.
Example
The Figure below shows the two banks of a river. A ferryboat operates between the two points P and Q that are directly opposite each other.
P Q
1.2 m s–1
90 m
The speed of flow of the river relative to the riverbanks
is 1.2 m s-1 in the direction shown. The speed of the ferry
boat in still water is 1.8 m s-1 in a direction perpendicular
to the river banks. The distance between P and Q is 90 m.
Clearly, if the ferryboat sets off from P directly towards Q, it will not land at Q.
It is left as an exercise for you to show that the speed of the ferryboat relative to the speed of the water is 2.2 m s -1 and
that it will land at a point 60 m downstream of Q. You should also show that, in order to land at Q, the ferry boat should leave P heading upstream at an angle of 56° to the riverbank and that the time taken to cross to Q is about 60 s.
For an aircraft, the rate of water flow becomes the wind speed and for a ship at sea it becomes a combination of wind, tide and current speeds.
The idea of relative measurement has far reaching consequences as will become clear to those of you who choose to study Special Relativity (Option D or in Option H.)
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2.2.1 Calculate the weight of a body using the expression W=mg.
2.2.2 Identify the forces acting on an object and draw free-body diagrams representing the forces acting.
2.2.3 Determine the resultant force in different situations
© IBO 2007
2.2.1 MASSANDWEIGHT
Before developing the ideas of Newtonian Mechanics, it is important to discuss the concepts of mass and weight. In everyday usage, mass and weight are often interchangeable; in physics there is a fundamental difference between the concepts.
So what is mass? Some textbooks will tell you that it is the ‘quantity of matter’ in a body. However, this is pretty meaningless for two reasons. It begs the question as to what is meant by ‘matter’ and also it gives no means of quantifying mass. In respect of the latter, if a physicist cannot measure something in the laboratory then that something does not belong to physics.
Mass is one of the fundamental properties of all matter and essentially measures a body’s inertia, that is, a body’s reluctance to change its state of motion. Moving bodies tend to keep on moving in the same direction and stationary bodies don’t start moving by themselves. The more massive a body, the more reluctant it is to change its state of motion. As we shall see in Section 2.2.8, it is quite possible to give a logical definition to mass that enables a physicist to quantify the concept. However, this definition does not tell you what mass is and why particles that make up bodies should have mass is one of the great unsolved mysteries of physics. The situation is further complicated by the fact that there are two types of mass. There is the mass we have just described, inertial mass and then there is the mass a body has that gives rise to the gravitational attraction between bodies. This is the gravitational mass of a body. We saw in Section 2.1.4 that experiment shows that the acceleration of free fall is independent of the mass of an object. This suggests that gravitational mass and inertial mass are in fact equivalent. (See section 2.2.8)
The equivalence of gravitational and inertial mass is one of the cornerstones of Einstein’s theory of general relativity and is discussed in more detail in Option H. The SI unit
of mass is the kilogram and the standard is the mass of a platinum alloy cylinder kept at the International Bureau of Weights and Measures at Sevres near Paris.
We shall see in section 2.2.8 that when we talk about the
weight of a body what we actually mean is the gravitational
force that the Earth exerts on the body. So weight is a force and since the force of gravity varies from place to place and also with height above the Earth’s surface, the weight of a body will also vary but it’s inertial and gravitational mass remains constant.
Exercise 2.2 (a)
1. This exercise is designed to help you distinguish between the concepts of mass and weight. Here are six different hypothetical activities
(a) You weigh yourself using bathroom scales. (b) You determine the mass of an object by
using a chemical beam balance. (c) You determine the density of lead. (d) You drop a brick on your foot. (e) You trap your fingers in a car door. (f) You design a suspension bridge.