Physical Sciences
CAPS
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Reg. No.: 2011/011959/07
Author: Miemie Pretorius
Cover illustration: Johann Strauss
Video recordings: Stellenbosch University, Telematic Services
Animations: Egbert Westra from Enjoythecompany
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MECHANICS
MECHANICS
Exercise 1: Resultant, Parallelogram, Triangle or Polygon
Exercise 2: Resultant is Zero if Object is Stationary or Moving at Constant Velocity
Exercise 3: The Angle Between Two Vectors
Exercise 4: Force Diagrams and Free-Body Diagrams
Exercise 5: Resolving a Vector into its Perpendicular Components
Exercise 6: Body on an Inclined Plane
Exercise 7: Contact Forces
Exercise 8: Newton’s First Law of Motion
Exercise 9: Newton’s Second Law of Motion: Law and Experiment
Exercise 10: Newton’s Second Law of Motion: Applications
Exercise 11: Newton’s Second Law of Motion: Velocity/Time Graphs 213
Exercise 12: Newton’s Second Law of Motion, Apparent Weight and “Lifts” 214
Exercise 13: Newton’s Second Law of Motion and Systems
Exercise 14: Newton’s Third Law of Motion
Exercise 15: Newton’s Universal Law of Gravity
Exercise 16: Newton’s Law of Gravity and Newton’s Third Law of Motion:
Exercise 17: Newton’s Law of Gravity on a Planet
Exercise 18: Terminal Velocity
MATTER AND MATERIALS
Exercise 19: Atomic Combinations: Forces, Bond Energy, Bond Length
Exercise 20: Atomic Combinations: Molecular Structure: Lewis Diagrams
Exercise 21: Atomic Combinations: Molecular Structure: Molecular Shape
Exercise 22: Atomic Combinations: Molecular Structure: Electronegativity
Exercise 23: Kinds of Particles and Intermolecular Forces
Exercise 24: Intermolecular Forces and the Physical State of Substances
Exercise 25: The Chemistry of Water
WAVES, SOUND AND LIGHT
Exercise 26: Background Knowledge
Exercise 27: Geometrical Optics: Refraction
Exercise 28: Critical Angle and Total Internal Reflection
Exercise 29: Applications of Reflection and Total Internal Reflection
Exercise 30: 2D and 3D Wave Fronts: Diffraction
MATTER AND MATERIALS
Exercise 31: Ideal Gases: The Kinetic Theory of Gases
Exercise 32: Ideal Gases: Definition
Exercise 33: Ideal Gases: Relationship between p and V (Boyle’s Law)
Exercise 34: Ideal Gases: Relationship between V and T (Charles’s Law)
Exercise 35: Ideal Gases: Relationship Between p and T (Guy Lussac’s Law)
Exercise 36: Ideal Gases: General Gas Law: p1V1/T1 = p2V2/T2
Exercise 37: Ideal Gases: General Gas Law: PV = nRT
Exercise 38: Ideal Gases: Deviations from Ideal Gas Behaviour
CHEMICAL CHANGE
Exercise 39: Quantitative Aspects of
Exercise
Exercise
Exercise
Exercise
ELECTRICITY
Exercise
SKILLS FOR PHYSICAL SCIENCES LEARNERS
Scientific Notation
Very large or very small numbers are preferablywritten in the scientific notation, e.g. a charge of 0,0000000326 C can be written as 3,26 x 10–8 C or a pressure of 350625 Pa can be written as 3,51 x 105 Pa.
Notice that in the scientific notation the number is written as a product of two numbers – the first number is between 1 and 10 and the second as 10n.
The first number is usually rounded off to two decimal places.
Please ensure that you know how to multiply and divide by numbers with exponents and how to use your calculator when dealing with exponents.
Conversion of Units
The 7 basic SI units:
Mass(m)is measured in kilogram(kg)
Time(t)is measured in second(s)
Length ( ) is measured in metre(m)
Electric current(I) is measured in ampere (A)
Amount of substance(n) is measured in mole (mol)
Temperature(T)is measured in kelvin(K)
Luminous intensity is measured in candela (cd)
All answers should be given in SI units, unless otherwise specified.
Mass is often given in gram.
In order to convert it to kg, it is useful to know that ‘kilo’ means ‘1000’ or 103
Therefore1 kg = 103 g and 1 g = 10–3 kg.
Time is often given in hours (h). Remember that 1 h is 60 minutes and 1 minute is 60 seconds.
Hence 1 h = 3600 s or 3,6 x 103 s and 1 s = 1/(3,6 x 103) h
Length is sometimes given in millimetre (mm) or centimetre (cm) or decimetre (dm) or kilometre (km).
In order to convert from one to the other, it is useful to understand the following diagram: For each space that you move to the right, you 10 or x10–1 ,
e.g. 5 mm = 5 x 10–1 cm = 5 x 10–2 dm = 5 x 10–3 m = 5 x 10–6 km, etc.
For each space that you move to the left, you x10, e.g. 5 m = 5 x 10 dm = 5 x 102 cm = 5 x 103 mm, etc.
If the conversion is for area, which is measured in m2, then for each space that you move to the right, you 102 or x10–2, e.g. 5 cm2 =5 x 10–2 dm2 = 5 x 10–4 m2, etc.
For each space you move to the left, you x102, e.g. 6 km2 = 6 x 106 m2, = 6 x 108 dm2 = 6 x 1012 mm2, etc.
If the conversion is for volume, which is measured in m3, then for each space that you move to the right, you 103 or x10–3, e.g. 250 cm3 = 250 x 10–3 dm3 = 250 10–6 m3, etc.
For each space you move to the left, you x103, e.g. 1 km3 = 1 x 109 m3 = 1 x 1015 cm3, etc.
Electric current is sometimes indicated as micro ampere ( A) or milli ampere (mA).
‘Micro’ means one millionth, or 10–6 and ‘milli’ means one thousandth, or 10–3 .
The scientific temperature scale that is used inScienceis the kelvin scale, but the celsius scale is popular for everyday use.
In order to convert from celsius to kelvin and vice versa, the following equation can be used:
T = t + 273, where T is the temperature in kelvin and t is the temperature in celsius.
What is ‘Rate’?
Rate is the change of a quantity per second
E.g. power is rate at which work is done or the amount of work done per second, i.e. (work done)/second.
Current is rate at which charge flows orthe amount of charge flowing past a point per second, i.e. (amount of charge)/second.
Directly and Inversely Proportional
If we say that quantity Ais directly proportional to quantity B, it means that if A doubles, B also doubles; if A increases 10 times, Balso increases 10 times, if A is halved, B is halved, etc.
This is written as A B, which we read as “A is directly proportional to B”.
To change this into a mathematical equation, we write it as A = kB, where k is a constant.
Example:
This means that it we measure the distance covered by different people for the same time, say 10 s, then the greater the speed of the person, the greater the distance covered by the person.
D v, which can be written as D = kv, where the constant k is thetime – the equation becomes
The graph of two quantities that are directly proportional, is always a straight line passing through the origin.
If we say that Pis inversely proportional to Q, it means that if P increases 2 times, Q decreases 2 times; if P increases 10 times, Q decreases 10 times, etc.
This is written as P 1/Q, which we read as “P is inversely proportional to Q”.
To change this into a mathematical equation, we write it as P = k/Q, where k is a constant andPQ = k.
The graph for two quantities that are inversely proportional is a hyperbola.
Graph of A and B, which are Graph of P and Q, which are directly proportional inversely proportional
Ratios
In Physical Sciences we are often requiredto divide a quantity into ratios.
Supposethree workers, X, Y and Z divide the profit of a business in a ratio of 1:3:5.
This means that for every R1 that X earns, Y earns R3 and Z earns R5, i.e. every R(1 + 3 + 5) = R9 is split up into R1, R3 and R5.
Hence X earns 1/9 of the profit, Y earns 3/9 of the profit and Z earns 5/9 of the profit.
Now it easy to calculate each one’s share for any profit, e.g. if the profit is R1800, X earns 1/9(R1800) = R200; Y earns 3/9(R1800) = R600 and Z earns 5/9(R1800) = R1000
Example: A potential difference of 24 V divides in a ratio of 2:2:4 across resistors P, Q and R.
Share of potential difference for P: 2/(2 + 2 + 4) = 2/8 2/8(24 V) = 6 V
Share of potential difference for Q: 2/(2 + 2 + 4) = 2/8 2/8(24 V) = 6 V
Share of potential difference for R: 4/(2 + 2 + 4) = 4/8 4/8(24 V) = 12 V
Practical Investigations
A practical investigation involves the relationship between two quantities
All other quantities have to be kept constant.
The investigative question should always relate the two quantities investigated, e.g. suppose we investigate the relationship between temperature and volume, then the question can be “What is the relationship between temperature and volume?” or “Does volume increase with temperature?”, etc.
The hypothesis should be an ‘answer’ tothe investigative question, e.g. “Volume is directly proportional to temperature” or “Volume increases as temperature increases”, etc.
The hypothesis may even be a wrong statement, but it has to ‘answer’ the investigative question. The results should be recorded in a table with suitable headings that contain units, e.g. the column that containsthe temperature should have the heading “Temperature (°C)” and the column containing the volume should have the heading “Volume (cm3)”.
The independent variable is the quantity that you change – suppose you decide you are going to take the volume at 0 °C, 10 °C, 20 °C, then temperature is the independent variable.
The dependent variable is the quantity that changes because of the change in the one that you control – in this case, the volume.
The constantvariables are those quantities that you will have to keep the same in order to have a fair investigation, i.e. you cannot have anything that will influence your readings except your two variables. E.g. you will have to work with the same substance – you cannot use hydrogen in one measurement and then iron in the next. You will also have to keep the mass of the substance constant – you cannot take 10 g of hydrogen in one measurement and 50 g of hydrogen in the next. You will also have to keep the pressure constant – one measurement cannot be taken at sea level and the other at high altitudes, etc. The graph should always have the two axes labelled properly, similar to the table, including units and then also a suitablescale.
The independent variable should always be on the horizontal or X-axis and the dependent variable on the vertical or Y-axis
The conclusion should state the relationship between the two variables.
Precautions have to be taken to exclude dangers and factors which could cause the investigation to fail.
A quantitative analysis involves measurements, e.g. when you determine the relationship between the volume of a gas and temperature, you have to measure temperature and volume.
A qualitative analysis involves a study, e.g. when you determine which materials conduct electricity; you only change the substances and observe in which cases the light bulb glows – no measurement required.
Graphs: Gradient and Area
In both Chemistry and Physics you often will come acrossgraphswhich you will have to interpret. At this stage there are two important skills to master that come in very useful: Lookat the gradient of the graph and the area included by the graph and the axes. In any graph the . This means, the gradient will be whatever is on the Y-axis divided by whatever is on the X-axis. Next look for an equation on the data sheet that contains these two quantities and rewrite the equation in such a way that the expression on the right hand side is the same as that of the gradient.
Example: In a velocity versus time graph,
Now look for an equation that contains the two variables, v and t, and see if you can rewrite the equation in such a way that v and t are the subject –the rest of the equation then represents the gradient. In our example we find a = ient represents a.
Models in Science
Models are used to represent something.
It tries to make a difficult concept understandable.
Most models do have some shortcomings, but as new discoveries are made,models keep on improving.
How to Solve Problems in Physical Sciences
As you read through the question, list all the quantities that are given,and very important, the quantity which is asked
Look for an equation under the relevant section on your data sheet, for an equation containing the quantities given and asked.
Sample
Whatever the gradient does, acceleration a does the same – the first part of the graph has a constant and negative gradient, hence the acceleration a will be constant and negative. In the second part of the graph the gradient is constant and positive; hence the acceleration is constant and positive. In any graph the area included by the graph and the axes is calculated as length times width. In our example it is v x t. Again see if you can find an equation that you can rearrange in such a way
Substitute the given values into the equation, but first ensure that your units are correct. Usually you have to use SI units, but in some instances you have to use other units, e.g. in n = m/M, the mass has to be in gram and not in the SI unit of kilogram!
MECHANICS
Resultant
The resultant of vectors is that single vector which has the same effect as all the vectors together. The resultant of two vectors that are not in a straight line can be determined in ONE of three ways:
1 Using a parallelogram/rectangle (tail-to-tail method), discussed below;
2 Using a triangle (head-to-tail method), discussed below;
3 Resolving the vectors into their perpendicular components, discussed on page 7.
NEVER MIX THE THREE METHODS – USE JUST ONE AT A TIME!!
A vector may be moved around until it is tail–to–tail or head–to–tail with another vector, provided its direction and magnitude remain the same.
Any one of the three methods can be used to determine the resultant and in all of these methods either a scale drawing or calculation can be used.
If the question specifically asks for a scale drawing, you are not allowed to use a calculation.
If the question states that you have to do a calculation, you are not allowed to do a scale drawing. If the question only requires you to determine the answer, without specifying a method, you may use a calculation or a scale drawing.
Parallelogram (Tail–to–Tail Method)
If a parallelogram is used, the two vectors are drawn as two adjacent sides of a parallelogram, starting at the same point
The diagonal, drawn from the same point, represents the resultant. Whenever this method is used, it is important to note that the two vectors, as well as the resultant, are all drawn starting from the same point. The parallelogram method can be used for only two vectors at a time.
Triangle (or Polygon) (Head–to–Tail Method)
If a triangle is used, the two vectors are placed head–to–tail, i.e. where the one arrow ends, the next one starts.
It does not matter in what order you take the vectors. The resultant is represented by the arrow drawn from the starting point of the first to the end point of the second vector, i.e. the arrow closing the triangle.
If we want to find the resultant of more than two vectors, the vectors are placed head-to-tail, in any order. The resultant is then the arrow drawn from the start of the first to the end of the last vector, i.e. the arrow closing the polygon.
In a vector diagram the resultant is the vector closing the polygon.
In a vector diagram the resultant can be recognised as the vector that is not head-to-tail (i.e. it is lying ‘the wrong way round’ – head-to-head and tail-to-tail).
If the vector diagram is a closed polygon, with all vectors lying head-to-tail, the resultant is zero.
In a right-angled triangle, the resultant can be calculated using the Law of Pythagoras (r2 = x2 + y2) and trigonometry
Exercise 1:
1 A student walks 4 km due north in one hour and 3 km due west in the next hour. What is his resultant displacement after two hours?
A 7,0 km B 5,0 km, 323º
C 2,5 km, 37º D 3,5 km, 323º (2)
2 An aircraft is flying directly north at constant airspeed in still air.
As the flight progresses, a wind arises, blowing from the west. What will be the effect on the speed and direction of the aircraft relative to the ground?
A No effect.
B Speed unchanged but direction changes.
C Speed increases but direction unchanged.
D Speed increases and direction changes.
3 Three horizontal forces of equal magnitude act on a box initially at rest as shown. The box
A accelerates towards the south-east.
B accelerates towards the south-west.
C accelerates towards the north-east.
D does not accelerate. (2)
4 In which ONE of the following vector diagrams is the resultant of the three vectors zero? (2)
5 According to the vector diagram
6 Vectors p, q and r form a vector diagram as indicated. Which statement regarding the relationship between the vectors is true?
(2)
7 In which ONE of the following vector diagrams is the resultant of the three vectors zero? (2)
8 Define the resultant of two vectors. (3)
9 Johannesburg is 1 500 km from Cape Town on a bearing of 045º (N 45º E).
Durban is 500 km from Johannesburg on a bearing of 150º (E 60º S). An aircraft on a test flight starts from Cape Town and flies to Johannesburg and then to Durban.
Find the magnitude of the resultant displacement of the aircraft for the total flight by means of an accurate scale vector diagram using a scale: 1 cm = 250 km.
Label the vectors. (8)
10 A fisherman in a boat first sails 3 km due south from the harbour, then he sails 4 km due east and then 2 km in a 45º N of E direction.
Use a scale drawing (in which 10 mm represents 1 km) to answer the following questions:
10.1 After these movements, how far will the fisherman be from the harbour? (9)
10.2 In what direction should he now sail to get back to the harbour in a straight line? (2)
11 A scale drawing in which vectors are used to represent four forces which act on an object forms a closed quadrangle as shown by the following diagram: What can be deduced by this vector diagram? Explain your answer. (4)
12 Use a scale drawing to represent the following vector equation to show its meaning properly: (Let 20 mm represent 1 vector unit.)
3 + 4 = 4 (4)
The Resultant is Zero if the Object is Stationary or Moving at Constant Velocity
When an object is at rest, the object is in equilibrium, i.e. the resultant is zero. When an object is moving at constant velocity, the body is also in equilibrium, i.e. resultant is zero. It is extremely important to immediately realise that whenever you see the expression constant velocity, you should know this implies the forces acting on the object have a resultant of zero. This knowledge is tested very frequently and is often skilfully camouflaged even in other sections of Physics. We will learn more about this in Newton’s First Law of Motion. When an object is in equilibrium under the influence of three (or four) forces, the three (or four) forces can be represented in magnitude and direction by the three (or four) sides of a closed triangle (or quadrangle), placed head-to-tail
To draw this triangle (or quadrangle), you can start with any force. Next any of the remaining forces can be moved around (keeping its direction) until it lies head-to-tail with the first force. Now move the third force until it lies head-to-tail with the second force, closing the triangle. Any unknown force or angle can now be calculated using trigonometry and Pythagoras if the triangle is a right-angled triangle or by a scale drawing.
Remember, an object is in equilibrium when it is at rest or moving at constant velocity
Please note that if you come across a closed vector diagram, with all the sides lying head-to-tail, it means the resultant is zero and the body is in equilibrium, i.e. at rest or moving at constant velocity.
Exercise 2:
1 The resultant of all the forces acting on a body, is zero if the body is
A falling freely from rest. B accelerating.
C moving in a circular path. D moving with constant velocity. (2)
2 A car moves along a horizontal road to the right at a constant velocity of 80 km h-1
Which ONE of the following diagrams best represents the forces acting on the car? (2)
3 A body moving at a CONSTANT VELOCITY on a horizontal plane, has a number of unequal forces acting on it. Which ONE of the following statements is TRUE?
A At least two of the forces must be acting in the same direction.
B The resultant of the forces is zero.
C Friction between the body and the plane causes a resultant force.
D The vector sum of the forces causes a resultant force which acts in the direction of motion. (2)
4 The diagram shows an object of weight W on a string. A horizontal force F is applied on the object so that the string makes an angle with the vertical when the object is at rest. The force exerted by the string is T. Which ONE of the following expressions is wrong?
A F + T + w = 0 B w = T cos
C T2 = w2 + F2 D tan (2)
5 A boy pushes a lawnmower of mass 40 kg with a constant velocity over a horizontal lawn. He exerts a push on the handle of 200 N at an angle of 20º to the horizontal. What is the resultant force acting on the mower? (2)
6 A mass m of weight W, is suspended from a light rope by means of a strong hook which is fixed at point X. The rope is strung between vertical poles as shown in the figure.
6.1 Say why a vector diagram of the forces acting at point X, plotted tail to head, will be a triangle. (3)
6.2 Draw a rough diagram of this triangle (in 6.1) of forces. Label each force and indicate the size of each angle. (6)
6.3 In which part of the rope is the force greater, in A or in B? (2)
6.4 The rope will break if the force in the rope exceeds 500 N. Calculate the maximum mass which the above arrangement can support. (Correct to one decimal place). (7)
7 A bell of mass 5 kg is supported by two rods bolted together at point P and attached to the wall as shown in the sketch. Ignore the weight of the rods in the questions which follow:
7.1 What is the magnitude of the downward force acting on the wall? (3)
7.2 Draw a labelled force diagram which shows all the forces acting on point P. (5)
7.3 Explain whether the forces acting at P are in equilibrium. (2)
7.4 Calculate the magnitude of the force exerted by the horizontal rod on P. (4)
7.5 If the angle between the rods were 45º instead of 30º, would the magnitude of force exerted by the horizontal rod at P be smaller, greater or the same? (2)
8 A car engine with a mass of 200 kg is lifted from its mountings by means of a chain and pulley system. In sketch A, the engine is suspended by the chain, hanging stationary. In sketch B, the engine is pulled sideways by a mechanic, using a rope. In the questions that follow, the masses of the chain and the rope can be ignored.
8.1 What is the magnitude of the tension in the chain (i.e. force exerted by the chain) in sketch A? (3)
8.2 Does the tension in the chain increase or decrease by pulling the engine sideways?
Explain your answer by referring to a properly labelled force diagram. (Do not use calculations.) (6)
8.3 Can the tension in the rope ever exceed that in the chain? (2)
9 A heavy crate, mass M, is lifted by means of a rope S and a pulley, so that it can be taken through a window to the first floor of a building. A second rope T, pulls the crate horizontally to the left so that it does not scratch the wall. At lunchtime the ropes are fixed and the crate is left to hang in equilibrium as shown in the diagram. The tension in the rope S is 6 500 N.
9.1 Now calculate
9.1.1 the mass of the crate. (4)
9.1.2 the tension in the rope T. (3)
9.2 Will the tension in the rope S increase if the crate is pulled further to the left? Explain your answer. (4)
The Angle Between Two Vectors
Whenever we talk about the angle between two vectors we refer to the angle measured between their tails and not between the head of one and the tail of the other.
The resultant of two vectors is a maximum when the angle between them is 0º and a minimum when the angle is 180º.
E.g.
This resultant is 7 N, because the angle between the vectors is 0º.
This resultant is 1 N because the angle between the vectors is 180º.
Any value between the minimum and maximum is possible, i.e. as the angle between them increases the resultant decreases.
DIAGRAMA
1 Which ONE of the following values of the angle between forces X and Z, gives rise to the smallest resultant?
A 0º B 60º C 90º D 120º (2)
2 Two forces, X and Y, can be replaced by a single force of magnitude 5 N. If the magnitude of X is 2 N, which ONE of the following can be the magnitude of force Y?
A 1 N B 2 N C 6 N D 10 N (2)
3 Two forces of constant magnitude act at the same point, as shown. If the angle between them increases from 10º to 80º, how will the magnitude of the resultant force change?
A Increases B Decreases
C Remains the same D Increases and then decreases (2)
4 An object experiences two successive displacements of 3 m and 5 m. Which ONE cannot be the magnitude of the resultant displacement (in m)?
A 2 B 4 C 8 D 10 (2)
5 Two forces of magnitude 5 N and 7 N act on an object. Which of the following cannot be a resultant of the two forces?
A 2 N B 9 N C 11 N D 13 N (2)
6 Two forces of 5 N and 7 N act on an object. They can be replaced by a single force which has the same effect as the other two. This force must have
A a magnitude of 2 N. B a magnitude of 12 N.
C a magnitude between 5 N and 7 N. D a magnitude between 2 N and 12 N. (2)
7 varies between 0º and 180º, the magnitude of the resultant force Fres …
A is always less than 20 N. B is always more than 60 N. C varies between 40 N and 80 N. D varies between 20 N and 60 N. (2)
Force Diagrams and Free-Body Diagrams
The force diagram for an object includes the object with all the forces acting on the object only, drawn as arrows indicating their directions. Note that a pushing force acts towards the object and a pulling force acts away from the object.
Example: A boy pushes a crate across a rough floor. In the free-body diagram for an object the object is indicated as a dot and all the forces are moved, whilst keeping their respective directions, until all the starting points start on the dot, i.e. all the forces act away from the dot.
1 A car moves along a horizontal road to the right at a constant velocity of 80 km h-1 Which ONE of the following diagrams best represents the forces acting on the car? (2)
Sample
Label each force in a full sentence and specify where possible the nature of the force – e.g. frictional force exerted by the plane on the object or gravitational force by the earth on the object, etc.
Only forces exerted on the object should be mentioned, therefore make certain each label reads: Force exerted by ….. on the object Should you mention forces acting on any other objects, you lose marks! In all situations the gravitational force of the earth on the object (weight) will be one of the forces. The relative magnitudes (not exactly according to scale) of the forces must also be indicated by the lengths of the arrows.
2 An object is suspended by means of a light string. The sketch shows a horizontal force F which pulls the object from the vertical position until it reaches an equilibrium position as shown. Which ONE of the following vector diagrams best represents all the forces acting on the object? (2)
3 A bullet is fired by a rifle as indicated in the sketch. If the bullet experiences air resistance, which diagram correctly indicates the forces acting on the bullet at the moment when it is in the position as indicated? (The forces are not drawn to scale.) (2)
4 A child sits on a swing. Her mother pulls the seat of the swing horizontally to the right as shown in the sketch. Which ONE of the following diagrams best represents the forces acting on the seat while the mother holds it? (2)
5 A boy pushes a lawnmower of mass 40 kg with a constant velocity over a horizontal lawn. He exerts a push on the handle of 200 N at an angle of 20º to the horizontal. On a copy of the sketch indicate (using arrows) and name all the forces acting on the mower while it is being pushed. (6)
6 A heavy crate, mass M, is lifted by means of a rope S and a pulley, so that it can be taken through a window to the first floor of a building. A second rope T, pulls the crate horizontally to the left so that it does not scratch the wall. At lunchtime the ropes are fixed and the crate is left to hang in equilibrium as shown in the diagram. The tension in the rope S is 6 500 N. Draw a free-body diagram of point O when O is in equilibrium. (3)
7 An elephant is dragging a tree trunk along a rough horizontal surface by means of an attached rope which makes an angle of 30º with the horizontal. The force in the rope is 4 000 N. Draw a force diagram of the log and on it indicate and label all the forces (other than the 4 000 N force) acting on the log. (6)
Resolving a Vector into its Perpendicular Components
A vector R can be resolved into its components, i.e. any two vectors, when combined, has as resultant the vector R. (Resolving a vector into two components is the ‘opposite’ of finding the resultant of two vectors.)
Several sets of components can be found for a specific vector, but we resolve a vector only on the Cartesian plane, i.e. the perpendicular X- and Y-axes.
The horizontal component Rx = R cos
The vertical component Ry = R sin where is the angle between the vector R and the X-axis
If the horizontal component Rx has a positive value, its direction is to the right along the positive X-axis and if it has a negative value, its direction is along the negative X-axis.
If the vertical component Ry has a positive value, its direction is upwards, in the direction of the positive Y-axis and if it has a negative value, it is downwards along the negative Y-axis.
The resultant of two vectors or any number of vectors, which are not perpendicular, can be determined by resolving each of the vectors into its X- and Y-components, including its sign.
All the X-components are added up to obtain the resultant X-component, Rx, and all the Ycomponents are added up to obtain the resultant Y-component, Ry.
These resultant X- and Y-components, Rx and Ry,are then combined, using the parallelogram method, to obtain the resultant. The Theorem of Pythagoras is used to determine its magnitude (R2 = Rx2 + Ry2) and trigonometry (tan x y) to obtain its direction.
If a body is in equilibrium (stationary or moving at constant velocity) the resultant X-component and the resultant Y-component are both zero
This means that the sum of components acting along the positive X-axis is equal in magnitude to the sum of components acting along the negative X-axis and the sum of components acting along the positive Y-axis is equal in magnitude to the sum of components acting along the negative Y-axis.
(Think of the total pull to the left must be equal to the total pull to the right and the total pull up must be equal to the total pull down, in order to keep the body in equilibrium.)
Note: Remember that sin and cos are co-functions, i.e. sin = cos (90º – ), hence Rx = R cos = R sin (90º – ) and Ry = R sin = R cos (90º – )
Exercise 5:
1 The following pair of numbers represents the X- and Y-components of a vector: 15 cos 30º; –15 cos 60º. The magnitude and direction of this vector is respectively
A 15; 30º above the positive X-axis B 15; 60º above the positive X-axis C 15; 30º below the positive X-axis D 15; 60º below the positive X-axis (2)
2 This sketch represents a bird flying in the direction 45º (measured clockwise from north).
The axis along which the component of its velocity is zero, is A AA B BB C CC D DD (2)
4 A load of weight W is suspended from two strings. F1 and F2 are the forces exerted by the strings on the load, in the directions shown in the figure. Which ONE of the following equations is valid for this situation?
A W2 = F12 + F22 B F1 sin 50º = F2 sin 30º C F1 cos 50º = F2 cos 30º D W = F1 + F2 (2)
5 As a result of the three forces F1, F2 and F3 acting on it, an object at point P is in equilibrium. Which ONE of the following statements is not true with reference to the three forces?
A The resultant of forces F1, F2 and F3 is zero.
Sample
3 A block of mass 4 kg is being pulled over a rough horizontal surface at a constant velocity, by a force of 100 N.
What is the magnitude of the frictional force in newton?
A 40 B 100 C 100 cos 30º D 100 sin 30º (2)
B Forces F1, F2 and F3 lie in the same plane.
C Force F3 is the resultant of forces F1 and F2
D The sum of the components of all the forces in any chosen direction is zero. (2)
6 Jabu and her sister Dudu hold a bag of potatoes at rest, using two ropes tied to the top of the bag as shown. The magnitude of the force which Dudu applies to the rope is greater than the magnitude of the force which Jabu applies on her rope. Which ONE of the following statements is true with reference to the magnitudes of the vertical and horizontal components of the forces applied by Jabu and Dudu?
7 A boy pushes a lawnmower of mass 40 kg with a constant velocity over a horizontal lawn. He exerts a push on the handle of 200 N at an angle of 20º to the horizontal. Calculate the magnitude of the horizontal frictional force exerted on the mower by the lawn. (5)
8 Explain the concept component of a vector with the aid of a sketch. (4)
9 An elephant is dragging a tree trunk along a rough horizontal surface by means of an attached rope which makes an angle of 30º with the horizontal. The force in the rope is 4 000 N.
9.1 Calculate the horizontal component of the force exerted by the rope on the tree trunk. (2)
9.2 Calculate also the vertical component of the force exerted by the rope on the tree trunk and explain why it serves a useful purpose. (4)
10 A heavy box, mass m, is lifted by means of a rope R which passes over a pulley fixed to a pole. A second rope S, tied to rope R at point P, exerts a horizontal force and pulls the box to the right. After lifting the box to a certain height, the box is held stationary as shown in the sketch. Ignore the masses of the ropes. The tension in the rope R is 5 850 N.
10.1 Draw the point P in your answer book and draw and label all the forces acting at P when P is in equilibrium. (3)
10.2 By resolving the force exerted by rope R into components, calculate the 10.2.1 magnitude of the force exerted by rope S (4) 10.2.2 mass, m, of the box. (4)
10.3 Will the tension in rope R INCREASE, DECREASE or REMAIN THE SAME if rope S is pulled further to the right (the length of rope R remains the same)? Give a reason for your choice. (3)
11 Tony would like to cross a gorge. He uses a cable suspended between two vertical steel poles, mounted on the rocks on either side of the gorge (refer to fig. 1). To keep pole P vertical and the pole
9 and cable in equilibrium, a stay wire is anchored at X. Halfway during his crossing, the tension in the cable near to pole P is 1 110 N and the cable makes an angle of 70º with the vertical at pole P (fig. 2). The stay wire breaks if its tension exceeds 2 000 N. Disregard the masses of the cable and the stay wire.
cable stay wire
11.1 Determine the magnitude of the horizontal component of the force which the cable exerts on the top of pole P. (4)
11.2 The horizontal component of the force exerted by the stay wire is of the same magnitude as the horizontal component the cable force. What is the magnitude of the horizontal component of the force exerted by the top of pole P on the cable and the stay wire separately? (2)
11.3 Determine the minimum angle which the stay wire can make with the vertical just before the stay wire breaks. (3)
11.4 If the pole has a mass of 150 kg, calculate the vertical force which the pole exerts on the rocks when the stay wire is anchored at the angle calculated in the previous question. (7)
12 Ziyanda keeps a wooden pole, PQ, vertically upright by two support wires, M and N, attached to its top end at P. In order to keep P stationary, support wire M, which makes an angle of 70º with the horizontal, exerts a force of 50 N on P. Support wire N makes an angle of 30º with the horizontal. The support wires and the pole are in the same vertical plane.
12.1 Calculate the magnitude and direction of the horizontal component of the force that support wire M exerts on P. (4)
12.2 State what is the magnitude and direction of the horizontal component of the force exerted by support wire N on point P, if the pole is to remain stationary. Give a reason for your answer. (3)
12.3 Calculate the magnitude of the vertical component of the force exerted on point P by:
12.3.1 Support wire N (3)
12.3.2 Support wire M (2)
12.4 Calculate the magnitude of the resultant force which the lower end of the pole, Q, exerts on the ground if the mass of the pole is 8 kg. (3)
13 An electric lamp of mass 1,2 kg hangs freely from an electric cable fixed to the horizontal ceiling of Thembi’s room (point R in FIGURE 1). To position the lamp directly over her table, Thembi tied a piece of string to the cable at point P and the other end to point Q on the ceiling, as shown in FIGURE 2. The mass of the string and the cable are both negligible. Thembi, who is a Physics student, wants to find out whether the tension in the electric cable is affected when the lamp is pulled aside.
To determine this, she measures the angle which the cable and the string make with the ceiling and finds them to be 70º and 30º respectively as shown. The tension in the string is 4,1 N. The following questions refer to FIGURE 2.
13.1 Point P is in equilibrium. Explain, using appropriate Physics concepts, what this means. (2)
13.2 The three forces acting at P can be resolved into their X- and Y-components.
Calculate the X- and the Y-components of 13.2.1 the force exerted by the string on point P (6) 13.2.2 the force exerted by the lamp on point P. (3)
13.3 Determine the tension in the cable PR using the components calculated in the previous question. (6)
14 Two learners wish to determine the resultant of three non-linear forces. They attach a sheet of white paper to a force board which is equipped with two pulleys. They knot 3 pieces of gut together and attach the other end of each piece to a mass piece. Two of the pieces of gut are passed over the pulleys while the third piece is allowed to hang. The system is then allowed to reach equilibrium and the position of the knot is marked on the white paper as well as the angles between the pieces of gut at the knot.
The following force diagram was obtained.
14.1 Determine the resultant of the three forces by resolving each force into its X- and Y-components.
Round off to 1 decimal place. (12)
14.2 What conclusion can be drawn about the equilibrium of the knot? (3)
15 Three puppies, Lady, Madat and Terror, are pulling a bone in different directions as indicated in the diagram.
15.1 Copy the table in your answer book and complete it. (12)
Force Fx = F cos Fy
Lady Madat
Terror
15.2 Determine the resultant of the forces. (5)
Body on an Inclined Plane
All the problems basically require that the weight of the object, W, which acts downward, is resolved into two components, i.e. the component WII parallel to the inclined plane and the component W perpendicular to the inclined plane. The angle, , between the weight, W, and the component W , will always be equal to the angle between the incline and the horizontal.
Component WII parallel to the inclined plane:
Component W perpendicular to the inclined plane:
It is calculated using equation WII = W sin , with the angle between the plane and horizontal. This component tends to move the object down the plane.
If some or other force prevents the block from sliding down the plane (e.g. friction), this force is equal in magnitude to this component and opposite in direction.
It is calculated using equation W = W cos , with the angle between the plane and horizontal. This component tends to move the object into the plane.
The normal force exerted by the plane on the block is equal in magnitude to this component and opposite in direction.
1 The diagram represents an object of weight W on a rough inclined plane of length L. The magnitude of the friction force F between the block and the plane is A W B W sin C W cos D W (2)
2 A block of wood of mass 3 kg is placed on an inclined plane as shown in the sketch and remains stationary.
2.1 Indicate all the forces acting on the stationary block by means of arrows and label them. (3)
2.2 Calculate the magnitude of the force R. (2)
2.3 Calculate the force which prevents the block from sliding down the slope. (6)
3 A block of mass 2 kg remains stationary on a plane inclined at 30º to the horizontal.
3.1 Draw a labelled force diagram (not to scale) which shows the block’s weight and the components of its weight parallel and perpendicular to the inclined plane. (5)
3.2 Calculate the values of the above two components of the block’s weight. (8)
3.3 What is the magnitude and direction of the friction force between the block and the plane? (3)
4 The figure shows a block of mass M held motionless on a frictionless plane by means of a light rope which is parallel with the plane. The plane is inclined at 35º to the horizontal.
4.1 Draw a force diagram for the block, showing all the forces acting on it. (6)
4.2 Draw a diagram showing how the gravitational force acting on the block can be resolved into components parallel with and perpendicular to the plane. (5)
4.3 If the force in the rope is 5 N, calculate the weight of the block. (4)
Contact Forces
Contact forces are applied between two objects that are in physical contact with each other, e.g. pushing forces, pulling forces, normal forces and frictional forces. Non-contact forces are forces that act over a distance, e.g. magnetic forces, gravitational forces, electrostatic forces.
The force exerted by a string on an object is always directed away from the object from the point where it is attached, because a string can only pull, not push. The tension in a light, inelastic string is the same everywhere, but the tension is different in two separate pieces of string.
inside the
Static forces of friction (forces of friction between two bodies trying to move over each other) can have a range of values from zero up to a maximum value of fsmax = s FN, where s is the static coefficient of friction for a static body on a surface. (Static frictional force fs FN and s is the constant of proportionality with no units.)
s = fsmax N (on data sheet)
The dynamic force of friction (force of friction when the bodies are moving over each other) for a surface is constant and is reasonably independent of the speed of motion.
Dynamic force of friction fk = k FN, where k is the dynamic coefficient of friction for a moving object. (Dynamic frictional force fk N and k is the constant of proportionality with no units.)
k = fk N (on data sheet)
The force of friction does not depend on area of contact of the surfaces if the normal force is constant. Remember that for an object on an inclined surface, the normal force (FN) is not equal to the weight (W) in magnitude, but FN = F = W cos
The resultant force for an object resting on an inclined plane, is zero.
The forces acting parallel to the plane, are the component of the weight pulling it down the incline, FII, and the static frictional force fsmax, which acts up the incline.
For equilibrium, fsmax = FII .......................[1]
But fsmax = s FN = s F = s mg cos .........[2]
FII = mg sin .........[3]
Substitute [2] and [3] into [1]: s mg cos = mg sin , hence s = (mg sin ) = tan .
The same argument can be used for an object sliding down the incline at constant velocity, but then k is obtained.
Exercise 7:
1.1 the type of force of which weight is an example. (1) 1.2 forces such as gravity, which act at a distance. (1) 1.3 the perpendicular component of the force exerted by a surface on an object in contact with the surface. (1)
2 A boy applies a force, F, to a crate as shown.
The crate moves at constant velocity, v. The crate is then turned on its end, reducing the surface area in contact with the floor by half. In this new orientation, he pushes the crate across the floor with the same constant velocity.
The force he now applies is ....
3 A wooden block of mass m
Each surface in contact with another experiences two contact forces: A normal force (FN) and a force of friction (f).
Normal forces are exerted at a right angle to any surface by both objects that are in contact. An object resting on a horizontal surface experiences a normal force equal to its weight in magnitude. If an object rests on an inclined plane, the magnitude of the normal force of the object on the plane can be calculated using W = W cos = mg cos , where is the angle between the plane and horizontal. The normal force by the plane on the object is therefore also W = W cos = mg cos
Forces of friction act along the surface between two bodies when they move or try to move over each other, and in a direction so as to oppose relative motion of the surfaces.
This is because of the interlocking of the irregularities in the surfaces, which impedes motion. Forces of friction depend on the nature of the surfaces (i.e. smoothness, indicated by the coefficient of friction ) and the normal force (indicated by FN).
Forces of friction exerted by two bodies on each other can be static or dynamic friction forces.
Which ONE of the following expressions represents the magnitude of the frictional force on the block?
4 A lady injured her back when she slipped and fell in a supermarket. She holds the owner of the supermarket accountable for her medical expenses. The owner claims that the floor covering was not wet and meets the accepted standards. He therefore cannot accept responsibility.
The matter eventually ends up in court. Before passing judgement, the judge approaches you, a science student, to determine whether the coefficient of static friction of the floor is a minimum of 0,5 as required. He provides you with a tile from the floor, as well as one of the shoes the lady was wearing on the day of the incident. 4.1 Write down an expression for the coefficient of static friction. (2)
4.2 Plan an investigation that you will perform to assist the judge in his judgement. Follow the steps outlined below to ensure that your plan meets the requirements:
4.2.1 Formulate an investigative question. (2)
4.2.2 Apparatus: List ALL the other apparatus, except the tile and the shoe, that you will need. (2)
4.2.3 A stepwise method: How will you perform the investigation? Include a relevant, labelled free-body diagram. (5)
4.2.4 Results: What will you record? (2)
4.2.5 Conclusion: How will you interpret the results to draw a conclusion? (2)
5 A bullet of mass 2 g is fired horizontally into a stationary wooden block of mass 1 kg. The bullet strikes the block at a velocity of 490 m s-1. The impact causes the block-bullet system to slide a distance of 20 cm from its original position before coming to rest. Assume the block experiences a constant frictional force.
5.1
Draw a free-body diagram of the block-bullet system. Show and label ALL the forces acting on the blockbullet system while it is moving. (3)
5.2 Assume, while the block-bullet system is moving, it experiences a constant acceleration of –2,4 m s-2. Calculate the coefficient of friction between the wooden block and the surface. (5)
6 A 10 kg box rests on a horizontal floor. The coefficient of static friction, μs = 0,4 and the coefficient of kinetic friction, μk = 0,3. Determine the force of friction acting on the box if a horizontal external applied force is exerted on it of magnitude 6.1 38 N 6.2 40 N (7)
7 Your friend wants a ride on his sled. If you are on flat ground, will you exert less force if you push him or pull him?
Explain your answer. (7)
8 The skier in the diagram descends the 30º slope. Assume the coefficient of dynamic friction is 0,1.
8.1 Draw the free-body diagram for the skier. (3)
8.2 Calculate her acceleration. (6)
8.3 Calculate the speed she will reach after 4 s. (4)
8.4 Suppose that the snow is slushy and the skier moves down the 30º slope at constant speed. What can you say about the coefficient of friction, μk? Explain. (5)
9 A block is given a push so that it slides up a ramp. After reaching its highest point, it slides back down. Why is the magnitude of its acceleration less on the descent than on the ascent? (8)
10 You can hold a book against a rough wall and prevent it from slipping down by pressing only horizontally. How can the application of a horizontal force keep an object from moving vertically? (4)
11 The coefficient of dynamic friction between a 35 kg box and the floor is 0,3.
11.1 What horizontal force is required to move the crate at a constant speed across the floor? (5) 11.2 What horizontal force is required if μk is zero? (2)
14 On a rainy day, you worry about parking your car in your driveway, which has an incline of 12º. The Jones’s driveway has an incline of 9º, and Bongi’s driveway across the street has one of 6º. The coefficient of static friction between the tyres and wet concrete is 0,15. Which driveway(s) will be safe to park? (8)
15 A crate on the bed of a truck slides forward when the driver decelerates from 40 km h-1 to rest in 3,5 s or less, but not if he decelerates in a longer time.
15.1 Calculate his acceleration. (6)
15.2 Calculate the maximum coefficient of static friction between the crate and the bed. (6)
16 A rollercoaster, mass m, reaches the top of the hill with a speed of 6 km h-1. It then descends the hill, which is 45 m long and at an average angle of 45º. Assume μk = 0,12.
Sample
12 A force of 40 N is required to start a 5,1 kg box moving across a horizontal concrete floor.
12.1 What is the coefficient of static friction between the box and the floor? (3)
12.2 If the 40 N force continues, the box accelerates at 0,7 m s-2. What is the coefficient of kinetic friction? (7)
13 A drawer, mass 20 kg, becomes stuck. The owner gradually pulls with more and more force. When the applied force reaches 8 N, the drawer suddenly opens, throwing all the contents to the floor. Find the maximum coefficient of static friction between the drawer and the cabinet. (4)
16.1 Calculate the frictional force between the rollercoaster and the surface of the hill. (4)
16.2 Calculate the net force acting on the rollercoaster. (3)
16.3 Calculate its speed when it reaches the bottom. (8)
17 An 18 kg box is released on a 37º incline and accelerates down the incline at 0,27 m s-2
17.1 Calculate the friction force impeding its motion. (8)
17.2 Calculate the coefficient of friction. (3)
18 An engineer has to design a supermarket, where several ramps have to connect different parts of the supermarket and where customers will have to push grocery carts up the ramps. A survey shows that almost no-one complains if the force required to push the trolley does not exceed 50 N. Will a 5º slope be too steep for a 30 kg grocery cart, full of groceries? Assume μk = 0,1. (10)
19 A 3 kg block slides at a constant velocity of 7 m s-1 along a horizontal surface. It then strikes a rough surface, causing it to experience a constant frictional force of 30 N. The block slides 2 m under the influence of this frictional force before it moves up a frictionless ramp inclined at an angle of 20º to the horizontal, as shown. The block moves a distance d up the ramp, before it comes to rest.
19.1 Show by calculation that the speed of the block at the bottom of the ramp is 3 m s-1. (5)
19.2 Draw a free-body diagram to show all the forces acting on the block in a direction parallel to the incline, whilst the block is sliding up the ramp. (2)
s-1
19.3 Calculate the distance, d, the block slides up the ramp. (5)
20 John applies a force F to help his friend in a wheelchair to move up a ramp of length 10 m and a vertical height of 1,5 m as shown in the diagram. The combined mass of his friend and the wheelchair is 120 kg. The frictional force between the wheels of the wheelchair and the surface of the ramp is 50 N. The rotational effects of the wheels of the wheelchair may be ignored. The wheelchair moves up the ramp at constant velocity.
h = 1,5 m F
10 m
20.1 What is the magnitude of the net force acting on the wheelchair as it moves up the ramp? Give a reason for your answer. (2)
20.2 Calculate the magnitude of force F exerted by John. (4)
21 A crate of cool drinks, mass 230 kg, is at rest on an incline that makes an angle of 30º with the horizontal.
21.1 Calculate the magnitude of the force that prevents the crate from sliding down the slope. (3)
21.2 Calculate the magnitude of the coefficient of static friction between the crate and the slope. (4)
Newton’s First Law of Motion
Newton’s First Law of Motion: An object continues in a state of rest or uniform velocity unless it is acted upon by an unbalanced (or net or resultant) force.
In other words, an object at rest tends to remain at rest and an object moving tends to continue moving at that constant velocity when the resultant force is zero (no unbalanced forces or resultant or net force acting on it) OR an object at rest remains at rest and a moving object continues moving at a constant velocity unless an unbalanced force acts upon it.
Applicable to all cases where Fnet = 0
Therefore, whenever a body is at rest or moving at a constant velocity, the resultant force is zero, i.e. all the forces in the X-direction and all the forces in the Y-direction must add up to zero. Also keep this in mind when you draw a force diagram or a free-body diagram. Inertia of a body is its tendency to resist any change in its motion. The inertia of an object is determined by its mass. (NB: Inertia is not a force – it is a property!) Make sure whether you have to explain your answer using Newton’s First Law or the property, inertia.
1.1 the resistance of an object to changing its position. (1)
2 Sipho, a five year old boy, stands in the coach of a passenger train while the train is moving at a high constant speed on the straight section of a line between two suburban stations. He jumps straight upward and lands on his feet again. The position where he lands will probably be ….. from where he jumped.
A the same position B slightly forward
C slightly backward D a noticeable distance backward (2)
3 A workman, standing on a scaffold, lowers an object of weight 300 N by means of a rope, at constant speed. If the weight of the rope is negligible, the force that the man exerts, is
A equal to 300 N
B constant and less than 300 N
C greater than 300 N D less than 300 N and decreasing. (2)
4 According to Newton’s First Law
A the acceleration of a body is directly proportional to the force causing the acceleration.
B the velocity of a body remains constant unless an unbalanced force acts upon it.
C the impulse on a body is the product of the mass and the change in velocity.
D the sum of the gravitational potential energy and the kinetic energy of a body is constant. (2)
5 Ziyanda places 4 two-rand coins, one above the other, on a horizontal table. Friction is negligible. She shoots another two-rand coin (X) which strikes the lowest coin horizontally. Coin X displaces and replaces the lowest coin, while all the other coins remain at the same place on the table. Which ONE of the following laws explains why the top three coins are not displaced?
A Newton’s First Law of Motion
B Newton’s Third Law of Motion.
C Law of Conservation of Momentum
D Law of Conservation of Energy. (2)
7 An object is moving to the right at constant velocity whilst a force F is acting on it. The magnitude of the frictional force acting on the object is
A zero.
C smaller than F.
8 If an object is at rest, it can be concluded that
A no external forces act on it.
B the net force on the object is zero.
B equal to F.
D larger than F. (2)
C the object is prevented from moving by friction.
D the object experiences a net force greater than zero. (2)
9 A driver of a motor places a book on the level dashboard of his car in front of him. He observes that the book slides towards the windscreen when the brakes are applied. Which of Newton’s laws best explains the motion of the book?
A First Law
Sample
6 A body slides along a frictionless, horizontal surface at constant velocity.
For which ONE of the following pairs are the magnitudes of both physical quantities zero?
A Displacement and momentum
C Displacement and resultant force
B Acceleration and momentum
D Acceleration and resultant force (2) X two-rand coins displaced coin
B Second Law
C Third Law D Law of Universal Gravitation (2)
10 While driving very fast along a slippery road, the driver of a car is dismayed to find that the car does not respond when he attempts to turn a corner.
10.1 Name and state a law of motion which can be used to explain why the car does not respond to the steering. (4)
10.2 From the law you have quoted above, state what is necessary in order to turn the car. (3)
11 The driver of an empty delivery van which is travelling at 18 m s-1 has to make an emergency stop. He applies the brakes. What property of matter is illustrated by the fact that a loose object in the van carries on moving after the van has stopped? (2)
12 The sketch shows a sheet of paper with a metal sphere at rest on it, lying on the smooth surface of a table. The sheet of paper is carefully pulled at constant velocity towards the right without the sphere moving relative to the paper. The moment that the force acting on the paper stops, the sphere starts to roll towards the right.
12.1 Which property of matter is illustrated by the sphere after the paper stops? (3)
12.2 Name and formulate the law that pertains to the behaviour of the ball after the paper stops. (4)
12.3 Explain why the wearing of safety belts by the occupants of a car, can save their lives during a collision. (6)
13 A car of mass 800 kg is being towed by a truck, using a tow rope. When the car is being towed along a level road at constant speed, the force exerted on the rope is 100 N. Explain why a force is needed to tow the car at a constant speed. (2)
14 Name and state the Physics law which applies to the following real-life situation: A bottle is placed on a sheet of paper. When the paper is jerked in a horizontal direction, the bottle remains in the same position. (3)
15 A lifebuoy is dropped from a hovering rescue helicopter to a man in difficulties off the Clifton beachfront. He is hoisted up together with the buoy at a constant speed. Ignore air friction and answer the following question: What is the magnitude of the upward force in the cable while the man is out of the water and being hoisted upwards? (3)
16 A flatbed truck carrying a concrete block near the back of its flat bed, as shown, is travelling due west along a straight level road. The truck collides head on with a concrete bridge support and stops. The concrete block slides into the back of the driver’s cab, and bends the cab inwards before stopping. Assume friction between the floor of the truck and the block to be negligible. Considering the stages of motion of the block on the truck, namely sliding along the bed of the truck and colliding with the back of the cab, explain briefly how Newton’s First Law of motion is illustrated during the motion of the block. (2)
Exercise 8:
17 The diagram shows a truck with a crate immediately behind the cab in position A. The truck suddenly accelerates forward and the crate slides towards the back of the truck coming to rest at the position indicated B.
17.1 Draw a force diagram, indicating and labelling the horizontal force(s) acting on the crate while the truck accelerates. (2)
17.2 Briefly explain why the crate slides towards the back of the truck. (4)
18 A dummy is placed on the driver’s seat of a car during an experiment to investigate the effects of collisions. The car starts from rest and accelerates towards a wall. The car collides with the wall and the dummy carries on moving forward and collides with the windscreen. Name and state the relevant law of motion which explains this situation. (4)
19 A boy, on frictionless roller-skates, is standing at rest on the back of a truck which is travelling at a constant velocity of 10 km h-1 as shown.
19.1 While the truck is moving, in which direction and with what speed must the boy move, relative to the truck, in order to be stationary relative to the ground? (2)
19.2 If the boy is initially at rest on the truck, in which direction, relative to the truck, will the boy be displaced (if at all) if the truck slows down? (2)
19.3 Name and state, in words, the law used in the previous question. (4)
20 A Boeing 747 accelerates on a runway.
20.1 Draw a labelled free body diagram, showing all the forces acting on the Boeing before take-off. (4)
20.2 The captain detects a problem with his aircraft during take-off. He applies brakes and brings the aircraft to a standstill just before the end of the runway. Passenger A sits in his seat without being buckled up. Describe the motion of passenger A at the moment that the pilot applies the brakes. (1)
20.3 Which law of motion was used to predict the motion in the previous question? Write down this law in words. (3)
Newton’s Second Law of Motion: Law and Experiment
Newton’s Second Law of Motion: An object experiencing an unbalanced/resultant/net force will accelerate. The magnitude of this acceleration is directly proportional to the magnitude of the force and inversely proportional to the mass of the object. The acceleration is in the direction of the unbalanced/resultant/net force.
It can be expressed by the equation Fnet = ma (on data sheet)
Acceleration is the rate of change of velocity. Know the experiment verifying the law. The graphs obtained from this experiment: When the mass (m) is kept constant, the following graph is obtained: Because it is a straight line, we can conclude that acceleration is directly proportional to resultant force, hence net
The solid line represents motion where friction is absent or has been compensated for (e.g. using an inclined plane) and the broken line represents motion where friction is present and has not been compensated for.
1.1 the rate of change of velocity. (1)
1.2 the quotient of the net force acting on a body and the acceleration produced in that direction. (1)
2 During a trolley experiment a pupil investigates the relationship between acceleration a and mass m of a trolley, with the resultant force F on the trolley kept constant.
Which ONE of the following graphs would best represent the outcome of such an investigation? (2)
3 In an investigation of Newton’s Second Law, graphs of acceleration a versus force F for two bodies X and Y are obtained. Which ONE of the following statements is true?
A The bodies have equal mass.
B Body X has the smaller mass.
C Body Y has the smaller mass.
D Mass does not affect the gradient of the curve. (2)
4 In an investigation of the relationship between acceleration (a) and force (F) for two objects (P and Q) moving on a frictionless surface, the following graph was obtained:
Which ONE of the following statements is true?
A Object Q has a smaller mass than object P.
B Object Q has a bigger mass than object P.
C The gradient of the graph is not affected by the mass of the object.
D Objects P and Q have equal mass. (2)
5 When a resultant force of 1 N acts on an object of 1 kg, the object …
A will move a distance of 1 meter before it comes to rest.
B moves a distance of 1 meter in 1 second.
C moves at a velocity of 1 m s-1.
Sample
We can find the frictional force by taking the difference between the two graphs on the force-axis (indicated by the double arrow).
The gradient = (From Fnet = ma)
When the force is kept constant, the following graphs are obtained:
From the graphs it is evident that acceleration is inversely proportional to the mass of the object, thus . The gradient of the straight- net = ma)
D accelerates at 1 m s-2. (2)
6 During an investigation of Newton's second law, trolleys of different masses were accelerated with the same applied force. To compensate for friction, the trolley track was slightly lifted and the angle between the trolley track and the horizontal was kept constant throughout the whole investigation. The dotted line on the sketch graph represents the expected results.
Point R is on the line, but points P and Q deviate slightly from the expected positions.
Which ONE of the following best explains why points P and Q are not on the straight line?
A The track was lifted too much, causing the trolleys to accelerate more than expected.
B The frictional force increases when the mass of the trolleys increases.
C The lifting of the track caused the lighter trolleys to accelerate more than expected.
D The frictional force for the lighter trolleys was more than expected. (2)
7 A pupil investigates how the acceleration of an object to which a constant force is applied, varies with changing mass. In order to do this, he accelerates first one, then two and lastly three identical trolleys at a time, down a friction-compensated slope as shown in the figure. He uses an elastic cord attached to a stick.
His results are as follows:
7.1 Plot the results on a sketch-graph of acceleration against mass. Label the axes. (6)
7.2 What possible mathematical relationship between acceleration and mass for constant force does the shape of the graph suggest? (3)
7.3 In order to confirm the relationship suggested above, it is desirable to plot a graph on different axes that would give a straight line.
7.3.1 Draw and label the new system of axes. (3)
7.3.2 Calculate the values needed and complete the graph. (6)
8 Susan conducts an experiment to investigate the relationship between the resultant force exerted on an object and the acceleration produced. To start off with, she sets up a runway so that a trolley will pull a tickertape through a ticker-timer when it travels down the slope.
Ticker-timer
8.1 Susan gives the trolley a brief push and then immediately switches on the ticker-timer. A section of the ticker-tape is shown below.
Ticker-tape Trolley Direction
8.1.1 What needs to be done to the slope in order to be able to call it friction compensated? (2)
8.1.2 How should the dots on the ticker-tape be spaced once the slope has been friction compensated? (2)
8.1.3 What is the magnitude of the resultant force on the trolley when it moves freely down the friction compensated slope? (2)
8.2 Once the slope has been friction compensated, Susan accelerates the trolley, first by means of one elastic cord, then repeats the process with two cords and finally with three elastic cords, all stretched to the same extent. The three sets of results are graphed on the velocity-time axes.
8.2.1 What does the shape of the graph and any one of the sets of results indicate about the relationship between acceleration and force? (4)
8.2.2 What does a comparison of the slopes of the three graphs suggest about the relationship between acceleration and force?
8.2.3 If Susan wishes to satisfy herself that the answer to 8.2.2 is correct,
A What type of graph should she plot, and (3)
B what should the slope of the graph be? (3)
9 A team of pupils investigates the relationship between the applied force, F, on a trolley and the acceleration a, of the trolley. The trolley is accelerated using stretched rubber bands along a plank, tilted to compensate for friction. The procedure is repeated using 1, 2 and 3 rubber bands which are parallel and extended to the same length. The trolley pulls a ticker tape through a ticker timer with a frequency of 50 Hz. The result for the procedure with 2 rubber bands is given in the sketch. On the strip of tickertape lines p, q, r, s and t are shown. The lines mark every FIFTH (5) ticker dot. The distances between the lines are indicated on the tape. Lines are drawn at every FIFTH ticker dot.
9.2 The experimental results obtained by the pupils are represented by the circular graph points shown. What relationship between acceleration and number of rubber bands used, is suggested by the graph? (2)
9.3 The pupils repeat the experiment, but using two trolleys, one stacked on top of the other. The results are represented by the x points on the graph. From the two graphs, what relationship can you infer between acceleration and mass, for a given applied force? Explain your reasoning. (4)
10 A length of ticker tape is placed through a ticker timer with a frequency of 20 Hz.
9.1 Calculate
9.1.1 the time taken for the trolley to move the 20 mm corresponding to the change in position from p to q on the tape. How does this compare with the time to move the 140 mm from s to t? (3)
9.1.2 the average velocity over interval qr. (3)
9.1.3 the average velocity over the interval st. (2)
9.1.4 the average acceleration of the trolley. (3)
The tape is attached to a trolley of mass 2 kg which is then pulled by a constant force.
The portion of the ticker tape shows the dots made by the ticker timer marked off in 10 dot intervals (PQ, QR and RS) and the distance covered for each 10 dot interval.
10.1 Calculate the average velocity of the trolley represented by the dots between P and Q. (4)
10.2 Calculate the average velocity of the trolley represented by the dots between R and S. (4)
10.3 Calculate the acceleration of the trolley. (4)
10.4 How would the spacing of the dots on the tape change if the mass of the trolley were doubled, using the same force? Explain your answer. (4)
11 Learners at a school were required to achieve the following outcome: Learners should be able to use graphical methods to determine how the acceleration of an object, to which a constant force is applied, varies with changing mass. In order to do this, Keshev conducted the following experiment: He first accelerated one, then two and lastly, three identical trolleys, mass 1 kg each, down a frictioncompensated track, as shown. Keshev used a metre ruler and an elastic cord, stretched to the same length and at the same angle for each of the three situations, in order to maintain a constant force.
metre ruler elastic cord
His results and the graph he drew, were as follows:
Mass of trolleys (kg) Acceleration (m s-2)
0,96
0,49
0,31
Sample
In order to obtain a relationship between acceleration and mass it is necessary to plot a graph which tends to produce a straight line.
11.1 Draw and label a new system of axes which should produce a straight line graph. Do the necessary calculations, plot the points and sketch the graph. (5)
11.2 State the relationship between the acceleration, a, and the mass, m. (2)
11.3 Determine the gradient of the graph sketched in the question above. (3)
11.4 What physical quantity does the gradient calculated represent? (2)
11.5 State, in words, Newton’s Second Law of Motion. (3)
12 Overloading of motor vehicles is one of the many causes of road accidents on our country’s roads. A Physical Sciences class investigated the relationship between force, mass, acceleration, stopping distance and time. Different groups investigated different relationships. One group did the investigation below. These learners used the following apparatus in their investigation: Four trolleys, a ticker timer, four lengths of ticker tape, a ramp (runway), a non-elastic string, a pulley, a 250 g mass piece. They set up the apparatus as shown: The runway was inclined to compensate for friction. The mass piece accelerated the trolley down the incline. The incline was kept constant for the duration of the experiment.
The mass was increased by stacking the trolleys one on top of the other after each reading. After analysis of the four ticker tapes, the following results were recorded:
2
4 1,9
12.1 Formulate an investigative question for this investigation. (2)
12.2 Copy and complete the table. Draw a graph of the dependent variable on the y-axis versus the independent variable on the x-axis. Indicate the following on your graph: Suitable heading, labels for the axes, appropriate scale, the four plotted points.
Draw a straight line through the LAST THREE points. (6)
12.3 Draw a conclusion from the graph. (2)
12.4 Extrapolate (extend) the line on the graph.
Give a possible reason why the first point does NOT lie on the line. (2)
12.5 Use your conclusion in question 12.3 to explain to motorists why overloading is dangerous. (3)
13 Learners set up a runway to investigate the relationship between mass and acceleration. The following are the results that were obtained during the investigation:
m (trolley mass piece)
(g) a (m s-2) ma (N)
200 10
400 5
13.1 Calculate the values of ma in the table. (2)
13.2 Give the scientific reason why learners have placed a book under the track. (2) 13.3 Which variable is kept constant during this investigation? (1)
13.4 Draw a sketch graph of acceleration (on the Y-axis) versus mass (X-axis). (3)
13.5 Write down the mathematical relationship between acceleration and mass from the learners’ conclusion based on the graph obtained. (2)
13.6 Explain the relationship that you have written in the previous question. (1)
13.7 Name the law that was investigated by the learners. (1)
4 They repeat this procedure by adding mass pieces one at a time and thus increasing the force acting on the trolley.
5 After analysing the tapes, they tabulate their results as shown: Force (N) Mass of trolley (kg) Acceleration (m s-2)
0,25 0,50 0,50
0,50 0,50 1,00
0,75 0,50 1,50
14.1 Formulate an investigative question that the learners would have for this investigation. (2)
14.2 Write down a possible hypothesis for this investigation. (2)
Sample
14 High school learners carry out an investigation to find the relationship between force and acceleration of an object. They use the following materials: Trolley and trolley track, ticker timer, ticker tape, four identical mass pieces, a string and a power source. Their method is as follows:
1 They place the trolley on the track and tilt the track so that the trolley is just about to move down the incline on its own.
2 They attach a ticker tape to the trolley and pass the tape through the ticker timer.
3 They attach a mass piece to the trolley as shown in the diagram below, switch the ticker timer on and allow the trolley to accelerate down the track.
14.3 Which variable did the learners keep constant in this investigation? (1)
14.4 Write down the independent variable. (1)
14.5 Write down the dependent variable. (1)
14.6 Deduce the relationship between the net force and the acceleration from the learners’ results. (2)
14.7 In this investigation, why would the force applied by the mass pieces be regarded as the net force?(4)
14.8 A block of mass 5 kg is pulled with a force of 16 N on a frictionless surface.
Calculate the acceleration of the block. (3)
15 A group of learners set up an arrangement as shown in the diagram to investigate the relationship between force and acceleration.
They kept the mass of the trolley constant.
The time, t, for the trolley to move a distance of 1 m from rest was determined for various values of the pulling force F. The acceleration is then calculated using the equation 2, where The following values were obtained and are recorded in the table:
The relation between force F (N) and acceleration a (m s-2) is shown in the graph.
15.1 Give a suitable hypothesis for this investigation (2)
15.2 What does the gradient of the graph represent?
(You do not have to calculate the gradient.) (1)
15.3 Draw a sketch of the trolley and draw arrows on your sketch to represent all the external forces acting on the trolley while it is accelerating along the bench. Label each force. (4)
15.4 Name the law which is investigated in this experiment. (1)
15.5 The mass of the trolley is now doubled and the whole investigation is repeated. What is the effect of the increased mass on the gradient of the graph?
(Use only REMAINS THE SAME, INCREASES or DECREASES as your answer.) (1)
Newton’s Second Law of Motion: Applications
Before applying Fnet = ma (on data sheet), make sure that Fnet is the resultant/net force and not the force applied (i.e. you have considered ALL the forces acting on the body)!!
Newton's Second Law of Motion: The equations Fnet = ma and Fnet = t are equivalent: Fnet = t = (pf – pi t = (mvf – mvi t = m(vf – vi t = ma.
When an object accelerates, the equation Fnet = ma can be applied separately in the x and y directions.
• All the examination preparation for physics and chemistry you need in small, digestible, bite-sized ‘quanta’.
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