DINALEDI

TEACHER TRAINING MATERIALS

PHYSICAL SCIENCES

MODULE 1

NOVEMBER 2006

The quantities used in Physics may be divided in

Unit 1: V

A vector is a physical quantity that has both magExamples of vectors: Displacement Force Momentum Velocity Acceleration A scalar is a physical quantity that has magnitudExamples of scalars Distance Mass Time Speed Volume Energy VECTOR REPRESENTATION An arrow is used to represent a vector graphicallymagnitude of the vector. All the examples of vectdescribed above. VECTOR ADDITION There are two main methods of vector addition, nmethod. Both methods give the same results. VECTORS AND THE CARTESIAN PLANE In order to give direction we make use of angles, when teaching vector representation. This allowsimportant for all vector calculations and represen RESULTANT When two or more vectors are added, we get a re When two or more forces act simultaneously at aat the same point that would produce the same e

Lesson 1 ectors and Scalars

MODULE 1

to two groups: vectors and scalars.

nitude and direction.

e (size) only.

. The length of the arrow would represent the ors stated above can be represented using the arrow

amely the head to tail method and the parallelogram

the Cartesian plane would be the best place to start us to indicate the frame of reference, which is tations.

sultant vector.

point, the resultant force is that single force, applied ffect.

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A resultant displacement is the single displacement that brings about the same change in position as all the other displacements together. EQUILIBRANT The equilibrant is equal in magnitude and opposite in direction to the resultant. NOTE: A body moves with constant velocity if it covers equal displacements in equal time intervals. Finding the Resultant Vector When two or more vectors, act simultaneously at the same point, the resultant vector (i.e. the vector sum) can be determined using the tail-to-head method or the tail-to- tail method. Head to Tail Method If the vectors are drawn using the tail to head method, then the resultant is the closing side of the vector diagram. Note:

1. If one is finding the resultant of two vectors, then the resultant is the closing side of the vector triangle.

2. If one is finding the resultant of three vectors, then the resultant is the closing side of the vector quadrilateral.

3. If one is finding the resultant of many vectors, then the resultant is the closing side of the vector polygon.

Example: A bird flies a distance x on a bearing of 400 and then flies a distance y on a bearing of 1100 Find the resultant vector. Solution: By Calculation: Make a sketch of the vector diagram using the Tail to Head method.

Sketch: N

N

N

1100

y

x θ

400 R

2

The magnitude of the resultant can be found by the theorem of Pythagoras if and only if one of the angles in the diagram is equal to 90°, otherwise the cosine or sine rule could be used to do calculations. Solution: By the graphical method Draw the vectors according to scale using the Tail to Head method. Note: The following is a rough sketch of the diagram that must be drawn to scale in order to obtain the answer graphically.

0° 0°

1100 90°

s1 s2 40° 180° θ R

90° 180° Tail-to-Tail Method (Parallelogram Method) If the given vectors are drawn using the parallelogram method, then the resultant is given by the diagonal of the parallelogram, where the given vectors are adjacent sides. Note: This method can be used to find the resultant of only two vectors at a time. Let us look at finding the resultant of the previous example using the parallelogram method. Solution:

Sketch: N

y

x

R 1100

400

y

The vectors must be drawn such that there is aNorth line at the tail of each vector. The tails ofsuccessive vectors should be drawn at the headof the preceding vector. The vectors can bedrawn in any order. All the North lines must beparallel to each other.

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The magnitude of the resultant can be found using the cosine rule: The direction of the resultant can be found using the sine rule or any other relevant trigonometric ratio. Both the magnitude and the direction of the resultant can be found using the graphical method, i.e. by drawing the above sketch according to scale.

TRIANGLE LAW FOR THREE FORCES IN EQUILIBRIUM:

When three forces are in equilibrium (i.e. the resultant is zero), they would form a closed vector triangle when drawn using the tail-to-head method.

Similarly, if the resultant of many vectors acting at a point is zero, then these vectors would form a closed vector polygon when drawn using the tail-to-head method.

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1. The sine rule is used when: You have any two sides and any known angle,

OR two angles and a known side.

cC

bB

aA sinsinsin

==

2. The cosine rule is used when:

You have two (2) sides and the included angle (between the known sides) c2 = a2 + b2 - 2ab.cos C

The above rules can be found on the information sheet. USING THE SINE RULE: When? If none of the angles in the closed vector triangle forms a right angle (90˚), (Side, side, angle.) or (angle, angle, side). QUESTION 1: Z Y 100º 30º X A hanging flower basket is suspended from an iron frame at point Z by two wooden struts, as shown in the diagram. The total mass of the flower basket is 1,5 kg. 1.1 Draw a rough-labelled sketch of all the vectors. 1.2 Determine by means of an accurate construction (scale: 10mm = 15 N) OR by calculation

the magnitude and direction of the force in struts X and Y. 1.3 Use the sine rule to calculate the magnitude and direction of USING THE COSINE RULE: (side, included angle, side) A pilot wants to fly at 200km/h on a bearing of 300˚ but the wind blows at 60km/h on the bearing 60˚. What velocity (v) must he fly relative to still air? The angle between R and W = 120° Put the wind head- to- head with the resultant. The angle between the resultant and the wind remains 120˚. Please note: in the diagrams below, angles and lines are not drawn to scale.

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v2 = R2 + w2 -2 R w cos 120°

60 k

m.h-1

120o 200 km.h -1

RW

θV

= 2002 + 602 – 2. 200. 60 cos 120° = 40000 + 3600 + 12000 = 55 600 v = 235.8 km. h-1 Now you have to work out the bearing. Show your calculation: Note: The addition of vectors is commutative (i.e. the order in which the vectors are drawn makes no difference) as is illustrated below.

120o RW

θV

300o

c

b

a

b

ac

R

b

ac

R

b a

c

Ror

or

What does the term “addition” of vectors mean? Think of vectors working in the same direction as well as in different directions. QUESTION 2: How would you explain the different methods of adding vectors? (Head-to-tail method, parallelogram method; adding vectors at 90°.) Vectors acting at 1800 to each other F1 = 110 N at 270° F2 = 100 N at 90°

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If these two vectors are placed head to tail then: rr 21 FFFR

r+=

= 10 N, 270° This shows that the resultant force of 10 N acts in the direction to the left. IMPORTANT: Always explain the direction of the answer when calculating mathematical answers of vectors for scientific problems. This is an example of vector addition. The resultant of any two vectors, R, has a maximum magnitude when the vectors have the same direction, i.e. the angle between the vectors is 0º. R has minimum magnitude when the vectors are acting in the opposite direction to each other i.e. the angle between the vectors is 180º. VECTORS ACTING ON AN OBJECT AT A RIGHT ANGLE (90°) When adding vectors at 90° it is necessary to use the theorem of Pythagoras (use the magnitudes of the vectors mathematically): hypotenuse2 = x2+ y2 THUS: Vector 22 = Vector 12 + Vector 32 Vector 2 (hypotenuse) Vector 1 (opposite) θ

Vector 3 (adjacent) The six trigonometric ratios could be used when solving problems involving right-angled triangles.

Sin θ = opposite side / hypotenuse Cos θ = adjacent side / hypotenuse Tan θ = opposite side / adjacent

RESOLUTION OF VECTORS: CALCULATING COMPONENTS OF VECTORS Any vector quantity can be resolved (split) into components along any axes. In school we often resolve vectors into components that are perpendicular to each other.

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QUESTION 3: A boy (mass 50 kg) flies a kite on an open field. The wind causes a pulling force in the rope of 1000 N and the boy feels this effect in his hands as he holds onto the rope attached to the kite. At the same time that he experiences a forward pull, he feels a lifting effect. This implies a horizontal and a vertical force together has a resultant force of 1000 N on the boy. Is there a risk that the boy might lift off the ground? Use θ = 30°. Fy F = 1000N θ Fx

Calculate the vertical and horizontal components as Fy___________________________________________________

_____________________________________________________

_____________________________________________________ Fx___________________________________________________

_____________________________________________________

_____________________________________________________Which component might endanger the boy to lift off the ground? ____________ Why?_______________________________________________________________

OBJECT ON AN INCLINED PLANE When an object is on a slope, the force due to gravity, Fg, acting on the object can be broken into two components, which are at right angles to each other namely the force acting parallel to the slope, and the force acting perpendicular to the slope. The parallel component, F║, is the force, which causes the object to slide down the slope, and the perpendicular component, F┴, is the force with which the object pushes onto the surface. Suppose our object has a mass of 20Kg

When an object is at rest on a level surface then the normal force is always equal and opposite of the weight of the object.

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On an inclined surface notice that... • The angle of the incline and the

angle of the weight away from the normal is the same size..

• The weight continues to act vertically downwards toward the centre of the earth.

• The normal force is equal and opposite to the component of the weight perpendicular .

©1998 Science Joy Wagon QUESTION 4: A box of mass 3 kg is placed on a smooth plank, which makes an angle of 25º with the ground. Determine the magnitude of 8.1 the weight of the box. 8.2 the force which causes the box to slide down the plank. 8.3 the force exerted by the box on the plank, measured perpendicularly to the plank. All the applications for vectors discussed in general in this unit are applicable to the calculations with velocity or with any other vector quantity. TERMINAL VELOCITY As an object falls from rest, its speed increases. As the object falls, it collides with air particles, causing friction between itself and the air. Air friction increases until its magnitude equals the magnitude of the objects weight. The object then reaches a constant speed of about 200km.h-1 after falling for about 10 seconds.

Force due to gravity

V (km h-1)

Force due to gravity

Force due to gravity

air friction air

friction

t (s)

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The speed of the falling object approaches a maximum as the air friction increases. Terminal velocity (maximum constant velocity) is reached when the air friction is equal in magnitude and opposite in direction to the force due to gravity on the falling object. General hints for solving vector problems Vector problems can be solved by constructions or by calculations. Solving by construction:

• Make a rough sketch. • Determine the magnitude of each vector according to the scale given. • Indicate north were applicable. • Label each vector correctly. • Distinguish between resultant and the components.

Solving by calculation:

• Make a rough sketch or planning diagram. • In your rough sketch draw the vectors using directed line segments. • Use sin, cos and tan ratios where applicable or use the sine- and cosine rules.

OTHER APPLICATIONS OF VECTORS AND DIAGRAMS: Velocity and displacement of a boat on a river. The following misconceptions so often cause the learners to interpret questions incorrectly. • The man aims to sail to a point across the river.

The man rows the boat such that the velocity of the boat in still water is 20 m.s-1 in still water. What would the resultant velocity of the boat be?

v =

20 m

.s-1

river

v v

c c

Rθ

river bank R > v

I II III I: In the diagram the first part of the diagram I shows the effort of the man rowing alone in still water

(v). If no current flows, he will reach the opposite bank as planned.

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II: Part II indicates that a current in the river (c) flowing at a right angle down stream to the right, will have such an effect on his rowing, that he will be off target and more to the right when he reaches the opposite bank.

III: The combined effect of the water (displacement, velocity or force) and the rowing (displacement,

velocity or force) has a resultant effect (R) in a direction θ away form the original intended direction. The equation form Pythagoras or a scale diagram will resolve the magnitude and direction of the resultant effect (displacement, velocity or force).

QUESTION 5: • Use the information above to find the magnitude of the resultant velocity of the boat if the man drifts

off in a direction 60° while he rows in a northern direction at 20 m.s-1 On the other hand, if he wants to reach a point directly across his starting point (that is the place where he should end up), he must compensate for the flow of the water (current velocity c). Let us assume the water flows to the west in this example, at a velocity of 5 m.s-1. The result of his motion, the Resultant, is in the NORTHERN direction, perpendicular to the direction of the water flow.

IV

river

R

river bank R < v

c

Rv

c

R

QUESTION 6: Find the magnitude and direction of the resultant velocity if the man rows at a velocity of 20 m.s-1, at an angle into the current to be “washed” across

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TUTORIAL ON LESSONS 1, 2 AND 3 VECTORS AND SCALARS

1 How is the bearing of a vector measured? 2 What is the bearing of a force which is acting in (a) an easterly direction;

(b) a westerly direction; (c) a south-westerly direction?

3 (a) Distinguish between distance and displacement. (b) Which of these is a vector quantity and which a scalar? 4 What is meant by the resultant of a number of vectors? 5. A force of 5 N and a force of 3 N act simultaneously on a body.

(a) As the angle between these forces is made smaller, what happens to the magnitude of their resultant?

6. A car travels 20km due north, and then 12km on a bearing of 90°. Use a scale drawing to find the

resultant displacement. 7. The sketch shows a weight of 600N held in equilibrium by strings T1 and T2.

Find T1 and T2. 30̊ T2 T1

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8. A bird sits on a clothes line. The weight of the bird causes the line to sag at an angle of 20°. The tension in the line is 5N to the left and right of the bird. What is the mass of the bird?

5N 5 N 9. A man exerts a force of 250 N at 35° to the horizontal to push a lawnmower across as lawn.

(a) What is the effective push on the mower horizontally? (b) If the mass of the mower is 100kg what is the resultant vertical force exerted by the mower

on the lawn?

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MODULE 1

Lesson 4 Unit 1: Forces and Newton’s first law

FORCES: Force is a vector quantity. Therefore always consider both the magnitude and direction of the forces when solving problems. Example: Label the forces on a girl standing on the ground. Normal force

Force of gravity (weight of girl) Downward force: Downward force of gravity that the earth exerts on the girl. Upward force: Normal force of the earth on the girl. Normal: This is the supporting force that a surface exerts on an object that it is placed on it. The Normal is always perpendicular to the surface. Forces are due to objects having interaction effects on each other. Every force is a push or a pull caused by the interaction between the two objects.

• Force is a vector quantity. • The unit of a force is Newton (N) • The symbol that we often use is F.

We study two types of forces, namely contact and non –contact forces: The interaction between objects in contact, gives rise to contact forces and the interaction between objects that are at a distance from each other gives rise to non-contact forces.

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Examples of contact and non-contact forces Contact Forces Non-contact Forces

Normal Gravitational force Friction Electrostatic force Man pushing box Magnetic force

Force exerted by the table on the object. Force exerted by the table on

the object

Force exerted by the earth on the object Force exerted by the earth on

the object. Note: A free-body diagram is a force diagram in which the object is represented by a point. QUESTION 1: Draw a free-body diagram for a shopping trolley pushed by a lady. Upward supporting force that the floor exerts on the trolley Backward push due to friction f Force X Downward pull by the earth on the trolley. (Attractive force of the earth on the trolley) QUESTION 2: Name the forward force X in the diagram. NOTE: Weight is the one force that is always present; no one can control the effect that gravity has on any mass piece. Use the force of gravity instead of the word “weight”. Weight always works perpendiculary downwards towards the centre of the earth (see Newton’s second law later).

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FRICTION ( Ff ) Friction normally acts against the direction of linear motion. Friction force is then the force opposing motion.

Frictional Force (f) direction of motion

Frictional forces act along the common surface between two bodies in the direction, which opposes their relative motion. The object has to touch the surface so that friction can be created. Remember that a friction force is the force that normally opposes motion; friction force acts on the same plane where linear motion takes place; objects have to be in contact with one another for a friction force to develop. It is also possible to eliminate friction, e.g. on smooth surfaces. QUESTION 3: Discuss the following experiment and give your opinion of the role that friction plays in the setup of the exercise. Aim: To find out what role friction plays when a force is applied on

the cell phone (as in the diagram). Apparatus: Place an object (e.g. a cell phone) on a sheet of paper that lies

on a smooth surface. Diagram: Fapplied Cell phone

Method: Exert a horizontal pushing force on the cell phone. Results: ____________________________________________________________________ (Observation) ____________________________________________________________________ (discuss the forces applicable): _________________________________________________________

Conclusion ___________________________________________________________________ NEWTON’S FIRST LAW: The law states: An object will remain in its state of rest or will continue to move at a constant velocity in a straight line unless a resultant force acts on it. Translated from Latin: Every body preserves its state of rest or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed thereon.

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Common experience seems to contradict this. We know that nothing continues to move forever without change. If the resultant force is equal to zero the following will be true:

• The forces are at equilibrium • There is no acceleration • A body is stationary or is moving at a constant velocity.

The phrase constant velocity implies that there is no acceleration.

Vconstant ⇒ a = 0 Inertia is the objects resistance to any change to its state of motion due to its mass. The bigger the mass, the bigger the object’s inertia will be. QUESTION 4: How would you show / demonstrate inertia to the learners using very basic materials you have in your class? List the examples and explain them. __________________________________________________________________________________ QUESTION 5: Why do we need safety belts in our cars? QUESTION 6: What causes whiplash? The teacher could demonstrate the following when explaining inertia:

1. Fill a glass beaker with water and place it on a piece of paper that is on a horizontal table. Briskly and swiftly slide the paper away from underneath the beaker.

2. Place a coin on a cardboard that is placed over a beaker/ cup. Exert a horizontal force on the cardboard in order to flick it away from the beaker.

3. Tie a mass piece with two strings: one free hanging at the bottom and the other attached to a fixed point above the mass piece. Slowly tug at the bottom string and observe. Repeat this demonstration, but swiftly jerk the bottom string off and observe.

QUESTION 7: Why do you get the different results for slow applied force and quick “plucking” force? QUESTION 8: When are forces in equilibrium? Give various examples. Use drawings where necessary. QUESTION 9: Read through the following scenario and explain what happens: A truck, with a ladder, which is free to move on its roof, speeds down the road in the rain. The driver does not see the car in front of him and he crashes into it. Explain what happens to the ladder and why.

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TUTORIAL ON LESSON 4

NEWTON I

1 You need to push a box with a force of 100N across a carpet at a constant speed. What is the magnitude of the frictional force exerted by the carpet on the box?

2 Explain, using Laws of Physics, what happens when a passenger standing in a bus is thrown

forward when the bus stops. 3 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.

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1.

Lesson 5 and 6 Unit 2: Newton’s second law

MODULE 1 NEWTON’S SECOND LAW IN THE SPECIAL CASE OF CONSTANT MASS Statement of Newton’s Second Law in the special case of constant mass, states that:

When a resultant force acts on an object it will accelerate in the direction of the resultant force. This acceleration is directly proportional to the resultant force and inversely proportional to the mass of the object.

Fresultant = mass x acceleration

Fres = ma APPLICATION OF NEWTON’S SECOND LAW. Apply Newton’s second law to the vertical motion with a helicopter. Show only the vertical forces) Note: The upward force is the thrust of the air on the blades.

Draw the free body diagram for the helicopter. Note: A force diagram is a picture of the object(s) under study, with all the forces acting on it (them) drawn in as arrows. In a free-body diagram, the object under study is drawn as a dot and all the forces acting on it are drawn as arrows.

Fup

mg w

Fup = mgFdown = mg

eightForce due to gravity

The forces on the helicopter are in equilibrium when the helicopter:

• hovers stationary in the air.

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• moves upwards with a constant speed. • moves downwards with a constant speed.

QUESTION 1: Draw a picture showing any other example of a situation from every day life, where Newton’s second law is applicable. Mass and weight According to the Encyclopedia Britannica, the mass of a body stays constant (unchanged), unless parts of the object are removed. This notion, expressed as the theory of conservation of mass, held that the mass of an object, or collection of objects, never changes, no matter how constituent parts rearrange themselves. If a body splits into pieces, the mass divides with the pieces so that the sum of the mass of the composite would be equal to the sum of the masses of constituent pieces together. This is true in general, but when Einstein did research in 1905, he found that mass is not always conserved. When he compared the mass of chemicals of a radioactive nature before and after experiments, the mass is not the same. This was a small change and eventually led to the discovery of other natural phenomena such as the wave properties of matter, matter turning into different forms of energy, etc. For the purpose of school level studies, the teacher does not have to go into so much detail. Weight (force due to gravity) is a force that a body of a certain mass experiences due to the presence of a second massive object, such as the Earth or the moon. Weight is thus a consequence of the universal law of gravitation: any two objects (or particles) in the universe, because of their masses, attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The distance must be measured from the centre of gravity of each object/particle. Thus more massive objects with larger masses weighing more in the same location will have a stronger gravitation force. The farther an object is from Earth, the smaller its weight; the closer to earth, the larger the weight. To make things easier for scientific calculations, scientists have agreed on a standardized average value for the gravity on the surface of the earth, 9.8 m.s-2, but in calculations it is acceptable to round the figure off to 10 m.s-2. The mass of that body is determined by the quantity of matter it is made up of. Mass is a quantitative measure of the inertia of the body. Inertia is thus a fundamental property of all matter. Inertia is a measure of the resistance that a body of matter offers to a change in its state of motion. By international agreement, the standard unit of mass is kilogram (kg). The original unit of mass is a platinum-iridium cylinder made of exactly one kilogram, commonly called the International Prototype Kilogram, is kept at the International Bureau of Weights and Measures in Sevres in France. QUESTION 2: In different situations, you may have different values for weight. How can this be? Explain. QUESTION 3: An astronomer of 80 Kg, travels through space to investigate another planet, planet X. Complete the table below to show the different measurements he will have if he measured his mass and weight.

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PLACE (where the quantity is measured

MASS (on a kilogram scale)

WEIGHT (on a Newton scale)

On the earth, just before he leaves In space, during his travels through the vacuum

On planet X that has double the gravity as the moon (moon’s g is 1/6 of earth’s g)

On the moon where he stops over for an experiment

APPLICATION OF NEWTON’S SECOND LAW TWO OBJECTS THAT ARE JOINED TOGETHER Given:

10 kg5 kg

F Fres = 45 N

F is a fraction of 45N = 105

5+

x 45

= 15N By taking the objects “apart” (separating them), the forces acting on each individual object can be identified.

10 kg15 N 45 N5 kg

15 N

For the 5 kg object: For the 10 kg object: Fr = ma Fr = Fright + Fleft = ma +15= 5 x a +45 + (-15) = 10 x a a = 15/5 30 = 10a = 3 m.s-2 a = 3 m.s-2 Both answers are positive and reinforce the answer for the acceleration of the obdirection choice with positive as “ to the right “ now shows that the positive answer is aright. It is imperative that the learners draw pictures to work from in their calculations. LINKED OBJECTS, VARIOUS GROUPS/SITUATIONS:

1. Linked objects on linear planes. 2. Pulley systems on a table with a pulley at the end of the table. 3. Pulley systems hanging from a ceiling.

Choose to the right as the positive direction

ject. The original cceleration to the

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1. Objects on a linear plane: Given: A 5 kg mass experiences a 2N friction force and a 10 g mass experiences 4N friction force when pulled by an applied total force of 30N in a certain direction.

10 kg5 kg

2 N

Ftotal

4 N

Note: The frictional force for the two masses is not the same. This is because

f = µkN The objects have to be in contact with a rough surface before a friction force starts its function of opposing the motion on the rough surface. It is possible to overcome friction, by applying a larger force to move the objects. Note: the force in the rope develops in both directions as the applied force pulls the rope tight, but acts in opposite directions for each object.

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For the 5 kg object: For the 10 kg object: FR = Fright + Fleft = ma FR = Fright + Fleft = ma FR = Ftension + Ff = 5 a FR = Ftotal + Ff + Ftension = 10 a Add direction for each force: FR = Ftension - Ff = 5 a FR = Ftotal - Ff - Ftension = 10 a FR = Ftension –2 = 5 x a +30 -4 - Ftension = 10 a

26 - Ftension = 10 a a = Ftension - 2 a = 26 - Ftension

5 10 ∴ Ftension - 2 = 26 - Ftension

5 10 ∴ 5 (26 - Ftension) = 10 (Ftension - 2) ∴ 130 - 5 Ftension = 10 Ftension - 20 ∴ 130 - 20 = 10 Ftension + 5 Ftension ∴ 110 = 15 Ftension ∴ Ftension = 110 / 15

= 7.333 N, to the left for the 10 kg and to the right for the 5 kg object.

Now solve the acceleration of the whole system (both equations should add up to the same acceleration). For the 5 g object: For the 10 kg object: a = Ftension - 2 or a = 26 - Ftension

5 10 = 7.333 – 2 /5 = 26 - 7.333 / 10 = 1 m.s-2 = 1 m.s-2 2. Pulleys on a table: Determine the magnitude of the forces in the string and the magnitude of the acceleration of the system. Given:

8 kg

2 kg

8 kg

2 kg

16 N

20 N

8 kg

2 kg

16 N

16 N

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a = total

res

mF = 20

10 = 2 m.s-2 in the direction of the movement, which is to the right for the 8 kg mass and

downwards for the 2 kg mass. Remember

• The forces are equal in magnitude on all points of the string, because it is one string. If the forces in the string are not equal in magnitude at all points then the string/rope would break or kink.

• The forces always act towards the pulley, or away from the object, because it is a pulling force in the string.

3. Pulleys attached to a ceiling Determine the magnitude of the forces in the string and the magnitude of the acceleration of the system. Put in weights

Given

4 kg 6 kg

Step 1

4 kg 6 kg

40 N60 N

Step 2

4 kg 6 kg

Fres = 20 N

Step 3

4 kg 6 kg

Fres = 20 N

X 20 = 8 N410 +8 N

40 + 8 = 48 N

Step 4

4 kg 6 kg

60 N40 N

• The heavier mass piece (6 kg) moves downwards.

4 kg 6 kg

48 N 48 N

• The lighter mass piece (4 kg) moves upwards.

a = Fres/mtotal =2010

= 2 m.s-2

• You have to add the accelerating force to the original weight. • The forces in the string have equal magnitudes on either side of the pulley. • Forces act towards the frictionless pulley.

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NB: Frictionless three-body-problem: Determine the magnitude of the forces on each string and the acceleration of the system. Given:

10 kg

4 kg

16 N

W = 100 N

6 kg

= m.g= (6 + 4) 10= 10 x 10= 100 N

W

The resultant downwards force = 100N as this force is equal to the weight of both hanging masses. The fraction of the resultant force for each mass is: 10/20 x 100 = 50N Going sideways to the right, leave out the ‘+’ or ’-‘for the horizontal motion. a = Fres/mtotal = 100/ 20 = 5 m.s-2 to the right and downwards.

10 kg

4 kg

16 N

100 N

6 kg

-20 N

-30 N

X 100420

X 100620

Going down is negative

Going down is negative

10 kg

4 kg

16 N

6 kg

-20 N

-30 N

70 - 20 =

100 - 30 =70 N70 N

50 N

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QUESTION 4: How would you explain the pulley systems to the learners? QUESTION 5: In your opinion what are the vital concepts the learners should remember when attempting linked object calculations?

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TUTORIAL ON LESSONS 5 AND 6

NEWTON II

1 A horizontal force of 20N accelerates a 4kg object, resting on a horizontal rough surface, at 4m.s-2. What is the force of friction opposing the applied force?

2 James tied two frictionless metal blocks together with a thread (T) and pulled them with

a spring balance on a horizontal frictionless surface. The reading on the spring balance is 7,0N. The blocks P and Q, have masses of 1kg and 2,5kg respectively

. 7,0 N

T P=1kg Q=2,5kg 2.1 Draw a force diagram for block P showing and labelling all the unbalanced forces. 2.2 Draw a force diagram for block Q showing and labelling all the unbalanced forces. 2.3 Calculate the acceleration of the blocks. Show all your calculations. 2.4 Calculate the magnitude of the force exerted by the thread T on each block. 3 The diagram shows a crane with a 120kg load hanging in a stationary position from its cable.

120 kg

3.1 Draw a force diagram, indicating and labeling all the forces acting on the 120kg load.

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3.2 What is the magnitude of the tension in the cable when the load hangs in a stationary position?

3.3 The crane operator changes the tension in the cable to 1380N. In which direction and with what type of motion will the load start moving?

3.4 Calculate the magnitude of the type of motion mentioned in 3.3. 4 A 3kg mass on a table is moved to the right by a 70N force, as shown. 70 N T

5 kg

3 kg

5 A 5 kg mass piece hangs from a rope which runs over a frictionless pulley. The rope,

in turn, is attached to a 3 kg mass. Ignore friction . 5.1 Draw two separate force diagrams showing and naming all the forces acting on each of

the masses 5.2 Apply Newton’s Second Law to each mass and give an equation for the resultant force

acting on 6.2 the 3kg mass 6.3 the 5kg mass 6.4 using these equations, calculate 5.3.1 the magnitude of the acceleration experienced by the masses. 5.3.2 The tension T in the string. 6 A weather rocket, mass 10kg, is launched vertically upwards. The engine fires for 5s,

accelerating the rocket upward at 30m.s-2 before the fuel is used up. 6.1 Draw a sketch of the rocket, showing and naming the forces acting on it while the

engine is firing. 6.2 Calculate the magnitude of the upward force exerted by the engine. 6.3 What is the acceleration of the rocket once the engine has stopped?

28

MODULE 1

Lesson 7 Unit 3: Newton’s third law and linked objects

NEWTON’S THIRD LAW For every action there is an equal but opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always act in pairs - equal and opposite action-reaction force pairs. Note: These action-reaction pairs of forces act on different bodies, therefore they cannot nullify each other to give a resultant of zero. If body A exerts a force on body B, then body B exerts a

force on body A that is equal in magnitude but opposite in direction.

A

A- F

BF

BF

If

then

29

QUESTION 1: Sketch the action-reaction pairs of forces for each of the following scenarios, using Newton’s third law: A book lies on a table. A girl pushes a boy standing on a skateboard by touching his shoulder. A parent pulls a child on a swing seat up to a certain height. A boy kicks a soccer ball with his soccer boot on.

30

TUTORIAL ON LESSON 7

NEWTON III

1. A book rests on a table. What force is the Newton reaction force to the weight of the

book? A the force exerted by the book on the table. B the force exerted by the table on the book.

C the force exerted by the earth on the book. D the force exerted by the book on the earth.

2. A bee flies into the windscreen of an on-coming car. Compared to the magnitude of

the force of the car on the bee, the magnitude of the force of the bee on the car during the collision is …

A zero

B the same C bigger D smaller

3. When an apple falls from a tree, it falls to the ground because of the gravitational force

between the apple and the earth. If F1 is the magnitude of the force exerted by the earth on the apple and F2 is the magnitude of the force exerted by the apple on the earth, then …

A F1 is smaller than F2. B F1 is greater than F2.

C F1 is equal to F2.

31

MODULE 1

Unit Graphs of motion Graphs of motion are a graphical representatThe graphs of motion are used in the tickcourse. SECTION 1: GRAPHS OF MOTION. Average Speed (vave) The average speed is equal to the total distdistance. If an object travels 3 km N in one hour and thkm in 2 hours with an average speed of 3,5 k Average Velocity The average velocity is the resultant changposition. If an object travels 3 km N in one hour and and this occurred in a time interval of 2 hour530. 2. Instantaneous Speed (v) or Velocity Instantaneous speed (or velocity) is th For an object travelling with constant accelemidpoint of the time interval.

V av = 2vu + =

23010 +

= 20 m.s-1

vave = total distance divided by the tota

vave = st

Lesson 8 and 9 3: Graphs of motion

ion of the total action a certain object has undergone. er timer calculation that will be discussed later on in the

ance travelled divided by the time taken to travel that total

en 4 km E in 1 hour, it would have covered a distance of 7 m.h-1.

e in position divided by the time taken for that change in

then 4 km E in 1 h, its resultant change in position is 5 km s. Its average velocity would be 2,5 km.h-1 , on a bearing of

. e speed (or velocity) at that instant.

ration the average speed is the instantaneous speed at the

30

10

vav

ts2 4

vm.s-1

l time

32

3. Distance (s) The area under a Speed/ time graph gives distance.

20

0 6 106 4 t

s

vm.s-1

s = area of the rectangle + area of the triangle = 20 x 6 + 0, 5(20 x 4) = 120 + 40 = 160 m.

4. Acceleration (a) Acceleration can be calculated from the slope of a velocity / time graph Acceleration is the rate of change in velocity.

50

10ts0 8

a = gradient = 12

12

ttvv

tv

−−

=∆∆

=30 -10 8 = 20 8 =2 5 m.s-1

To find the acceleration of a body we need:

1. Two velocities, the initial velocity and the final velocity. 2. Subtract the initial velocity from the final velocity to find the change in velocity (v – u). 3. Divide the change in velocity by the time it took for the change in velocity to take place.

33

The definition for acceleration is: the rate of change in velocity.

THERE ARE THREE MAIN POSSIBILITIES FOR ACCELERATION: 1. The constant resultant force is in the same direction as the initial speed (u). This will cause

speeding up, going faster or acceleration.

+

t

F

+

t

a

t

v

speed up

slope

t

s

no limit

curve

+ F+ u (initial speed)

then “a” has the same shape as F

2. The constant resultant force is in the opposite direction to the initial speed (u). This will

cause the object to slow down, i.e. t

v

slope

slow down

t

s

curve

taper off

- F+ u (initial speed)

then “a” has the same shape as F-

t

F

-

t

a

3. The resultant force is equal to zero Newton.

This will mean that the object does not accelerate, a = 0 m.s-2

00 =⇒= aFres The object moves at constant speed or the object could be at rest. The vector sum of all the forces equals zero.

34t

a

t

v

constant

flat

t

sslopezero

+

Make sure that you fully understand the graphs below. It is the section of the work most learners battle the most with.

WHAT WILL HAPPEN IF YOU THROW AN OBJECT UPWARDS? The only force acting on the object while it is in the air is weight. This force remains constant and is directed vertically downwards. Let the initial velocity (u) upwards be positive

t

v

Slow down,stop, turn,speed up

t

s

stop, turn-

t

F

constantweight

-

t

a

same as F

force acceleration speed distance

On the way up the object moves against the force of gravity and slows down. At the top it stops and changes direction. The object moves with gravity on the way down and the object‘s speed increases.

35

GETTING THE VELOCITY: TIME GRAPH FROM THE SPEED: TIME GRAPH. The place on the graph where the object stops and change direction fold the speed graph downwards over the bending point

t

v

stop, turn

speed

t

vvelocity

+ 0 -

Make your own notes as you go along. GET THE CORRESPONDING CHANGE IN POSITION (OR THE DISPLACEMENT) : TIME GRAPH FROM THE DISTANCE GRAPH. Where the object stops and change direction fold the distance graph downward (“over the point”).

t

s

stop, turn

distance

change in position /displacement

t

s

A scalar like speed is a physical quantity that can never have a negative value. The minus sign for the same motion’s vector is just an indication that the motion is in the opposite direction as in the reference. DROPPING AN OBJECT THAT BOUNCES The only force working on the object while it is in the air is weight downwards (ignoring air friction). Choose downwards as positive. As the object hits the floor a brief motion in the negative direction occurs.

t

s

t

v

stop

velocity

t

s

stop,turn

change in position /displacement

t

F

- - t

a

same as F

+

t

v

Slow down,stop, turn,speed up

speed

stop turn

distance

36

QUESTION 1 Write down what you think is very important when starting out the graphs of motion. QUESTION 2: What are the different types of graphs of motion? Write them down. It is very important that the learners can draw the particular graphs given different information/graphs/problems. It is also important that learners are able to recognize which graph is relevant to which sort of problem. QUESTION 3: Write down the key word and concepts you would use in teaching the graphs of motion. QUESTION 4: What TYPE of activities would you let the learners engage in, in an effort to teach the graphs of motion? It is important to know that for any type of motion the learners can draw a graph. They must always be aware that in an exam question they could be asked to draw a graph that is purely based on the skills of application. QUESTION 5: When drawing a graph we always put the time interval on the x-axis, why? (Discuss the different variables for an experiment.) Some help with problem solving: Learners are sometimes confused about what gradients and areas of graphs represent. It is useful to consider the correct units are obtained when gradients and areas are calculated. 1. Consider the following graphs for uniform velocity: Calculation of the gradient involves calculating the change along the y-axis divided by the change along the x-axis. 2. For the displacement/ time graph the unit of s÷t, or ∆y/∆x, is m.s-1.

This is the unit for velocity. 3. For the velocity/time graph the unit of the gradient is v÷t, or ∆y/∆x,

m.s-2, which is also the unit for acceleration. NOTE: that the calculation of the gradient of an acceleration/time graph does not have to be calculated. Calculation of the area involves multiplying the y value by the x value. 1. Area under the velocity/time graph has the unit m

This is the unit for displacement. 2. Area under the acceleration /time graph has the unit m.s-1

This is the unit for velocity. NOTE: the calculation of the area under the displacement /time graph is meaningless. The area under a velocity/time graph represents the change in displacement. The area under an acceleration/time graph represents the change in velocity. Note: Always use the term negative acceleration and do not use the term deceleration. Negative acceleration does not necessarily imply slowing down while deceleration implies slowing down.

37

Remember that the equation that the learners use to calculate the gradient on a graph is:

Gradient = xy

∆∆

When you’re trying to determine the gradient of a curved graph you draw the tangent (straight lines touching the curve at a certain point) of the graph. The tangent to the curve must be drawn such that it corresponds to the appropriate x-value on the x-axis.The tangent can be used to get the ∆x and ∆y values for an interval. The ratio of the fairly accurate ∆x and ∆y values you can be calculated and used as the gradient equation. Summary: When you find that your learners do not use the calculations in the correct format, this suggested explanation might be useful to use to find the gradient or the area under a graph. gradient

Use the ∆x and ∆y as xy

∆∆ s

(m) ∆y = ∆s ∆x = ∆t t(s)

E.g. vts

xy

=∆∆

=∆∆

QUESTION 7: Given the speed vs time graph draw the corresponding:

a) distance- time graph b) displacement- time graph.

Use the correct distance and displacement values and make sure that the time intervals are correct.

20 50 60 65 75 80

10

20

Car stops and turns

Distance = area under graph

speed

38

TUTORIAL ON LESSONS 8 AND 9

GRAPHS OF MOTION

1. The following information can be obtained from a velocity-time graph: A The displacement from the gradient

B The displacement from the area between the curve and the horizontal axis C The acceleration from the area between the curve and the time axis D The velocity from the gradient 2. The displacement of an object is proportional to

A the area under a displacement-time graph B the gradient of a velocity-time graph C the area under an acceleration-time graph D the area under a velocity-time graph

3. The velocity-time graph below represents the motion of a vehicle: v (m.s-1) 30 20 10 0 10 20 30 40 50 t (s) 3.1 Describe the motion of the vehicle represented by the above graph. 3.2 Determine the acceleration of the vehicle in the first 10s. 3.3 What is the velocity of the vehicle during the next 20s? 3.4 Find the acceleration of the vehicle in the last 20s. 3.5 Determine the distance covered by the vehicle in the 50s.

39

Applications TICKER TIMERS A practical investigation to consider using very b THE TISSUE PAPER EXPERIMENT. To investigate the acceleration produced by a cof five learners. You will need the following materials per group:1. A roll of toilet paper. 2. A Koki-pen (permanent marker). 3. A stopwatch.

The procedure / method: Let the 5 learners assume the following roles: Learner 1: To hold the tissue paper with two forefingers frofingers. Learner 2: To hold the Koki-pen ready to make dots as theLearner 3: To be a timekeeper, ready with the stopwatch. Learner 4: Ready to walk with the tissue paper. Learner 5: Gives instructions and records the results. Learner 1: Must hold the tissue roll, while learner 4 getsforce as he/she walks along. Learner 5: Must give instructions to learner 2, 3 and 4 to start walking for about 4m. Learner 3: Must start recording the time from learner 4‘s Stationary position until he/ she reaches the 4mLearner 2: Will be making dots on the tissue paper as he/sBefore the learners can record their results thgroup understands their role. Record your results as follow: Total time taken: _______________Total distance walked: _______________Total dots made: _______________ Analyse the tissue paper as follows: 1. Mark the dots- A, B, C, D etc.

MODULE 1

Lesson 10

of vectors: Ticker timers

asic equipment.

onstant force on a given mass. Learners work in groups

m both hands so that the paper sheets can roll over the

tissue paper passes him/her.

ready to pull out the roll sheets, applying a constant

start their respective tasks. For example learner 4 must

mark.

he goes along. ey must practice the experiment until every body in the

_____________________________________________ _____________________________________________ _____________________________________________

40

2. Measure the distances between the consecutive points on the paper. Record your observation in the table below: Distance (m) between points

Time taken in seconds(s)

Velocity (m.s-1)

Acceleration (m.s-2)

A - B

B - C

C - D

Why is this important? __________________________________________________________________________________ Calculations:

1. Calculate the ratio of total dots : total time taken as follows: Total dots

Total time

2. What is the time taken to make one dot?

3. Plot the following graphs: a. Distance versus time. b. Velocity versus time. c. Acceleration versus time.

Discussion: What information about the motion can you deduce from the graphs you have plotted? Learners have difficulty plotting the different graphs of motion as well as describing motion from a depicted graph. How would you overcome this problem? Write down the key word and concepts you would use to explain ticker timers and incorporating it with the graphs of motion. It is highly effective to create a fixed format for your learners to use when attempting a ticker timer problem. Colour makes difficult problems allot more manageable. SAMPLE OF TICKER TIMER FORMAT Very important learners must work in SI units. We often use ticker timers to study motion. There are several terms we need to understand that are used in these studies. Instead of using a watch or clock to measure time, we use a ticker timer. The ticker timer makes dots on tape and the time that passes between the dots are constant (time lapses between dots made and are when the spaces appear). Results gathered from ticker timer problems and the graphs of the motion should be used in unison. The frequency is the number of taps or dots made per second on the ticker tape. Unit: hertz (Hz). Ticker timers usually work at 50 hertz but these settings can vary according to the setting of the available apparatus. In our calculations the frequency is usually a given.

41

The period (T) is the time between consecutive dots or taps on the ticker tape. T = 1 /f so if the frequency = 50 Hz then T = 1/ f

= 1/ 50 = 0, 02 s

12 mm 24 mm 36 mm A B C D . . . . Given the tape AD with a period of 0,2s v a-b = ∆ s / ∆ t = 0.012m / 0.2s = 0.06m/s v c-d = ∆ s / ∆ t = 0.036m / 0.2s = 0.18m/s Put the average speed in the middle of the time interval, because the average velocity is really reached in the middle of the interval to make it the instantaneous velocity at the centre of the interval for uniform acceleration. Calculating instantaneous velocity: How is instantaneous velocity calculated if one, for instance, wants the instantaneous velocity at point B? When working with interval AC, point B will be exactly in the middle of the interval AC. Working with an interval such as AC, it is important to use the displacement from A to C, as well as the time from A to C. In the example ribbon above, interval AC will have ∆s = displacement A to C = 12 + 24 = 36 mm, 0.036 m (in SI units), and ∆t = time A to C = 2 x 0.02 s = 0.04 s.

vB = ∆ s/ ∆ t = 0.036 / 0.04 = 0.9 m/s Work with two intervals, e.g. interval AB and interval BC, but the intervals do not always have to be adjacent to each other. If details for intervals AB and CD are known, any two of these intervals will yield the same acceleration if the ribbon moved with uniform acceleration.

∆sAB = displacement AB = 12 = 0.012 m (in SI units), and ∆tAB = time A to B = 0.02 s.

vAB = ∆ s/ ∆ t = 0.012 / 0.02 = 0.6 m/s

∆sCD = displacement AB = 36 = 0.036 m (in SI units), and ∆tCD = time A to B = 0.02 s. vCD = ∆ s/ ∆ t = 0.036 / 0.02 = 1.8 m/s

a = ∆v ÷∆t = vCD - vAB = (1.8 - 0.6) ÷ 2 (0.02) = 25 m.s-1. Interesting: If the time interval between the consecutive dots is 0.1s and ∆ s is the distance between the dots, the acceleration is a tenth of the magnitude of the ∆s. (E.g. if ∆ s = 4 mm, a = 0,4 m.s -2 ).

8 m m 1 2 m m 1 6 m m

0 . 1 s 0 . 1 s 0 . 1 s

4 m m 4 m ms s

42

Learners can work out the following in this sequence: 1. Frequency (f) - 1/ seconds, unit is Hertz (Hz). 2. Period (T) - 1/ frequency, unit seconds (s). 3. Time for an interval - the number of spaces multiplied by the period 4. Work out the average velocity: v = ∆s /∆t. It is a very important concept and learners usually battle with this concept. If learners are asked to work out the average velocity for an interval, it is the displacement (s) covered in that specific interval only, divided by the total time (t) taken to undergo the displacement in that interval. Learners have to, at least, work out 2 average velocities, so that they can subtract an initial velocity (u) from a final velocity (v): ∆v = v – u in the next calculation for acceleration. 5. Work out the acceleration: change in velocity divided the time it took to undergo the change.

a = tv

∆∆

a = v – u ∆t

If your school has the correct equipment, you can make proper ticker-timer ribbons for the learners to work from. Use the trolleys and transformers etc. The suppliers usually make up complete kits that your school can purchase. It is a good idea to invest in at least one ticker-timer kit. Note: Design a practical session where the learners have the opportunity to do the experiment on their own as well as in a group. Groups should not be bigger than 6 learners per group, although the ideal is to have 4 to 5 members in a group.

43

TUTORIAL ON LESSON 10

TICKER-TIMERS 1 A paper tape, fixed to a moving trolley, is pulled through a ticker-timer. The following

table lists the distances which the trolley travelled in each time interval of 5 ticks : T = time number of intervals

s (mm) 0 25 75 125 175 225 275 t (s) 0 0.1 0.2 0.3 0.4 0.5 0.6

1.1 What is the period of the ticker- timer? 1.2 What is the frequency of the time? 1.3 What is the average velocity during the first time interval? 1.4 What is the average velocity during the second time interval? 1.5 What is the average acceleration for these two time intervals? Multiple Choice Questions 1. The SI unit of period is A hertz B second C hour D kilohertz 2. The distance between 10 dots on a strip of ticker-tape which is pulled through a ticker-

timer (which can make 50 dots per second), represents the ……. during a time interval. A displacement B velocity C acceleration D time elapsed 3. The lengths of all the 10 dot intervals on a strip of ticker-tape are all the same (and

greater than zero). This means that the object's (to which the tape was attached) A acceleration is greater than zero B velocity is zero C position does not change D velocity is constant 4. The lengths of successive 10 dot intervals on a strip of ticker-tape increase. This

means that the object to which the tape was attached has A zero acceleration B a decreasing C constant velocity D an increasing velocity

44

Equ For the purpose of this training discussion, theuniform acceleration when using the equations The equations that the learners will find on the i

1. v = Δs Δt

2. v = u + at

3. s = ( u + v). t

2

4. v2 = u2 + 2 as 5. s = ut + ½ at2

QUESTION 1: The learners can also use the equation used infor motion for changing velocity? Explain why. Learners should always use the equations just a

s = ut + ½ at2

It is not a good idea to use this equation to calmathematical problem for most of the learners. QUESTION 2: How would you suggest learners calculate timeAn object freely falls from rest from a building 60 It is very important that the learners choose theor run the possibility to make mistakes.

MODULE 1

Lesson 11

ations of motion

study of motion is restricted to straight-line motion with of motion.

nformation sheets are:

the ticker timer calculations. Which equations are used

s they are given. For example:

culate the value of the time, since t and t2 seem to be a

taken for a motion in the following example? m high.

appropriate equation so that they don’t waste any time

45

Study the table below: Symbol used in equations

Description Unit Vector

u Initial velocity; velocity at time o seconds, when the displacement is o meters.

m.s-1 Yes

v Final velocity; the velocity after time ‘t’ passes and the displacement ‘s’ pass.

m.s-1 Yes

s Displacement; the direction (usually + or – for straight line motion) and the distance of the moving objects from its starting point after time ‘t’. in these equations we use ‘s’ in similar ways as ∆s.

m Yes

a Acceleration; the change in velocity during a certain time.

m.s-2 Yes

t Time; time that passes during which velocity changes and the objects displacement changes by s.

s No

v

Average velocity

tsv

∆∆

=

m.s-1 Yes

∆v Change of velocity ∆v = v – u.

m.s-1 yes

Before learners start a calculation it is advised that they draw a picture or diagram as this could help them to succeed in solving the problem. QUESTION 3: Write down the steps you will use to teach the learners how to solve the problems where equations of motion are involved. QUESTION 4: Read through the problem below: A motor car, exceeding the speed limit, crosses a speed trap on a long straight road at 72 km.hr-1 and continues at this speed. As the speedster passes the speed trap, a traffic cop sets out after her on his motor bike, accelerating uniformly from rest at 2 m.s-2 until he eventually overtakes the motorist.

1. After 10s how far is the person ahead? 2. How many seconds after starting off, will the traffic cop overtake the motorist?

Remember to work in SI units! Firstly make a sketch indicating all given Solve the problem

46

Tips on solving the problem: For each of the persons doing the motion you can: 1. Write down the given information 2. Draw a picture 3. Place the information in the appropriate places in the drawing. Before answering the question the learners should write down “stuva” s = displacement t = time u = initial velocity v = final velocity a = acceleration AND Draw a picture of the motion for the cop and the motorist. Use a rough graph if it is easier to “see” the movement of the

1. They should fill in the given information in the appropriate places. 2. They should choose an equation from the info sheet that is applicable for the motion. 3. They must convert to SI units if necessary. 4. They must substitute the values in their appropriate places. 5. Then they can solve the calculation.

QUESTION 5: Write down the activities you would engage the learners in to illustrate the concepts of motion. It is very important that the learners work through as many examples as the time available allow them to. The learners should be exposed to calculations that are very basic right to the calculations that are the most difficult. QUESTION 6: What sort of assessment would you use to make sure that the learners are competent? Note that these equations are valid for rectilinear motion at constant acceleration. They can be applied to motion on the horizontal- or vertical plane. There are 3 equations of motion are: v = u + a t (mentions t but not s) s = u t + ½ a t2 (mentions t and also s, but not v) v2= u2 + 2a s (no mention of t) Since each of the equations above has four of the five factors (s, t, u, v and a) in it, they can be used when three factors are known. u = initial speed or velocity v = final speed or velocity a = acceleration t = time You will find the above mentioned equations on your information sheet A car that is speeding up. The acceleration and velocity vectors will either both be positive or will both be negative. QUESTION 7: A car accelerates from 20 m.s-1 at 3 m.s-2 for 12 seconds. (I) How fast does it travel after 12 s? (II) How far does it travel?

47

A car that is slowing down The acceleration and velocity vectors will have opposite directions. QUESTION 8: If a car would travel at 20 m.s-1 and would apply is bakes for 5 second and decreases its velocity at 3 m.s-2.

(I) How fast will it travel after 5 seconds? (II) How far will it travel while slowing down?

If there is no mention of time in the problem. QUESTION 9: A plane slows down at 6 m.s-2 over a distance of 2 500 m until it reaches a speed of 200 m. (I) Calculate the initial speed. (II) How long did it take the plane to make the change in its velocity? Some common mistakes made with calculations: The equation for the circumference of a circle is 2πr. The area of a circle is calculated with the formula πr 2, but it is not necessary to calculate the area for examples where an athlete runs around a track. Very often though, the learners use the incorrect equation. Take a minute to point this common mistake out to them during the lesson. If the learners are asked to calculate the average speed, they should use the equation total distance/ total time; for average velocity the equation is total displacement/ total time. Projectile motion at an angle: QUESTION 10 Vertical component: fire a projectile at 100 m.s-1 at an angle of 70° to the vertical. How high will the object go vertically upwards? Horizontal component- fire a projectile at 100 m.s-1 at an angle of 70°to the vertical. How far will the object travel horizontally if it is in the air for 6 seconds?

48

TUTORIAL ON LESSON 11

EQUATIONS OF MOTION

1. A skier moves with uniform speed of 10 m.s-1 over level snow, then accelerates at 2

m.s-2 down a long, straight slope. Calculate 1.1 his speed after travelling down the slope for 5 sec 1.2 the distance travelled down the slope in 5 sec 1.3 his speed at the bottom of the slope if its length is 200 m 2. An aircraft starts from rest and accelerates down the runway prior to take-off. It moves

600 m in 12 s. Find 2.1 acceleration the craft undergoes 2.2 take-off velocity 3. A truck breaks down on the open road. The driver places warning triangles 80 m from

either end of the truck. A car is approaching at 80 km.hr-1 and applies its brakes immediately on seeing the warning triangles. Its brakes can decelerate the vehicle at 3 m.s-2. Can the car stop in time ?

4. A car travelling at 30 m.s-1 is brought uniformly to rest in a distance of 50 m. Calculate 4.1 the acceleration 4.2 the time taken to come to rest 5. Car A is travelling at a constant velocity of 28 m.s-1. It passes car B at the instant that

car B starts accelerating from rest with constant acceleration of 4 m.s-2. Calculate the time taken for car B to catch up with car A.

49

MODULE 1

Lesson 12 and 13 Newton’s universal gravitational law and free fall

The universal law and constant: This is a small portion of the work. Grade 11 learners are expected to substitute into equations to solve the unknown, but for grade 12 learners the emphasis is usually on the understanding and application of inverse and direct proportionality. The UNIVERSAL GRAVITATIONAL LAW states:

Any two objects or particles in the universe will attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square distance between the gravitational centres of the objects.

The equation used in this section is:

F = G 221

rmm

QUESTION 1: Using the equation how would you derive the unit for the universal gravitational constant? The equation is usually supplied on the information sheet. The universal constant (G) has a value of 6, 7 x 10 -11 N m2 kg2 The value for the universal constant is also available on the information sheet. QUESTION 2: List a few activities that learners could engage in to explain the law and the constant. QUESTION 3: What are the key words and concepts in this section? Acceleration due to gravity If we stand on the surface of the earth and drop an object it will accelerate downwards, at an average of 10 m.s-2 as explained in the earlier lessons. From the previous knowledge the learners gathered from Newton’s second law they must know that the resultant force causes the acceleration. The force given by the weight of the body, due to the gravity on any planet, is:

F resultant = m a We use this idea when we say a mass of 50 kg has a weight of 500N.

50

This is the way the learners should calculate the weight of objects: Forcegravity = Mass x acceleration F gravity = m x a [a = g = 10 m.s-2] ∴ F gravity = m x g REMEMBER:

• Weight always acts vertically downwards towards the center of the earth (PERPENDICULAR TO THE SURFACE).

• In free fall, you are weightless. • Should you fall off a building you do not feel your weight. • Weigh due to gravitation is always present; you must always take weight into consideration!

F

Earth

QUESTION 4: An object moves upward at 5 m.s-1 when thrown into the air, turns around and falls back into your hand after a few seconds. What is the value for each of the factors at the following positions? (Use “up” and “down” to indicate directions for the vectors.) FACTOR POSITION VALUE (unit) VELOCITY Leaving hand ACCELERATION Highest point TIME TAKEN Back in hand The relationship between ‘g’, the mass and radius of the earth.

If we use the equation, F = G 221

rmm

to calculate the weight (F gravity) of an object on the surface of

the earth, we substitute

• m1 with the mass of the object. • m2 with the mass of the earth (M). • r with the radius of the earth.

F gravity = G 221

rMm

If you replace the F gravity with mg and divide it by m, we get the following equation:

g = G 2rM

QUESTION 5: Will the force of gravity (weight) of an object be the same on the moon? Use the above equation(s) and discussion to explain your answer.

51

QUESTION 6: Which teaching resources can be used to make the teaching/ learning experience a dynamic one? QUESTION 7: The learners could be asked to derive the unit of an equation. How would you teach this to the learners? The inverse square law The inverse square law is evident in both Newton’s Law of Universal Gravitation and Coulomb’s Law. Is the inverse square law evident in other equations in physics? QUESTION 9: The learners often have difficulty understanding and applying the square, why do you think this is? Is it possible to overcome the problem by planning correctly? Gravitational calculations The following are examples of calculations that you can expect to come across when doing this section of the work with the learners. QUESTION 10: The force of gravity on an astronaut on the surface of the earth is F gravity. The astronaut launches into space at an altitude above the surface of the earth equal to the radius of the earth. What will the force of gravity on the astronaut at this height be? QUESTION 11: What are the problem-solving steps you would apply to the above problems in an effort to solve them? In your opinion, is it necessary to draw a picture before starting the calculation? Speak to your colleagues to find what works for them. Objects that are in free fall (ignore air resistance)

a = g = 10 m.s-2 g = acceleration due to the gravitational force exerted by the center of the earth on the center of an object. It takes any object of any mass and volume 1 s to travel 5 m from rest. During the 2nd second, that same object will move another 15 m, giving it a total displacement of 20 m after 2 s. Know the following table - it will come in very handy for any free falling object subjected to the gravitational field on earth:

t v s 1 10 5 2 20 20 3 30 45 4 40 80 5 50 125

Dropping an object from rest: After I second it falls 5m and reaches a speed of 10m/s After 2 seconds it falls a total of 20m and reaches a speed of 20 m/s After 5 second it falls a total of 125m and reaches a speed of 50m/s WHEN YOU FIRE AN OBJECT UPWARDS AT 50 m.s-1 When will the object pass the 120m mark? Let the object drop from rest for 5 seconds to reach a speed of 50 m/s. The object will have covered 125m.

52

Consider only the vertical motion The velocity of the body (at the top) is instantaneously zero v = 0 m.s-1

v = 0 m.s-1

125 m

5 s

v = 50 m.s-1 v = 50 m.s-1

It takes 1 second to fall 5 m The object reaches the 120m mark 1 second before and after it was at the top.

The times are: 5 - 1 = 4s at point A and 5 + 1 = 6 s at point B.

120 m

A

5 s

125 m

B 5 m

FIRE AN OBJECT UPWARDS OFF A CLIFF: Fire an object upwards at 40 m/s off the top of a cliff that is 10m high. How long doe sit take to hit the bottom of the cliff? (i) Consider the upward motion. Let the object drop from rest for 4s until it reaches a speed of 40m/s. thus it will fall 80m. It will thus cover 80m to get to the top of its flight.

cliff

100 m

V = 40 m.s-1

4 s

V = 0 m.s-1

V = 40 m.s-1

80 m

4 s

53

(ii) Consider the downward motion.

The object must fall 80m + 100m = 180m before it will hit the ground.

cliff

4 s

V = 0 m.s-1

80 m

100 m

80 m

180 m

The time the object uses to fall 180m s = 5 t2 180 = 5 t2 t 2= 36 t = 6 s The total time of flight = 4s (up) + 6s (down) = 10 s. If you have to convert km.h-1 to m.s-1: Take the value given, multiply with 1 000 to convert the “kilo” and divide 602 because every hour has 3 600 seconds. (1 000 ÷ 602 = 0.2777777, which means if you multiply the given by 0.2777777, or divide the given by 3.6, the answer will be the same.) QUESTION 12: Convert the following: 3,6 km.h -1 to m.s-1: __________________________________________ 10 m.s-1 to km.h -1: ___________________________________________ 72 km.h -1 to m.s-1: ___________________________________________ When you are obliged to work with SI units this conversion will come in very handy. IF THE INITIAL VELOCITY (u) IS UNKNOWN: A man fires an object off a building that is 120m high. If it takes the object 12s to hit the bottom of the building, what was the initial speed upwards? Regard the initial velocity (u) as positive; thus upwards is + s is a vector; it is the change in position from the start to the end, going downwards.

54

c l i f f

a = - 1 0 m . s - 2

- 1 2 0 m

- 1 2 0 m s

0u

QUESTION 13: Find the initial velocity of the object. Regard u as positive so upwards is positive. s is a vector; it is the change in position from the start to the end going downwards. Acceleration is downwards (-) due to the gravitational pull of the earth, as upwards has been chosen as the positive direction. Free falling objects QUESTION 14: The radius of the earth is 6400 km. An astronaut orbiting 200 km above the earth’s surface is 6600 km from the centre of the earth. Yet the TV broadcasts of astronauts’ in earth orbit show them as weightless! How would you explain the above statement? QUESTION 15: In fact, the astronauts are not weightless, but they experience a feeling of weightlessness because they are in free fall. Why does the space capsule not fall onto the earth? Remember in a free fall situation:

1. The resultant force is equal to the weight! 2. If you were to use mass in your equation you have to work with the equations associated with

Newton’s laws. 3. If however there is no mass in the calculation you may use the equations of motion.

QUESTION 16: When is an object in a state of free fall?

55

TUTORIAL ON LESSON 12 AND 13

UNIVERSAL GRAVITATION AND FALLING BODIES

1 A stone is dropped from a bridge and hits the water 150m below. 1.1 How long does the stone take to reach the water? 1.2 With what speed does the stone strike the water?

2 A ball is thrown vertically upwards. It takes 6s for the ball to return to the thrower’s

hand. 2.1 Calculate the maximum height reached by the ball. 2.2 With what speed does it return to the thrower’s hand?

3 A mass of 1kg on the earth is attracted with a force of 9,8N. The radius of the

earth is 6,4 x 10+6m, calculate the mass of the earth.

4 A man on earth weighs 750N. What will be the force of gravity on him in a satellite orbiting the earth at a distance of twice the radius of the earth, above the surface of the earth?

5 A girl has a mass of 50kg. What gravitational force would be exerted on her on a

planet, which has a mass equal to that of the earth and a radius 3 times, that of the earth?

56

Ty Types of energy Gravitational potential energy (Ep) Potential energy is stored up energy by virtue of t E Unit: joule (J) or N.m h: is the vertical height above the zero point g: is the gravitational acceleration due to the eart Example 1: Calculate the work done to raise a 50kg mass thrW = m g h = 50 x 10 x 6 = 3 000J. Kinetic energy (Ek) Kinetic energy is by the virtue of the motion of th EUnit: joule (J) or N.m V: is the velocity of the object Example 2: Calculate the work done to stop a 50kg frictionlesEk = ½ m v 2 = ½ 50 x 4 = 100 J Make sure you know all the concepts and theirabove shown calculations look relatively easy but Work done (W) Work is the product of the force acting and the diss is the displacement moved against friction.

friction force

F cos θ = component of the force F in the directio W

Lesson 14 pes of energy

MODULE 1

he position of the object in the gravitational field.

p = m g h

h’s pull.

ough 6m vertically.

e body

k = ½ m v 2

s mass that moves horizontally at 2 m.s-1

applications, before you attempt any calculation. The they are the vital building blocks for the work to follow!

W= F s

placement in the direction of the force.

m

s

θ

Fstring

n of the object’s movement = F cos θ. s

57

• Work done is a transfer of ordered energy, on a big scale, by moving directed force (with no

temperature change) • Heat is a transfer of disordered energy by molecules doing work randomly on a microscopic

scale, and raising the temperature. • Heat is a process not a ‘thing’. Heat is energy in transit. • Thermal energy = Random Ep = Ek of the molecules. • Work is only done if the object moves against some resistance while remaining in contact with

the object exerting he force. • Spin a stone on a string in a circle • Inertia makes it move tangentially

W = F cos 90˚. s = 0 • No work is done at 90 ˚ to a force

F

s Question 1: What types of energy will the learners encounter in their studies? Write down the activities you would engage the learners in to illustrate each of the types of energy in the table below.

Type of energy activities

Potential energy:

Kinetic energy:

Question 2: Discuss and write down the common experience of the learner’s problems concerning energy. Question 3: What is the difference between potential and kinetic energy? Looking at the different equations for the energy types, how would you explain the differences to the learners? Write down the units for potential and kinetic energy. Potential energy: ___________________________ Kinetic energy: ___________________________ Question 5: How would you calculate the work done by a certain object? Question 7: Write down different situations in our every day lives where energy is applicable.

58

THERThe fof the THERAll ththe s Thesat the QuesDevedown QuesWhat

MODULE 1

Unit 5: Types of force situa

E IS FRICTION orce then is doing work against friction and force is used to overcome friction.

- F

E IS NO FRICTIO e force is used to do useful work.. There peed and kinetic energy.

e force situations link up with energy. The lse sections alone, but everything in physic

tion 4: lop a short practical you would engage th.

tion 6: are common misconceptions the learners

Lesson 15 tions

this could cause an increase in internal energy. Some

F

is no change in temperature but there is an increase in

Fres

earners should always be aware of the fact that we look s will and can link up in one way or another.

e learners in to illustrate the energy concept. Write it

have concerning energy?

59

MODULE 1

Unit 5: Energy ENERGY CALCULATIONS DERIVING THE ENERGY EQUATIONS v 2 = 2 a s if u = 0m v 2 = 2 m a s multipm v 2 = m a s = F s 2 But m a s = m g h for a vertical rise h, subject to m v Three equivalent formulae for energy and work d If there is an increase in height, then work done W = If there is a change in speed, the work done = ch W Unit of work = Joule J = Unit of energy. A joule is the amount of work done when a fodirection of the force. Look at the examples that follow, they would acould be applied. Example 3: On a horizontal road a 2400kg truck is traveling a 9600N force. How far does it move while slowi W = ∆ Ep + ∆ Ek F s = ½ m ∆ v 2 on a horizontal road 9600 s = ½ (2400) (20 -0) 2 s = 480 000/ 9 600 s = 50m Work done = energy transferred from one systeWe can say that energy is the ability to do work.

Lesson 16 calculations and concepts

, starting from rest. ly both sides by m

the acceleration due to the earth’s gravitational pull, g

2 /2 = mgh = F s

one

= change in potential energy

∆Ep = m g ∆h

ange in kinetic energy = ∆Ek = m ∆v 2

rce of 1 N moves the object a distance of 1m in the

ssist you in understanding how this section of the work

at 20m.s -1 and applies its brakes, it is brought to rest by ng down?

m to another

60

LAW OF CONSERVATION OF ENERGY. The total mechanical energy within a closed system remains constant i.e. the total Ep + Ek remains constant.

Ep + Ek (at A) = Ep+ Ek (at B) Ep + Ek = mechanical energy.

Example 4: A bead on a frictionless sliding wire- Determine the speed at point B The energy equation ignores thEpath between points A and B.

Ep + Ek (at A) = Ep+ Ek (at B)

6 m4 m

3 m.s-1

A

B

mgh + ½ m v 2 = mgh + ½ m v 2 10 x 6 + ½ (3) 2 = 10 x 4 + ½ v 2 v 2 = 49 v = 7 m.s -1

Example 5: Below is a bead sliding along a frictionless surface, Determine its speed at point B if it slides up the ramp.

1 m

B

A (measure vertical height from A)

6 m.s-1

Ep + Ek (at A) = Ep+ Ek (at B) mgh + ½ m v 2 = mgh + ½ m v 2 0 + ½ (6) 2 = 10 x 1 + ½ v 2 v 2 = 16 v = 4 m.s -1

Example 6: a pendulum If a pendulum bob rises through a vertical height of 5m determine the maximum horizontal speed at the lowest point of its swing.

61

5 m

v

B

Bob is stationary at A

A

Ep + Ek (at A) = Ep+ Ek (at B) v = 0 at A h = 0 at B Ek = 0 at A Ep = 0 at B

Ep (At top, A) = Ek (at bottom, B) mgh = ½ m v 2 v 2 = 20 h

= 20 x 5 = 100 v = 10 m.s -1 Example 7: Cliff problems A man drops a 50kg rock off a vertical cliff that is 45m high (1) Calculate the speed of the rock at the bottom of the cliff.

cliffA (v = 0 m.s-1) object is stationary at A

B (h = 0)

45 mLoss in Ep = gain in Ek

Ep + Ek (at A) = Ep+ Ek (at B)

Ep (at top, A) = Ek (at bottom, B) mgh = ½ m v 2 v 2 = 20 h

= 20 x 45 v = 30 m/s The total energy at any height at any time = Ep + Ek Ep at top = mgh = 50 x 10 x 45 = 22500 J Ek at bottom = 22500 J (2) Calculate the speed when the rock is 25m above the ground. Ep at 25m = mgh = 50 x 10 x 25 = 12500 J Ep + Ek = 12500 J = Total energy at 25m Thus Ek = 22500 - 12500 = 10 000 ½ 50 v 2 = 10 000/ 25 = 400 v = 20 m.s -1 In this section you would have to explain the energy calculations and how they are derived.

62

Question 7: A man drops a 50kg rock off a vertical cliff that is 45m high; calculate the speed of the rock at the bottom of the cliff. Analyse the problem and solve it. Draw the situation to illustrate an understanding of the problem. Energy concepts: There are some energy concepts that have proven to be more difficult than the energy calculations that focus on linear planes. The concepts are:

1. Non conservation of energy and 2. Curved pathways.

1. Non-conservation of energy Mechanical energy is not conserved when friction is present, as work has to be done to compensate for the friction. Ep + Ek (start) = Ep + Ek (finish) + W friction Friction reduces the available energy. Friction transfers the energy out of the system in the form of heat. Some mechanical energy is converted by friction into internal energy and the system will get hotter. 2. Curved paths The equations of motion only work on straight lines of motion. Whenever we do work we simply transfer energy from one system to another Energy lost by one system will be gained by other system. Work done on a system = change in mechanical energy Question 8: Explain how you would introduce the following situations to the learners. Non conservation of energy: Curved pathways: Question 9: What sort of activities would you use to illustrate the two situations in question 8?

63

Ch

Power and Efficiency Power is the ra the rate

Unit of power: Watt (W) A Watt is the rate of working or energy tr Example: A car moves at a constant speed of 20 m/s the frictional force (F), if it has an 88 kW mo

• Modify this formula by dividing both sid

W/t = F s/ t • For constant speed

P = F v (the useful formula)

88 000 = F 20 F = 88 000/ 20 = 4400N

Lesson 18 apter 5: Power

MODULE 1

te of doing work

or

of energy transfer

P = W/t

ansfer of one joule per second.

thus the frictional force is equal to the driving force. Find tor.

W = F s

es by time t

64

EFFICIENCY EFFECIENCY OF AN ELECTRIC MOTOR A 100 W electric motor takes 1 minute to lift a 50 kg mass through 6m. Calculate the efficiency of the motor. Work done by motor: W = P t

= 100 x 60 motor = 6000 J

50 kg

Work done on mass: W = ∆Ep = m g ∆h

= 50 x 10 x 6 = 3000 J

Efficiency = energy output/ energy input x 100% Efficiency = work done on mass / work done by motor x 100%

= 3000 / 6000 x 100 = 50%

Make sure that you can do the above mentioned calculations before attempting similar problems in a class situation.

Read through the problem below: A car moves at a constant speed of 20 m/s thus the frictional force equals the driving force. Find the frictional force (F), if it has an 88 kW motor. The solution: W = F s Modify this formula by dividing both sides by time t

W/t = F s/ t For constant speed P = F v (the useful formula) 88 000 = F 20 F = 88 000/ 20

= 4400N Question 1: Why is it necessary to consider efficiency? Question 2: Define efficiency: Question 3: What teaching resources do you intend on using in an attempt to illustrate the above concepts?

65

F

FRICTION EFFECTS AIR FRICTION SLOWS FALLING OBJECTS If a 20 kg object falls for 5s and hits the ground

(i) Calculate the work done to overco(ii) What average force stops the box

Answers:

(i) Without air resistance v = 10 t = 10 x 5 = 50 m/s Ek = ½ m v 2 = ½ 20 (50) 2 = 25 000 J With air resistance: v = 40 m/s Ek = ½ m v 2 = ½ 20 (40) 2 = 16 000 J Energy used to convert overcome air f W = ∆ Ek = 25 000 - 16 000 = 9 000 J

Some mechanical energy is transferred to the amolecules move freely and randomly but more e

(ii) W = ∆ Ek = F s 16000 = F x 0.4 F = 16000 / 0.4 = 40 000 N

Question1: From previous knowledge we know that frictioenergy used/ applied? Question 2: Consider the problem given: A 20 kg object falls for 5s and hits the ground atWithout air friction: v = 50 m. sWith air friction: v = 40 m. s 1. Calculate the work done to overcome the air 2. What average force stops the box if it makes How would you solve the above problem?

Lesson 19 riction effects

MODULE 1

at 40 m.s -1 me the air friction. if it makes a 40 cm dent in the ground?

riction ir as heat energy by working against the friction. The air nergetically.

n has an effect on motion, but how does it affect the

40 m/s -1 -1

friction. a 40 cm dent in the ground?

66

EN Graphs of energy ENERGY GRAPHS TO GET AN Ek GRAPH FROM AN Ep GRAPH Ek + Ep = a constant Given: an Ep graph Fold it over its horizontal centre line to get the E

h

Ep

Note: temperature is determined, on the atomatoms. The most learners experience problems with gpossible. In doing this they will gain confidencproblems become. If they are still struggling with graphs after this swith you, the educator so that their hiccup can bThe learners should not be afraid to ask a quenever know; the question just asked could be thYou are the only one that knows whether your le Question 3: How do you derive a kinetic energy graph from Note: temperature is determined, on the anatoatoms.

MODULE 1

LESSON 20 ERGY GRAPHS

k graph.

Ek

h

ic scale, by the random kinetic energy of the object’s

raph calculations. They must do as many examples as e in your own abilities and you will find the easier the

ection, encourage the learners to make an appointment e sorted out before it becomes a major problem. stion, there is no such thing as a stupid question. You e exact same one fellow class mates are struggling with. arners are able to do this or not.

a potential energy graph? mic scale, by the random kinetic energy of the object’s

67

Question 4: What are the steps you would follow to teach the energy graphs and equations to the learners for exam purposes? Question 6: What are the teaching resources you can implement in an effort to illustrate clearly to the learners the concepts of graphs?

68

Momentum MOMENTUM p = m.v Def: momentum is the product of mass and veloUnit: kg. m.s -1 Momentum is a vector. Always remember direction. Example 1: Calculate the momentum of a 10kg object movin p = m.v = 10.8 = 80 kg. m.s -1 north. Example 2: Calculate the velocity of an object if it has a measterly direction v = p/m = 40/4 = 10 m.s -1 east. The examples are an easy way of showing how Question 1: How would you introduce the learners to the con Momentum sums are basic sums and the learnexam papers. This is a very important section of the work, aneasy. Momentum is the product of mass and velocity. The unit for momentum is: kg. m.s-1

Momentum is a vector and therefore it is very im

Calculate momentum through

p = Question 2: What sort of activities would you engage the lea Newton combined the two quantities (mass represents the product of the two quantities. Note: momentum and inertia must not be confuof an object to stay in the state of motion it is,

Lesson 21 Momentum

MODULE 1

city.

g at 8m/s due north.

ass of 4kg and if it has momentum of 40kg.m/s in an

the concepts work.

cept of momentum?

ers are not likely to find such an easy question in their

d compared with Newton’s laws of motion it is relatively

portant that the learners remember direction.

the equation:

m. v

rners in to illustrate momentum?

and velocity) into a measurement of motion, which

sed with each other! The inertia of a body is a property while momentum is a way to compare the motion of

69

bodies. Bodies will therefore always posses inertia, even if it’s motionless, while its momentum then will be equal to zero. In a soccer game it’s important for the players of each team to stop their opponent from reaching their goal. Which of the following players will be the most difficult to stop? Player 1: mass of 125kg running with a velocity of 8 m.s-1 Player 2: mass of 80kg running with a velocity of 10 m.s-1 Question 7: How does mass influence the momentum? Question 8: Player 1 will be more difficult to stop, why? Remember: always use examples that are relevant to the learners and to which they can relate. Question 9: What assessment criteria would you use to assess your learners? Question 10: Why would you use the above mentioned assessment criteria?

70

Unit 6: C Change in momentum. When the velocity of a body changes, the chang Change in momentum = final momentum – inIn symbols: ∆p = m v –m u = m (v – u) The change in momentum takes place when duchanges due to external factors. Rememberecommended to use the initial direction of mremember that if an answer ends up with a nanswer. In the event that an object bounces bac Change in momentum: ∆p = m (v – u)

• ∆p is the change in momentum. • The mass does not change • The velocity changes • The change in velocity = v – u

There are certain rules you can follow that woustick to these rules you are very unlikely to make Rule 1: Always take the direction of initial veRule 2: if the object bounces back then the fi Look at the examples that follow Example 3: If a 4kg object hits a wall at 6m/s and bounces b u = + 6 m.s -1 v = - 3 m.s -1 ∆ p = m (v – u) = 4 [–(+6)] = 4 (- 9) = - 36 kg. m.s -1 towards the wall = 36 kg. m.s -1away from the wall. Look at the example below and solve the probleA ball with a mass of 200g strikes a racket at riga velocity of 9 m.s-1

Question1: Calculate the change in momentum of the ball.

Lesson 22 hange in momentum

MODULE 1

e in momentum is represented by the equation:

itial momentum

ring motion the velocity, in either one of the directions, r: it is very important to remember direction. It is

otion as the positive direction. The learners should egative, they have to clarify and explain the negative k the learners should take the final velocity as negative.

ld allow you to solve these problems effectively. If you a mistake during the calculation.

locity (u) as positive. nal velocity (v) is negative.

ack at 3m/s calculate the change in momentum

m: ht angles with a velocity of 12 m.s-1 and rebounds with

71

Remember: that you have to choose a direction and it will definitely help to draw a picture of the motion. The learners must always explain the negative sign should it occur in their calculation. Question 3: Provide appropriate examples you would use in explaining the change of momentum. Question 4: What are the teaching resources you could use to illustrate change in momentum? The trolley experiments can also be used to investigate the change of momentum. Learners should be encouraged to do the experiments individually as well as in groups. In the previous lessons we have discussed the working of the trolleys. It was used in the motion section or in the ticker timer parts. Please refer to the mentioned sections. Another important law comes into play during this section of the work; it is the conservation of momentum. The principle of conservation of momentum states: The total linear momentum of an isolated system is conserved in magnitude and direction. Question 5: Write down the activities you would engage the learners in to illustrate the conservation and change in momentum. The learners should be encouraged to use the trolley experiments’ results to derive the law of conservation of momentum. Question 7: What are common misconceptions the learners experience within the above mentioned section?

72

Newton’s Newton’s second law restated. It is imperative that the learners know and undelinked. As is momentum and Newton’s seconddirectly proportional to the resultant force actingm. a, but the acceleration can be substituted awill look as follow

Fres = m

Thus we can define the resultant force as the Fres = m (v –u)/ t Thus we can define a force as the rate of chang Example 4: A 0,2 kg ball travels with a velocity of 20 m.s –1 acalculate the force on the ball if the contact time Fres = m (v –u)/ t =0,2 [30 – (+20)]/0,01 = - 1000 N The shorter the time in contact is, the bigger theWhy is this important? Because it is the introduction to the concept of imLook at the example below. A 2kg ball travels with a velocity of 20 m.s-1and 1. Calculate the force applied on the ball if the cont Remember: the shorter the time the ball is in cobe.

Lesson 23 second law restated.

MODULE 1

rstand that all the principles and concepts in physics are law. The rate of change of momentum of a body is

on it, and is in the direction of the resultant force. Fres = ccording to the definition of acceleration. The equation

tuv )( −

rate of change of momentum.

e of momentum with time.

nd hits an object and returns with a velocity of 30 m.s -1 is 0,01 s.

resultant force

pulse!

hits an object. The ball returns with a velocity, of 30 m.s-

act time is 0,001 s.

ntact with the force, the bigger the resultant force must

73

Impu Impulse and Collisions Impulse is the change in momentum. Note: thecollision impulse is give to both bodies. Impulse Impulse = change in momentum Ft = m (v – u) Note: there are two units: N.s or kg. m.s -1 Impulse could be explained by using examples: Example: A 0,2 kg ball travels with a velocity of 20 m.s -1acalculate the force on the ball if the contact time F = m (v –u)/ t =0, 2 [30 – (+20)]/0, 01 = - 1000 N Ft = m (v – u) = 0, 2 (- 30 – 20) = - 10 kg. m.s -1OR Ft = 1000 x 0, 01 = - 10 N.s Impulse is given to both bodies during a collision COLLISIONS The law of the conservation of momentum is verThere are different types of collisions the learner

1. Elastic collisions. 2. Inelastic collisions (normal everyday co3. Completely inelastic collisions.

Elastic collisions conserve KINETIC ENERGY • No kinetic energy is lost • Perpetual motion • Difference in 2 objects' speed before collision e• Elastic collisions only happen when atoms and Inelastic Collisions (normal everyday collisionsSome kinetic energy is "lost" as:

• Noise energy • Thermal energy caused by friction • Elastic Ep of deformation

Lesson 24 lse and collisions

MODULE 1

re are two units for impulse: N.s or kg.m.s-1. During a is for the higher grade learners.

nd hits an object and returns with a velocity of 30 m.s -1 is 0,01 s.

.

y important in this section. s will have to deal with.

llisions)

quals difference in 2 objects' speed after collision molecules bounce around.

)

74

• Raising the internal energy of the surroundings • Speed slows down e.g. the object bounces back at a slower speed off a fixed surface

Completely inelastic collisions • Objects stick together after Impact • A lot of kinetic energy is "lost” (Converted into other forms of energy) Question1: Impulse can be deduced from Newton’s second law. Write down how you would deduce the formula for impulse from Newton’s second law. F∆t is the impulse of the force which causes the change in velocity of the body and it is equal to the change in momentum of the body. Impulse is a vector therefore the direction of the motion must always be taken in to consideration. Before starting a problem the learners should choose a direction as positive the opposite direction will obviously be the negative then. The change in momentum of a body depends on the force (F) acting on the body as well as the time (∆t) the force acts on the body. Use relative and applicable examples of impulse to explain the concept. Learners so often forget the direction it is important that you stress direction. Collisions The law of the conservation of momentum is very important in this section. There are different types of collisions the learners will have to deal with. Elastic collisions. Inelastic collisions (normal everyday collisions). Completely inelastic collisions. Question 2: Write down the differences between the types collisions. Write down the activities you would engage the learners in to illustrate each of the collisions in the table below.

Type of collision Activities Elastic Inelastic Completely inelastic

Remember: It is advised that the learners draw a picture of the motion before and after the collisions. They should also write down the given information in a structured way so that they are sure what is expected from them. Then the learners will also encounter a situation where an explosion is taking place. An explosion, in addition to its usual meaning, describes any situation in which two stationary moves apart because of the equal and opposite forces they exert on each other. Question 7: Which equation should the learners use in the explosion calculations? Question 9: How does road and car safety link up with momentum?

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Momentum and

Before any momentum calculation is attempted in the right manner There are certain givens thare not given to them directly. Principle of conservation of linear momentumThe total momentum in a closed system remains However, kinetic energy is not conserved in macroscopic collisions. Somehow gas particles losing energy. The basic content that the learners have to know1. Collisions lose no momentum Momentum before collision = momentum after c m1u1 + m2u2 = m1v1 + m2v2

collide

u1 u2

m1 m2

Example 1: (let to the right be +) BEFORE AN INELASTIC COLLISION AF

Detekg if the 20 kg

30 kg 20 kg

Before an inelastic collision

6 m.s -1 - 10 m.s -1

Lesson 25 more difficult calculations

MODULE 1

it should firstly be analysed so that it can be approached at the learners should know off by heart. These givens

constant in both magnitude and direction.

an inelastic collision. Kinetic energy is always lost in bounce around forever on the microscopic scale without

:

ollision

bounce apart

v1 v2

m1 m2

TER THE COLLISION

rmine the velocity of the 30 stands still after the collision.

After the collision

30 kg 20 kg

V = ?0 m.s -1

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momentum before collision = momentum after collision m1u1 + m2u2 = m1v1 + m2v2 30(6) + 20(-10) = 30v + 0

v = 180 – 200/30 = -20/30 = - 0, 67 m.s -1 to the left

Example 2: (let to the right be +)

Determine the velocity of the 25 kg and the 10 kg mass if they stick together

after the collision.

25 kg 10 kg

before collision

20 m.s -1 - 10 m.s -1

after collision

25 kg 10 kg

V = ?

Momentum before collision = momentum after collision m1u1 + m2u2 = (m1 + m2) v 25.20 + 10 (-10) = (25 + 10) v

v = 500 – 100/ 35

= 400/35 = 11, 4 m/s to the right Kinetic energy is not conserved in an inelastic collision

Ek before collision = 0, 5 m1u12 + 0, 5 m2u22 = 0, 5 (25)202 + 0, 5 (10)102

= 5000 + 500 = 5500J

Ek after collision = 0, 5 (m1 +m2) v2 = 0, 5 (35) (11, 4)2

=2274, 3 J 5500 - 2274, 3 = 3225, 7 J. This is the kinetic energy that is converted into noise energy, thermal energy, elastic Ep of deformation and raising the internal energy of the surroundings. • Kinetic energy is always "lost" in a macroscopic collision • Somehow gas particles bounce around forever on the microscopic atomic scale without losing

energy

Ft = m (v-u)

The impulse of a force is the product of the resultant force acting and the time for which it acts.

An explosion is a negative collision • Explosions are called super elastic collisions • There is an increase in kinetic energy • No momentum is lost in an explosion

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MORE CHALLENGING COLLISION CALCULATIONS Example 3: (let to the right be +) A 10 g bullet travels horizontally at 2000 m.s -1 passing through a 2 kg pendulum, which then moves horizontally at 4 m.s -1. Calculate the final speed of the bullet. Only consider horizontal motion.

m1u1 + m2u2 = m1v1 + m2v2 0, 01 x 2000 + 0 = 0,01v + 2 x 4

2 kgu

0,01 kg

before collision

2000 m.s -1

after collision

2 kg

0,01v = 20 – 8 0,01v = 12 v = 1200 m.s -1 These are some of the difficult collision sums you will come across. Work through them systematically. It is very important that you draw yourself a sketch of the before and after situation! Example 4: (let to the right be +) A 200 kg railway truck moves horizontally at 20 m.s -1, when 800kg of grain drops vertically on the truck. Calculate the final horizontal speed of the loaded truck. • Only consider horizontal motion • While they fall the grain have no horizontal motion.

200 kg

20 m.s-1

800 kg

800 kg

Momentum before collision = momentum after collision

m1u1 + m2u2 = (m1 + m2) v

200 x 20 + 0 = (200 + 800) v

v = 4000/ 1000

v = 4 m.s -1

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Example 5: (let to the right be +)

From a stationary 40 kg boat a 60 kg man dives at 4 m/s to the right and simultaneously an 80 kg man dives at 2m-s-1 to the left. Determine the velocity of the boat as it reacts to them diving off

Mom

before explosion = mom after explosion

40 ve kg

80 kg 60 kg

locity of boat, v2 ?

80 kg 60 kg

-2 m.s -1 -4 m.s -1

40 kg

(m1 + m2 + m3) u = m1v1 + m2v2 + m3v3 0 = 80 (-2) + 40 v2 + 60 (4) v2 = 160 – 240 /40 v2 = - 80/40 = 2 m.s -1 ,to the left.

Now that the learners can analyse momentum problems here are some difficult collision examples. Please take time to attempt the questions. Question1: A 20g bullet travels horizontally at 2000 m.s-1 passing through a 2kg pendulum which then moves horizontally at 4 m.s-1. Calculate the final speed of the bullet. Question 2: A 20g bullet travels horizontally and embeds in a stationary 1,9kg pendulum, which rises vertically to a height of 5m. Calculate the initial speed of the bullet. Question 3: A 220kg truck moved at a velocity of 20 m.s-1. 850kg of grain is dropped vertically on the truck. Calculate the speed of the loaded truck. Question 4: Two boys are standing on a 40kg stationary boat. The one boy1 has a mass of 40kg and boy2 a mass of 85kg. Simultaneously the two boys dive off the boat, boy1 with a velocity of4 m.s-1 and boy2 with a velocity of 2 m.s-1. Determine the velocity of the boat as it reacts to the boys diving off. Question 5: Two trolleys are connected with a spring. The spring is released and the trolleys move in different directions but strike buffers at the same time. Each of the trolleys has a mass of 500g. Trolley A is 320mm from the buffer; Trolley B has a piece of metal on it and is 80mm from the buffer. 1. Deduce an equation from the principle of the conservation of momentum, from which you can calculate the mass of the piece of metal on trolley B in terms of the displacements the trolleys experience before striking the buffers. 2. Calculate the mass of the piece of metal Tip: you cannot solve this problem without making a drawing. Question 6: After completing the above examples what sort of difficulties can you identify?

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You can overcome these difficulties by introducing the learners to examples from various resources. Build your learners self-confidence and they will be able to solve the difficult calculations in the end. ASSESSMENT TIPS – Physics section Note: this is a summary of the most common problems learners’ experience. There are a lot more problems, which have not been covered in this summary.

1. Naming of forces acting on an object: The use of correct terminology, as well as knowing the directions of forces is important. As a result they cannot apply the knowledge gained to solve problems. 2. The stating of laws: The learners have difficulty in stating the laws precisely. They have the tendency of omitting the key words in laws. And they tend to use poor language. 3. Equations of motion: The learners tend to compare ‘a’ for horizontal motion and ‘g’ for vertical motion when applying equations of motion.

4. Terminology: It is obvious that some educators are not insistent on the use of proper terminology. Hence, learners use terminology they make up themselves and is so doing loose allot of marks, especially when stating laws.

Note: The grade 12-learner preparations for the assessment are critical to our learners; on the next page follow some tips on preparation and completing assessment tasks. Read through them and think of other tips to share with your learners. It is very important that your learners are mentally and emotionally prepared for the exams ahead. More assessment tips

5. Work through as much past assessment tasks as possible This will help the learners to identify patterns, identify the level of the questions asked and test their readiness for assessment. 6. Manage time effectively.

a. Read every question carefully. b. When your time for a question is up move on! c. Number all the questions clearly. d. Answer clearly. e. If you don’t know the answers, try telling the assessor what you know. Don’t leave

open spaces.

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