Further Mathematics Lesson Note SS2 Second Term
Further Mathematics SS2 – Edudelight.com
SCHEME OF WORK
FURTHER MATHEMATICS
SS 2
- Review of 1st term’s work and conic section: definition of circles and part of circle.
- Equation of circle given centre and radius.
- General Equation of a circle
- Finding centre and radius of a given circle.
- Finding equation of a circle given the end point of the diameter
- Equation of circle passing through 3 points
Equation of tangent to a circle
Length of tangent to a circle
- Statistics
(i) Probability (ii) Classical Frequential and axiomatic approaches to probability
(b) Sample space and event space
(c) Mutually exclusive, independent and conditional events
(d) Conditional probability
(e) Probability trees
- Permutations (i) Permutation on arrangement (ii) Cyclic Permutation (iii) arrangement of identical object. (iv) Arrangement in which repetitions are allowed.
- Combination
Introduction to combination on selection
Conditional arrangements and selection
Problems involving arrangement and selection
Problems
- Review of the 1st half term’s & periodic test.
- Dynamics (i) Newton’s Law of motion
Motion along inclined plane
Motion of connected particles
- Work, power and energy
Impulse and momentum
- Projectiles
Trajectory of projectiles
Greatest height reached projectiles conts
Time and flight
Range
Projection along inclined plane
- Introduction to operation research inventory model
Concept of inventory
Definition of important terms in inventory holding list
Demand ordering list etc
Computation of optimal quantity
12 & 13 revision of second term’s work and preparation of examination
WEEK ONE
REVISION O LAST TERM WORK
WEEK 2
CIRCLE
A Circle has been defined as the locus of points equidistant from a fixed point. A circle is completely specified if we know.
i) the Centre ii) the radius
Equation of a circle centre (a, b) radius r
| P (a, b) |
| Q |
| r |
| R |
| Q |
| b |
| X – A |
| a |
The above diagram shows a circle centre (a, b) radius r considers PQR.
PR = x – a
RQ = y – b
Since PQR is a right angled triangle, we have PR2 + RQ2 = PQ2
| Y – b |
| Q |
| r |
Hence (x – a)2 + (y – b)2 = r2 (1)
| x – a |
| P (a, b) |
| R |
Equation (1) is the equation of a circle centre (a, b) radius r.
| y |
If the centre of the circle is the origin (O, O) equation (1) becomes x2 + y2 = r2
| x |
| O |
| X N |
| y |
| r |
| P(x, y) |
Alternatively, we can consider OPN
ON2 + NP2 = OP2
X2 + y2 = r2
Example 1:
Find the equation of the circle centre (-3, -2), radius 2 units
Solution
From the formular, equation of a circle centre (a, b) and radius r,
(x – a)2 + (y – b)2 = r2
a = 3, b = -2 r = 2
(x – 3)2 + (y – (-2)2
(x – 3)2 + (y + 2)2 = 4
X2 – 3x – 3x + 9 + y2 + 2y + 2y = 4
X2 – 6x + 9 + y2 + 4y + 4 = 4
X2 + y2 – 6x + 4y + 9 = 0
Example 2
Find the equation of the circle centre origin, and radius 3.
Solution
X2 + y2 = 32 since the equation is at the centre of the origin
X2 + y2 = 9.
THE GENERAL EQUATION OF A CIRCLE
Recall that the equation of the circle centre (a1 b) and radius r is
(x – a)2 + (y – b)2 = r2
Upon expansion, we have
X2 – 2ax + a2 + y2 – 2by + b2 = r2
X2 + y2– 2ax– 2by+ a2+ b2 – r2 = 0
The above equation can be written as
X2 + y2 + 2gx + 2fy + c = 0
Where a = -g, b = -f and c = a2 + b2 – r2
The equation
Is called the general equation of the circle. You will observe that the general equation of a circle:
- Is a second degree equation in x and y
- Has coefficient of x2 and y2 being equal.
- Has no xy term
So, given the centre and radius of a circle, we can easily find the general equation of a circle. On the other hand, given the general equation of a circle, we can find the radius and centre of the circle, by the method of completing the squares.
EQUATION OF THE CIRCLE WHICH IS DESCRIBED ON THE LINE JOINING (X1, Y1) AND (X2, Y2) AS DIAMETER
| A (X1, Y1) |
| B (X2, Y2) |
| P(X, Y) |
Let A and B be the points (x1, Y1) and (x2, Y2) respectively.
Join AB. Let P(x, y). be any point on the circle, Join AP, BP A= (x1, y1)
< APB = 900 P = (x, y)
Slope of A = Gradient of AP =
Slope BP = Gradient of BP =
Since AP is to BP, That is m, XM2 = -1
= -1
(y – y1) (y – y2) = – (x – x1) (x – x2)
(x – x1) (y – y2) + (y – y1) (y – y2) = 0
This is the required equation of the circle with (x1, y1) and (x2, y2) as the coordinates of the ends of a diameter.
Example 4
Find the equation of the circle whose diameter has the end point (-5, 1) and (3, -7)
Solution
Take (x1, y1) = (-5, 1) and (x2, y2) = (3, 7)
Equation of the circle
(x – x1) (x – x2) + (y – y1) (y – y2) = 0
(x + 5) (x – 3) + (y – 1) (y + 7) = 0
X2 + y2 + 2x + 6y – 22 = 0
WRAP UP AND ASSESSMENT
The equation of a circle centre origin, radius r is x2 + y2 = r2
The equation of a circle centre (a, b) radius r is (x – a)2 + (y – b)2 = r2. The general equation of a circle is of the form x2 + y2 + 2gx + 2fy + c = 0. The general equation of a circle (i) is a second degree equation in x and y (ii) has coefficients of x2 and y2 being equal (iii) has no xy term.
1. Find the equations of the following circles
a. Centre (o, o) radius 7
b. Centre (o, o) radius
c. Centre (2, -3) radius 4
2. Find the equation of the circle passing through (3, -2), (4, 5) and (-6, 3).
3. Find the equation of the circle whose centre is the point (2, 3) and which passes through the intersection of the lines 3x -2y – 1 = 0 and 4x + y -27 = 0.
4. Find the centers and radii of the following circles.
X2 + y2 – 2x – 4y + 1 = 0; 4x2 + 4y2– 4x – 12y + 1 = 0
5. Find the equation of the circle whose centre is the same as that of the circle x2 + y2 – 6x + 2y + 4 = 0 and which passes through the point (7, 4)
Solution
x2 + y2 – 6x + 2y + 4 = 0
From completing the square;
X2 – 6x + 9 + y2 + 2y + 1 = -4 + 9 + 1
(x – 3)2 + (y + 1)2 = 6
(x – a)2 + (y –b)2 = r2
-a = -3 y – b = y + 1
a = 3 -b = 1
b = 1
(a, b) = (3, -1) is the centre which passes through (7, 4)
r2 = cp2
r2 = (3 – 7)2 + (-1 -4)2
(-4)2 + (-1 + 4)2
r2 – (-4)2 + (5)2
r2 = 16 + 25
r2 = 41
C= a2 + b2 – r2
= (-3)2 + (1)2 – 41
9 + 1 – 41 = -31
Hence the equation of the circle
(x – 3)2 + (y + 1)2 = 41
(x – 3) (x – 3) + (y + 1) (y + 1) = 41
X2 – 3x – 3x + 9 + y2 + y + y + 1 = 41
X2 – 6x + 9 + y2 + 2y + 1 = 41
X2 + y2 – 6x + 2y = 41 – 10
X2 + y2 – 6x + 2y = 31
X2 + y2 – 6x + 2y – 31 =
OR
From the General equation of a circle C = -31
X2 + y2 + 2gx + 2fy + c = 0
X2 + y2 + 2(-3)x + x(1)y – 31 = 0
X2 + y2 – 6x + 2y – 31 = 0
TICKET OUT
1. Find the equation of the circle which passes through the points (1, -2), (4, -3) and has its centre on the line 3x + 4y + 10 = 0
2. Find the equation of the circle circumscribing the triangle formed by the lines 3x + y -5 = 0, x + y + 1 = 0 and 2x + y – 4 = 0
Further Mathematics SS2 – Edudelight.com
WEEK 3
CIRCLE EQUATION OF TANGENT TO A CIRCLE AND LENGTH OF TANGENT
| T (x1, y1) |
Let the equation of the cir
X2 + y2 + 2gx + 2fy + c = 0
At (x1, y1)
X12 + y12 + 2gx, + 2fy + c = 0
C = -(x12 + y12 + 2gx1 + 2fy1) ………….(1)
From.
X2 + y2 + 2gx + 2fy + c = 0
2x + 2y + 2g + 2f = 0
X + y + g + f = 0
(y + f) + x + g = 0
(y + f) = -(x + g)
=
At (x1, y1), =
The equation of the tangent is
=
(y – y1) (y1 + f) = – (x1 + g) (x – x1)
Yy1 + yf1 – y12 – y1f = -(x1x + x12 + xg +x1g)
Yy1 + yf – y12 – y1f = -x1x + x12 – xg + x1g
Yy1 + yf + x1x + xg = y12 + y1f + x12 + x1g
Xx1 + yy1 + xg + yf = x12 + y12 + x1g + y1f……………….(2)
Note: the equation of tangent at x1, y1) to the circle x2 + y2 = a2 is given as xx1 + yy1 = a2
Adding x1g + y1f to both sides of (2), we have
xx1 + yy1 + xg + x1g + yf + y1f = x12 + y12 + 2x1g + 2y1f
xx1 + yy1 + (x + x1)g + (y + y1)f = x12 + y12 + 2x1g + 2y1f
From (1), x12 + y12 + 2x1g + 2y1f = -c
Xx1 + yy1 + (x + x1)g + (y + y1)f + c = 0
Hence the equation of the tangent to the circle
X2 + y2 + 2gx + 2fy + c = 0, at the point (x1, y1) on the circle is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
Example 5
Find the equations of the tangent to the circle 2x2 + 2y2 – 2x – 5y + 3 = 0 at (1, 1)
Solution
The equation of the circle can be rewritten as
X2 + y2 – x – +
X2 + y2 + 2gx + 2fy + c = 0 General Equation of a circle
2gx = -x 2fy = C =
2g = -1 f =
g =
The equation of tangent at the point (1, 1) is
X(1) + y(1) + (x + 1) (y + 1) = 0
X + y – – – + = 0
4x + 4y – 2x – 2 – 5y – 5 + 6 = 0
4x + 4y – 2y – 5y – 1 = 0
+2x – y – 1 = 0; 2x – y – 1 = 0 is the equation of the tangent at the point (x1, y1) = (1, 1)
| T |
| t |
LENGTH OF A TANGENT
| r Q (a, b) |
| P |
| Q (X1, Y1) |
Consider the diagram
Above
Since OTQ is a right-angled triangle
OQ2 = OT2 + QT2
P2 = r2 + t2
t2 = p2 – r2
Also P2 = (a – x1)2 + (b – y1)2
Example 6
Find the length of the tangent to the circle x2 + y2 – 2x – 4y – 4 = 0 from the point (8, 10)
Solution
x2 + y2 – 2x – 4y – 4 = 0
x2– 2x + y2 – 4y = – 4
x2– 2x + 1+ y2 – 4y + 4 = 4 + 5
(x – 1)2 + (y – 2)2 = 9
(x – 1)2 + (y – 2)2 = 32
Hence (1, 2) is the centre of the circle with radius 3 units. Then we have
P2 = (1 – 8)2 + (2 – 10)2
P2 (-7)2 + (-8)2
P2 = 49 + 64
P2 = 113
r2 = 9
t2 = p2 – r2
= 113 – 9
t2 = 104
t = is the length of the tangent to the circle x2 + y2 – 2x – 4y – 4 = 0 from the point (8, 10)
WRAP UP AND ASSESSMENT
The equation of the tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 at the point (x1, y1) on the circle is xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
A. Find the equation of the tangents to the following circles at the given points on the circles. (i) x2 + y2 + 2x – 3y – 13 = 0 at (1, -2)
(ii) 3x2 + 3y2 – 8x – 6y – 61 = 0 at (4, 5)
B. Find the lengths of the tangents of the following circles from the given points.
a) x2 + y2 + 5x + 4y – 20 = 0 at (2, 3)
b) x2 + y2 – 3x + 2y – 10 = 0 at (-4, 1)
TICKET OUT
C. Find the equation of the tangents to the following circles at the given points.
a. x2 + y2 = 169 (12, -5)
b. x2 + y2 = 10 at the point whose abscissa is 1.
c. x2 + y2 – 4x + 2y + 3 = 0 at (1, -2)
3. Find the equation of the circle which has centre c(3, 1) and which touches the line 5x – 12y + 10 = 0. Find the equation of the equation of the tangent to the circle at the point
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WEEK 4
PROBABILITY
Probability is a branch of Mathematics dealing with random experiments in the tossing of a coin, we cannot predict which side of the coin would show, hence we say, this is a Random experiment. The ratio of number of success, a particular side of the coin shows up to the total number of trials becomes the probability of that particular side, either the head or the tail.
Hence probability becomes which is also known as the Relative frequency of the number of times a head shows up.
SAMPLE SPACE AND EVENT SPACE
An outcome is any result of an experiment in probability. If we cannot predict before hand, the outcome of an experiment, the experiment is called a random experiment. The set of all possible outcomes of any random experiment will be called a sample space and it will be denoted by S. The probability of an event E denoted Pr (E) is defined as
Pr (E) =
Note: the probability of an impossible event is zero. Since n( ) = 0
The probability of any event E is therefore a number which satisfies the inequality 0 Pr (E) 1.
The sample space to which we assign each of the sample elements, equal probability is called an equiprobable sample space. The sample elements are said to be equiprobable or equally likely.
Example 1.
If a fair coin is tossed once, what is the probability of obtaining: a) a head b) a tail?
Solution
Since the events head and tail are equally likely, then as the total probabilities is equal to unity (1) we have Pr (Head) = Pr (Tail) = P
But Pr (H) + Pr (T) = 1
P + P = 1
2p = 1
P =
(i) Pr (H) = and (ii) Pr (T) =
Example 2
If a fair die is tossed, find the probability of obtaining a (i) 6 (ii) 3
Solution
In fairness, each face of a die has an equal chance of showing uppermost when tossed i.e
Pr (6) = Pr (5) = Pr (4) = Pr (3) = Pr (2) = Pr (1) = P
But P + P + P + P + P + P = 1
6P = 1
P =
Pr(6) = Pr(3) =
Working definition of probability
This definition is the simplified form of the classical definition. If an event A has n exhaustive ways of occurring, and out of these, k ways are favorable, we write.
Pr(A) = =
SET NOTATION IN PROBABILITY
The definition of probability or indeed, the concept of probability is better understood by expressing it in the language of sets. A if x E µ is a universal set and E denotes member of, the probability of the event x is given by Pr (x) =
A set whose members represent all possible outcomes of an observation or experiment is called a sample space. An event is a subset of the sample space.
Fundamental Results of Probability Set Functions
If A and B denote events and Ac, reads complement of A or not A and Pr (A) means probability of the event A. then the following hold:
- Pr(A) + Pr(Ac) = 1
- If A1, A2, A3 …..A is a set of exhaustive, mutually exclusive events, the.
Pr(A1), + Pr (A2) + Pr (A3) + …… + Pr (Ak) – 1
- O Pr (A) 1
At no occasion must the probability of an event be greater than 1 or less than zero.
- If two events A and B are mutually exclusive then
- Pr (A U B) = Pr (A) + Pr(B). (Addition Law)
- Pr (A n B) = 0
Note: if A and B are not mutually exclusive then Pr (A U B) = Pr(A) + Pr (B) – Pr(A n B)
- If two events A and B are independent then Pr (A n B) = Pr (A) x Pr (B) (Multiplication Law)
- If µ and denote the sample space and the null set respectively, then
- Pr( ) =1
- Pr ( ) = 0
WRAP UP AND ASSESSMENTS
Pr (A) = where A C U and µ is the total sample space.
When two events A and B are mutually exclusive Pr (A U B) = Pr (A) + Pr(B). i.e A and B cannot occur together.
A box contains 6 green balls, 4 red balls and 5 yellow balls. If three balls are drawn successively at random from the box, find the probability that they are drawn in the order green, red and yellow if each ball is (a) replaced (b) not replaced
TICKET OUT
Box x contains 7 electric bulbs of which 3 are defective and box y contains 5 bulbs of which 2 are defective A bulb is drawn at random from each box, what is the probability.
(i) Both are non-defective
(ii) one is defective and the other is not?
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WEEK FIVE
CONDITIONAL PROBABILITY AND TREE PROBABILITY
The ability to solve a problem successfully in the topic “probability” depends largely on being able to determine when events are mutually exclusive independent or dependent. The next important aspect is that the sample space must be known and the events in the sample space must be equiprobable.
With conditional probability, we shall link the probability of an event A to another event B. given the conditions surrounding the events A and event B, it is possible to compute the probability of A given B has already occurred. This is called the conditional probability of A given B and written P (A/B)
Definition
If A and B are any two events defined in a finite sample space, then
P(A/B) =
Suppose µ is a finite sample space and A, B C µ , then P(B) = and P(A n B) =
P (A/B) = P(A n B) / Pr(B)
= provided P(B) 0
TREE DIAGRAM
Tree diagram method can also be used to solve problems on probability involving independent and dependent event. Each of the branches of the tree diagram represents different possibilities in which probability can be written or formed. (Examples illustrated in the classroom)
WRAP UP AND ASSESSMENT
If two events A and B are said to be dependent then P(A n B) = P(A) x P(B/A), where P(B/A) means the probability of B given that A has occurred.
If three cards are chosen from a park without replacement, what is the probability of obtaining?
(a) at least 2 spades
(b) at most two spades?
TICKET OUT
A coin is weighed in such a way that the probability of a head is and the probability of a tail is . If a head appears then a letter is selected at random from among (a, b, c, d, e, f); and if a tail, a letter is selected at random from among (a, b, c, d).
i. Find the probability that a vowel is selected
ii. if a vowel was selected, what is the probability that a tail appeared on the coin?
2. Two boxes labeled X and Y contains rings. Box X contains 4 gold, 3 diamond and 3 brass rings. Box Y contains 6 gold, 2 diamonds and 1 brass ring. A box is selected at random and a ring is picked from it. Calculate the probability that the selection is
i. a gold from X
ii. a diamond ring
iii. either a gold or diamond
iv. if a gold is selected, what is the probability it came from box X?
3. If X and Y are events, given that P(X) = Pr (Y) = and P(X U Y) =
Find (i) P C x/y) ii) P(y/x) iii. P(X n Yc) iv. P(X/Yc).
4. Four boys participated in a competition in which their respective chances of winning prizes are , , and . What is the probability that at most two of them win prizes?
5. A bag contains 6 red and 4 black balls, if three balls are selected at random from the bag, one after the other and without replacement find the probability that (i) they are of the same colour (ii) one black and two red balls are selected.
6. the probability that a player will score a penalty kick is 0.8. if he takes 5 penalty kicks calculate, correct (4 d.p), the probability he will score.
(i) no goal
(ii) at most two goals.
WEEK SIX
PERMUTATION
Suppose we are interested in the number of different arrangements of three men in a line. In order to be able to distinguish between the different arrangement, let us label the three men a, b and c.
The problem reduces to finding the number of ways of arranging the letters a, b and c. here, order is of paramount importance as abc and cba for example are two different arrangement altogether.
Consider the diagram below
| c |
3 2 1
| b |
| a c b |
| a b c |
| a |
| a |
| a |
| a |
| a |
| a |
| b |
| b |
| b |
| c |
| c |
| c |
| c |
| c |
| c a b |
| c b a |
| b c a |
| b a c |
The diagram above is known as the tree diagram abc bac cab ac bca cba, we observe here that there are 6 different arrangements.
In general, the number of different arrangements of n different objects is equal to the products n x (n -1) x (n -2) x ………. X 4 x 3 x 2 x 1, this product can be written n! for short
n! is read n – factorial
This, n! = n(n-1) (n-2) (n-3) x ……… x 4 x 3 x 2 x 1.
In general, if an operation can be performed in p-ways and another operation can be performed in q-ways, then the number of different ways of performing the two operations one after the other is p x q ways.
The arrangement of objects taking into account the different orders or arrangement is called PERMUTATION.
Suppose, we are only interested in the number of ways the first and second positions can be taken by 4 people in a race, assuming there is no tie.
The first position can be taken in 4 ways by any of the 4 athletes. The second position can be taken in 3 ways by any of the remaining 3 athletes. This arrangement is called the Permutation of 4 people taking 2 at a time and is designated. This 4P2 = 4 x 3
Hence in general:
nPr = is the permutation of n objects taking r at a time.
Recall that n! = n(n-1) (n -2) x ….. x 3 x 2 x 1 = n(n-1)!
If we put n = 1 then 1! = 1 x 0!
So in our factorial notation, we define
O! = 1
CYCLIC PERMUTATION
In cyclic permutation, we are concerned about arrangement of things about a circular object. The number of permutations of n people round a circular table is obtained by fixing one person and permitting the remaining (n-1) about the fixed person. The number of ways of doing this is 1(n -1)! Ways.
In general, the number of ways of arranging fixed objects rounds a circular ring which can be turned over =
PERMUTATION INVOLVING INDISTIGUISHABLE OBJECTS
The number of ways of permuting n objects taking n at a time with n1, objects alike, n2 objects alike ……nj objects alike n1 + n2 + n3 + ………. nj = n is given by
CONDITIONAL PERMUTATION
Sometimes restrictions can be placed on the order of arrangements of objects that are to be permuted. If that happen, the permutation is said to be conditional.
WRAP UP AND ASSESSMENT
The arrangement of objects taking into account the different orders of arrangement is called PERMUTATION. nPr = . The number of ways of arranging n-fixed objects round a circulating which can be turned over is
Find the number of ways the letters of the word FURTHER can be permuted.
Find the number of permutations of the letter of the word MATHEMATICS.
Find the number of ways 6 people can be seated in a round table, if two particular friends must sit next to each other.
TICKET OUT
Find the number of ways of arranging the letters of the word ABAKALIKI.
WEEK SEVEN
MID TERM EXAMINATION
WEEK EIGHT
COMBINATION
Combination, otherwise called selection, is a set of quantities chosen from a given set where attention is not paid on the order of the quantities within the set.
The number of combinations or selections of n unlike things taken r at a time is
Notation: The notation for combination is nCr
Thus, nCr =
npr = r! ncr.
It can be shown too that ncr = nCn –r
For if nCr = then by replacing r by n – r we have
nCn-r = (n –r)!
= nCr
WRAP UP AND ASSESSMENT
The number of ways r-objects can be selected from n-objects without much regard being paid to the order of arrangement is denoted nCr. nCr = =
A woman wishes to pick 2 parts of black shoes from 5 parts of black shoes and 3 packs of white shoes from 7 parts of white shoes. Find the number of ways, the woman can make her selections. Ex 5 No 13, 12, 19.
TICKET OUT
A committee consisting of 3 men and 5 women is selected from 5 men and 10 women. Find how many ways this committee can be formed.
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WEEK 9
Newton’s Laws of Motion
The First law of Motion
It states that “Everybody continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by external impressed forces.
The tendency of a body to remain in its state of rest or uniform motion in a straight line is called inertia.
The Second Law of Motion
The law states that the rate of change of momentum of a body is proportional to the applied force and is in the devotion of the force.
By Newton’s second Law
F
| F =ma |
F ma
F = kma, when k = 1,
The Newton’s Third law
Action and reaction are equal and opposite
Example
A boy sits on a log. The mass of the log is 8kg and the weight of the boy is 55N. what is the reaction of the ground on the log on which the body is sitting? (Take g = 9.8m/s2)
Solution
By newton’s second law F=ma
Weight of the log = 8 x 9.8 = 78.4N
Weight of the boy and log = 78.4 + 55 = 133.4N
By newton’s third law, the ground will exert and equal but opposite force on the log in which the body is sitting.
Hence the reaction force R = 133.4N
Exercise
A load of mass 80kg is placed in a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards.
- At constant velocity
- With acceleration of 2m/s2 1 take g = 9.8m/s2
Motion Along an inclined plane
Example: an object whose weight is 10kg is placed on a smooth plane inclined at 300 to the horizontal find
- The acceleration of the object as its moves down the plane.
- The velocity attained after 3 seconds if
- It states from rest
| F |
It moves with an initial velocity of 5ms-1 (take g = 10m/s2)
Solution
| 30 |
| mg |
| Mg cos 30 |
| Mg sin 30 |
- When the body is moving down. The plane the net force mg sin 300 = ma
a = g sin 300 = 10 x 0.5
a = 5m/s2
- t = 3, a= 5m/s2, u = o
from v = u + at
v = o + 5 x 3
v = 15 m/s
ii. if u = 5m/s
v = 5 + 5(3)
= 5 + 15
V = 20 m/s
Motion of connected particles
Example: Two particles whose masses are 15kg and 12kg respectively are attached to the ends of a light in extensible string. The string passes over a light frictionless pulley and the masses hang freely. The system is released from rest when the 15kg mass is 32m above the floor. Find
- the tension in the string
- the line taken by the 15kg mass to reach the floor 1 take g = 10m/s2)
Solution
| 15g |
| a |
| a |
| T |
| T |
| 12g |
Let the acceleration be a
By newton’s second law F = ma
For the 15kg mass
15g – T = 15a………………….(1)
For the 12kg mass
T – 12g = 12a …………………(2)
- + (2) gives
3g = 27a
A =
From (2) T = 12a + 12g = 12 x + 12 x 10 =
ii. Let t be the time taken by the 15kg mass to reach the floor, 3 the distance cover. Since the body starts from rest u = 0, S =
t2 =
t2 =
t2 =
WEEK 10
Work, Power and Energy
The work done by a constant force F is defined as the product of the force and the distance W = /F/d
Example: The resistance to the motion of a cart being push by a man is 220N. if the man pushed the cart a distance of 12km, how much work has he done against the resistance?
Solution
F = /w/d = 220N x 12000m = 2640000J
Power
Power is the rate of doing work. The unit of power is watt (w)
Exercise:
On the level, a car develops a power of 60kw if the resistance to motion is 900n, what is the maximum speed of the car?
Working at the same power and with the same resistance operating, what would be the maximum speed possible up an inclined of the car is 800kg?
What is the acceleration at the time when the car is moving up the inclined plane at 40m/s? (Take g = 10m/s)
Energy
| Ek = |
Kinetic Energy
, m = mass of body
V = velocity of body
| Ep =mgh |
Potential Energy
M = mass, g = gravity, h = height
Law of Conservation of Energy
It state that energy cannot be created nor destroyed. It can only be changed from one form to another.
| V = |
When a particle falls from a height it, gain in K.E is equal to loss in P.E
mgH =
Impulse and Momentum
The momentum of a body is the product of mass and acceleration. Newton’s second law states
That F =
| Ft = mv – mu |
Ft is called impulse and the S.I unit is Ns.
Conservation of Momentum
It states that if there are no external forces, the total momentum before collision of two particles is equal to the total momentum of the two particles after collision.
Consider two particles of mass M1 and M2 which collide with initial velocities U1 and U2 respectively. Let us assume that the particles have final velocities V1 and V2 respectively
M1 U1 + M2U2 = M1V1 + M2V2
Example
A force FN acts on a body of mass 2kg travelling 45m/s for 0.5s. if the final velocity of the body is 6.5m/s. Find F.
Solution
U =4.5 m/s, V = 6.5 m/s, m =2kg, t =0.5s
Impulse Ft = mv – mu
F =
F = F = 8N
Exercise: A bullet of mass 0.03kg is fired with a velocity of 248m/s into a block of wood of mass 0.9kg. find the common velocity of the bullet and the block when the bullet is embedded.
Assignment
A bullet is fired with a speed of 500m/s into a block of wood of mass 0.5kg and becomes embedded in it. If it gives the block a speed of 15 m/s. Find correct to two significant figures. The mass of the bullet.
WEEK 11
PROJECTILES
Motion Under gravity
- V = u + at
- V2 = u2 + 2as
- S = ut + at2
Where u is the initial velocity, V is the final velocity, S the distance travelled, a the acceleration and t the time.
Analogous Equations
Suppose a particle is projected with an initial velocity u of an angle to the horizontal
| H |
| x |
| U cos |
| x |
| U |
| U cos |
| A |
| R |
| B |
| vy |
Recall
V = u +at
S = ut + at2
ax = o, ay = -g
For a particle under gravity we have the following
Vx = ucos Vy = u sin – gt
Sx = ucos Sy = ucos – gt2
Where Sx = horizontal distance
Sy = vertical distance
Vx = horizontal component of velocity
Vy = vertical
R = the range
Example
A particle is projected with an initial velocity of 42m/s at an angle 50o to the horizontal. After 2 seconds. Find
- The vertical component of the velocity
- The horizontal component of the velocity
- The magnitude of the velocity
- The vertical distance travelled
- The horizontal distance travelled (take g = 9.8 m/s2
Solution
| 42 m/s |
| x |
| 50 |
- Vy = u sin -gt
= 42 sin 50o – 9/8 x 2
= 12.57m/s
- Vx = u cos
= 42 cos 50o
=27m/s
- V =
=
= 29.78m/s
- Sy = ut sin – gt2
=42 x 2 sin 50o – x 9.8 x 22
=44.75m
- Sx = Utcos
=42 x 2cos 500
=53.99m
The Greatest Height Reached
Vy2 u2sin2 – 2gsy, at maximum height Vy = 0
O = u2 sin 2 – 2gSy
| hmax = |
Sy =
Time Taken to Reach the Greatest Height
Vy = Usin – gt
| t = |
O = u sin – gt
Example: A bullet is fired to the horizontal. If the initial velocity of the bullet is 43m/s. Find
- The greatest height reached
- The time taken to reach the greatest height
| 43 m/s |
Solution
| 36o |
- h =
h =
h = 32.59m
- t =
Time of Flight
It is the time taken for the particle to return to it initial level
Sy = ut sin gt2, but Sy = O
O = utSin – gt2
t (u sin – gt) = 0
| T = |
t = o or t =
but t = o is the initial time, Hence the time of flight denoted T is
Range: If is the horizontal distance covered when the particle returns to its original level
Sx = u tcos
R = uT cos , where T=
R = ucos x
R = called the range.
The Trajectory of a Projectile
Let y be the vertical distance and x the horizontal distance
X = ut cos ……………..(1)
Y = utSin – ………………(2)
From (1) t = ……………….(3)
Substituting t = into (2) gives
Y = u sin –
Y = xtan – ……………………(4)
Hence the path or trajectory of a projectile is a parabola.
Exercise:
A particle is projected at an angle to the horizontal with an initial velocity u. if the maximum horizontal distance travelled is 20m and the greates height reached is 10m. Find u and
Assignment: New further Mathematics Project Page 273 Q 15, 16 and 17.
