WAEC · 2009 · Nov/Dec · Further Maths · Paper 2

Question 1

Indices, logarithms & surds

(a)
Solve $2^{2x + 2} - 9(2^x) = -2$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Write $2^{2x + 2} = 2^2 \times (2^x)^2$
$= 4(2^x)^2$ .
Let $y = 2^x$ .
The equation becomes $4y^2 - 9y + 2 = 0$ .
Factorise: $(4y - 1)(y - 2) = 0$ , so $y = \frac14$ or $y = 2$ .
$2^x = 2 = 2^1$ gives $x = 1$ .
$2^x = \frac14 = 2^{-2}$ gives $x = -2$ .
So $x = 1$ or $x = -2$ .

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Question 2

Functions

Given that $f(x) = \dfrac{x}{2}$ , $g(x) = \dfrac{5x}{x^2 - 4}$ , $x \ne \pm 2$ , and $h(x) = (x - 1)^{\frac12}$ , where $x \in \mathbb R$ , the set of real numbers, find:

(a)
$g \circ f(x)$ ;
(b)
the inverse of $h(x)$ .

Worked solution (try it first)

(a)

$g \circ f(x)$ means $g(f(x))$ : put $f(x) = \frac{x}{2}$ in place of $x$ in $g$ .
$g\left(\frac{x}{2}\right) = \dfrac{5 \cdot \frac{x}{2}}{\left(\frac{x}{2}\right)^2 - 4}$
$= \dfrac{\frac{5x}{2}}{\frac{x^2}{4} - 4}$ .
Multiply the top and bottom by 4: $\dfrac{10x}{x^2 - 16}$ .
So $g \circ f(x) = \dfrac{10x}{x^2 - 16}$ , $x \ne \pm 4$ .

(b)

Let $y = (x - 1)^{\frac12}$ and make $x$ the subject.
Square both sides: $y^2 = x - 1$ , so $x = y^2 + 1$ .
Swap the letters: $h^{-1}(x) = x^2 + 1$ (for $x \ge 0$ ).

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Question 3

Polynomials & quadratic roots

(a)
The roots of the equation $3x^2 + 4x - 5 = 0$ are $\alpha$ and $\beta$ . Find the equation whose roots are $\left(\dfrac{1 + \alpha}{\beta}\right)$ and $\left(\dfrac{1 + \beta}{\alpha}\right)$ .

Worked solution (try it first)

(a)

From $3x^2 + 4x - 5 = 0$ : $\alpha + \beta = -\frac43$ and $\alpha\beta = -\frac53$ .
Sum of the new roots: $\dfrac{1 + \alpha}{\beta} + \dfrac{1 + \beta}{\alpha} = \dfrac{\alpha + \alpha^2 + \beta + \beta^2}{\alpha\beta}$ .
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$= \frac{16}{9} + \frac{10}{3}$
$= \frac{46}{9}$ .
So the sum is $\dfrac{-\frac{12}{9} + \frac{46}{9}}{-\frac53} = \dfrac{34}{9} \times \left(-\dfrac35\right)$
$= -\dfrac{34}{15}$ .
Product of the new roots: $\dfrac{(1 + \alpha)(1 + \beta)}{\alpha\beta} = \dfrac{1 + (\alpha + \beta) + \alpha\beta}{\alpha\beta}$
$= \dfrac{1 - \frac43 - \frac53}{-\frac53}$
$= \dfrac{-2}{-\frac53}$
$= \dfrac65$ .
The equation is $x^2 - (\text{sum})x + \text{product} = 0$ : $x^2 + \frac{34}{15}x + \frac65 = 0$ .
Multiply by 15: $15x^2 + 34x + 18 = 0$ .

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Question 4

Sequences, series & binomial expansion

(a)
Write down the first three terms of the binomial expansion of $(1 + 3x)^{\frac13}$ in ascending powers of $x$ .
(b)
Use the expansion in (a) to find, correct to three decimal places, the value of $28^{\frac13}$ .

Worked solution (try it first)

(a)

$(1 + a)^n = 1 + na + \dfrac{n(n - 1)}{2!}a^2 + \ldots$ with $n = \frac13$ and $a = 3x$ .
Second term: $\frac13(3x) = x$ .
Third term: $\dfrac{\frac13\left(-\frac23\right)}{2}(3x)^2 = -\frac19 \times 9x^2$
$= -x^2$ .
So $(1 + 3x)^{\frac13} = 1 + x - x^2 + \ldots$

(b)

Write $28 = 27\left(1 + \frac{1}{27}\right)$ , so $28^{\frac13} = 3\left(1 + \frac{1}{27}\right)^{\frac13}$ .
Match $3x = \frac{1}{27}$ : $x = \frac{1}{81}$ .
$28^{\frac13} \approx 3\left(1 + \frac{1}{81} - \frac{1}{6561}\right)$
$= 3(1 + 0.012346 - 0.000152)$
$= 3(1.012193)$ .
$= 3.03658$ , which is $3.037$ to 3 decimal places.

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Question 5

The table shows the marks obtained by a group of students.

Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79
Number of students	2	5	9	15	18	14	10	7

(a)
If a student is selected at random from the group, find the probability that this student scored at most 69 marks.
(b)
Calculate the median of the distribution.

Worked solution (try it first)

(a)

The total number of students is $2 + 5 + 9 + 15 + 18 + 14 + 10 + 7 = 80$ .
At most 69 marks means every class except 70–79: $80 - 7 = 73$ students.
Probability $= \dfrac{73}{80}$ .

(b)

The median is the $\frac{80}{2} = 40$ th mark.
Cumulative frequencies: $2, 7, 16, 31, 49, \ldots$ , so the median class is 40–49.
Median $= L_1 + \left(\dfrac{\frac{N}{2} - F}{f_m}\right)c$ with $L_1 = 39.5$ , $F = 31$ , $f_m = 18$ and $c = 10$ .
Median $= 39.5 + \dfrac{40 - 31}{18} \times 10$
$= 39.5 + 5$
$= 44.5$ .

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Question 6

Probability & distributions

The chances of three hunters hitting a target are $\frac13$ , $\frac23$ and $\frac15$ respectively. If they fire independently at the target, find the probability that

(a)
at least one of them hits the target;
(b)
only two of them hit the target.

Worked solution (try it first)

(a)

The probabilities of missing are $1 - \frac13 = \frac23$ , $1 - \frac23 = \frac13$ and $1 - \frac15 = \frac45$ .
The hunters fire independently, so $P(\text{none hits}) = \frac23 \times \frac13 \times \frac45$
$= \frac{8}{45}$ .
$P(\text{at least one}) = 1 - \frac{8}{45}$
$= \frac{37}{45}$ .

(b)

First two hit, third misses: $\frac13 \times \frac23 \times \frac45 = \frac{8}{45}$ .
First and third hit, second misses: $\frac13 \times \frac13 \times \frac15 = \frac{1}{45}$ .
Second and third hit, first misses: $\frac23 \times \frac23 \times \frac15 = \frac{4}{45}$ .
Add the three cases: $\frac{8 + 1 + 4}{45} = \frac{13}{45}$ .

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Question 7

Vectors

The position vectors of points $P$ and $R$ are $(4\mathbf i - 3\mathbf j)$ and $(8\mathbf i + \mathbf j)$ respectively. Find:

(a)
the unit vector in the direction of $(4\mathbf i - 3\mathbf j)$ ;
Separate values with commas, e.g. 3, −2
(b)
correct to one decimal place, the acute angle between the two vectors.

Worked solution (try it first)

(a)

The magnitude is $|4\mathbf i - 3\mathbf j| = \sqrt{4^2 + 3^2}$ , which is $\sqrt{25} = 5$ .
Divide the vector by its length: $\frac15(4\mathbf i - 3\mathbf j) = \frac45\mathbf i - \frac35\mathbf j$ .

(b)

Dot product: $(4\mathbf i - 3\mathbf j)\cdot(8\mathbf i + \mathbf j) = 32 - 3$
$= 29$ .
$|8\mathbf i + \mathbf j| = \sqrt{64 + 1}$
$= \sqrt{65}$ .
$\cos\theta = \dfrac{29}{5\sqrt{65}}$
$= 0.7194$ .
So $\theta = 44.0^\circ$ (1 d.p.).

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A body of mass 4 kg hangs from a fixed point $O$ by a light inextensible string. It is pulled aside by a horizontal force $F$ N and rests in equilibrium with the string inclined at an angle of $60^\circ$ to the downward vertical. [Take $g = 10\text{ m s}^{-2}$ ] Calculate, correct to two decimal places, the:

(a)
magnitude of $F$ ;
(b)
tension in the string.

Worked solution (try it first)

(a)

The weight is $mg = 4 \times 10 = 40$ N, acting downwards.
Three forces act on the body: the weight, the horizontal force $F$ and the tension $T$ along the string, at $60^\circ$ to the downward vertical.
Resolve horizontally: $T\sin 60^\circ = F$ .
Resolve vertically: $T\cos 60^\circ = 40$ .
Divide the first equation by the second: $\tan 60^\circ = \dfrac{F}{40}$ , so $F = 40\tan 60^\circ = 40\sqrt3 = 69.28$ N (2 d.p.).

(b)

From the vertical equation: $T = \dfrac{40}{\cos 60^\circ}$
$= \dfrac{40}{0.5}$
$= 80.00$ N.

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Question 9

(a)
Express $\dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)}$ in partial fractions.
(b)
If $y = \sqrt{x + 1}$ , determine the value of $x$ for which $\dfrac{dy}{dx} = y$ .

Worked solution (try it first)

(a)

The factor $x^2 + 1$ doesn't factorise, so write $\dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)} = \dfrac{P}{x - 1} + \dfrac{Qx + R}{x^2 + 1}$ .
Multiply through by the denominator: $4x^2 - x + 3 = P(x^2 + 1) + (Qx + R)(x - 1)$ .
Put $x = 1$ : $4 - 1 + 3 = 2P$ , so $P = 3$ .
Compare $x^2$ terms: $4 = P + Q$ , so $Q = 1$ .
Compare constants: $3 = P - R$ , so $R = 0$ .
So the partial fractions are $\dfrac{3}{x - 1} + \dfrac{x}{x^2 + 1}$ .

(b)

$y = (x + 1)^{\frac12}$ , so $\dfrac{dy}{dx} = \frac12(x + 1)^{-\frac12}$
$= \dfrac{1}{2\sqrt{x + 1}}$ .
Set $\dfrac{1}{2\sqrt{x + 1}} = \sqrt{x + 1}$ and multiply by $2\sqrt{x + 1}$ : $1 = 2(x + 1)$ .
So $x + 1 = \frac12$ and $x = -\frac12$ .

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Question 10

(a)(i)
An exponential sequence is given by $18, 2, \frac29, \ldots$ Find an expression for the $n$ th term;
(a)(ii)
the sum of the first $n$ terms.
(b)(i)
Find the equation of the tangent to the curve $y = -x^2 + x + 1$ at the point $(2, -1)$ .
(b)(ii)
Find the intercepts of the tangent in (b)(i) with the axes.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

The first term is $a = 18$ and the common ratio is $r = \frac{2}{18} = \frac19$ .
$T_n = ar^{n-1}$
$= 18\left(\frac19\right)^{n-1}$ , which is also $162\left(\frac19\right)^n$ .

(ii)

$r < 1$ , so use $S_n = \dfrac{a(1 - r^n)}{1 - r}$
$= \dfrac{18\left(1 - (\frac19)^n\right)}{\frac89}$ .
$\dfrac{18}{\frac89} = \dfrac{18 \times 9}{8}$
$= \dfrac{81}{4}$ , so $S_n = \frac{81}{4}\left(1 - \left(\frac19\right)^n\right)$ .

(b)(i)

Differentiate: $\dfrac{dy}{dx} = -2x + 1$ .
At $x = 2$ the gradient is $-4 + 1 = -3$ .
Tangent through $(2, -1)$ : $y + 1 = -3(x - 2)$ , which gives $y + 3x - 5 = 0$ .

(ii)

On the $x$ -axis, $y = 0$ : $3x = 5$ , so $x = \frac53$ .
The intercept is $\left(\frac53, 0\right)$ .
On the $y$ -axis, $x = 0$ : $y = 5$ .
The intercept is $(0, 5)$ .

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Question 11 ✱

(a)(i)
The polynomial $f(x) = 2x^3 + px^2 + qx + r$ is divisible by $(2x^2 - 7x + 3)$ . It has a remainder of $-36$ when it is divided by $(x + 1)$ . Find the values of the constants $p$ , $q$ and $r$ ;
Separate values with commas, e.g. 3, −2
(a)(ii)
the zeros of $f(x)$ .
Separate values with commas, e.g. 3, −2
(b)
Find the truth set of $x^2 - 3x + 2 < 0$ .
Show the answer
$\{x : 1 < x < 2\}$

Worked solution (try it first)

(a)(i)

$2x^2 - 7x + 3 = (2x - 1)(x - 3)$ , so $f(\frac12) = 0$ and $f(3) = 0$ .
$f(\frac12) = 0$ : $\frac14 + \frac{p}{4} + \frac{q}{2} + r = 0$ , so $p + 2q + 4r = -1$ .
$f(3) = 0$ : $54 + 9p + 3q + r = 0$ , so $9p + 3q + r = -54$ .
Remainder theorem, $f(-1) = -36$ : $-2 + p - q + r = -36$ , so $p - q + r = -34$ .
Solve the three equations: $p = -11$ , $q = 17$ and $r = -6$ .
Check: $f(-1) = -2 - 11 - 17 - 6 = -36$ ✓.

(ii)

$f(x) = 2x^3 - 11x^2 + 17x - 6$
$= (2x^2 - 7x + 3)(x - 2)$
$= (2x - 1)(x - 3)(x - 2)$ .
So the zeros are $x = \frac12$ , $2$ and $3$ .

(b)

Factorise: $(x - 1)(x - 2) < 0$ .
The product is negative only when the brackets have opposite signs, which is between the roots.
The truth set is $\{x : 1 < x < 2\}$ .

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Question 12 ✱✱

(a)(i)
Two linear transformations are defined by $P : (x, y) \to (-2x + 3y,\ 5x - 4y)$ and $T : (x, y) \to (-3x - 4y,\ 6x - 5y)$ . Find the inverse of $T$ ;
Separate values with commas, e.g. 3, −2
(a)(ii)
the image of $(1, 3)$ under the transformation $P$ .
Separate values with commas, e.g. 3, −2
(b)
Find the volume, in terms of $\pi$ , of the solid formed when the area enclosed by the lines $y = 1$ , $2y = x + 1$ and $x = 0$ is rotated through two right angles about the $y$ -axis.

Worked solution (try it first)

(a)(i)

The matrix of $T$ is $\begin{pmatrix} -3 & -4 \\ 6 & -5 \end{pmatrix}$ , with determinant $(-3)(-5) - (-4)(6) = 15 + 24 = 39$ .
Swap the leading diagonal, change the signs of the others and divide by 39: $T^{-1} = \frac{1}{39}\begin{pmatrix} -5 & 4 \\ -6 & -3 \end{pmatrix}$ .
So $T^{-1} : (x, y) \to \left(\dfrac{-5x + 4y}{39},\ \dfrac{-6x - 3y}{39}\right) = \left(\dfrac{-5x + 4y}{39},\ \dfrac{-2x - y}{13}\right)$ .

(ii)

The matrix of $P$ is $\begin{pmatrix} -2 & 3 \\ 5 & -4 \end{pmatrix}$ .
Multiply by $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ : $\begin{pmatrix} -2 + 9 \\ 5 - 12 \end{pmatrix} = \begin{pmatrix} 7 \\ -7 \end{pmatrix}$ .
The image is $(7, -7)$ .

(b)

The region is the triangle with corners $(0, \frac12)$ , $(0, 1)$ and $(1, 1)$ .
Along it, $x = 2y - 1$ for $\frac12 \le y \le 1$ .
Two right angles is half a turn, so the volume is half of $\pi\int x^2\,dy$ : $V = \dfrac{\pi}{2}\displaystyle\int_{1/2}^{1} (2y - 1)^2\,dy$ .
Expand and integrate: $\dfrac{\pi}{2}\left[\frac43y^3 - 2y^2 + y\right]_{1/2}^{1} = \dfrac{\pi}{2}\left(\frac13 - \frac16\right)$ .
So $V = \dfrac{\pi}{2} \times \dfrac16$
$= \dfrac{\pi}{12}$ cubic units.

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Question 13

(a)
If ${}^nP_3 = 10\left({}^{n-2}P_2\right)$ , find $n$ .
Separate values with commas, e.g. 3, −2
(b)(i)
Four students are to be selected from 4 boys and 6 girls to represent a school in a competition. If there is no restriction, in how many ways can they be selected?
(b)(ii)
Calculate the probability of having an equal number of boys and girls.

Worked solution (try it first)

(a)

${}^nP_3 = \dfrac{n!}{(n - 3)!}$
$= n(n - 1)(n - 2)$ and ${}^{n-2}P_2 = \dfrac{(n - 2)!}{(n - 4)!}$
$= (n - 2)(n - 3)$ .
So $n(n - 1)(n - 2) = 10(n - 2)(n - 3)$ .
Divide by $n - 2$ (it isn't zero, since $n \ge 4$ ): $n(n - 1) = 10(n - 3)$ .
Expand: $n^2 - n = 10n - 30$ , so $n^2 - 11n + 30 = 0$ .
Factorise: $(n - 5)(n - 6) = 0$ , so $n = 5$ or $n = 6$ .
Both work: ${}^5P_3 = 60 = 10 \times {}^3P_2$ and ${}^6P_3 = 120 = 10 \times {}^4P_2$ .

(b)(i)

Order doesn't matter, so choose 4 of the 10 students: ${}^{10}C_4 = 210$ ways.

(ii)

Equal numbers means 2 boys and 2 girls: ${}^4C_2 \times {}^6C_2 = 6 \times 15 = 90$ ways.
Probability $= \dfrac{90}{210} = \dfrac37$ .

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Question 14

Statistics & correlation

The age distribution of final year students in a polytechnic is shown in the table below.

Age (years)	20–24	25–29	30–34	35–39	40–44
Number of students	40	45	36	14	5

Calculate, correct to one decimal place, the:

(a)
mean age;
(b)
standard deviation of the ages.

Worked solution (try it first)

(a)

Class marks: $22, 27, 32, 37, 42$ .
Multiply by the frequencies: $fx = 880, 1215, 1152, 518, 210$ .
$\sum f = 140$ and $\sum fx = 3975$ .
Mean $= \dfrac{\sum fx}{\sum f}$
$= \dfrac{3975}{140}$
$= 28.39$ , which is $28.4$ years (1 d.p.).

(b)

$fx^2 = 19\,360, 32\,805, 36\,864, 19\,166, 8820$ , so $\sum fx^2 = 117\,015$ .
Variance $= \dfrac{\sum fx^2}{\sum f} - \bar x^2$
$= \dfrac{117\,015}{140} - \left(\dfrac{3975}{140}\right)^2$
$= 835.82 - 806.15$
$= 29.67$ .
Standard deviation $= \sqrt{29.67} = 5.4$ years (1 d.p.).

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Question 15

In a test, eight students obtained the following marks in Biology and Physics.

Student	A	B	C	D	E	F	G	H
Biology	76	52	63	48	84	36	28	70
Physics	75	78	28	45	56	71	54	58

(a)
(i) Calculate, correct to two decimal places, Spearman's rank correlation coefficient of the distribution. (ii) Comment on your result in (a)(i).
(b)(i)
Three-digit numbers are to be formed from 1, 2, 3, 4 and 5. If repetition is not allowed, how many numbers can be formed?
(b)(ii)
What is the probability of selecting an odd number from the numbers formed in (b)(i)?

Worked solution (try it first)

(a)(i)

Rank the Biology marks (highest = 1): A 2, B 5, C 4, D 6, E 1, F 7, G 8, H 3.
Rank the Physics marks: A 2, B 1, C 8, D 7, E 5, F 3, G 6, H 4.
Differences $d$ : $0, 4, -4, -1, -4, 4, 2, -1$ , so $\sum d^2 = 0 + 16 + 16 + 1 + 16 + 16 + 4 + 1$
$= 70$ .
$\rho = 1 - \dfrac{6\sum d^2}{n(n^2 - 1)}$
$= 1 - \dfrac{6 \times 70}{8 \times 63}$
$= 1 - \dfrac{420}{504}$
$= 0.17$ (2 d.p.).

(ii)

$\rho$ is positive but close to 0, so there is only a weak positive correlation between the marks in the two subjects.

(b)(i)

There are 5 choices for the first digit, 4 for the second and 3 for the third: $5 \times 4 \times 3 = 60$ numbers.

(ii)

An odd number ends in 1, 3 or 5: 3 choices for the last digit, then 4 and 3 for the others, giving $3 \times 4 \times 3 = 36$ .
Probability $= \dfrac{36}{60} = \dfrac35$ .

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Question 16 ✱

Vectors

(a)
Find the unit vector along the resultant of the vectors $(3\mathbf i - 2\mathbf j)$ and $(\mathbf i + 5\mathbf j)$ .
Separate values with commas, e.g. 3, −2
(b)
The position vectors of points $P$ , $Q$ , $R$ and $S$ are $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ , $\begin{pmatrix} 10 \\ 4 \end{pmatrix}$ , $\begin{pmatrix} 3 \\ 12 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ respectively. (i) Show that $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{RS}$ . (ii) Calculate, correct to one decimal place, angle $PSQ$ .

Worked solution (try it first)

(a)

Resultant: $(3\mathbf i - 2\mathbf j) + (\mathbf i + 5\mathbf j) = 4\mathbf i + 3\mathbf j$ .
Its magnitude is $\sqrt{4^2 + 3^2} = 5$ , so the unit vector is $\frac15(4\mathbf i + 3\mathbf j) = \frac45\mathbf i + \frac35\mathbf j$ .

(b)(i)

$\overrightarrow{PQ} = \begin{pmatrix} 10 \\ 4 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}$
$= \begin{pmatrix} 12 \\ 1 \end{pmatrix}$ and $\overrightarrow{RS} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} - \begin{pmatrix} 3 \\ 12 \end{pmatrix}$
$= \begin{pmatrix} 1 \\ -12 \end{pmatrix}$ .
$\overrightarrow{PQ}\cdot\overrightarrow{RS} = 12 \times 1 + 1 \times (-12)$
$= 0$ , so $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{RS}$ .

(ii)

$\overrightarrow{SP} = \begin{pmatrix} -2 - 4 \\ 3 - 0 \end{pmatrix}$
$= \begin{pmatrix} -6 \\ 3 \end{pmatrix}$ and $\overrightarrow{SQ} = \begin{pmatrix} 10 - 4 \\ 4 - 0 \end{pmatrix}$
$= \begin{pmatrix} 6 \\ 4 \end{pmatrix}$ .
$\overrightarrow{SP}\cdot\overrightarrow{SQ} = -36 + 12$
$= -24$ .
$|\overrightarrow{SP}| = \sqrt{45}$
$= 3\sqrt5$ and $|\overrightarrow{SQ}| = \sqrt{52}$
$= 2\sqrt{13}$ .
$\cos\angle PSQ = \dfrac{-24}{3\sqrt5 \times 2\sqrt{13}}$
$= \dfrac{-4}{\sqrt{65}}$
$= -0.4961$ .
The cosine is negative, so the angle is obtuse: $\angle PSQ = 119.7^\circ$ (1 d.p.).

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Question 17

A body of mass 3 kg resting on a smooth surface is acted upon by forces $P(10\text{ N}, 045^\circ)$ , $Q(12\text{ N}, 180^\circ)$ , $R(15\text{ N}, 215^\circ)$ and $S(8\text{ N}, 090^\circ)$ . Calculate, correct to one decimal place, the:

(a)
resultant force acting on the body;
(b)
acceleration with which the body begins to move;
(c)
time it takes to cover the first 4 metres.

Worked solution (try it first)

(a)

Resolve each force into east ( $x$ ) and north ( $y$ ) components, using the bearing $\theta$ : $x = F\sin\theta$ , $y = F\cos\theta$ .
East: $10\sin 45^\circ + 12\sin 180^\circ + 15\sin 215^\circ + 8\sin 90^\circ = 7.071 + 0 - 8.604 + 8$
$= 6.467$ .
North: $10\cos 45^\circ + 12\cos 180^\circ + 15\cos 215^\circ + 8\cos 90^\circ = 7.071 - 12 - 12.287 + 0$
$= -17.216$ .
Magnitude: $|\mathbf R| = \sqrt{6.467^2 + 17.216^2}$
$= \sqrt{338.21}$
$= 18.4$ N (1 d.p.).

(b)

Newton's second law, $F = ma$ : $a = \dfrac{18.39}{3} = 6.1$ m s $^{-2}$ (1 d.p.).

(c)

The body starts from rest, so $s = ut + \frac12at^2$ with $u = 0$ : $4 = \frac12(6.13)t^2$ .
$t^2 = \dfrac{8}{6.13} = 1.305$ , so $t = 1.1$ s (1 d.p.).

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Question 18 ✱

(a)
A body of mass 8 kg is placed on a smooth plane inclined at an angle of $40^\circ$ to the horizontal. Find the magnitude of the force: (i) acting perpendicular to the plane; (ii) acting along the plane, required to keep the body in equilibrium. [Take $g = 10\text{ m s}^{-2}$ ]
Separate values with commas, e.g. 3, −2
(b)(i)
A particle of mass 800 g is moving in a straight line with a velocity of $(5\mathbf i + 3\mathbf j)$ m s $^{-1}$ . It is acted upon by a force that changes the velocity to $(5\mathbf i + 12\mathbf j)$ m s $^{-1}$ . Find the impulse.
(b)(ii)
If the force acted for 2 s, find the acceleration of the particle.

Worked solution (try it first)

(a)

The weight is $mg = 8 \times 10 = 80$ N, acting vertically down.
Resolve it along and perpendicular to the plane.

(i)

Perpendicular to the plane: $80\cos 40^\circ = 61.28$ N.
This is balanced by the normal reaction.

(ii)

Down the plane: $80\sin 40^\circ = 51.42$ N.
The plane is smooth, so a force of $51.42$ N up the plane is needed for equilibrium.

(b)(i)

Impulse = change in momentum $= m(\mathbf v - \mathbf u)$ , with $m = 0.8$ kg.
$\mathbf v - \mathbf u = (5\mathbf i + 12\mathbf j) - (5\mathbf i + 3\mathbf j)$
$= 9\mathbf j$ , so the impulse is $0.8 \times 9\mathbf j = 7.2\mathbf j$ N s.

(ii)

Impulse $= Ft$ , so $F = \dfrac{7.2}{2} = 3.6$ N (in the $\mathbf j$ direction).
$a = \dfrac{F}{m} = \dfrac{3.6}{0.8} = 4.5$ m s $^{-2}$ , in the direction of $\mathbf j$ .

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