1.05q Trig in context: vectors, kinematics, forces

9

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0f58de6c-aba7-4a79-a962-c23be3ee0aa9-4_433_476_264_872} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} In Fig. 1, \(O A B\) is a sector of a circle with centre \(O\) and radius \(r\). \(A X\) is the tangent at \(A\) to the arc \(A B\) and angle \(B A X = \alpha\).
   1. Show that angle \(A O B = 2 \alpha\).
   2. Find the area of the shaded segment in terms of \(r\) and \(\alpha\).
2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0f58de6c-aba7-4a79-a962-c23be3ee0aa9-4_451_503_1162_861} \captionsetup{labelformat=empty} \caption{Fig. 2}
   \end{figure} In Fig. 2, \(A B C\) is an equilateral triangle of side 4 cm . The lines \(A X , B X\) and \(C X\) are tangents to the equal circular \(\operatorname { arcs } A B , B C\) and \(C A\). Use the results in part (a) to find the area of the shaded region, giving your answer in terms of \(\pi\) and \(\sqrt { } 3\).
   [0pt] [6]

3 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at \(\alpha\) to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all \(\beta\). \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} In the following, all lengths are in metres.

1. Find AC in terms of \(\alpha\), and hence show that \(\mathrm { GF } = 10 \sec \alpha \tan \beta\).
2. Show that \(\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )\). $$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$ Similarly, it can be shown that \(\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }\). [You are not required to derive this result.]
   You are now given that \(\alpha = 45 ^ { \circ }\) and that \(\tan \beta = t\).
3. Find CE and CD in terms of \(t\). Hence show that \(\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }\).
4. Show that \(\mathrm { GF } = 10 \sqrt { 2 } t\). For a certain value of \(\beta , \mathrm { DE } = 2 \mathrm { GF }\).
5. Show that \(t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }\). Hence find this value of \(\beta\).

2 In a game of rugby, a kick is to be taken from a point P (see Fig. 7). P is a perpendicular distance \(y\) metres from the line TOA. Other distances and angles are as shown. \begin{figure}[h] \end{figure}

1. Show that \(\theta = \beta - \alpha\), and hence that \(\tan \theta = \frac { 6 y } { 160 + y ^ { 2 } }\). Calculate the angle \(\theta\) when \(y = 6\).
2. By differentiating implicitly, show that \(\frac { \mathrm { d } \theta } { \mathrm { d } y } = \frac { 6 \left( 160 - y ^ { 2 } \right) } { \left( 160 + y ^ { 2 } \right) ^ { 2 } } \cos ^ { 2 } \theta\).
3. Use this result to find the value of \(y\) that maximises the angle \(\theta\). Calculate this maximum value of \(\theta\). [You need not verify that this value is indeed a maximum.]

7 A simple pendulum consists of a light inextensible string of length 0.8 m and a particle \(P\) of mass \(m \mathrm {~kg}\). The pendulum is hanging vertically at rest from a fixed point \(O\) when \(P\) is given a horizontal velocity of \(0.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that, in the subsequent motion, the maximum angle between the string and the downward vertical is 0.107 radians, correct to 3 significant figures.
2. Show that the motion may be modelled as simple harmonic motion, and find the period of this motion.
3. Find the time after the start of the motion when the velocity of the particle is first \(- 0.2 \mathrm {~ms} ^ { - 1 }\) and find the angular displacement of \(O P\) from the downward vertical at this time.

4 A particle is connected to a fixed point by a light inextensible string of length 2.45 m to make a simple pendulum. The particle is released from rest with the string taut and inclined at 0.1 radians to the downward vertical.

1. Show that the motion of the particle is approximately simple harmonic with period 3.14 s , correct to 3 significant figures. Calculate, in either order,
2. the angular speed of the pendulum when it has moved 0.04 radians from the initial position,
3. the time taken by the pendulum to move 0.04 radians from the initial position.

14 A particle \(P\) is projected from the point \(O\), at the top of a vertical wall of height \(H\) above a horizontal plane, with initial speed \(V\) at an angle \(\alpha\) above the horizontal. At time \(t\) the coordinates of the particle are \(( x , y )\) referred to horizontal and vertical axes at \(O\).

1. Express \(x\) and \(y\) as functions of \(t\). Let \(\theta\) be the angle \(O P\) makes with the horizontal at time \(t\).
2. (a) Show that $$\tan \theta = \tan \alpha - \frac { g } { 2 V \cos \alpha } t$$ (b) Show that when the particle attains its greatest height above the point of projection, where \(O P\) makes an angle \(\beta\) with the horizontal, $$\tan \beta = \frac { 1 } { 2 } \tan \alpha .$$ (c) If the particle strikes the ground where \(O P\) makes an angle \(\beta\) below the horizontal, show that $$H = \frac { 3 V ^ { 2 } \sin ^ { 2 } \alpha } { 2 g }$$

A new design for a company logo is to be made from two sectors of a circle, \(ORP\) and \(OQS\), and a rhombus \(OSTR\), as shown in the diagram below. \includegraphics{figure_7} The points \(P\), \(O\) and \(Q\) lie on a straight line and the angle \(ROS\) is \(\theta\) radians. A large copy of the logo, with \(PQ = 5\) metres, is to be put on a wall.

1. Show that the area of the logo, \(A\) square metres, is given by $$A = \frac{25}{8}(\pi - \theta + 2\sin\theta)$$ [4 marks]
2. 1. Show that the maximum value of \(A\) occurs when \(\theta = \frac{\pi}{3}\) Fully justify your answer. [6 marks]
   2. Find the exact maximum value of \(A\) [2 marks]
3. Without further calculation, state how your answers to parts (b)(i) and (b)(ii) would change if \(PQ\) were increased to 10 metres. [2 marks]

A particle is projected with velocity \(V\) at an angle \(\alpha\) to the horizontal up a plane inclined at \(\beta\) to the horizontal, where \(\alpha > \beta\).

1. Show that the time of flight is \(\frac{2V \sin(\alpha - \beta)}{g \cos \beta}\). [3]
2. Show that the range on the inclined plane is \(\frac{2V^2 \sin(\alpha - \beta) \cos \alpha}{g \cos^2 \beta}\). [4]
3. If the particle strikes the plane at right angles, prove that \(\tan \alpha = \cot \beta + 2 \tan \beta\). [5]