Question 9 - A-Level Maths

Pre-U Pre-U 9795/2 2011 June — Question 9 9 marks

Exam Board	Pre-U
Module	Pre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year	2011
Session	June
Marks	9
Topic	Vectors Introduction & 2D
Type	Closest approach of two objects
Difficulty	Standard +0.3 This is a standard closest approach problem requiring conversion of bearings to vectors, relative velocity/position vectors, and using the scalar product condition for minimum distance. All steps are routine applications of well-established techniques with no novel insight required. The multi-part structure guides students through the solution methodically, making it slightly easier than average.
Spec	1.10b Vectors in 3D: i,j,k notation 1.10h Vectors in kinematics: uniform acceleration in vector form 6.02i Conservation of energy: mechanical energy principle

9 At noon a vessel, $A$, leaves a port, $O$, and travels at $10 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ on a bearing of $042 ^ { \circ }$. Also at noon a second vessel, $B$, leaves another port, $P , 13 \mathrm {~km}$ due north of $O$, and travels at $15 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ on a bearing of $090 ^ { \circ }$. Take $O$ as the origin and $\mathbf { i }$ and $\mathbf { j }$ as unit vectors east and north respectively.

Express the velocity vector of $A$ relative to $B$ in the form $a \mathbf { i } + b \mathbf { j }$, where $a$ and $b$ are constants to be determined.
Express the position vector of $A$ relative to $B$, at time $t$ hours after the vessels have left port, in terms of $t , \mathbf { i }$ and $\mathbf { j }$.
Explain why the scalar product of the vectors in parts (i) and (ii) is zero when the two vessels are closest together.
Find the time at which the two vessels are closest together. $10 A$ and $B$ are two points 6 m apart on a smooth horizontal surface. A particle, $P$, of mass 0.5 kg is attached to $A$ by a light elastic string of natural length 2 m and modulus of elasticity 20 N , and to $B$ by a light elastic string of natural length 1 m and modulus of elasticity 10 N , such that $P$ is between $A$ and $B$.
Find the length $A P$ when $P$ is in equilibrium. $P$ is held at the point $C$, where $C$ is between $A$ and $B$ and $A C = 4.5 \mathrm {~m} . P$ is then released from rest. At time $t$ seconds after being released, the displacement of $P$ from the equilibrium position is $y$ metres.
Show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } = - 40 y$$
Find the time taken for $P$ to reach the mid-point of $A B$ for the first time. \includegraphics[max width=\textwidth, alt={}, center]{963c0834-fe49-480b-9bb5-1ace4254641a-6_750_1187_258_479} Two particles, $P$ and $Q$, are projected simultaneously from the same point on a plane inclined at $\alpha$ to the horizontal. $P$ is projected up the plane and $Q$ down the plane. Each particle is projected with speed $V$ at an angle $\theta$ to the plane. Both particles move in a vertical plane containing a line of greatest slope of the inclined plane and you are given that $\alpha + \theta < \frac { 1 } { 2 } \pi$ (see diagram).
Show that the range of $P$, up the plane, is given by $$\frac { 2 V ^ { 2 } \sin \theta } { g \cos ^ { 2 } \alpha } ( \cos \theta \cos \alpha - \sin \theta \sin \alpha ) .$$
Write down a similar expression for the range of $Q$, down the plane.
Given that the range up the plane is a quarter of the range down the plane and that $\alpha = \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$, find $\theta$.

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(i) $\mathbf{v}_A = (10\sin 42°)\mathbf{i} + (10\cos 42°)\mathbf{j}$ and $\mathbf{v}_B = 15\mathbf{i}$

$_A\mathbf{v}_B = (10\sin 42° - 15)\mathbf{i} + (10\cos 42°)\mathbf{j}$ (Accept $-8.31\mathbf{i} + 7.43\mathbf{j}$) M1A1 [2]

(ii) $\mathbf{r}_A = (10\sin 42°)t\mathbf{i} + (10\cos 42°)t\mathbf{j}$ and $\mathbf{r}_B = 15t\mathbf{i} + 13\mathbf{j}$

$_A\mathbf{r}_B = (10\sin 42° - 15)t\mathbf{i} + (10\cos 42°\ t - 13)\mathbf{j}$

(Accept $-8.31t\mathbf{i} + [7.43t - 13]\mathbf{j}$) M1A1 [2]

(iii) By considering relative motion, one object is taken to be at rest and the other moving relative to it. Hence, when closest together, the relative position vector is perpendicular to the relative velocity vector; (i.e. dot product = 0). B1 [1]

(iv) $_A\mathbf{r}_{B} \cdot {_A}\mathbf{v}_B = (10\sin 42° - 15)^2 t + 10\cos 42°(10\cos 42°\ t - 13) = 0$ M1A1

$\Rightarrow (100\sin^2 42° - 300\sin 42° + 225)t + 100\cos^2 42°\ t - 130\cos 42° = 0$

$\Rightarrow t = \dfrac{130\cos 42°}{(325 - 300\sin 42°)} = 0.777$ or 47 minutes (i.e. 1247) M1A1 [4]

(i) $\mathbf{v}_A = (10\sin 42°)\mathbf{i} + (10\cos 42°)\mathbf{j}$ and $\mathbf{v}_B = 15\mathbf{i}$

$_A\mathbf{v}_B = (10\sin 42° - 15)\mathbf{i} + (10\cos 42°)\mathbf{j}$ (Accept $-8.31\mathbf{i} + 7.43\mathbf{j}$) **M1A1** [2]

(ii) $\mathbf{r}_A = (10\sin 42°)t\mathbf{i} + (10\cos 42°)t\mathbf{j}$ and $\mathbf{r}_B = 15t\mathbf{i} + 13\mathbf{j}$

$_A\mathbf{r}_B = (10\sin 42° - 15)t\mathbf{i} + (10\cos 42°\ t - 13)\mathbf{j}$
(Accept $-8.31t\mathbf{i} + [7.43t - 13]\mathbf{j}$) **M1A1** [2]

(iii) By considering relative motion, one object is taken to be at rest and the other moving relative to it. Hence, when closest together, the relative position vector is perpendicular to the relative velocity vector; (i.e. dot product = 0). **B1** [1]

(iv) $_A\mathbf{r}_{B} \cdot {_A}\mathbf{v}_B = (10\sin 42° - 15)^2 t + 10\cos 42°(10\cos 42°\ t - 13) = 0$ **M1A1**

$\Rightarrow (100\sin^2 42° - 300\sin 42° + 225)t + 100\cos^2 42°\ t - 130\cos 42° = 0$

$\Rightarrow t = \dfrac{130\cos 42°}{(325 - 300\sin 42°)} = 0.777$ or 47 minutes (i.e. 1247) **M1A1** [4]

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9 At noon a vessel, $A$, leaves a port, $O$, and travels at $10 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ on a bearing of $042 ^ { \circ }$. Also at noon a second vessel, $B$, leaves another port, $P , 13 \mathrm {~km}$ due north of $O$, and travels at $15 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ on a bearing of $090 ^ { \circ }$. Take $O$ as the origin and $\mathbf { i }$ and $\mathbf { j }$ as unit vectors east and north respectively.\\
(i) Express the velocity vector of $A$ relative to $B$ in the form $a \mathbf { i } + b \mathbf { j }$, where $a$ and $b$ are constants to be determined.\\
(ii) Express the position vector of $A$ relative to $B$, at time $t$ hours after the vessels have left port, in terms of $t , \mathbf { i }$ and $\mathbf { j }$.\\
(iii) Explain why the scalar product of the vectors in parts (i) and (ii) is zero when the two vessels are closest together.\\
(iv) Find the time at which the two vessels are closest together.\\
$10 A$ and $B$ are two points 6 m apart on a smooth horizontal surface. A particle, $P$, of mass 0.5 kg is attached to $A$ by a light elastic string of natural length 2 m and modulus of elasticity 20 N , and to $B$ by a light elastic string of natural length 1 m and modulus of elasticity 10 N , such that $P$ is between $A$ and $B$.\\
(i) Find the length $A P$ when $P$ is in equilibrium.\\
$P$ is held at the point $C$, where $C$ is between $A$ and $B$ and $A C = 4.5 \mathrm {~m} . P$ is then released from rest. At time $t$ seconds after being released, the displacement of $P$ from the equilibrium position is $y$ metres.\\
(ii) Show that

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } = - 40 y$$

(iii) Find the time taken for $P$ to reach the mid-point of $A B$ for the first time.\\
\includegraphics[max width=\textwidth, alt={}, center]{963c0834-fe49-480b-9bb5-1ace4254641a-6_750_1187_258_479}

Two particles, $P$ and $Q$, are projected simultaneously from the same point on a plane inclined at $\alpha$ to the horizontal. $P$ is projected up the plane and $Q$ down the plane. Each particle is projected with speed $V$ at an angle $\theta$ to the plane. Both particles move in a vertical plane containing a line of greatest slope of the inclined plane and you are given that $\alpha + \theta < \frac { 1 } { 2 } \pi$ (see diagram).\\
(i) Show that the range of $P$, up the plane, is given by

$$\frac { 2 V ^ { 2 } \sin \theta } { g \cos ^ { 2 } \alpha } ( \cos \theta \cos \alpha - \sin \theta \sin \alpha ) .$$

(ii) Write down a similar expression for the range of $Q$, down the plane.\\
(iii) Given that the range up the plane is a quarter of the range down the plane and that $\alpha = \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$, find $\theta$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2011 Q9 [9]}}

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Q1 3 Q2 8 Q3 10 Q7 3 Q9 9 Q12 9

Pre-U Pre-U 9795/2 2011 June — Question 9 9 marks

(i) \(\mathbf{v}_A = (10\sin 42°)\mathbf{i} + (10\cos 42°)\mathbf{j}\) and \(\mathbf{v}_B = 15\mathbf{i}\)

\(_A\mathbf{v}_B = (10\sin 42° - 15)\mathbf{i} + (10\cos 42°)\mathbf{j}\) (Accept \(-8.31\mathbf{i} + 7.43\mathbf{j}\)) M1A1 [2]

(ii) \(\mathbf{r}_A = (10\sin 42°)t\mathbf{i} + (10\cos 42°)t\mathbf{j}\) and \(\mathbf{r}_B = 15t\mathbf{i} + 13\mathbf{j}\)

\(_A\mathbf{r}_B = (10\sin 42° - 15)t\mathbf{i} + (10\cos 42°\ t - 13)\mathbf{j}\)

(Accept \(-8.31t\mathbf{i} + [7.43t - 13]\mathbf{j}\)) M1A1 [2]

(iii) By considering relative motion, one object is taken to be at rest and the other moving relative to it. Hence, when closest together, the relative position vector is perpendicular to the relative velocity vector; (i.e. dot product = 0). B1 [1]

(iv) \(_A\mathbf{r}_{B} \cdot {_A}\mathbf{v}_B = (10\sin 42° - 15)^2 t + 10\cos 42°(10\cos 42°\ t - 13) = 0\) M1A1

\(\Rightarrow (100\sin^2 42° - 300\sin 42° + 225)t + 100\cos^2 42°\ t - 130\cos 42° = 0\)

\(\Rightarrow t = \dfrac{130\cos 42°}{(325 - 300\sin 42°)} = 0.777\) or 47 minutes (i.e. 1247) M1A1 [4]

This paper (6 questions)