Question 7 - A-Level Maths

Edexcel M1 — Question 7

Exam Board	Edexcel
Module	M1 (Mechanics 1)
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors 3D & Lines
Type	Kinematics with position vectors
Difficulty	Moderate -0.3 This is a standard M1 kinematics question using position vectors. Parts (a)-(c) involve routine calculations: finding velocity from displacement, writing position vector equations, and algebraic manipulation to show a given result. Part (d) requires solving a quadratic equation. While multi-step, all techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec	1.10a Vectors in 2D: i,j notation and column vectors 1.10b Vectors in 3D: i,j,k notation 1.10d Vector operations: addition and scalar multiplication 1.10e Position vectors: and displacement 1.10f Distance between points: using position vectors 1.10h Vectors in kinematics: uniform acceleration in vector form

7. Two ships $P$ and $Q$ are travelling at night with constant velocities. At midnight, $P$ is at the point with position vector $( 20 \mathbf { i } + 10 \mathbf { j } ) \mathrm { km }$ relative to a fixed origin $O$. At the same time, $Q$ is at the point with position vector $( 14 \mathbf { i } - 6 \mathbf { j } ) \mathrm { km }$. Three hours later, $P$ is at the point with position vector $( 29 \mathbf { i } + 34 \mathbf { j } ) \mathrm { km }$. The ship $Q$ travels with velocity $12 \mathbf { j } \mathrm {~km} \mathrm {~h} ^ { - 1 }$. At time $t$ hours after midnight, the position vectors of $P$ and $Q$ are $\mathbf { p } \mathrm { km }$ and $\mathbf { q } \mathrm { km }$ respectively. Find

the velocity of $P$, in terms of $\mathbf { i }$ and $\mathbf { j }$,
expressions for $\mathbf { p }$ and $\mathbf { q }$, in terms of $t$, i and $\mathbf { j }$. At time $t$ hours after midnight, the distance between $P$ and $Q$ is $d \mathrm {~km}$.
By finding an expression for $\overrightarrow { P Q }$, show that $$d ^ { 2 } = 25 t ^ { 2 } - 92 t + 292$$ Weather conditions are such that an observer on $P$ can only see the lights on $Q$ when the distance between $P$ and $Q$ is 15 km or less. Given that when $t = 1$, the lights on $Q$ move into sight of the observer,
find the time, to the nearest minute, at which the lights on $Q$ move out of sight of the observer.
1. In taking off, an aircraft moves on a straight runway $A B$ of length 1.2 km . The aircraft moves from $A$ with initial speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It moves with constant acceleration and 20 s later it leaves the runway at $C$ with speed $74 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find
2. the acceleration of the aircraft,
3. the distance $B C$.
4. Two small steel balls $A$ and $B$ have mass 0.6 kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when they collide directly. Immediately before the collision, the speed of $A$ is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the speed of $B$ is $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Immediately after the collision, the direction of motion of $A$ is unchanged and the speed of $B$ is twice the speed of $A$. Find
5. the speed of $A$ immediately after the collision,
6. the magnitude of the impulse exerted on $B$ in the collision.
\begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{94d9432d-1723-4549-ad5e-d4be0f5fd083-018_282_707_278_699}
\end{figure}

Show mark scheme Show mark scheme source

Question 7:

Part (a)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$t=0$ gives $\mathbf{v} = \mathbf{i} - 3\mathbf{j}$	B1
$\text{speed} = \sqrt{1^2 + (-3)^2}$	M1	$\sqrt{\text{sum of squares of components}}$
$= \sqrt{10} = 3.2$ or better	A1

Part (b)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$t=2$ gives $\mathbf{v} = (-3\mathbf{i} + 3\mathbf{j})$	M1	Clear attempt to substitute $t=2$
Bearing is $315°$	A1

Part (c)(i)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$1 - 2t = 0 \Rightarrow t = 0.5$	M1 A1	N.B. If two solutions offered by equating both $\mathbf{i}$ and $\mathbf{j}$ components to zero, give M0

Part (c)(ii)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$-(3t-3) = -3(1-2t)$, i.e. $\frac{1-2t}{3t-3} = \pm\frac{-1}{-3}$	M1 A1	Must be equation in $t$ only; correct equation (with $+$ sign)
Solving for $t$	DM1	Dependent on first M1
$t = \frac{2}{3},\ 0.67$ or better	A1

## Question 7:

### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $t=0$ gives $\mathbf{v} = \mathbf{i} - 3\mathbf{j}$ | B1 | |
| $\text{speed} = \sqrt{1^2 + (-3)^2}$ | M1 | $\sqrt{\text{sum of squares of components}}$ |
| $= \sqrt{10} = 3.2$ or better | A1 | |

### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $t=2$ gives $\mathbf{v} = (-3\mathbf{i} + 3\mathbf{j})$ | M1 | Clear attempt to substitute $t=2$ |
| Bearing is $315°$ | A1 | |

### Part (c)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1 - 2t = 0 \Rightarrow t = 0.5$ | M1 A1 | N.B. If two solutions offered by equating both $\mathbf{i}$ and $\mathbf{j}$ components to zero, give M0 |

### Part (c)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-(3t-3) = -3(1-2t)$, i.e. $\frac{1-2t}{3t-3} = \pm\frac{-1}{-3}$ | M1 A1 | Must be equation in $t$ only; correct equation (with $+$ sign) |
| Solving for $t$ | DM1 | Dependent on first M1 |
| $t = \frac{2}{3},\ 0.67$ or better | A1 | |

Show LaTeX source

7. Two ships $P$ and $Q$ are travelling at night with constant velocities. At midnight, $P$ is at the point with position vector $( 20 \mathbf { i } + 10 \mathbf { j } ) \mathrm { km }$ relative to a fixed origin $O$. At the same time, $Q$ is at the point with position vector $( 14 \mathbf { i } - 6 \mathbf { j } ) \mathrm { km }$. Three hours later, $P$ is at the point with position vector $( 29 \mathbf { i } + 34 \mathbf { j } ) \mathrm { km }$. The ship $Q$ travels with velocity $12 \mathbf { j } \mathrm {~km} \mathrm {~h} ^ { - 1 }$. At time $t$ hours after midnight, the position vectors of $P$ and $Q$ are $\mathbf { p } \mathrm { km }$ and $\mathbf { q } \mathrm { km }$ respectively. Find\\
(a) the velocity of $P$, in terms of $\mathbf { i }$ and $\mathbf { j }$,\\
(b) expressions for $\mathbf { p }$ and $\mathbf { q }$, in terms of $t$, i and $\mathbf { j }$.

At time $t$ hours after midnight, the distance between $P$ and $Q$ is $d \mathrm {~km}$.\\
(c) By finding an expression for $\overrightarrow { P Q }$, show that

$$d ^ { 2 } = 25 t ^ { 2 } - 92 t + 292$$

Weather conditions are such that an observer on $P$ can only see the lights on $Q$ when the distance between $P$ and $Q$ is 15 km or less. Given that when $t = 1$, the lights on $Q$ move into sight of the observer,\\
(d) find the time, to the nearest minute, at which the lights on $Q$ move out of sight of the observer.

\begin{enumerate}
  \item In taking off, an aircraft moves on a straight runway $A B$ of length 1.2 km . The aircraft moves from $A$ with initial speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It moves with constant acceleration and 20 s later it leaves the runway at $C$ with speed $74 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find\\
(a) the acceleration of the aircraft,\\
(b) the distance $B C$.
  \item Two small steel balls $A$ and $B$ have mass 0.6 kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when they collide directly. Immediately before the collision, the speed of $A$ is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the speed of $B$ is $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Immediately after the collision, the direction of motion of $A$ is unchanged and the speed of $B$ is twice the speed of $A$. Find\\
(a) the speed of $A$ immediately after the collision,\\
(b) the magnitude of the impulse exerted on $B$ in the collision.\\

  \item 
\end{enumerate}

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
  \includegraphics[alt={},max width=\textwidth]{94d9432d-1723-4549-ad5e-d4be0f5fd083-018_282_707_278_699}
\end{center}
\end{figure}

\hfill \mbox{\textit{Edexcel M1  Q7}}

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