OCR M4 2016 June — Question 2 9 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2016
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeClosest approach of two objects
DifficultyStandard +0.3 This is a standard M4 relative velocity question requiring vector subtraction to find relative velocity, then using perpendicular distance from a line for closest approach. The multi-part structure and bearing conversions add some computational work, but the methods are routine textbook applications with no novel problem-solving required. Slightly easier than average due to straightforward setup.
Spec1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.02a Kinematics language: position, displacement, velocity, acceleration3.02h Motion under gravity: vector form

\includegraphics{figure_2} Boat \(A\) is travelling with constant speed 7.9 m s\(^{-1}\) on a course with bearing 035°. Boat \(B\) is travelling with constant speed 10.5 m s\(^{-1}\) on a course with bearing 330°. At one instant, the boats are 1500 m apart with \(B\) on a bearing of 125° from \(A\) (see diagram).
  1. Find the magnitude and the bearing of the velocity of \(B\) relative to \(A\). [5]
  2. Find the shortest distance between \(A\) and \(B\) in the subsequent motion. [2]
  3. Find the time taken from the instant when \(A\) and \(B\) are 1500 m apart to the instant when \(A\) and \(B\) are at the point of closest approach. [2]

(i)
AnswerMarks Guidance
\(w^2 = 7.9^2 + 10.5^2 - 2(7.9)(10.5)\cos(30 + 35)\)M1 Use of cosine rule
\(w = 10.1\)A1 10.12658...
\(\frac{\sin\theta}{7.9} = \frac{\sin(30 + 35)}{10.12658...}\) or \(\frac{\sin\alpha}{10.5} = \frac{\sin(30 + 35)}{10.12658...}\)M1 Use of sine rule with their \(w\)
\(\theta = 45.0\) or \(\alpha = 70.0\)A1 44.99406... (\(\beta = 15.005...\)) or (\(\gamma = 74.994...\))
Bearing = \(330 - \theta = 285\) or \(180 + 35 + \alpha = 285\)A1 285.0059...
[5]
(ii)
AnswerMarks Guidance
Shortest distance = \(d = 1500\sin(44.99406... - 25)\)M1 \(1500\sin\beta\)
\(d = 513\) (3 sf)A1 512.8841...
[2]
(iii)
AnswerMarks Guidance
\(t = \frac{1500\cos(44.99406... - 25)}{10.12658...}\)M1 Use of \(s = ut\) with their \(w\) and \(\theta\)
\(= 139\) (3 sf)A1 139.1972... (\(\beta\) consistent with (ii))
[2]
**(i)**

$w^2 = 7.9^2 + 10.5^2 - 2(7.9)(10.5)\cos(30 + 35)$ | M1 | Use of cosine rule

$w = 10.1$ | A1 | 10.12658...

$\frac{\sin\theta}{7.9} = \frac{\sin(30 + 35)}{10.12658...}$ or $\frac{\sin\alpha}{10.5} = \frac{\sin(30 + 35)}{10.12658...}$ | M1 | Use of sine rule with their $w$

$\theta = 45.0$ or $\alpha = 70.0$ | A1 | 44.99406... ($\beta = 15.005...$) or ($\gamma = 74.994...$)

Bearing = $330 - \theta = 285$ or $180 + 35 + \alpha = 285$ | A1 | 285.0059...

| [5] |

**(ii)**

Shortest distance = $d = 1500\sin(44.99406... - 25)$ | M1 | $1500\sin\beta$

$d = 513$ (3 sf) | A1 | 512.8841...

| [2] |

**(iii)**

$t = \frac{1500\cos(44.99406... - 25)}{10.12658...}$ | M1 | Use of $s = ut$ with their $w$ and $\theta$

$= 139$ (3 sf) | A1 | 139.1972... ($\beta$ consistent with (ii))

| [2] |
\includegraphics{figure_2}

Boat $A$ is travelling with constant speed 7.9 m s$^{-1}$ on a course with bearing 035°. Boat $B$ is travelling with constant speed 10.5 m s$^{-1}$ on a course with bearing 330°. At one instant, the boats are 1500 m apart with $B$ on a bearing of 125° from $A$ (see diagram).

\begin{enumerate}[label=(\roman*)]
\item Find the magnitude and the bearing of the velocity of $B$ relative to $A$. [5]
\item Find the shortest distance between $A$ and $B$ in the subsequent motion. [2]
\item Find the time taken from the instant when $A$ and $B$ are 1500 m apart to the instant when $A$ and $B$ are at the point of closest approach. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR M4 2016 Q2 [9]}}