| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Kinematics with position vectors |
| Difficulty | Moderate -0.3 This is a straightforward M1 kinematics question requiring equating position vectors to find when they meet (basic algebraic manipulation) and comparing speeds using magnitude of velocity vectors. The calculations are routine with no conceptual challenges, making it slightly easier than average but still requiring proper method. |
| Spec | 1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Either \(-2 + 8t = 7t\) or \(t = 10 - 4t\) | M1 | Forming an equation for \(t\). Accept vector equation. May be implied by statement \(t = 2\). |
| \(t = 2\) | A1 | |
| Substituting \(t = 2\) in both expressions | B1 | e.g. showing \(t = 2\) satisfies both equations or a vector equation. |
| They meet at \((14, 2)\) | B1 | Accept \(\begin{pmatrix}14\\2\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Ashok's speed is \(\sqrt{8^2 + 1^2} = \sqrt{65}\) | B1 | |
| Kumar's speed is \(\sqrt{7^2 + (-4)^2} = \sqrt{65}\) km h\(^{-1}\) | B1 | |
| They both walk at the same speed | B1 | CAO from correct speeds. SC1 for finding both velocities correctly but neither speed. |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Either $-2 + 8t = 7t$ or $t = 10 - 4t$ | M1 | Forming an equation for $t$. Accept vector equation. May be implied by statement $t = 2$. |
| $t = 2$ | A1 | |
| Substituting $t = 2$ in **both** expressions | B1 | e.g. showing $t = 2$ satisfies both equations or a vector equation. |
| They meet at $(14, 2)$ | B1 | Accept $\begin{pmatrix}14\\2\end{pmatrix}$ |
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## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Ashok's speed is $\sqrt{8^2 + 1^2} = \sqrt{65}$ | B1 | |
| Kumar's speed is $\sqrt{7^2 + (-4)^2} = \sqrt{65}$ km h$^{-1}$ | B1 | |
| They both walk at the same speed | B1 | CAO from correct speeds. SC1 for finding both velocities correctly but neither speed. |
---3 The map of a large area of open land is marked in 1 km squares and a point near the middle of the area is defined to be the origin. The vectors $\binom { 1 } { 0 }$ and $\binom { 0 } { 1 }$ are in the directions east and north.
At time $t$ hours the position vectors of two hikers, Ashok and Kumar, are given by:
$$\begin{array} { l l }
\text { Ashok } & \mathbf { r } _ { \mathrm { A } } = \binom { - 2 } { 0 } + \binom { 8 } { 1 } t , \\
\text { Kumar } & \mathbf { r } _ { \mathrm { K } } = \binom { 7 t } { 10 - 4 t } .
\end{array}$$
(i) Prove that the two hikers meet and give the coordinates of the point where this happens.\\
(ii) Compare the speeds of the two hikers.
\hfill \mbox{\textit{OCR MEI M1 2015 Q3 [7]}}