| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | 3D vector motion problems |
| Difficulty | Moderate -0.3 This is a straightforward vector mechanics question requiring only direct application of F=ma, then standard constant acceleration formulas (v=u+at, s=ut+½at²). All steps are routine calculations with no problem-solving insight needed, making it slightly easier than average despite being multi-part. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Net force \(= (-80\mathbf{k}) + (-\mathbf{i}+16\mathbf{j}+72\mathbf{k}) = (-\mathbf{i}+16\mathbf{j}-8\mathbf{k})\) N | M1 | Sum forces |
| \(\mathbf{a} = \frac{1}{8}(-\mathbf{i}+16\mathbf{j}-8\mathbf{k}) = (-\frac{1}{8}\mathbf{i}+2\mathbf{j}-\mathbf{k})\) m s\(^{-2}\) | A1 | Divide by mass 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with \(t=4\) | M1 | |
| \(= 4(\mathbf{i}-4\mathbf{j}+3\mathbf{k}) + 8(-\frac{1}{8}\mathbf{i}+2\mathbf{j}-\mathbf{k})\) | A1 | |
| \(= (3\mathbf{i}+0\mathbf{j}-5\mathbf{k})\) m, i.e. \((3,0,-5)\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \( | OA | = \sqrt{3^2+0^2+(-5)^2} = \sqrt{34} \approx 5.83\) m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Horizontal distance \(= \sqrt{3^2+0^2} = 3\) m | M1 | |
| \(\sin\theta = \frac{5}{\sqrt{34}}\) or \(\tan\theta = \frac{5}{3}\) | ||
| \(\theta = \arctan\left(\frac{5}{3}\right) \approx 59.0°\) below horizontal | A1 | cao |
# Question 6:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Net force $= (-80\mathbf{k}) + (-\mathbf{i}+16\mathbf{j}+72\mathbf{k}) = (-\mathbf{i}+16\mathbf{j}-8\mathbf{k})$ N | M1 | Sum forces |
| $\mathbf{a} = \frac{1}{8}(-\mathbf{i}+16\mathbf{j}-8\mathbf{k}) = (-\frac{1}{8}\mathbf{i}+2\mathbf{j}-\mathbf{k})$ m s$^{-2}$ | A1 | Divide by mass 8 |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ with $t=4$ | M1 | |
| $= 4(\mathbf{i}-4\mathbf{j}+3\mathbf{k}) + 8(-\frac{1}{8}\mathbf{i}+2\mathbf{j}-\mathbf{k})$ | A1 | |
| $= (3\mathbf{i}+0\mathbf{j}-5\mathbf{k})$ m, i.e. $(3,0,-5)$ | A1 | cao |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $|OA| = \sqrt{3^2+0^2+(-5)^2} = \sqrt{34} \approx 5.83$ m | B1 | ft their position vector |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Horizontal distance $= \sqrt{3^2+0^2} = 3$ m | M1 | |
| $\sin\theta = \frac{5}{\sqrt{34}}$ or $\tan\theta = \frac{5}{3}$ | | |
| $\theta = \arctan\left(\frac{5}{3}\right) \approx 59.0°$ below horizontal | A1 | cao |
---6 A rock of mass 8 kg is acted on by just the two forces $- 80 \mathbf { k } \mathrm {~N}$ and $( - \mathbf { i } + 16 \mathbf { j } + 72 \mathbf { k } ) \mathrm { N }$, where $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane and $\mathbf { k }$ is a unit vector vertically upward.\\
(i) Show that the acceleration of the rock is $\left( - \frac { 1 } { 8 } \mathbf { i } + 2 \mathbf { j } - \mathbf { k } \right) \mathrm { ms } ^ { - 2 }$.
The rock passes through the origin of position vectors, O , with velocity $( \mathbf { i } - 4 \mathbf { j } + 3 \mathbf { k } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and 4 seconds later passes through the point A .\\
(ii) Find the position vector of A .\\
(iii) Find the distance OA .\\
(iv) Find the angle that OA makes with the horizontal.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI M1 2007 Q6 [8]}}