| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.8 This is a straightforward application of Newton's second law (F=ma) in vector form and basic kinematics with constant acceleration. Part (i) requires simple multiplication of mass by acceleration vector, and part (ii) uses the standard SUVAT equation s = ut + ½at² with vectors. Both parts are direct recall and substitution with no problem-solving or conceptual challenges, making this easier than average for A-level mechanics. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{v} = \mathbf{i} + (3-2t)\mathbf{j}\) | M1 | Differentiating \(\mathbf{r}\). Allow 1 error. Could use const accn. |
| A1 | ||
| \(\mathbf{v}(4) = \mathbf{i} - 5\mathbf{j}\) | F1 | Do not award if \(\sqrt{26}\) is given as vel (accept if \(v\) given and \(v\) given as well called speed or magnitude). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{a} = -2\mathbf{j}\) | B1 | Diff \(\mathbf{v}\). FT their \(\mathbf{v}\). Award if \(-2\mathbf{j}\) seen & isw. |
| Using N2L \(\mathbf{F} = 1.5 \times (-2\mathbf{j})\) | M1 | Award for \(1.5 \times (\pm \text{ their } \mathbf{a} \text{ or } a)\) seen. |
| \(-3\mathbf{j}\) N | A1 | cao. Do not award if final answer is not correct. [Award M1 A1 for \(-3\mathbf{j}\) WW] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 2+t\) and \(y = 3t - t^2\) | B1 | Must have both but may be implied. |
| Substitute \(t = x-2\), so \(y = 3(x-2)-(x-2)^2\) \([= (x-2)(5-x)]\) | B1 | cao. isw. Must see the form \(y = \ldots\) |
## Question 6:
**Part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = \mathbf{i} + (3-2t)\mathbf{j}$ | M1 | Differentiating $\mathbf{r}$. Allow 1 error. Could use const accn. |
| | A1 | |
| $\mathbf{v}(4) = \mathbf{i} - 5\mathbf{j}$ | F1 | Do not award if $\sqrt{26}$ is given as vel (accept if $v$ given and $v$ given as well called speed or magnitude). |
**Part (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{a} = -2\mathbf{j}$ | B1 | Diff $\mathbf{v}$. FT their $\mathbf{v}$. Award if $-2\mathbf{j}$ seen & isw. |
| Using N2L $\mathbf{F} = 1.5 \times (-2\mathbf{j})$ | M1 | Award for $1.5 \times (\pm \text{ their } \mathbf{a} \text{ or } a)$ seen. |
| $-3\mathbf{j}$ N | A1 | cao. Do not award if final answer is not correct. [Award M1 A1 for $-3\mathbf{j}$ WW] |
**Part (iii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 2+t$ and $y = 3t - t^2$ | B1 | Must have both but may be implied. |
| Substitute $t = x-2$, so $y = 3(x-2)-(x-2)^2$ $[= (x-2)(5-x)]$ | B1 | cao. isw. Must see the form $y = \ldots$ |
---6 An object of mass 5 kg has a constant acceleration of $\binom { - 1 } { 2 } \mathrm {~ms} ^ { - 2 }$ for $0 \leqslant t \leqslant 4$, where $t$ is the time in seconds.\\
(i) Calculate the force acting on the object.
When $t = 0$, the object has position vector $\binom { - 2 } { 3 } \mathrm {~m}$ and velocity $\binom { 4 } { 5 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Find the position vector of the object when $t = 4$.
\hfill \mbox{\textit{OCR MEI M1 Q6 [5]}}