| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| 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 F=ma in vector form (part i) and basic kinematics with constant acceleration (part ii). Both parts require only direct formula application with no problem-solving insight or multi-step reasoning, making it easier than average for A-level mechanics. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{F} = 5\begin{pmatrix}-1\\2\end{pmatrix} = \begin{pmatrix}-5\\10\end{pmatrix}\) so \(\begin{pmatrix}-5\\10\end{pmatrix}\) N | M1 | Use of N2L in vector form |
| A1 | Ignore units. Award 2 for answer seen. SC1 for \(\sqrt{125}\) or equiv seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{s} = \begin{pmatrix}-2\\3\end{pmatrix} + 4\begin{pmatrix}4\\5\end{pmatrix} + \frac{1}{2}\times4^2\times\begin{pmatrix}-1\\2\end{pmatrix}\) | M1 | Use of \(\mathbf{s} = t\mathbf{u} + 0.5t^2\mathbf{a}\) or integration of \(\mathbf{a}\). Allow \(\mathbf{s}_0\) omitted. If integrated need to consider \(\mathbf{v}\) when \(t=0\) |
| A1 | Correctly evaluated; accept \(\mathbf{s}_0\) omitted | |
| \(\mathbf{s} = \begin{pmatrix}6\\39\end{pmatrix}\) so \(\begin{pmatrix}6\\39\end{pmatrix}\) m | B1 | Correctly adding \(\mathbf{s}_0\) to a vector (FT). Ignore units. NB \(\begin{pmatrix}8\\36\end{pmatrix}\) seen scores M1 A1 |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{F} = 5\begin{pmatrix}-1\\2\end{pmatrix} = \begin{pmatrix}-5\\10\end{pmatrix}$ so $\begin{pmatrix}-5\\10\end{pmatrix}$ N | M1 | Use of N2L in vector form |
| | A1 | Ignore units. Award 2 for answer seen. SC1 for $\sqrt{125}$ or equiv seen |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{s} = \begin{pmatrix}-2\\3\end{pmatrix} + 4\begin{pmatrix}4\\5\end{pmatrix} + \frac{1}{2}\times4^2\times\begin{pmatrix}-1\\2\end{pmatrix}$ | M1 | Use of $\mathbf{s} = t\mathbf{u} + 0.5t^2\mathbf{a}$ or integration of $\mathbf{a}$. Allow $\mathbf{s}_0$ omitted. If integrated need to consider $\mathbf{v}$ when $t=0$ |
| | A1 | Correctly evaluated; accept $\mathbf{s}_0$ omitted |
| $\mathbf{s} = \begin{pmatrix}6\\39\end{pmatrix}$ so $\begin{pmatrix}6\\39\end{pmatrix}$ m | B1 | Correctly adding $\mathbf{s}_0$ to a vector (FT). Ignore units. NB $\begin{pmatrix}8\\36\end{pmatrix}$ seen scores M1 A1 |
---3 An object of mass 5 kg has a constant acceleration of $\binom { - 1 } { 2 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$ 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 2008 Q3 [5]}}