| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.3 This is a straightforward application of Newton's second law in vector form (F=ma) requiring vector addition and scalar multiplication. Part (i) involves simple algebraic manipulation to find the unknown force, and part (ii) uses constant acceleration kinematics (v = u + at). While it requires competence with 3D vectors, the problem-solving is routine with no conceptual challenges beyond standard M1 content. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}-1\\14\\-8\end{pmatrix} + \begin{pmatrix}3\\-9\\10\end{pmatrix} + \mathbf{F} = 4\begin{pmatrix}-1\\2\\4\end{pmatrix}\) | M1 | N2L. Allow sign errors in applying N2L. Do not condone \(\mathbf{F} = mg\mathbf{a}\). Allow one given force omitted. |
| M1 | Attempt to add \(\begin{pmatrix}-1\\14\\-8\end{pmatrix}\) and \(\begin{pmatrix}3\\-9\\10\end{pmatrix}\) | |
| \(\mathbf{F} = \begin{pmatrix}-6\\3\\14\end{pmatrix}\) | A1 | Two components correct |
| A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{v} = \begin{pmatrix}-3\\3\\6\end{pmatrix} + 3\begin{pmatrix}-1\\2\\4\end{pmatrix} = \begin{pmatrix}-6\\9\\18\end{pmatrix}\) so \(\begin{pmatrix}-6\\9\\18\end{pmatrix} \text{ m s}^{-1}\) | M1 | \(\mathbf{v} = \mathbf{u} + t\mathbf{a}\) with given \(\mathbf{u}\) and \(\mathbf{a}\). Could go via \(\mathbf{s}\). If integration used, require arbitrary constant (need not be evaluated). |
| A1 | cao isw | |
| speed is \(\sqrt{(-6)^2 + 9^2 + 18^2} = 21 \text{ m s}^{-1}\) | M1 | Allow \(-6^2\) even if interpreted as \(-36\). Only FT their \(\mathbf{v}\). |
| F1 | FT their \(\mathbf{v}\) only. [Award M1 F1 for 21 seen WWW] |
# Question 3:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}-1\\14\\-8\end{pmatrix} + \begin{pmatrix}3\\-9\\10\end{pmatrix} + \mathbf{F} = 4\begin{pmatrix}-1\\2\\4\end{pmatrix}$ | M1 | N2L. Allow sign errors in applying N2L. Do not condone $\mathbf{F} = mg\mathbf{a}$. Allow one given force omitted. |
| | M1 | Attempt to add $\begin{pmatrix}-1\\14\\-8\end{pmatrix}$ and $\begin{pmatrix}3\\-9\\10\end{pmatrix}$ |
| $\mathbf{F} = \begin{pmatrix}-6\\3\\14\end{pmatrix}$ | A1 | Two components correct |
| | A1 | cao |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = \begin{pmatrix}-3\\3\\6\end{pmatrix} + 3\begin{pmatrix}-1\\2\\4\end{pmatrix} = \begin{pmatrix}-6\\9\\18\end{pmatrix}$ so $\begin{pmatrix}-6\\9\\18\end{pmatrix} \text{ m s}^{-1}$ | M1 | $\mathbf{v} = \mathbf{u} + t\mathbf{a}$ with given $\mathbf{u}$ and $\mathbf{a}$. Could go via $\mathbf{s}$. If integration used, require arbitrary constant (need not be evaluated). |
| | A1 | cao isw |
| speed is $\sqrt{(-6)^2 + 9^2 + 18^2} = 21 \text{ m s}^{-1}$ | M1 | Allow $-6^2$ even if interpreted as $-36$. Only FT **their** $\mathbf{v}$. |
| | F1 | FT their $\mathbf{v}$ only. [Award M1 F1 for 21 seen WWW] |
---3 The three forces $\left( \begin{array} { r } - 1 \\ 14 \\ - 8 \end{array} \right) \mathrm { N } , \left( \begin{array} { r } 3 \\ - 9 \\ 10 \end{array} \right) \mathrm { N }$ and $\mathbf { F } \mathrm { N }$ act on a body of mass 4 kg in deep space and give it an acceleration of $\left( \begin{array} { r } - 1 \\ 2 \\ 4 \end{array} \right) \mathrm { m } \mathrm { s } ^ { - 2 }$.\\
(i) Calculate $\mathbf { F }$.
At one instant the velocity of the body is $\left( \begin{array} { r } - 3 \\ 3 \\ 6 \end{array} \right) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
(ii) Calculate the velocity and also the speed of the body 3 seconds later.
\hfill \mbox{\textit{OCR MEI M1 2010 Q3 [8]}}