| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find force using F=ma |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring two differentiations of a position vector (standard M1 technique) followed by direct application of F=ma. The algebra is simple (quadratic terms only), and both parts are routine textbook exercises with no problem-solving or insight required. Slightly easier than average due to its mechanical nature. |
| Spec | 1.07i Differentiate x^n: for rational n and sums3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differentiate: \(\mathbf{v} = 2t\,\mathbf{i} + (5-4t)\,\mathbf{j}\) | M1, A1 | At least 1 component correct. Award for RHS seen |
| Differentiate: \(\mathbf{a} = 2\mathbf{i} - 4\mathbf{j}\) | M1, F1 | Do not award if \(\mathbf{i}\) and \(\mathbf{j}\) lost in v. At least 1 component correct. FT from their 2-component v |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{F} + 12\mathbf{j} = 4(2\mathbf{i} - 4\mathbf{j})\) | M1 | N2L. Allow \(\mathbf{F} = mg\,\mathbf{a}\). No extra forces. Allow \(12\mathbf{j}\) omitted |
| A1 | Allow wrong signs otherwise correct with their vector a | |
| \(\mathbf{F} = 8\mathbf{i} - 28\mathbf{j}\) | A1 | cao |
## Question 3:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate: $\mathbf{v} = 2t\,\mathbf{i} + (5-4t)\,\mathbf{j}$ | M1, A1 | At least 1 component correct. Award for RHS seen |
| Differentiate: $\mathbf{a} = 2\mathbf{i} - 4\mathbf{j}$ | M1, F1 | Do not award if $\mathbf{i}$ and $\mathbf{j}$ lost in **v**. At least 1 component correct. FT from their 2-component **v** |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{F} + 12\mathbf{j} = 4(2\mathbf{i} - 4\mathbf{j})$ | M1 | N2L. Allow $\mathbf{F} = mg\,\mathbf{a}$. No extra forces. Allow $12\mathbf{j}$ omitted |
| | A1 | Allow wrong signs otherwise correct with their vector **a** |
| $\mathbf{F} = 8\mathbf{i} - 28\mathbf{j}$ | A1 | cao |
---3 The position vector, $r$, of a particle of mass 4 kg at time $t$ is given by
$$\mathbf { r } = t ^ { 2 } \mathbf { i } + \left( 5 t - 2 t ^ { 2 } \right) \mathbf { j }$$
where $\mathbf { i }$ and $\mathbf { j }$ are the standard unit vectors, lengths are in metres and time is in seconds.\\
(i) Find an expression for the acceleration of the particle.
The particle is subject to a force $\mathbf { F }$ and a force $12 \mathbf { j } \mathbf { N }$.\\
(ii) Find $\mathbf { F }$.
\hfill \mbox{\textit{OCR MEI M1 Q3 [7]}}