| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| 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 mechanics question testing basic vector manipulation with forces and Newton's second law. Part (a) is trivial recall, parts (b)-(c) involve standard vector addition and F=ma calculations, and part (d) requires understanding equilibrium. All steps are routine textbook exercises with no problem-solving insight required. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Weight \(= -2g\mathbf{j}\) N \([= -19.6\mathbf{j}\) N\(]\) | B1 | 2.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Horizontal force \(3\mathbf{i}\) | B1 | 3.3 |
| Newton's second law: \((-4\mathbf{i} + 12\mathbf{j}) + 3\mathbf{i} - 19.6\mathbf{j} = 2\mathbf{a}\) | M1 | 1.1a |
| \(\mathbf{a} = -0.5\mathbf{i} - 3.8\mathbf{j}\) m s\(^{-2}\) | A1 | 1.1b |
| Alternative: Horizontal: \(-4 + 3 = 2a_x\); Vertical: \(12 - 19.6 = 2a_y\) | B1, M1 | |
| \(\mathbf{a} = -0.5\mathbf{i} - 3.8\mathbf{j}\) m s\(^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{v} = \mathbf{u} + \mathbf{a}t = 5\mathbf{i} + (-0.5\mathbf{i} - 3.8\mathbf{j}) \times 4\) | M1 | 1.1a |
| \(= 3\mathbf{i} - 15.2\mathbf{j}\) m s\(^{-1}\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Constant velocity is equilibrium, so \((\mathbf{i} + 7.6\mathbf{j})\) N | B1 | 1.1b |
## Question 12(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Weight $= -2g\mathbf{j}$ N $[= -19.6\mathbf{j}$ N$]$ | B1 | 2.5 | Allow equivalent column vectors in all part questions |
**Total: [1]**
## Question 12(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Horizontal force $3\mathbf{i}$ | B1 | 3.3 | May be implied by correct resultant force |
| Newton's second law: $(-4\mathbf{i} + 12\mathbf{j}) + 3\mathbf{i} - 19.6\mathbf{j} = 2\mathbf{a}$ | M1 | 1.1a | Must be vector equation. Allow one missing or incorrect force |
| $\mathbf{a} = -0.5\mathbf{i} - 3.8\mathbf{j}$ m s$^{-2}$ | A1 | 1.1b | ISW if the magnitude is also given |
| **Alternative:** Horizontal: $-4 + 3 = 2a_x$; Vertical: $12 - 19.6 = 2a_y$ | B1, M1 | | 3 N force used in horizontal equation and not vertical. Considers motion in two directions. Allow one missing or incorrect force |
| $\mathbf{a} = -0.5\mathbf{i} - 3.8\mathbf{j}$ m s$^{-2}$ | A1 | | ISW if the magnitude is also given |
**Total: [3]**
## Question 12(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{v} = \mathbf{u} + \mathbf{a}t = 5\mathbf{i} + (-0.5\mathbf{i} - 3.8\mathbf{j}) \times 4$ | M1 | 1.1a | Using suvat equation(s) leading to a vector $\mathbf{v}$. Do not award if scalar added to vector |
| $= 3\mathbf{i} - 15.2\mathbf{j}$ m s$^{-1}$ | A1 | 1.1b | Mark final answer. Must be vector $\mathbf{v}$ and not speed. FT their vector acceleration |
**Total: [2]**
## Question 12(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Constant velocity is equilibrium, so $(\mathbf{i} + 7.6\mathbf{j})$ N | B1 | 1.1b | Cao. ISW if the magnitude is also given |
**Total: [1]**
---12 In this question the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are horizontal and vertically upwards respectively.
A particle has mass 2 kg .
\begin{enumerate}[label=(\alph*)]
\item Write down its weight as a vector.
A horizontal force of 3 N in the $\mathbf { i }$ direction and a force $\mathbf { F } = ( - 4 \mathbf { i } + 12 \mathbf { j } ) \mathrm { N }$ act on the particle.
\item Determine the acceleration of the particle.
\item The initial velocity of the particle is $5 \mathbf { i } \mathrm {~ms} ^ { - 1 }$.
Find the velocity of the particle after 4 s .
\item Find the extra force that must be applied to the particle for it to move at constant velocity.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2023 Q12 [7]}}