| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.8 This is a straightforward application of Newton's second law with vectors. Part (a) requires simple recall that weight = -mg j. Part (b) involves adding three force vectors and dividing by mass—purely mechanical calculation with no problem-solving insight required. Below average difficulty for A-level mechanics. |
| Spec | 1.10b Vectors in 3D: i,j,k notation3.03d Newton's second law: 2D vectors3.03f Weight: W=mg |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((W=)\ -2.5g\mathbf{j}\ \text{N}\ (= -24.5\mathbf{j}\ \text{N})\) | B1 [1] | 1.2 – Condone missing units. Must be negative. Allow any correct vector notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Newton's second law: \(-24.5\mathbf{j}+(3\mathbf{i}-2\mathbf{j})+(-\mathbf{i}+18\mathbf{j}) = 2.5\mathbf{a}\) | M1 | 1.1a – Attempt to use N2L with \(m=2.5\). Condone missing weight or incorrect weight vector. Must be sum of forces \(= 2.5\mathbf{a}\) |
| \(\mathbf{a} = \frac{1}{2.5}(2\mathbf{i}-8.5\mathbf{j}) = (0.8\mathbf{i}-3.4\mathbf{j})\ \text{m s}^{-2}\) | A1 [2] | 1.1 – Cao. Do not award if the magnitude of acceleration given as acceleration |
## Question 5:
**(a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(W=)\ -2.5g\mathbf{j}\ \text{N}\ (= -24.5\mathbf{j}\ \text{N})$ | B1 [1] | 1.2 – Condone missing units. Must be negative. Allow any correct vector notation |
**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Newton's second law: $-24.5\mathbf{j}+(3\mathbf{i}-2\mathbf{j})+(-\mathbf{i}+18\mathbf{j}) = 2.5\mathbf{a}$ | M1 | 1.1a – Attempt to use N2L with $m=2.5$. Condone missing weight or incorrect weight vector. Must be sum of forces $= 2.5\mathbf{a}$ |
| $\mathbf{a} = \frac{1}{2.5}(2\mathbf{i}-8.5\mathbf{j}) = (0.8\mathbf{i}-3.4\mathbf{j})\ \text{m s}^{-2}$ | A1 [2] | 1.1 – Cao. Do not award if the magnitude of acceleration given as acceleration |
---5 In this question, the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are horizontal and vertically upwards respectively.
A particle has mass 2.5 kg .
\begin{enumerate}[label=(\alph*)]
\item Write the weight of the particle as a vector.
The particle moves under the action of its weight and two external forces ( $3 \mathbf { i } - 2 \mathbf { j }$ ) N and $( - \mathbf { i } + 18 \mathbf { j } ) N$.
\item Find the acceleration of the particle, giving your answer in vector form.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2019 Q5 [3]}}