| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| 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 vector mechanics question requiring basic application of Newton's second law. Part (a) involves recognizing that zero horizontal acceleration means horizontal forces sum to zero (k + 4 = 0), and part (b) is simple constant acceleration kinematics. Both parts are routine calculations with no problem-solving insight required, making it easier than average. |
| Spec | 3.02e Two-dimensional constant acceleration: with vectors3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Component in \(\mathbf{i}\) direction zero \(\Rightarrow k = -4\) | B1 [1] | Allow \((-4\mathbf{i} + 5\mathbf{j})\) seen instead of \(k\); do not allow \(k = -4\mathbf{i}\) or similar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Weight \(= -0.8g\mathbf{j}\) | B1 | Allow if seen |
| N2L: \(5\mathbf{j} + 3\mathbf{j} - 0.8g\mathbf{j} = 0.8\mathbf{a}\) | M1 | Condone missing weight |
| \([\Rightarrow \mathbf{a} = 0.2\mathbf{j}]\) | M1 | Using their \(\mathbf{a}\) in a vector suvat equation(s) |
| \(\mathbf{v} = (4\mathbf{i} + 7\mathbf{j}) + 0.2\mathbf{j} \times 10\), velocity is \((4\mathbf{i} + 9\mathbf{j})\ \text{m s}^{-1}\) | A1 [4] | Must be in correct vector form; accept fully correct column vector |
| Alternative: Acceleration is vertical, consider only vertical motion | M1 | Applying N2L in 1-dimension; condone missing weight |
| \(5 + 3 - 0.8g = 0.8a\), \([\Rightarrow a = 0.2]\) | B1 | Including the weight (consistent sign convention) |
| \(v = u + at = 7 + 0.2 \times 10 = 9\), so velocity is \((4\mathbf{i} + 9\mathbf{j})\ \text{m s}^{-1}\) | M1, A1 [4] | Using suvat in vertical direction; must be in vector form; accept fully correct column vector |
## Question 11:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Component in $\mathbf{i}$ direction zero $\Rightarrow k = -4$ | B1 [1] | Allow $(-4\mathbf{i} + 5\mathbf{j})$ seen instead of $k$; do not allow $k = -4\mathbf{i}$ or similar |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Weight $= -0.8g\mathbf{j}$ | B1 | Allow if seen |
| N2L: $5\mathbf{j} + 3\mathbf{j} - 0.8g\mathbf{j} = 0.8\mathbf{a}$ | M1 | Condone missing weight |
| $[\Rightarrow \mathbf{a} = 0.2\mathbf{j}]$ | M1 | Using their $\mathbf{a}$ in a vector suvat equation(s) |
| $\mathbf{v} = (4\mathbf{i} + 7\mathbf{j}) + 0.2\mathbf{j} \times 10$, velocity is $(4\mathbf{i} + 9\mathbf{j})\ \text{m s}^{-1}$ | A1 [4] | Must be in correct vector form; accept fully correct column vector |
| **Alternative:** Acceleration is vertical, consider only vertical motion | M1 | Applying N2L in 1-dimension; condone missing weight |
| $5 + 3 - 0.8g = 0.8a$, $[\Rightarrow a = 0.2]$ | B1 | Including the weight (consistent sign convention) |
| $v = u + at = 7 + 0.2 \times 10 = 9$, so velocity is $(4\mathbf{i} + 9\mathbf{j})\ \text{m s}^{-1}$ | M1, A1 [4] | Using suvat in vertical direction; must be in vector form; accept fully correct column vector |
---11 In this question, the unit vector $\mathbf { i }$ is horizontal and the unit vector $\mathbf { j }$ is vertically upwards.
A particle of mass 0.8 kg moves under the action of its weight and two forces given by ( $k \mathbf { i } + 5 \mathbf { j }$ ) N and $( 4 \mathbf { i } + 3 \mathbf { j } ) \mathrm { N }$. The acceleration of the particle is vertically upwards.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $k$.
Initially the velocity of the particle is $( 4 \mathbf { i } + 7 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$.
\item Find the velocity of the particle 10 seconds later.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2019 Q11 [5]}}