| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: equilibrium (find unknowns) |
| Difficulty | Moderate -0.8 This is a straightforward application of Newton's laws in vector form with standard equilibrium and F=ma calculations. Part (a) requires simple vector addition with weight, and part (b) is direct application of F=ma. No problem-solving insight needed, just routine mechanical application of basic principles. |
| Spec | 1.10b Vectors in 3D: i,j,k notation3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors3.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Weight \(\begin{pmatrix}0\\-1.5g\end{pmatrix}\) | B1 (AO 1.2) | Allow seen or implied by correct answer |
| Equilibrium equation: \(\begin{pmatrix}0\\-1.5g\end{pmatrix}+\begin{pmatrix}4\\-2\end{pmatrix}+\mathbf{F}_2=\mathbf{0}\) | M1 (AO 1.1a) | FT their weight; allow sign errors |
| \(\mathbf{F}_2 = \begin{pmatrix}-4\\2+1.5g\end{pmatrix} = \begin{pmatrix}-4\\16.7\end{pmatrix}\) N | A1 [3] (AO 1.1) | Must be vector; allow \(\mathbf{i}\)-\(\mathbf{j}\) form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Newton's second law: \(\begin{pmatrix}0\\-1.5g\end{pmatrix}+\begin{pmatrix}4\\-2\end{pmatrix}+\begin{pmatrix}2\\20\end{pmatrix}=m\mathbf{a}\) | M1 (AO 1.1a) | Addition of vectors; allow if weight missing but other two forces and acceleration seen in equation; allow wrong weight only if given as a vector |
| \(\mathbf{a} = \frac{1}{1.5}\begin{pmatrix}6\\3.3\end{pmatrix} = \begin{pmatrix}4\\2.2\end{pmatrix}\) m s\(^{-2}\) | A1 [2] (AO 1.1) | Must be vector; any correct form |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Weight $\begin{pmatrix}0\\-1.5g\end{pmatrix}$ | B1 (AO 1.2) | Allow seen or implied by correct answer |
| Equilibrium equation: $\begin{pmatrix}0\\-1.5g\end{pmatrix}+\begin{pmatrix}4\\-2\end{pmatrix}+\mathbf{F}_2=\mathbf{0}$ | M1 (AO 1.1a) | FT their weight; allow sign errors |
| $\mathbf{F}_2 = \begin{pmatrix}-4\\2+1.5g\end{pmatrix} = \begin{pmatrix}-4\\16.7\end{pmatrix}$ N | A1 [3] (AO 1.1) | Must be vector; allow $\mathbf{i}$-$\mathbf{j}$ form |
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## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Newton's second law: $\begin{pmatrix}0\\-1.5g\end{pmatrix}+\begin{pmatrix}4\\-2\end{pmatrix}+\begin{pmatrix}2\\20\end{pmatrix}=m\mathbf{a}$ | M1 (AO 1.1a) | Addition of vectors; allow if weight missing but other two forces and acceleration seen in equation; allow wrong weight only if given as a vector |
| $\mathbf{a} = \frac{1}{1.5}\begin{pmatrix}6\\3.3\end{pmatrix} = \begin{pmatrix}4\\2.2\end{pmatrix}$ m s$^{-2}$ | A1 [2] (AO 1.1) | Must be vector; any correct form |
---4 In this question, the $x$ and $y$ directions are horizontal and vertically upwards respectively.\\
A particle of mass 1.5 kg is in equilibrium under the action of its weight and forces $\mathbf { F } _ { 1 } = \binom { 4 } { - 2 } \mathrm {~N}$\\
and $\mathbf { F } _ { 2 }$. and $\mathbf { F } _ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the force $\mathbf { F } _ { 2 }$.
The force $\mathbf { F } _ { 2 }$ is changed to $\binom { 2 } { 20 } \mathrm {~N}$.
\item Find the acceleration of the particle.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q4 [5]}}