| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Linear combination of vectors |
| Difficulty | Easy -1.3 This is a straightforward vector addition question requiring only basic arithmetic with components, followed by standard interpretation of the zero vector in two physical contexts. The calculation is routine and the interpretations (equilibrium for forces, returning to start for displacement) are standard textbook applications with no problem-solving required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement3.03a Force: vector nature and diagrams |
| Answer | Marks |
|---|---|
| M1 | \(p + q = 28\mathbf{i} - 3.5\mathbf{j}\) |
| M1 | \(28\mathbf{i} - 3.5\mathbf{j} = k(8\mathbf{i} - \mathbf{j})\) |
| B1 | Or equivalent. \(k\) may be implied by going straight to 3.5 |
| A1 | \(k = 3.5\) (So they are parallel) |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | \(p + q = 28\mathbf{i} - 3.5\mathbf{j}\) | |
| M1 | \(p + q: \tan\theta = \frac{-3.5}{28} \Rightarrow \theta = -7.13°\) | |
| M1 | \(8\mathbf{i} - \mathbf{j}: \tan\theta = \frac{-1}{8} \Rightarrow \theta = -7.13°\) | Comparing the ratio of the components in each of the two vectors is sufficient, using any consistent sign convention. The angle does not need to be worked out, nor does \(\tan\) have to be seen. |
| A1 | So they are parallel | Both ratios the same and correct |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | \(3p + 10q = (36+160)\mathbf{i} + (-15 + 15)\mathbf{j}\) | |
| B1 | \(= 196\mathbf{i}\) | Or equivalent explanation. May be shown on a diagram |
| B1 | Zero \(\mathbf{j}\) component so horizontal |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | The horizontal component must be zero | |
| B1 | \(12k + 3 \times 16 = 0 \Rightarrow k = -4\) | Substituting \(k = -4\) and showing \(\mathbf{i}\) component is zero is acceptable |
| B1 | \(w = -24.5\mathbf{j}\) | Award for 24.5 seen |
| B1 | \(mg = 24.5 \Rightarrow m = 2.5\) | Award for 2.5 seen. FT from their weight |
| B1 | The mass is 2.5 kg |
# Question 1
## (i)
M1 | $p + q = 28\mathbf{i} - 3.5\mathbf{j}$
M1 | $28\mathbf{i} - 3.5\mathbf{j} = k(8\mathbf{i} - \mathbf{j})$
B1 | Or equivalent. $k$ may be implied by going straight to 3.5
A1 | $k = 3.5$ (So they are parallel)
**Alternative**
B1 | $p + q = 28\mathbf{i} - 3.5\mathbf{j}$
M1 | $p + q: \tan\theta = \frac{-3.5}{28} \Rightarrow \theta = -7.13°$
M1 | $8\mathbf{i} - \mathbf{j}: \tan\theta = \frac{-1}{8} \Rightarrow \theta = -7.13°$ | Comparing the ratio of the components in each of the two vectors is sufficient, using any consistent sign convention. The angle does not need to be worked out, nor does $\tan$ have to be seen.
A1 | So they are parallel | Both ratios the same and correct
[3]
## (ii)
B1 | $3p + 10q = (36+160)\mathbf{i} + (-15 + 15)\mathbf{j}$
B1 | $= 196\mathbf{i}$ | Or equivalent explanation. May be shown on a diagram
B1 | Zero $\mathbf{j}$ component so horizontal
[2]
## (iii)
B1 | The horizontal component must be zero
B1 | $12k + 3 \times 16 = 0 \Rightarrow k = -4$ | Substituting $k = -4$ and showing $\mathbf{i}$ component is zero is acceptable
B1 | $w = -24.5\mathbf{j}$ | Award for 24.5 seen
B1 | $mg = 24.5 \Rightarrow m = 2.5$ | Award for 2.5 seen. FT from their weight
B1 | The mass is 2.5 kg
[3]1 The vectors $\mathbf { P } , \mathbf { Q }$ and $\mathbf { R }$ are given by
$$\mathbf { P } = 5 \mathbf { i } + 4 \mathbf { j } , \quad \mathbf { Q } = 3 \mathbf { i } - 5 \mathbf { j } , \quad \mathbf { R } = - 8 \mathbf { i } + \mathbf { j }$$
\begin{enumerate}[label=(\roman*)]
\item Find the vector $\mathbf { P } + \mathbf { Q } + \mathbf { R }$.
\item Interpret your answer to part (i) in the cases\\
(A) $\mathbf { P } , \mathbf { Q }$ and $\mathbf { R }$ represent three forces acting on a particle,\\
(B) $\mathbf { P } , \mathbf { Q }$ and $\mathbf { R }$ represent three stages of a hiker's walk.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 Q1 [3]}}