| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of two vector forces (direction/magnitude conditions) |
| Difficulty | Standard +0.3 This is a straightforward M1 mechanics question requiring basic vector addition, understanding of parallel vectors, and Newton's second law. Part (a) is simple trigonometry, part (b) requires equating components of parallel vectors (standard technique), and part (c) applies F=ma with magnitude calculation. All steps are routine applications of standard methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03n Equilibrium in 2D: particle under forces3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\tan\theta = \frac{2}{1} \Rightarrow \theta = 63.4°\) | M1 A1 | |
| angle is \(153.4°\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((4+p)\mathbf{i} + (q-5)\mathbf{j}\) | B1 | |
| \((q-5) = -2(4+p)\) | M1 A1 | |
| \(2p + q + 3 = 0\ *\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(q = 1 \Rightarrow p = -2\) | B1 | |
| \(\Rightarrow \mathbf{R} = 2\mathbf{i} - 4\mathbf{j}\) | M1 | |
| \(\Rightarrow | \mathbf{R} | = \sqrt{2^2 + (-4)^2} = \sqrt{20}\) |
| \(\sqrt{20} = m \cdot 8\sqrt{5}\) | M1 A1 f.t. | |
| \(\Rightarrow m = \frac{1}{4}\) | A1 cao | (7) [14] |
## Question 6:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{2}{1} \Rightarrow \theta = 63.4°$ | M1 A1 | |
| angle is $153.4°$ | A1 | **(3)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(4+p)\mathbf{i} + (q-5)\mathbf{j}$ | B1 | |
| $(q-5) = -2(4+p)$ | M1 A1 | |
| $2p + q + 3 = 0\ *$ | A1 | **(4)** |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $q = 1 \Rightarrow p = -2$ | B1 | |
| $\Rightarrow \mathbf{R} = 2\mathbf{i} - 4\mathbf{j}$ | M1 | |
| $\Rightarrow |\mathbf{R}| = \sqrt{2^2 + (-4)^2} = \sqrt{20}$ | M1 A1 f.t. | |
| $\sqrt{20} = m \cdot 8\sqrt{5}$ | M1 A1 f.t. | |
| $\Rightarrow m = \frac{1}{4}$ | A1 cao | **(7) [14]** |
---6. Two forces, $( 4 \mathbf { i } - 5 \mathbf { j } ) \mathrm { N }$ and $( p \mathbf { i } + q \mathbf { j } ) \mathrm { N }$, act on a particle $P$ of mass $m \mathrm {~kg}$. The resultant of the two forces is $\mathbf { R }$. Given that $\mathbf { R }$ acts in a direction which is parallel to the vector ( $\mathbf { i } - 2 \mathbf { j }$ ),
\begin{enumerate}[label=(\alph*)]
\item find the angle between $\mathbf { R }$ and the vector $\mathbf { j }$,
\item show that $2 p + q + 3 = 0$.
Given also that $q = 1$ and that $P$ moves with an acceleration of magnitude $8 \sqrt { } 5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, (c) find the value of $m$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2009 Q6 [14]}}