| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of two vector forces (direction/magnitude conditions) |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question requiring basic vector addition and Newton's second law. Part (a) uses the parallel vector condition (ratio of components equal), which is a standard technique. Part (b) applies F=ma to find acceleration, then uses constant acceleration to find velocity and speed. All steps are routine with no problem-solving insight required, making it easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((4\mathbf{i} - 2\mathbf{j}) + (2\mathbf{i} + q\mathbf{j}) = (6\mathbf{i} + (q-2)\mathbf{j})\) | M1A1 | First M1 for adding forces; First A1 for \((6\mathbf{i} + (q-2)\mathbf{j})\) seen or implied |
| \(6 = 2(q-2)\) — ratio 2:1 | DM1 | Dependent on first M1, for using 'parallel to \((2\mathbf{i}+\mathbf{j})\)' to obtain equation in \(q\) only |
| \(q = 5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(6\mathbf{i} + 3\mathbf{j} = 1.5\mathbf{a}\) | M1 | First M1 for resultant force \(= 1.5\mathbf{a}\) |
| \(\mathbf{a} = (4\mathbf{i} + 2\mathbf{j})\ \text{m s}^{-2}\) | A1 | |
| \(\mathbf{v} = \mathbf{u} + \mathbf{a}t = (-2\mathbf{i}+4\mathbf{j}) + 2(4\mathbf{i}+2\mathbf{j})\) | M1 | Second M1 for \((-2\mathbf{i}+4\mathbf{j}) + 2\times(\text{their } \mathbf{a})\); M0 if force used instead of \(\mathbf{a}\) |
| \(= 6\mathbf{i} + 8\mathbf{j}\) | A1ft | Follow through on their velocity at \(t=2\) |
| \(\text{speed} = \sqrt{6^2 + 8^2}\) | M1 | Third M1 for finding magnitude of their velocity at \(t=2\) |
| \(= 10\ \text{m s}^{-1}\) | A1 |
# Question 2:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $(4\mathbf{i} - 2\mathbf{j}) + (2\mathbf{i} + q\mathbf{j}) = (6\mathbf{i} + (q-2)\mathbf{j})$ | M1A1 | First M1 for adding forces; First A1 for $(6\mathbf{i} + (q-2)\mathbf{j})$ seen or implied |
| $6 = 2(q-2)$ — ratio 2:1 | DM1 | Dependent on first M1, for using 'parallel to $(2\mathbf{i}+\mathbf{j})$' to obtain equation in $q$ only |
| $q = 5$ | A1 | |
## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $6\mathbf{i} + 3\mathbf{j} = 1.5\mathbf{a}$ | M1 | First M1 for resultant force $= 1.5\mathbf{a}$ |
| $\mathbf{a} = (4\mathbf{i} + 2\mathbf{j})\ \text{m s}^{-2}$ | A1 | |
| $\mathbf{v} = \mathbf{u} + \mathbf{a}t = (-2\mathbf{i}+4\mathbf{j}) + 2(4\mathbf{i}+2\mathbf{j})$ | M1 | Second M1 for $(-2\mathbf{i}+4\mathbf{j}) + 2\times(\text{their } \mathbf{a})$; M0 if force used instead of $\mathbf{a}$ |
| $= 6\mathbf{i} + 8\mathbf{j}$ | A1ft | Follow through on their velocity at $t=2$ |
| $\text{speed} = \sqrt{6^2 + 8^2}$ | M1 | Third M1 for finding magnitude of their velocity at $t=2$ |
| $= 10\ \text{m s}^{-1}$ | A1 | |
**N.B. In (b), if scalars used throughout: M0A0M0A0M0A0**
---\begin{enumerate}
\item Two forces $( 4 \mathbf { i } - 2 \mathbf { j } ) \mathrm { N }$ and $( 2 \mathbf { i } + q \mathbf { j } ) \mathrm { N }$ act on a particle $P$ of mass 1.5 kg . The resultant of these two forces is parallel to the vector $( 2 \mathbf { i } + \mathbf { j } )$.\\
(a) Find the value of $q$.
\end{enumerate}
At time $t = 0 , P$ is moving with velocity $( - 2 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
(b) Find the speed of $P$ at time $t = 2$ seconds.\\
\hfill \mbox{\textit{Edexcel M1 2014 Q2 [10]}}