| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.3 This is a straightforward M1 vector mechanics question requiring routine application of F=ma, constant acceleration equations, and basic vector operations. Part (a) is a 'show that' with the answer given, parts (b-c) are standard bookwork, and part (d) requires showing parallel vectors but with the time specified. No problem-solving insight needed, just methodical application of standard techniques. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors3.02e Two-dimensional constant acceleration: with vectors3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{F} = m\mathbf{a}\): \(3\mathbf{i} - 2\mathbf{j} = 0.5\mathbf{a}\) | M1 | Use of \(\mathbf{F} = m\mathbf{a}\) |
| \(\mathbf{a} = 6\mathbf{i} - 4\mathbf{j}\) | A1 | Correct answer |
| \( | \mathbf{a} | = \sqrt{6^2 + (-4)^2} = 2\sqrt{13}\) ms\(^{-2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\): \(\mathbf{v} = (\mathbf{i} + 3\mathbf{j}) + 2(6\mathbf{i} - 4\mathbf{j})\) | M1A1 ft | M1 for \((\mathbf{i}+3\mathbf{j}) + (2 \times \text{their } \mathbf{a})\); ft for correct expression |
| \(= 13\mathbf{i} - 5\mathbf{j}\) ms\(^{-1}\) | A1 | isw if they find speed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Distance \(= 2 | \mathbf{v} | = 2\sqrt{4+1} = 2\sqrt{5} = 4.47\) m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(t=3.5\), velocity of \(P\) is \((\mathbf{i}+3\mathbf{j})+3.5(6\mathbf{i}-4\mathbf{j}) = 22\mathbf{i}-11\mathbf{j}\) | M1A1 ft | M1 for \((\mathbf{i}+3\mathbf{j})+(3.5 \times \text{their } \mathbf{a})\) or their (b) \(+(1.5 \times \text{their } \mathbf{a})\); ft for correct expression of form \(a\mathbf{i}+b\mathbf{j}\) |
| \(22\mathbf{i}-11\mathbf{j} = 11(2\mathbf{i}-\mathbf{j})\) | A1 | Given conclusion reached correctly, oe |
# Question 5:
## Part 5a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{F} = m\mathbf{a}$: $3\mathbf{i} - 2\mathbf{j} = 0.5\mathbf{a}$ | M1 | Use of $\mathbf{F} = m\mathbf{a}$ |
| $\mathbf{a} = 6\mathbf{i} - 4\mathbf{j}$ | A1 | Correct answer |
| $|\mathbf{a}| = \sqrt{6^2 + (-4)^2} = 2\sqrt{13}$ ms$^{-2}$ | M1A1 | Allow $\sqrt{6^2 + 4^2}$; given answer |
## Part 5b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{v} = \mathbf{u} + \mathbf{a}t$: $\mathbf{v} = (\mathbf{i} + 3\mathbf{j}) + 2(6\mathbf{i} - 4\mathbf{j})$ | M1A1 ft | M1 for $(\mathbf{i}+3\mathbf{j}) + (2 \times \text{their } \mathbf{a})$; ft for correct expression |
| $= 13\mathbf{i} - 5\mathbf{j}$ ms$^{-1}$ | A1 | isw if they find speed |
## Part 5c:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Distance $= 2|\mathbf{v}| = 2\sqrt{4+1} = 2\sqrt{5} = 4.47$ m | M1A1 | M1 for $2\sqrt{2^2+(-1)^2}$ or $\sqrt{4^2+(-2)^2}$; A1 for $2\sqrt{5}$ or $\sqrt{20}$ or 4.5 or 4.47 or better |
## Part 5d:
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $t=3.5$, velocity of $P$ is $(\mathbf{i}+3\mathbf{j})+3.5(6\mathbf{i}-4\mathbf{j}) = 22\mathbf{i}-11\mathbf{j}$ | M1A1 ft | M1 for $(\mathbf{i}+3\mathbf{j})+(3.5 \times \text{their } \mathbf{a})$ or their (b) $+(1.5 \times \text{their } \mathbf{a})$; ft for correct expression of form $a\mathbf{i}+b\mathbf{j}$ |
| $22\mathbf{i}-11\mathbf{j} = 11(2\mathbf{i}-\mathbf{j})$ | A1 | Given conclusion reached correctly, oe |
---5. A particle $P$ of mass 0.5 kg is moving under the action of a single force $( 3 \mathbf { i } - 2 \mathbf { j } ) \mathrm { N }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the acceleration of $P$ is $2 \sqrt { 13 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
At time $t = 0$, the velocity of $P$ is $( \mathbf { i } + 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\item Find the velocity of $P$ at time $t = 2$ seconds.
Another particle $Q$ moves with constant velocity $\mathbf { v } = ( 2 \mathbf { i } - \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\item Find the distance moved by $Q$ in 2 seconds.
\item Show that at time $t = 3.5$ seconds both particles are moving in the same direction.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q5 [12]}}