| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | October |
| Marks | 9 |
| 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 mechanics question requiring application of Newton's second law in vector form (F=ma) to find constants, then using constant acceleration equations. The steps are routine: sum forces in i and j components, equate to mass times acceleration, solve simultaneous equations, then apply v=u+at. No novel problem-solving or geometric 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 notation3.03d Newton's second law: 2D vectors3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((-10\mathbf{i}+a\mathbf{j})+(b\mathbf{i}-5\mathbf{j})+(2a\mathbf{i}+7\mathbf{j}) = 3(3\mathbf{i}+4\mathbf{j})\) | M1 | Apply \(\mathbf{F}=m\mathbf{a}\); need all terms; allow \(m\) instead of 3 |
| \(a - 5 + 7 = 12 \Rightarrow a = 10\) | M1 A1 | Equating coefficients of \(\mathbf{j}\); M0 if 0 instead of \(m\mathbf{a}\) |
| \(-10 + b + 2a = 9 \Rightarrow b = -1\) | M1 A1 (5) | Equating coefficients of \(\mathbf{i}\); M0 if 0 instead of \(m\mathbf{a}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(20\mathbf{i}+20\mathbf{j} = \mathbf{u}+4(3\mathbf{i}+4\mathbf{j})\) | M1 | Apply \(\mathbf{v}=\mathbf{u}+t\mathbf{a}\); need all terms; must be vector \(\mathbf{u}\) |
| \(\mathbf{u} = (8\mathbf{i}+4\mathbf{j})\) | A1 | |
| \(u = \sqrt{8^2+4^2} = \sqrt{80} = 8.9\) (or better) | M1 A1 (4) | Independent M1 for finding magnitude; A1 for \(\sqrt{80}\) or 8.9 or better |
## Question 2:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|---------|
| $(-10\mathbf{i}+a\mathbf{j})+(b\mathbf{i}-5\mathbf{j})+(2a\mathbf{i}+7\mathbf{j}) = 3(3\mathbf{i}+4\mathbf{j})$ | M1 | Apply $\mathbf{F}=m\mathbf{a}$; need all terms; allow $m$ instead of 3 |
| $a - 5 + 7 = 12 \Rightarrow a = 10$ | M1 A1 | Equating coefficients of $\mathbf{j}$; M0 if **0** instead of $m\mathbf{a}$ |
| $-10 + b + 2a = 9 \Rightarrow b = -1$ | M1 A1 (5) | Equating coefficients of $\mathbf{i}$; M0 if **0** instead of $m\mathbf{a}$ |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|---------|
| $20\mathbf{i}+20\mathbf{j} = \mathbf{u}+4(3\mathbf{i}+4\mathbf{j})$ | M1 | Apply $\mathbf{v}=\mathbf{u}+t\mathbf{a}$; need all terms; must be vector $\mathbf{u}$ |
| $\mathbf{u} = (8\mathbf{i}+4\mathbf{j})$ | A1 | |
| $u = \sqrt{8^2+4^2} = \sqrt{80} = 8.9$ (or better) | M1 A1 (4) | Independent M1 for finding magnitude; A1 for $\sqrt{80}$ or 8.9 or better |
---2. [In this question $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane.]
Three forces, $( - 10 \mathbf { i } + a \mathbf { j } ) \mathrm { N } , ( b \mathbf { i } - 5 \mathbf { j } ) \mathrm { N }$ and $( 2 a \mathbf { i } + 7 \mathbf { j } ) \mathrm { N }$, where $a$ and $b$ are constants, act on a particle $P$ of mass 3 kg . The acceleration of $P$ is $( 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ and the value of $b$.
At time $t = 0$ seconds the speed of $P$ is $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and at time $t = 4$ seconds the velocity of $P$ is $( 20 \mathbf { i } + 20 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$
\item Find the value of $u$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2016 Q2 [9]}}