| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | When moving parallel to given vector |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring differentiation of position to find velocity, setting velocity components proportional to find T, then differentiating again for acceleration. The algebra is routine and the method is standard textbook material, making it slightly easier than average. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Differentiate \(\mathbf{p}\) to obtain \(\mathbf{v}\) | M1 | |
| \(\mathbf{v} = (3t^2 - 9t - 24)\mathbf{i} + (-3t^2 + 6t + 12)\mathbf{j}\) | A1 | |
| Equate coefficients and obtain quadratic in \(T\): \(3T^2 - 9T - 24 = -3T^2 + 6T + 12\) | DM1 | Dependent on preceding M1 |
| \(6T^2 - 15T - 36 = 0\) | ||
| Solve for \(T\): \(3(2T+3)(T-4) = 0\) | M1 | Independent. Solve a 3 term quadratic in \(T\) |
| \(T = 4\) | A1 | |
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Differentiate \(\mathbf{v}\) to obtain \(\mathbf{a}\) | M1 | |
| \(\mathbf{a} = (6t-9)\mathbf{i} + (-6t+6)\mathbf{j}\) | A1 | |
| Use their \(T\): \(\mathbf{a} = (6T-9)\mathbf{i} + (-6T+6)\mathbf{j} = 15\mathbf{i} - 18\mathbf{j}\) | DM1 | Dependent on preceding M1 |
| Use Pythagoras: \( | \mathbf{a} | = \sqrt{15^2 + 18^2}\) |
| \(= \sqrt{549} = 23.4\) (m s\(^{-2}\)) | A1 | 23.4 or better |
| (5 marks, 10 marks total for Q4) |
# Question 4:
## Part 4a:
| Working | Mark | Notes |
|---------|------|-------|
| Differentiate $\mathbf{p}$ to obtain $\mathbf{v}$ | M1 | |
| $\mathbf{v} = (3t^2 - 9t - 24)\mathbf{i} + (-3t^2 + 6t + 12)\mathbf{j}$ | A1 | |
| Equate coefficients and obtain quadratic in $T$: $3T^2 - 9T - 24 = -3T^2 + 6T + 12$ | DM1 | Dependent on preceding M1 |
| $6T^2 - 15T - 36 = 0$ | | |
| Solve for $T$: $3(2T+3)(T-4) = 0$ | M1 | Independent. Solve a 3 term quadratic in $T$ |
| $T = 4$ | A1 | |
| **(5 marks)** | | |
## Part 4b:
| Working | Mark | Notes |
|---------|------|-------|
| Differentiate $\mathbf{v}$ to obtain $\mathbf{a}$ | M1 | |
| $\mathbf{a} = (6t-9)\mathbf{i} + (-6t+6)\mathbf{j}$ | A1 | |
| Use their $T$: $\mathbf{a} = (6T-9)\mathbf{i} + (-6T+6)\mathbf{j} = 15\mathbf{i} - 18\mathbf{j}$ | DM1 | Dependent on preceding M1 |
| Use Pythagoras: $|\mathbf{a}| = \sqrt{15^2 + 18^2}$ | M1 | |
| $= \sqrt{549} = 23.4$ (m s$^{-2}$) | A1 | 23.4 or better |
| **(5 marks, 10 marks total for Q4)** | | |
---\begin{enumerate}
\item At time $t$ seconds $( t \geqslant 0 )$, a particle $P$ has position vector $\mathbf { r }$ metres with respect to a fixed origin $O$, where
\end{enumerate}
$$\mathbf { r } = \left( t ^ { 3 } - \frac { 9 } { 2 } t ^ { 2 } - 24 t \right) \mathbf { i } + \left( - t ^ { 3 } + 3 t ^ { 2 } + 12 t \right) \mathbf { j }$$
At time $T$ seconds, $P$ is moving in a direction parallel to the vector $\mathbf { - i } - \mathbf { j }$\\
Find\\
(a) the value of $T$,\\
(b) the magnitude of the acceleration of $P$ at the instant when $t = T$.
\hfill \mbox{\textit{Edexcel M2 2016 Q4 [10]}}