| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring differentiation of position vectors to find velocity and acceleration, then using Pythagoras to find speed. The steps are routine: differentiate twice, substitute t=1.5, solve a simple equation for c, then calculate acceleration magnitude. Slightly easier than average due to mechanical application of standard techniques with no conceptual challenges. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\mathbf{v} = (18 - 12t^2)\mathbf{i} + 2ct\mathbf{j}\) | M1 A1 A1 | |
| \(t = \frac{3}{2}\): \(\mathbf{v} = -9\mathbf{i} + 3c\mathbf{j}\) | M1 | |
| \( | \mathbf{v} | = 15 \Rightarrow 9^2 + (3c)^2 = 15^2\) |
| \(\Rightarrow (3c)^2 = 144 \Rightarrow c = 4\) | A1 | |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\mathbf{a} = -24t\mathbf{i} + 8\mathbf{j}\) | M1 | |
| \(t = \frac{3}{2}\): \(\mathbf{a} = -36\mathbf{i} + 8\mathbf{j}\) | M1 | |
| A1 \(\checkmark\) | ||
| (3) |
## Question 3:
### Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $\mathbf{v} = (18 - 12t^2)\mathbf{i} + 2ct\mathbf{j}$ | M1 A1 A1 | |
| $t = \frac{3}{2}$: $\mathbf{v} = -9\mathbf{i} + 3c\mathbf{j}$ | M1 | |
| $|\mathbf{v}| = 15 \Rightarrow 9^2 + (3c)^2 = 15^2$ | M1 | |
| $\Rightarrow (3c)^2 = 144 \Rightarrow c = 4$ | A1 | |
| | **(6)** | |
### Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\mathbf{a} = -24t\mathbf{i} + 8\mathbf{j}$ | M1 | |
| $t = \frac{3}{2}$: $\mathbf{a} = -36\mathbf{i} + 8\mathbf{j}$ | M1 | |
| | A1 $\checkmark$ | |
| | **(3)** | |
---3.A particle $P$ moves in a horizontal plane.At time $t$ seconds,the position vector of $P$ is $\mathbf { r }$ metres relative to a fixed origin $O$ ,and $\mathbf { r }$ is given by
$$\mathbf { r } = \left( 18 t - 4 t ^ { 3 } \right) \mathbf { i } + c t ^ { 2 } \mathbf { j } ,$$
where $c$ is a positive constant.When $t = 1.5$ ,the speed of $P$ is $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ .Find
\begin{enumerate}[label=(\alph*)]
\item the value of $c$ ,
\item the acceleration of $P$ when $t = 1.5$ . $\mathbf { r }$ metres relative to a fixed origin $O$ ,and $\mathbf { r }$ is given by $$\begin{aligned} \mathbf { r } = \left( 18 t - 4 t ^ { 3 } \right) \mathbf { i } + c t ^ { 2 } \mathbf { j } , \\ \text { where } c \text { is a positive constant.When } t = 1.5 \text { ,the speed of } P \text { is } 15 \mathrm {~m} \mathrm {~s} ^ { - 1 } \text { .Find } \end{aligned}$$ (a)the value of $c$ ,
3.A particle $P$ moves in a horizontal plane.At time $t$ seconds,the position vector of $P$ is D啨\\
(b)the acceleration of $P$ when $t = 1.5$ .
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2005 Q3 [9]}}