Edexcel M2 2015 January — Question 3 12 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2015
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (vectors)
TypeDistance between two positions
DifficultyStandard +0.3 This is a straightforward mechanics question requiring differentiation of position vectors to find velocity and acceleration, then applying the parallel condition. All steps are routine M2 techniques with no novel problem-solving required. The 'show that' part guides students to the answer, and parts (b)-(d) are direct applications of standard formulas. Slightly above average only due to the multi-part nature and vector component manipulation.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration
  1. At time \(t\) seconds \(( t \geqslant 0 )\) a particle \(P\) has position vector \(\mathbf { r }\) metres, with respect to a fixed origin \(O\), where
$$\mathbf { r } = \left( \frac { 1 } { 8 } t ^ { 4 } - 2 \lambda t ^ { 2 } + 5 \right) \mathbf { i } + \left( 5 t ^ { 2 } - \lambda t \right) \mathbf { j }$$ and \(\lambda\) is a constant. When \(t = 4 , P\) is moving parallel to the vector \(\mathbf { j }\).
  1. Show that \(\lambda = 2\)
  2. Find the speed of \(P\) when \(t = 4\)
  3. Find the acceleration of \(P\) when \(t = 4\) When \(t = 0 , P\) is at the point \(A\). When \(t = 4 , P\) is at the point \(B\).
  4. Find the distance \(A B\).
Question 3:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{v} = \left(\frac{1}{2}t^3 - 4\lambda t\right)\mathbf{i} + (10t - \lambda)\mathbf{j}\)M1 Differentiate \(\mathbf{r}\); attempt seen for one or both components (at least one power going down)
A1\(\mathbf{i}\) component correct
A1\(\mathbf{j}\) component correct; allow if M1 earned here but \(\mathbf{j}\) component not seen in (a) but then seen correct in (b)
\(\frac{1}{2}t^3 - 4\lambda t = 0\) when \(t=4\): \(\frac{64}{2} - 16\lambda = 0\)DM1 Dependent on first M1; set \(\mathbf{i}\) component of their \(\mathbf{v}\) equal to zero; allow with no \(\mathbf{j}\) component or incorrect \(\mathbf{j}\) component
\(\lambda = 2\)A1 Given answer; allow with no \(\mathbf{j}\) component or incorrect \(\mathbf{j}\) component
[5]
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(t=4,\ \lambda=2\): speed \(= 38\) (m s\(^{-1}\))B1 Must be a scalar, not a vector
[1]
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mathbf{a} = \left(\frac{3t^2}{2} - 4\times2\right)\mathbf{i} + 10\mathbf{j}\)M1 Differentiate \(\mathbf{v}\)
\(= 16\mathbf{i} + 10\mathbf{j}\) ISWA1 CSO
[2]
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(t=0\): \(\mathbf{r} = 5\mathbf{i}\)B1
\(t=4\): \(\mathbf{r} = -27\mathbf{i} + 72\mathbf{j}\)B1
Distance \(= \sqrt{32^2 + 72^2}\) (m)M1 Use Pythagoras to find \(
\(= 8\sqrt{97} = 78.8\) (m)A1 78.7908..., \(8\sqrt{97}\)
[4] 
## Question 3:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = \left(\frac{1}{2}t^3 - 4\lambda t\right)\mathbf{i} + (10t - \lambda)\mathbf{j}$ | M1 | Differentiate $\mathbf{r}$; attempt seen for one or both components (at least one power going down) |
| | A1 | $\mathbf{i}$ component correct |
| | A1 | $\mathbf{j}$ component correct; allow if M1 earned here but $\mathbf{j}$ component not seen in (a) but then seen correct in (b) |
| $\frac{1}{2}t^3 - 4\lambda t = 0$ when $t=4$: $\frac{64}{2} - 16\lambda = 0$ | DM1 | Dependent on first M1; set $\mathbf{i}$ component of their $\mathbf{v}$ equal to zero; allow with no $\mathbf{j}$ component or incorrect $\mathbf{j}$ component |
| $\lambda = 2$ | A1 | **Given answer**; allow with no $\mathbf{j}$ component or incorrect $\mathbf{j}$ component |
| **[5]** | | |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=4,\ \lambda=2$: speed $= 38$ (m s$^{-1}$) | B1 | Must be a scalar, not a vector |
| **[1]** | | |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{a} = \left(\frac{3t^2}{2} - 4\times2\right)\mathbf{i} + 10\mathbf{j}$ | M1 | Differentiate $\mathbf{v}$ |
| $= 16\mathbf{i} + 10\mathbf{j}$ ISW | A1 | CSO |
| **[2]** | | |

### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0$: $\mathbf{r} = 5\mathbf{i}$ | B1 | |
| $t=4$: $\mathbf{r} = -27\mathbf{i} + 72\mathbf{j}$ | B1 | |
| Distance $= \sqrt{32^2 + 72^2}$ (m) | M1 | Use Pythagoras to find $|\mathbf{r}_4 - \mathbf{r}_0|$ for $\mathbf{r}_0\neq\mathbf{0}$, $\mathbf{r}_4\neq\mathbf{0}$ |
| $= 8\sqrt{97} = 78.8$ (m) | A1 | 78.7908..., $8\sqrt{97}$ |
| **[4]** | | |

---
\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}

$$\mathbf { r } = \left( \frac { 1 } { 8 } t ^ { 4 } - 2 \lambda t ^ { 2 } + 5 \right) \mathbf { i } + \left( 5 t ^ { 2 } - \lambda t \right) \mathbf { j }$$

and $\lambda$ is a constant.

When $t = 4 , P$ is moving parallel to the vector $\mathbf { j }$.\\
(a) Show that $\lambda = 2$\\
(b) Find the speed of $P$ when $t = 4$\\
(c) Find the acceleration of $P$ when $t = 4$

When $t = 0 , P$ is at the point $A$. When $t = 4 , P$ is at the point $B$.\\
(d) Find the distance $A B$.\\

\hfill \mbox{\textit{Edexcel M2 2015 Q3 [12]}}
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