Edexcel M2 2011 January — Question 6 12 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2011
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeVector form projectile motion
DifficultyModerate -0.3 This is a standard M2 projectile motion question using vector notation. Part (a) requires applying SUVAT equations with constant acceleration (routine derivation), parts (b-c) involve basic substitution and differentiation, and parts (d-e) require solving for when velocity components satisfy tan(45°)=1. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average for A-level.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model

  1. \hspace{0pt} [In this question, the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are in a vertical plane, \(\mathbf { i }\) being horizontal and \(\mathbf { j }\) being vertically upwards.]
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c5760fa5-3c7f-4e29-87a2-b3b4145b9361-12_689_1042_360_459} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} At time \(t = 0\), a particle \(P\) is projected from the point \(A\) which has position vector 10j metres with respect to a fixed origin \(O\) at ground level. The ground is horizontal. The velocity of projection of \(P\) is \(( 3 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\), as shown in Figure 3. The particle moves freely under gravity and reaches the ground after \(T\) seconds.
  1. For \(0 \leqslant t \leqslant T\), show that, with respect to \(O\), the position vector, \(\mathbf { r }\) metres, of \(P\) at time \(t\) seconds is given by $$\mathbf { r } = 3 t \mathbf { i } + \left( 10 + 5 t - 4.9 t ^ { 2 } \right) \mathbf { j }$$
  2. Find the value of \(T\).
  3. Find the velocity of \(P\) at time \(t\) seconds \(( 0 \leqslant t \leqslant T )\). When \(P\) is at the point \(B\), the direction of motion of \(P\) is \(45 ^ { \circ }\) below the horizontal.
  4. Find the time taken for \(P\) to move from \(A\) to \(B\).
  5. Find the speed of \(P\) as it passes through \(B\).

Question 6:
Part (a):
AnswerMarks Guidance
AnswerMarks Guidance
Using \(s = ut + \frac{1}{2}at^2\)M1 Method must be clear
\(\mathbf{r} = (3t)\mathbf{i} + (10 + 5t - 4.9t^2)\mathbf{j}\)A1 A1 Answer given
Part (b):
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{j}\) component \(= 0\): \(10 + 5t - 4.9t^2\)M1
quadratic formula: \(t = \frac{5 \pm \sqrt{25 + 196}}{9.8} = \frac{5 \pm \sqrt{221}}{9.8}\)DM1
\(T = 2.03\)(s), \(2.0\) (s), positive solution onlyA1
Part (c):
AnswerMarks Guidance
AnswerMarks Guidance
Differentiating the position vector (or working from first principles)M1
\(\mathbf{v} = 3\mathbf{i} + (5 - 9.8t)\mathbf{j}\) (ms\(^{-1}\))A1
Part (d):
AnswerMarks Guidance
AnswerMarks Guidance
At \(B\) the \(\mathbf{j}\) component of velocity is the negative of the \(\mathbf{i}\) component: \(5 - 9.8t = -3\)M1
\(8 = 9.8t\)A1
\(t = 0.82\)
Part (e):
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{v} = 3\mathbf{i} - 3\mathbf{j}\), speed \(= \sqrt{3^2 + 3^2} = \sqrt{18} = 4.24\) (m s\(^{-1}\))M1A1
# Question 6:

## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $s = ut + \frac{1}{2}at^2$ | M1 | Method must be clear |
| $\mathbf{r} = (3t)\mathbf{i} + (10 + 5t - 4.9t^2)\mathbf{j}$ | A1 A1 | Answer given |

## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{j}$ component $= 0$: $10 + 5t - 4.9t^2$ | M1 | |
| quadratic formula: $t = \frac{5 \pm \sqrt{25 + 196}}{9.8} = \frac{5 \pm \sqrt{221}}{9.8}$ | DM1 | |
| $T = 2.03$(s), $2.0$ (s), positive solution only | A1 | |

## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiating the position vector (or working from first principles) | M1 | |
| $\mathbf{v} = 3\mathbf{i} + (5 - 9.8t)\mathbf{j}$ (ms$^{-1}$) | A1 | |

## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $B$ the $\mathbf{j}$ component of velocity is the negative of the $\mathbf{i}$ component: $5 - 9.8t = -3$ | M1 | |
| $8 = 9.8t$ | A1 | |
| $t = 0.82$ | | |

## Part (e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{v} = 3\mathbf{i} - 3\mathbf{j}$, speed $= \sqrt{3^2 + 3^2} = \sqrt{18} = 4.24$ (m s$^{-1}$) | M1A1 | |

---
\begin{enumerate}
  \item \hspace{0pt} [In this question, the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in a vertical plane, $\mathbf { i }$ being horizontal and $\mathbf { j }$ being vertically upwards.]
\end{enumerate}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{c5760fa5-3c7f-4e29-87a2-b3b4145b9361-12_689_1042_360_459}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

At time $t = 0$, a particle $P$ is projected from the point $A$ which has position vector 10j metres with respect to a fixed origin $O$ at ground level. The ground is horizontal. The velocity of projection of $P$ is $( 3 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, as shown in Figure 3. The particle moves freely under gravity and reaches the ground after $T$ seconds.\\
(a) For $0 \leqslant t \leqslant T$, show that, with respect to $O$, the position vector, $\mathbf { r }$ metres, of $P$ at time $t$ seconds is given by

$$\mathbf { r } = 3 t \mathbf { i } + \left( 10 + 5 t - 4.9 t ^ { 2 } \right) \mathbf { j }$$

(b) Find the value of $T$.\\
(c) Find the velocity of $P$ at time $t$ seconds $( 0 \leqslant t \leqslant T )$.

When $P$ is at the point $B$, the direction of motion of $P$ is $45 ^ { \circ }$ below the horizontal.\\
(d) Find the time taken for $P$ to move from $A$ to $B$.\\
(e) Find the speed of $P$ as it passes through $B$.

\hfill \mbox{\textit{Edexcel M2 2011 Q6 [12]}}