Edexcel M1 2008 January — Question 2 8 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2008
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSUVAT in 2D & Gravity
TypeFree fall: time or distance
DifficultyModerate -0.8 This is a straightforward two-stage SUVAT problem with standard bookwork calculations. Part (a) uses s=ut+½at² with u=0, part (b) uses v=u+at, and part (c) requires applying SUVAT again for the free-fall phase. All steps are routine applications of memorized formulae with no problem-solving insight required, making it easier than average but not trivial due to the two-stage nature.
Spec3.02a Kinematics language: position, displacement, velocity, acceleration3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form

2. A firework rocket starts from rest at ground level and moves vertically. In the first 3 s of its motion, the rocket rises 27 m . The rocket is modelled as a particle moving with constant acceleration \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Find
  1. the value of \(a\),
  2. the speed of the rocket 3 s after it has left the ground. After 3 s , the rocket burns out. The motion of the rocket is now modelled as that of a particle moving freely under gravity.
  3. Find the height of the rocket above the ground 5 s after it has left the ground.

Question 2:
Part (a)
AnswerMarks Guidance
WorkingMarks Notes
\(27 = 0 + \frac{1}{2}a \cdot 3^2 \Rightarrow a = 6\)M1 A1 (2)
Part (b)
AnswerMarks Guidance
WorkingMarks Notes
\(v = 6 \times 3 = 18 \text{ ms}^{-1}\)M1 A1 f.t. (2)
Part (c)
AnswerMarks Guidance
WorkingMarks Notes
From \(t=3\) to \(t=5\): \(s = 18 \times 2 - \frac{1}{2} \times 9.8 \times 2^2\)M1 A1 f.t.
Total ht. \(= s + 27 = 43.4\text{ m},\ 43\text{ m}\)M1 A1 (4) Total: 8
## Question 2:

### Part (a)
| Working | Marks | Notes |
|---------|-------|-------|
| $27 = 0 + \frac{1}{2}a \cdot 3^2 \Rightarrow a = 6$ | M1 A1 (2) | |

### Part (b)
| Working | Marks | Notes |
|---------|-------|-------|
| $v = 6 \times 3 = 18 \text{ ms}^{-1}$ | M1 A1 f.t. (2) | |

### Part (c)
| Working | Marks | Notes |
|---------|-------|-------|
| From $t=3$ to $t=5$: $s = 18 \times 2 - \frac{1}{2} \times 9.8 \times 2^2$ | M1 A1 f.t. | |
| Total ht. $= s + 27 = 43.4\text{ m},\ 43\text{ m}$ | M1 A1 (4) | **Total: 8** |

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2. A firework rocket starts from rest at ground level and moves vertically. In the first 3 s of its motion, the rocket rises 27 m . The rocket is modelled as a particle moving with constant acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find
\begin{enumerate}[label=(\alph*)]
\item the value of $a$,
\item the speed of the rocket 3 s after it has left the ground.

After 3 s , the rocket burns out. The motion of the rocket is now modelled as that of a particle moving freely under gravity.
\item Find the height of the rocket above the ground 5 s after it has left the ground.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2008 Q2 [8]}}