OCR M2 2007 June — Question 3 8 marks

Exam BoardOCR
ModuleM2 (Mechanics 2)
Year2007
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeEngine power on road constant/variable speed
DifficultyStandard +0.3 This is a straightforward application of the work-energy principle and power formula (P = Fv). Part (i) uses work done = change in KE with constant power, requiring simple algebra. Part (ii) applies P = Fv then F = ma. Both parts are standard M2 techniques with no resistance forces to complicate the problem, making it slightly easier than average.
Spec6.02k Power: rate of doing work6.02l Power and velocity: P = Fv

3 A rocket of mass 250 kg is moving in a straight line in space. There is no resistance to motion, and the mass of the rocket is assumed to be constant. With its motor working at a constant rate of 450 kW the rocket's speed increases from \(100 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) to \(150 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a time \(t\) seconds.
  1. Calculate the value of \(t\).
  2. Calculate the acceleration of the rocket at the instant when its speed is \(120 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

Question 3(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(WD = \frac{1}{2}\times250\times150^2 - \frac{1}{2}\times250\times100^2\)M1
\(1\,560\,000\)A1 \(1\,562\,500\)
\(450\,000 = 1\,560\,000/t\)M1
\(3.47\)A1 4
Question 3(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(F = 450\,000/120\)M1
\(3750\)A1
\(3750 = 250a\)M1
\(15\ \text{ms}^{-2}\)A1 4
## Question 3(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $WD = \frac{1}{2}\times250\times150^2 - \frac{1}{2}\times250\times100^2$ | M1 | |
| $1\,560\,000$ | A1 | $1\,562\,500$ |
| $450\,000 = 1\,560\,000/t$ | M1 | |
| $3.47$ | A1 **4** | |

## Question 3(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 450\,000/120$ | M1 | |
| $3750$ | A1 | |
| $3750 = 250a$ | M1 | |
| $15\ \text{ms}^{-2}$ | A1 **4** | |

---
3 A rocket of mass 250 kg is moving in a straight line in space. There is no resistance to motion, and the mass of the rocket is assumed to be constant. With its motor working at a constant rate of 450 kW the rocket's speed increases from $100 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $150 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a time $t$ seconds.\\
(i) Calculate the value of $t$.\\
(ii) Calculate the acceleration of the rocket at the instant when its speed is $120 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

\hfill \mbox{\textit{OCR M2 2007 Q3 [8]}}