| Exam Board | OCR MEI |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Rocket/thrust problems (mass decreasing) |
| Difficulty | Challenging +1.8 This is a challenging Further Mechanics question requiring derivation of the rocket equation from first principles using variable mass dynamics, followed by numerical application involving logarithms. While the rocket equation is a standard M4 topic, the derivation requires careful application of momentum principles with changing mass, and part (ii) involves non-trivial algebraic manipulation of logarithmic equations. This exceeds typical A-level difficulty but is standard for Further Maths M4. |
| Spec | 6.03a Linear momentum: p = mv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((m + \delta m)(v + \delta v) - \delta m(v - u) - mv = -mg\delta t\) | M1* | Attempt at momentum equation; 4 terms (allow one slip or sign errors) |
| NB \(\delta m < 0\) | A1 | Condone wrong sign of \(\delta m\) |
| \(m\frac{\delta v}{\delta t} + u\frac{\delta m}{\delta t} + \delta m\frac{\delta v}{\delta t} = -mg\) | M1 dep* | Simplify and divide by \(\delta t\) |
| \(m\frac{dv}{dt} + u\frac{dm}{dt} = -mg\) | A1 | Complete argument including sign of \(\delta m\) correct; if differential equation stated rather than derived then SC B3 |
| \(m = m_0 - kt\) | B1 | Seen or implied |
| \(\frac{dv}{dt} = \frac{uk}{m_0 - kt} - g\) | M1 | Substituting their \(m_0 - kt\) into their three term differential equation which contains two derivatives |
| \(v = \int\left(\frac{uk}{m_0 - kt} - g\right)dt\) | M1 | Separation of variables; must be of correct form \(v = \int\left(\frac{A}{B - Dt} - g\right)dt\) and attempt to integrate |
| \(= -u\ln(m_0 - kt) - gt (+c)\) | A1 | |
| When \(t = 0\), \(v = 0 \Rightarrow c = u\ln m_0\) | ||
| \(v = u\ln\left(\frac{m_0}{m_0 - kt}\right) - gt\) | E1 | |
| [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t = 24\) sec | B1 | |
| \(7910 = 3000\ln\left(\frac{m_0}{m_0 - 2400}\right) - 9.8 \times 24\) | M1 | Substitute into given equation and attempt to solve |
| \(m_0 = 2570.14\ldots\) so \(2570\) kg (3sf) | A1 | |
| [3] |
# Question 1:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(m + \delta m)(v + \delta v) - \delta m(v - u) - mv = -mg\delta t$ | M1* | Attempt at momentum equation; 4 terms (allow one slip or sign errors) |
| NB $\delta m < 0$ | A1 | Condone wrong sign of $\delta m$ |
| $m\frac{\delta v}{\delta t} + u\frac{\delta m}{\delta t} + \delta m\frac{\delta v}{\delta t} = -mg$ | M1 dep* | Simplify and divide by $\delta t$ |
| $m\frac{dv}{dt} + u\frac{dm}{dt} = -mg$ | A1 | Complete argument including sign of $\delta m$ correct; if differential equation stated rather than derived then SC B3 |
| $m = m_0 - kt$ | B1 | Seen or implied |
| $\frac{dv}{dt} = \frac{uk}{m_0 - kt} - g$ | M1 | Substituting their $m_0 - kt$ into their three term differential equation which contains two derivatives |
| $v = \int\left(\frac{uk}{m_0 - kt} - g\right)dt$ | M1 | Separation of variables; must be of correct form $v = \int\left(\frac{A}{B - Dt} - g\right)dt$ and attempt to integrate |
| $= -u\ln(m_0 - kt) - gt (+c)$ | A1 | |
| When $t = 0$, $v = 0 \Rightarrow c = u\ln m_0$ | | |
| $v = u\ln\left(\frac{m_0}{m_0 - kt}\right) - gt$ | E1 | |
| **[9]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t = 24$ sec | B1 | |
| $7910 = 3000\ln\left(\frac{m_0}{m_0 - 2400}\right) - 9.8 \times 24$ | M1 | Substitute into given equation and attempt to solve |
| $m_0 = 2570.14\ldots$ so $2570$ kg (3sf) | A1 | |
| **[3]** | | |
---1 A rocket is launched vertically upwards from rest. The initial mass of the rocket, including fuel and payload, is $m _ { 0 }$ and the propulsion system ejects mass at a constant mass rate $k$ with constant speed $u$ relative to the rocket. The only other force acting on the rocket is its weight. The acceleration due to gravity is constant throughout the motion.
At time $t$ after launch the speed of the rocket is $v$.\\
(i) Show that while mass is being ejected from the rocket $v = u \ln \left( \frac { m _ { 0 } } { m _ { 0 } - k t } \right) - g t$.
The rocket initially has 2400 kg of fuel which is ejected at a constant rate of $100 \mathrm {~kg} \mathrm {~s} ^ { - 1 }$ with constant speed $3000 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ relative to the rocket.\\
(ii) Given that the rocket must reach a speed of $7910 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ before releasing its payload, find the maximum possible value of $m _ { 0 }$.
\hfill \mbox{\textit{OCR MEI M4 2015 Q1 [12]}}