| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2004 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Rocket/thrust problems (mass decreasing) |
| Difficulty | Challenging +1.8 This is a challenging M5 variable mass problem requiring application of the rocket equation (momentum conservation with variable mass), integration of a logarithmic form, and careful algebraic manipulation. While the structure is guided by parts (a) and (b), it demands understanding beyond standard mechanics and involves non-trivial calculus—significantly harder than typical A-level questions but standard for Further Maths M5. |
| Spec | 3.02g Two-dimensional variable acceleration4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts6.02a Work done: concept and definition6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \((m+\delta m)(v+\delta v) + (-\delta m)(v-u) - mv = -mg\delta t\) | M1 A2 (−1 each error) | |
| \(m\frac{dv}{dt} + u\frac{dm}{dt} = -mg\) | A1 | |
| \(m = M(1-kt) \Rightarrow \frac{dm}{dt} = -kM\) | B1 | |
| \(M(1-kt)\frac{dv}{dt} + u(-kM) = -M(1-kt)g\) | M1 | |
| \(\frac{dv}{dt} = \frac{ku}{1-kt} - g\) (*) | A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = \int_0^{1/3k} \frac{ku}{1-kt} - g\; dt\) | M1 | |
| \(= \left[-u\ln(1-kt) - gt\right]_0^{1/3k}\) | A1 | |
| \(u\ln\left(\frac{3}{2}\right) - \frac{g}{3k}\) | A1 | (3 marks) |
## Question 5:
### Part (a)
$(m+\delta m)(v+\delta v) + (-\delta m)(v-u) - mv = -mg\delta t$ | M1 A2 (−1 each error) |
$m\frac{dv}{dt} + u\frac{dm}{dt} = -mg$ | A1 |
$m = M(1-kt) \Rightarrow \frac{dm}{dt} = -kM$ | B1 |
$M(1-kt)\frac{dv}{dt} + u(-kM) = -M(1-kt)g$ | M1 |
$\frac{dv}{dt} = \frac{ku}{1-kt} - g$ (*) | A1 | (7 marks)
### Part (b)
$v = \int_0^{1/3k} \frac{ku}{1-kt} - g\; dt$ | M1 |
$= \left[-u\ln(1-kt) - gt\right]_0^{1/3k}$ | A1 |
$u\ln\left(\frac{3}{2}\right) - \frac{g}{3k}$ | A1 | (3 marks)
---5. A rocket is launched vertically upwards under gravity from rest at time $t = 0$. The rocket propels itself upward by ejecting burnt fuel vertically downwards at a constant speed $u$ relative to the rocket. The initial mass of the rocket, including fuel, is $M$. At time $t$, before all the fuel has been used up, the mass of the rocket, including fuel, is $M ( 1 - k t )$ and the speed of the rocket is $v$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { k u } { 1 - k t } - g$.
\item Hence find the speed of the rocket when $t = \frac { 1 } { 3 k }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2004 Q5 [10]}}