| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Rocket/thrust problems (mass decreasing) |
| Difficulty | Challenging +1.8 This is a variable mass rocket problem from M5 requiring application of the rocket equation (momentum conservation with variable mass), integration, and interpretation of physical constraints. While the setup is standard for M5, it requires careful handling of the mass function, differentiation, and the multi-step reasoning to connect the constraint that the rocket moves upwards to the inequality for U. This is more challenging than typical M5 questions but follows a recognizable pattern for this topic. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(-mg\,\delta t = (m+\delta m)(v+\delta v) + \delta m(U-v) - mv\) | M1 A2 | |
| \(-mg = m\dfrac{dv}{dt} + U\dfrac{dm}{dt}\) | A1 | |
| \(m = M_0\!\left(1 - \frac{1}{2}t\right) \Rightarrow \dfrac{dm}{dt} = -\frac{1}{2}M_0\) | B1 | |
| \(-M_0 g\!\left(1-\frac{1}{2}t\right) = M_0\!\left(1-\frac{1}{2}t\right)\dfrac{dv}{dt} - \frac{1}{2}M_0 U\) | M1 | |
| \(\dfrac{U}{(2-t)} - 9.8 = \dfrac{dv}{dt}\) | A1 | *(7)* |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\dfrac{dv}{dt} > 0\) when \(t=0 \Rightarrow \dfrac{U}{2} - 9.8 > 0\) | M1 | |
| \(\Rightarrow U > 19.6\) | A1 | *(2)* |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(v = \displaystyle\int \frac{U}{(2-t)} - 9.8\; dt\) | M1 | |
| \(= -U\ln(2-t) - 9.8t + C\) | A1 | |
| \(t=0, v=0\): \(0 = -U\ln 2 + C \Rightarrow C = U\ln 2\) | M1 | |
| \(v = U\ln\dfrac{2}{(2-t)} - 9.8t\) | ||
| \(t=1\): \(v = U\ln 2 - 9.8\) | M1 A1 | Total: 14 |
## Question 4:
### Part (a):
| Working | Marks | Notes |
|---|---|---|
| $-mg\,\delta t = (m+\delta m)(v+\delta v) + \delta m(U-v) - mv$ | M1 A2 | |
| $-mg = m\dfrac{dv}{dt} + U\dfrac{dm}{dt}$ | A1 | |
| $m = M_0\!\left(1 - \frac{1}{2}t\right) \Rightarrow \dfrac{dm}{dt} = -\frac{1}{2}M_0$ | B1 | |
| $-M_0 g\!\left(1-\frac{1}{2}t\right) = M_0\!\left(1-\frac{1}{2}t\right)\dfrac{dv}{dt} - \frac{1}{2}M_0 U$ | M1 | |
| $\dfrac{U}{(2-t)} - 9.8 = \dfrac{dv}{dt}$ | A1 | ***(7)*** |
### Part (b):
| Working | Marks | Notes |
|---|---|---|
| $\dfrac{dv}{dt} > 0$ when $t=0 \Rightarrow \dfrac{U}{2} - 9.8 > 0$ | M1 | |
| $\Rightarrow U > 19.6$ | A1 | ***(2)*** |
### Part (c):
| Working | Marks | Notes |
|---|---|---|
| $v = \displaystyle\int \frac{U}{(2-t)} - 9.8\; dt$ | M1 | |
| $= -U\ln(2-t) - 9.8t + C$ | A1 | |
| $t=0, v=0$: $0 = -U\ln 2 + C \Rightarrow C = U\ln 2$ | M1 | |
| $v = U\ln\dfrac{2}{(2-t)} - 9.8t$ | | |
| $t=1$: $v = U\ln 2 - 9.8$ | M1 A1 | **Total: 14** |
---4. At time $t = 0$ a rocket is launched from rest vertically upwards. The rocket propels itself upwards by expelling burnt fuel vertically downwards with constant speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$ relative to the rocket. The initial mass of the rocket is $M _ { 0 } \mathrm {~kg}$. At time $t$ seconds, where $t < 2$, its mass is $M _ { 0 } \left( 1 - \frac { 1 } { 2 } t \right) \mathrm { kg }$, and it is moving upwards with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { U } { ( 2 - t ) } - 9.8 .$$
\item Hence show that $U > 19.6$.
\item Find, in terms of $U$, the speed of the rocket one second after its launch.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2008 Q4 [14]}}