| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| 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 with resistance forces, followed by solving a non-trivial differential equation with variable coefficients. Part (a) requires careful force analysis including thrust and resistance, while part (b) demands integration techniques with substitution and manipulation of exponential/logarithmic forms—significantly harder than standard mechanics but follows established M5 patterns. |
| Spec | 4.10a 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 particles |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Mass at time \(t\): \(m(t) = M - \lambda t\) | B1 | |
| Thrust force \(= \lambda U\) (rate of mass ejection × relative speed) | B1 | |
| Newton's second law: \((M-\lambda t)\frac{dv}{dt} = \lambda U - kv\) | M1 A1 | |
| \(\frac{dv}{dt} = \frac{\lambda U - kv}{M - \lambda t}\) | A1 | Rearranging to required form |
| Full justification of signs and thrust derivation | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dv}{dt} + \frac{k}{M-\lambda t}v = \frac{\lambda U}{M-\lambda t}\) | M1 | Writing as linear ODE |
| Integrating factor: \(e^{\int \frac{k}{M-\lambda t}dt} = e^{-\frac{k}{\lambda}\ln(M-\lambda t)} = (M-\lambda t)^{-k/\lambda}\) | M1 A1 | |
| \(\frac{d}{dt}\left[v(M-\lambda t)^{-k/\lambda}\right] = \lambda U (M-\lambda t)^{-k/\lambda - 1}\) | M1 | |
| Integrating: \(v(M-\lambda t)^{-k/\lambda} = \frac{\lambda U}{k}(M-\lambda t)^{-k/\lambda} + C\) | M1 A1 | |
| At \(t=0\), \(v=0\): \(C = -\frac{\lambda U}{k}M^{-k/\lambda}\) | M1 | |
| \(v = \frac{\lambda U}{k}\left\{1 - \left(1 - \frac{\lambda t}{M}\right)^{k/\lambda}\right\}\) | A1 |
# Question 6:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Mass at time $t$: $m(t) = M - \lambda t$ | B1 | |
| Thrust force $= \lambda U$ (rate of mass ejection × relative speed) | B1 | |
| Newton's second law: $(M-\lambda t)\frac{dv}{dt} = \lambda U - kv$ | M1 A1 | |
| $\frac{dv}{dt} = \frac{\lambda U - kv}{M - \lambda t}$ | A1 | Rearranging to required form |
| Full justification of signs and thrust derivation | M1 A1 | |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dv}{dt} + \frac{k}{M-\lambda t}v = \frac{\lambda U}{M-\lambda t}$ | M1 | Writing as linear ODE |
| Integrating factor: $e^{\int \frac{k}{M-\lambda t}dt} = e^{-\frac{k}{\lambda}\ln(M-\lambda t)} = (M-\lambda t)^{-k/\lambda}$ | M1 A1 | |
| $\frac{d}{dt}\left[v(M-\lambda t)^{-k/\lambda}\right] = \lambda U (M-\lambda t)^{-k/\lambda - 1}$ | M1 | |
| Integrating: $v(M-\lambda t)^{-k/\lambda} = \frac{\lambda U}{k}(M-\lambda t)^{-k/\lambda} + C$ | M1 A1 | |
| At $t=0$, $v=0$: $C = -\frac{\lambda U}{k}M^{-k/\lambda}$ | M1 | |
| $v = \frac{\lambda U}{k}\left\{1 - \left(1 - \frac{\lambda t}{M}\right)^{k/\lambda}\right\}$ | A1 | |
---6. A rocket-driven car moves along a straight horizontal road. The car has total initial mass $M$. It propels itself forwards by ejecting mass backwards at a constant rate $\lambda$ per unit time at a constant speed $U$ relative to the car. The car starts from rest at time $t = 0$. At time $t$ the speed of the car is $v$. The total resistance to motion is modelled as having magnitude $k v$, where $k$ is a constant.
Given that $t < \frac { M } { \lambda }$, show that
\begin{enumerate}[label=(\alph*)]
\item $\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { \lambda U - k v } { M - \lambda t }$,
\item $v = \frac { \lambda U } { k } \left\{ 1 - \left( 1 - \frac { \lambda t } { M } \right) ^ { \frac { k } { \lambda } } \right\}$.\\
(6)\\
(Total 13 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2005 Q6 [13]}}