| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2003 |
| 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 derivation of the rocket equation with resistance, then solving a first-order ODE with specific parameters. Part (a) demands careful application of Newton's second law to a variable mass system (non-standard), while part (b) requires solving a linear ODE with variable coefficients—both significantly above routine A-level but standard for Further Maths M5. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| (a) \((m + \delta m)(v + \delta v) + (-\delta m)(v - U) - mv = -kv\delta t\) | M1 A1 A1 |
| \(\Rightarrow m\frac{dv}{dt} + U\frac{dm}{dt} = -kv\) | A1 |
| \(m = M - \lambda t\) | B1 |
| \(\Rightarrow (M - \lambda t)\frac{dv}{dt} = \lambda U - kv\) | M1 |
| \(\Rightarrow \frac{dv}{dt} = \frac{\lambda U - kv}{M - \lambda t}\) (*) | A1 cso (7) |
| (b) Separating variables: \(\int \frac{dv}{U - v} = \int \frac{10}{M - 10t} dt\) or equivalent | M1 |
| Integrating: \(\ln(U - v) = \ln(M - 10t) + c\) | M1 A1 |
| Using limits correctly: \(\int_0^t = [\,]_0^t\) applied or \(t = 0, v = 0\) to find "\(c\)" \(\left[c = \ln\left(\frac{U}{M}\right)\right]\) | M1 |
| Complete method to find \(v\): \(\left[\ln\left(\frac{U}{U-v}\right) = \ln\left(\frac{M}{M-10t}\right)\right]\) | M1 |
| \(v = \frac{10Ut}{M}\) | A1 (6) |
**(a)** $(m + \delta m)(v + \delta v) + (-\delta m)(v - U) - mv = -kv\delta t$ | M1 A1 A1 |
$\Rightarrow m\frac{dv}{dt} + U\frac{dm}{dt} = -kv$ | A1 |
$m = M - \lambda t$ | B1 |
$\Rightarrow (M - \lambda t)\frac{dv}{dt} = \lambda U - kv$ | M1 |
$\Rightarrow \frac{dv}{dt} = \frac{\lambda U - kv}{M - \lambda t}$ (*) | A1 cso (7) |
**(b)** Separating variables: $\int \frac{dv}{U - v} = \int \frac{10}{M - 10t} dt$ or equivalent | M1 |
Integrating: $\ln(U - v) = \ln(M - 10t) + c$ | M1 A1 |
Using limits correctly: $\int_0^t = [\,]_0^t$ applied or $t = 0, v = 0$ to find "$c$" $\left[c = \ln\left(\frac{U}{M}\right)\right]$ | M1 |
Complete method to find $v$: $\left[\ln\left(\frac{U}{U-v}\right) = \ln\left(\frac{M}{M-10t}\right)\right]$ | M1 |
$v = \frac{10Ut}{M}$ | A1 (6) |
**Total: (13 marks)**
---4. A rocket-driven car propels itself forwards in a straight line on a horizontal track by ejecting burnt fuel backwards at a constant rate $\lambda \mathrm { kg } \mathrm { s } ^ { - 1 }$ and at a constant speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$ relative to the car. At time $t$ seconds, the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the total resistance to the motion of the car has magnitude $k v \mathrm {~N}$, where $k$ is a positive constant. When $t = 0$ the total mass of the car, including fuel, is $M \mathrm {~kg}$. Assuming that at time $t$ seconds some fuel remains in the car,
\begin{enumerate}[label=(\alph*)]
\item show that
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { \lambda U - k v } { M - \lambda t }$$
\item find the speed of the car at time $t$ seconds, given that it starts from rest when $t = 0$ and that $\lambda = k = 10$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2003 Q4 [13]}}