| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2010 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable mass problems (mass increasing) |
| Difficulty | Challenging +1.8 This is a challenging M5 variable mass problem requiring derivation of a differential equation from first principles using F=d(mv)/dt with time-varying mass (sphere volume), then solving a first-order linear ODE with integrating factor. Requires multiple sophisticated techniques and careful algebraic manipulation, significantly above average A-level difficulty but standard for Further Maths M5. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(\frac{dr}{dt} = \lambda \Rightarrow r = \lambda t + a\) | B1 | |
| \((m+\delta m)(v+\delta v) - mv = mg\,\delta t\) | M1 A1 | |
| \(\frac{dv}{dt} + \frac{v}{m}\frac{dm}{dt} = g\) | DM1 A1 | |
| \(\frac{dm}{dt} = 4\pi r^2(\rho)\lambda\) | A1 (B1) | |
| \(\frac{dv}{dt} + \frac{v}{4\pi r^3\rho}\cdot 4\pi r^2\rho\lambda = g \Rightarrow \frac{dv}{dt} + \frac{3v\lambda}{r} = g\) | DM1 | |
| \(\frac{dv}{dt} + \frac{3v\lambda}{\lambda t+a} = g\) * | A1 | Total: (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(R = e^{\int\frac{3\lambda}{\lambda t+a}dt} = e^{3\ln(\lambda t+a)} = e^{\ln(\lambda t+a)^3} = (\lambda t+a)^3\) | M1 A1 | |
| \(v(\lambda t+a)^3 = g\int(\lambda t+a)^3\,dt\) | DM1 | |
| \(v(\lambda t+a)^3 = \frac{1}{4\lambda}g(\lambda t+a)^4\) | A1 | |
| \(t=0, v=0 \Rightarrow C = -\frac{1}{4\lambda}ga^4\) | DM1 | |
| \(\lambda t + a = 3a\) | DM1 | |
| \(v = \frac{1}{4\lambda}g(3a) - \frac{1}{4\lambda}\frac{ga^4}{27a^3} = \frac{20ag}{27\lambda}\) | A1 | Total: (7), Q Total: 15 |
## Question 5:
### Part (a):
| Working/Answer | Mark | Notes |
|---|---|---|
| $\frac{dr}{dt} = \lambda \Rightarrow r = \lambda t + a$ | B1 | |
| $(m+\delta m)(v+\delta v) - mv = mg\,\delta t$ | M1 A1 | |
| $\frac{dv}{dt} + \frac{v}{m}\frac{dm}{dt} = g$ | DM1 A1 | |
| $\frac{dm}{dt} = 4\pi r^2(\rho)\lambda$ | A1 (B1) | |
| $\frac{dv}{dt} + \frac{v}{4\pi r^3\rho}\cdot 4\pi r^2\rho\lambda = g \Rightarrow \frac{dv}{dt} + \frac{3v\lambda}{r} = g$ | DM1 | |
| $\frac{dv}{dt} + \frac{3v\lambda}{\lambda t+a} = g$ * | A1 | **Total: (8)** |
### Part (b):
| Working/Answer | Mark | Notes |
|---|---|---|
| $R = e^{\int\frac{3\lambda}{\lambda t+a}dt} = e^{3\ln(\lambda t+a)} = e^{\ln(\lambda t+a)^3} = (\lambda t+a)^3$ | M1 A1 | |
| $v(\lambda t+a)^3 = g\int(\lambda t+a)^3\,dt$ | DM1 | |
| $v(\lambda t+a)^3 = \frac{1}{4\lambda}g(\lambda t+a)^4$ | A1 | |
| $t=0, v=0 \Rightarrow C = -\frac{1}{4\lambda}ga^4$ | DM1 | |
| $\lambda t + a = 3a$ | DM1 | |
| $v = \frac{1}{4\lambda}g(3a) - \frac{1}{4\lambda}\frac{ga^4}{27a^3} = \frac{20ag}{27\lambda}$ | A1 | **Total: (7), Q Total: 15** |\begin{enumerate}
\item A raindrop falls vertically under gravity through a cloud. In a model of the motion the raindrop is assumed to be spherical at all times and the cloud is assumed to consist of stationary water particles. At time $t = 0$, the raindrop is at rest and has radius $a$. As the raindrop falls, water particles from the cloud condense onto it and the radius of the raindrop is assumed to increase at a constant rate $\lambda$. A time $t$ the speed of the raindrop is $v$.\\
(a) Show that
\end{enumerate}
$$\frac { \mathrm { d } v } { \mathrm {~d} t } + \frac { 3 \lambda v } { ( \lambda t + a ) } = g$$
(b) Find the speed of the raindrop when its radius is $3 a$.\\
\hfill \mbox{\textit{Edexcel M5 2010 Q5 [15]}}