| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Standard +0.3 This is a straightforward application of Newton's second law to form a differential equation, followed by solving a first-order linear vector DE with given initial conditions. The solution method (integrating factor or recognizing the standard form) is routine for Further Maths students, and the calculation at t=ln 4 is designed to give clean numbers. Slightly above average difficulty due to vector notation and the mechanics context, but still a standard textbook exercise. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2\frac{d\mathbf{v}}{dt} = -19.6\mathbf{k} - \mathbf{v}\) | M1 | M1 for use of \(\mathbf{F} = m\mathbf{a}\) vertically with usual rules |
| \(\frac{d\mathbf{v}}{dt} + 0.5\mathbf{v} = -9.8\mathbf{k}\) | A1 (2) | A1 for printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Aux Eqn: \(m + 0.5 = 0\) | M1 | First M1 for forming auxiliary equation |
| \(\mathbf{v} = \mathbf{A}e^{-0.5t}\) (CF) | A1 | First A1 for correct CF |
| \(\mathbf{v} = -19.6\mathbf{k}\) (PI) | B1 | B1 for correct PI |
| \(\mathbf{v} = \mathbf{A}e^{-0.5t} - 19.6\mathbf{k}\) (GS) | M1 | Second M1 for \(\mathbf{v} =\) their CF + their PI |
| \(t=0,\ \mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 11.6\mathbf{k}) \Rightarrow \mathbf{A} = (4\mathbf{i} - 6\mathbf{j} + 31.2\mathbf{k})\) | M1 | Third M1 for use of initial conditions to find constant |
| \(\mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 31.2\mathbf{k})e^{-0.5t} - 19.6\mathbf{k}\) | A1 | Second A1 for correct particular solution |
| \(t = \ln 4,\ \mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 31.2\mathbf{k})\frac{1}{2} - 19.6\mathbf{k}\) | M1 | Fourth M1 for substituting \(t = \ln 4\) |
| \(= (2\mathbf{i} - 3\mathbf{j} - 4\mathbf{k})\) | A1 (8) | Third A1 for correct answer |
# Question 2:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\frac{d\mathbf{v}}{dt} = -19.6\mathbf{k} - \mathbf{v}$ | M1 | M1 for use of $\mathbf{F} = m\mathbf{a}$ vertically with usual rules |
| $\frac{d\mathbf{v}}{dt} + 0.5\mathbf{v} = -9.8\mathbf{k}$ | A1 (2) | A1 for printed answer |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Aux Eqn: $m + 0.5 = 0$ | M1 | First M1 for forming auxiliary equation |
| $\mathbf{v} = \mathbf{A}e^{-0.5t}$ (CF) | A1 | First A1 for correct CF |
| $\mathbf{v} = -19.6\mathbf{k}$ (PI) | B1 | B1 for correct PI |
| $\mathbf{v} = \mathbf{A}e^{-0.5t} - 19.6\mathbf{k}$ (GS) | M1 | Second M1 for $\mathbf{v} =$ their CF + their PI |
| $t=0,\ \mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 11.6\mathbf{k}) \Rightarrow \mathbf{A} = (4\mathbf{i} - 6\mathbf{j} + 31.2\mathbf{k})$ | M1 | Third M1 for use of initial conditions to find constant |
| $\mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 31.2\mathbf{k})e^{-0.5t} - 19.6\mathbf{k}$ | A1 | Second A1 for correct particular solution |
| $t = \ln 4,\ \mathbf{v} = (4\mathbf{i} - 6\mathbf{j} + 31.2\mathbf{k})\frac{1}{2} - 19.6\mathbf{k}$ | M1 | Fourth M1 for substituting $t = \ln 4$ |
| $= (2\mathbf{i} - 3\mathbf{j} - 4\mathbf{k})$ | A1 (8) | Third A1 for correct answer |
---2. [In this question, $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane and $\mathbf { k }$ is a unit vector vertically upwards.]
A particle of mass 2 kg moves under the action of a constant gravitational force $- 19.6 \mathbf { k } \mathrm {~N}$. The particle is subject to a resistive force $- \mathbf { v }$ newtons, where $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ is the velocity of the particle at time $t$ seconds.
\begin{enumerate}[label=(\alph*)]
\item By writing down an equation of motion of the particle, show that $\mathbf { v }$ satisfies the differential equation
$$\frac { \mathrm { d } \mathbf { v } } { \mathrm {~d} t } + 0.5 \mathbf { v } = - 9.8 \mathbf { k }$$
When $t = 0 , \mathbf { v } = ( 4 \mathbf { i } - 6 \mathbf { j } + 11.6 \mathbf { k } )$
\item Find $\mathbf { v }$ when $t = \ln 4$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2017 Q2 [10]}}