| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.3 This is a standard mechanics application of second-order differential equations with given solution form. Part (a) is straightforward force equation setup (F=ma), parts (b)-(c) involve routine substitution into the given solution form and differentiation. The critical damping context and solution form are provided, requiring only algebraic manipulation and calculus rather than independent problem-solving or derivation of the auxiliary equation. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4\frac{d^2x}{dt^2} = -9x - 12v\) | M1A1 | Using F = ma |
| \(= -9x - 12\frac{dx}{dt}\) | M1 | |
| Hence \(4\frac{d^2x}{dt^2} + 12\frac{dx}{dt} + 9x = 0\) ** | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Auxiliary eqn: \(4m^2 + 12m + 9 = 0\) | B1 | |
| \((2m+3)^2 = 0,\ m = -\frac{3}{2},\ \lambda = \frac{3}{2}\) | B1 | |
| \(t=0,\ x=4 \Rightarrow B=4\) | B1 | |
| \(t=0,\ \dot{x} = e^{-\lambda t}(-\lambda(At+B)+A) = 0 \Rightarrow -6 + A = 0,\ A = 6\) | B1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dot{x} = e^{-\frac{3}{2}t}\left(-\frac{3}{2}(6t+4)+6\right) = -9te^{-\frac{3}{2}t}\) | M1A1 | |
| \(\ddot{x} = e^{-\frac{3}{2}t}\left(-9 - (-9t)\times\frac{3}{2}\right)\) | M1 | |
| Acceleration \(= 0\) when \(t = \frac{2}{3}\); at which time \(v = -6e^{-1}\), so max speed \(= 6/e \approx 2.21 \ \text{m s}^{-1}\) (3sf) | A1, A1 | (5) Total: 13 |
## Question 6:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4\frac{d^2x}{dt^2} = -9x - 12v$ | M1A1 | Using F = ma |
| $= -9x - 12\frac{dx}{dt}$ | M1 | |
| Hence $4\frac{d^2x}{dt^2} + 12\frac{dx}{dt} + 9x = 0$ ** | A1 | (4) |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Auxiliary eqn: $4m^2 + 12m + 9 = 0$ | B1 | |
| $(2m+3)^2 = 0,\ m = -\frac{3}{2},\ \lambda = \frac{3}{2}$ | B1 | |
| $t=0,\ x=4 \Rightarrow B=4$ | B1 | |
| $t=0,\ \dot{x} = e^{-\lambda t}(-\lambda(At+B)+A) = 0 \Rightarrow -6 + A = 0,\ A = 6$ | B1 | (4) |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dot{x} = e^{-\frac{3}{2}t}\left(-\frac{3}{2}(6t+4)+6\right) = -9te^{-\frac{3}{2}t}$ | M1A1 | |
| $\ddot{x} = e^{-\frac{3}{2}t}\left(-9 - (-9t)\times\frac{3}{2}\right)$ | M1 | |
| Acceleration $= 0$ when $t = \frac{2}{3}$; at which time $v = -6e^{-1}$, so max speed $= 6/e \approx 2.21 \ \text{m s}^{-1}$ (3sf) | A1, A1 | (5) **Total: 13** |
---\begin{enumerate}
\item A particle $P$ of mass 4 kg moves along a horizontal straight line under the action of a force directed towards a fixed point $O$ on the line. At time $t$ seconds, $P$ is $x$ metres from $O$ and the force towards $O$ has magnitude $9 x$ newtons. The particle $P$ is also subject to air resistance, which has magnitude $12 v$ newtons when $P$ is moving with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(a) Show that the equation of motion of $P$ is
\end{enumerate}
$$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 12 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 x = 0$$
It is given that the solution of this differential equation is of the form
$$x = \mathrm { e } ^ { - \lambda t } ( A t + B )$$
When $t = 0$ the particle is released from rest at the point $R$, where $O R = 4 \mathrm {~m}$.
Find,\\
(b) the values of the constants $\lambda , A$ and $B$,\\
(c) the greatest speed of $P$ in the subsequent motion.\\
\hfill \mbox{\textit{Edexcel M4 2011 Q6 [13]}}