| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.3 This is a standard M2 variable force question requiring chain rule differentiation (v dv/dx), Newton's second law, separation of variables, and limit analysis. All techniques are routine for M2 students, with straightforward algebra and no novel insights required. Slightly above average difficulty due to multiple parts and the need to connect several standard methods. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = (8-2x)(-2) = -16+4x\) | B1 | Any correct form |
| \(-R = 0.25(-16+4\times1)\) | M1 | For using Newton's second law and substituting for \(x\) |
| Magnitude of the force is 3 N | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int dt = \int \dfrac{dx}{8-2x}\) | B1 | |
| M1* | For attempting to integrate | |
| \(t = -\frac{1}{2}\ln(8-2x)\) \((+C)\) | A1 | |
| (\(C = \frac{1}{2}\ln8\)) | M1*dep | For using \(x=0\) when \(t=0\) to find \(C\) |
| \(2t = \ln\dfrac{8}{8-2x} \Rightarrow e^{2t} = \dfrac{8}{8-2x}\) | M1*dep | For converting to exponential form |
| \(x = 4(1-e^{-2t})\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t \geq 0 \Rightarrow\) | M1 | |
| \(0 < e^{-2t} \leq 1 \Rightarrow 0 \leq 1-e^{-2t} < 1\) | ||
| \(\Rightarrow 0 \leq x < 4\) | A1 | [2] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = (8-2x)(-2) = -16+4x$ | B1 | Any correct form |
| $-R = 0.25(-16+4\times1)$ | M1 | For using Newton's second law and substituting for $x$ |
| Magnitude of the force is 3 N | A1 | **[3]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int dt = \int \dfrac{dx}{8-2x}$ | B1 | |
| | M1* | For attempting to integrate |
| $t = -\frac{1}{2}\ln(8-2x)$ $(+C)$ | A1 | |
| ($C = \frac{1}{2}\ln8$) | M1*dep | For using $x=0$ when $t=0$ to find $C$ |
| $2t = \ln\dfrac{8}{8-2x} \Rightarrow e^{2t} = \dfrac{8}{8-2x}$ | M1*dep | For converting to exponential form |
| $x = 4(1-e^{-2t})$ | A1 | **[6]** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t \geq 0 \Rightarrow$ | M1 | |
| $0 < e^{-2t} \leq 1 \Rightarrow 0 \leq 1-e^{-2t} < 1$ | | |
| $\Rightarrow 0 \leq x < 4$ | A1 | **[2]** |7 A particle of mass 0.25 kg moves in a straight line on a smooth horizontal surface. A variable resisting force acts on the particle. At time $t \mathrm {~s}$ the displacement of the particle from a point on the line is $x \mathrm {~m}$, and its velocity is $( 8 - 2 x ) \mathrm { m } \mathrm { s } ^ { - 1 }$. It is given that $x = 0$ when $t = 0$.\\
(i) Find the acceleration of the particle in terms of $x$, and hence find the magnitude of the resisting force when $x = 1$.\\
(ii) Find an expression for $x$ in terms of $t$.\\
(iii) Show that the particle is always less than 4 m from its initial position.
\hfill \mbox{\textit{CAIE M2 2005 Q7 [11]}}