| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance with other powers |
| Difficulty | Standard +0.8 This question requires setting up F=ma with a non-standard resistance force (proportional to 1/v rather than v or v²), then applying the chain rule v(dv/dx) = a to obtain a separable differential equation. While the integration itself is straightforward (v³ term), the setup requires understanding of variable force mechanics and the connection between acceleration forms, which goes beyond routine application. The numerical calculation in part (ii) is standard once the relationship is established. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using Newton's second law with \(a = v\dfrac{dv}{dx}\) | M1 | |
| \(-\dfrac{1}{3v} = 0.2v\dfrac{dv}{dx}\) | A1 | |
| \(3v^2\dfrac{dv}{dx} = -5\) from correct working | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Separating variables and attempting to integrate | M1 | |
| \(v^3 = (A) - 5x\) | A1 | |
| Using \(x = 0\) and \(v = 4\) to find \(A\), then substituting \(x = 7.4\) (or equivalent using limits) | M1 | |
| \(v = 3\) | A1 |
# Question 4:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Using Newton's second law with $a = v\dfrac{dv}{dx}$ | M1 | |
| $-\dfrac{1}{3v} = 0.2v\dfrac{dv}{dx}$ | A1 | |
| $3v^2\dfrac{dv}{dx} = -5$ from correct working | A1 | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Separating variables and attempting to integrate | M1 | |
| $v^3 = (A) - 5x$ | A1 | |
| Using $x = 0$ and $v = 4$ to find $A$, then substituting $x = 7.4$ (or equivalent using limits) | M1 | |
| $v = 3$ | A1 | |
---4 A particle of mass 0.2 kg moves in a straight line on a smooth horizontal surface. When its displacement from a fixed point on the surface is $x \mathrm {~m}$, its velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The motion is opposed by a force of magnitude $\frac { 1 } { 3 v } \mathrm {~N}$.\\
(i) Show that $3 v ^ { 2 } \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 5$.\\
(ii) Find the value of $v$ when $x = 7.4$, given that $v = 4$ when $x = 0$.
\hfill \mbox{\textit{CAIE M2 2003 Q4 [7]}}