| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - non-gravitational context |
| Difficulty | Challenging +1.2 This is a variable force mechanics problem requiring Newton's second law in the form v(dv/dx) = a, followed by separating variables and integrating. While it involves non-standard resistance (proportional to v²/x²), the mathematical steps are straightforward: apply F=ma, separate variables (1/v integrates to ln v), and use initial conditions. More challenging than routine kinematics but standard for M2 level with clear signposting across two parts. |
| 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 | Marks | Guidance |
| \(0.25v\frac{\mathrm{d}v}{\mathrm{d}x} = -kv^2x^{-2} \rightarrow v\frac{\mathrm{d}v}{\mathrm{d}x} = -4kv^2x^{-2}\) | B1 | AG |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int\frac{\mathrm{d}v}{v} = -4k\int x^{-2}\,\mathrm{d}x\) | M1 | Attempt to integrate |
| \(\ln v = \frac{4k}{x}\ (+c)\) | A1 | |
| \(x = 0.8,\ v = 3\) hence \(c = \ln 3 - 5k\) | A1 | Finds \(c\) |
| \(\ln v = \frac{4k}{x} + \ln 3 - 5k\) | M1 | |
| \(v = 3^{\left(\frac{4k}{x}-5k\right)}\) | A1 | |
| Total | 5 |
**Question 3(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.25v\frac{\mathrm{d}v}{\mathrm{d}x} = -kv^2x^{-2} \rightarrow v\frac{\mathrm{d}v}{\mathrm{d}x} = -4kv^2x^{-2}$ | B1 | AG |
| **Total** | **1** | |
**Question 3(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int\frac{\mathrm{d}v}{v} = -4k\int x^{-2}\,\mathrm{d}x$ | M1 | Attempt to integrate |
| $\ln v = \frac{4k}{x}\ (+c)$ | A1 | |
| $x = 0.8,\ v = 3$ hence $c = \ln 3 - 5k$ | A1 | Finds $c$ |
| $\ln v = \frac{4k}{x} + \ln 3 - 5k$ | M1 | |
| $v = 3^{\left(\frac{4k}{x}-5k\right)}$ | A1 | |
| **Total** | **5** | |3 A smooth horizontal surface has two fixed points $O$ and $A$ which are 0.8 m apart. A particle $P$ of mass 0.25 kg is projected with velocity $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ horizontally from $A$ in the direction away from $O$. The velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when the displacement of $P$ from $O$ is $x \mathrm {~m}$. A force of magnitude $k v ^ { 2 } x ^ { - 2 } \mathrm {~N}$ opposes the motion of $P$.\\
(i) Show that $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 4 k v ^ { 2 } x ^ { - 2 }$.\\
(ii) Express $v$ in terms of $k$ and $x$.\\
\hfill \mbox{\textit{CAIE M2 2019 Q3 [6]}}