| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable force (position x) - find velocity |
| Difficulty | Standard +0.8 This is a multi-part variable force question requiring application of F=ma with v dv/dx, solving a differential equation, and analyzing motion constraints. Part (i) is straightforward force equation setup, but parts (ii)-(iii) require careful analysis of acceleration conditions and integration to find velocity constraints, which is above-average difficulty for M2 level. |
| Spec | 1.08h Integration by substitution3.02a Kinematics language: position, displacement, velocity, acceleration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.2v\frac{dv}{dx} = 0.09\sqrt{x} - 0.3\) | M1 | Use Newton's Second Law horizontally |
| \(v\frac{dv}{dx} = 0.45\sqrt{x} - 1.5\) | A1 | AG |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0 = 0.45x^{\frac{1}{2}} - 1.5\) | M1 | Equate acceleration to zero |
| \(x = \frac{100}{9}\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int v\, dv = \int(0.45x^{\frac{1}{2}} - 1.5)dx\) | M1 | Attempt to integrate |
| \(\frac{v^2}{2} = \frac{0.45}{\frac{3}{2}}x^{\frac{3}{2}} - 1.5x(+c) = 0.3x^{\frac{3}{2}} - 1.5x(+c)\) | A1 | |
| \(0.3\left(\frac{100}{9}\right)^{\frac{3}{2}} - 1.5\left(\frac{100}{9}\right) + c = 0\) | M1 | |
| \(c = \frac{50}{9}\) | A1 | |
| \(x = 0\), \(\frac{v^2}{2} > \frac{50}{9}\) so \(v > \frac{10}{3}\) | A1 | |
| Total: 5 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.2v\frac{dv}{dx} = 0.09\sqrt{x} - 0.3$ | M1 | Use Newton's Second Law horizontally |
| $v\frac{dv}{dx} = 0.45\sqrt{x} - 1.5$ | A1 | AG |
| **Total: 2** | | |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0 = 0.45x^{\frac{1}{2}} - 1.5$ | M1 | Equate acceleration to zero |
| $x = \frac{100}{9}$ | A1 | |
| **Total: 2** | | |
---
## Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int v\, dv = \int(0.45x^{\frac{1}{2}} - 1.5)dx$ | M1 | Attempt to integrate |
| $\frac{v^2}{2} = \frac{0.45}{\frac{3}{2}}x^{\frac{3}{2}} - 1.5x(+c) = 0.3x^{\frac{3}{2}} - 1.5x(+c)$ | A1 | |
| $0.3\left(\frac{100}{9}\right)^{\frac{3}{2}} - 1.5\left(\frac{100}{9}\right) + c = 0$ | M1 | |
| $c = \frac{50}{9}$ | A1 | |
| $x = 0$, $\frac{v^2}{2} > \frac{50}{9}$ so $v > \frac{10}{3}$ | A1 | |
| **Total: 5** | | |
---6 A particle $P$ of mass 0.2 kg is projected horizontally from a fixed point $O$ on a smooth horizontal surface. When the displacement of $P$ from $O$ is $x \mathrm {~m}$ the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A horizontal force of variable magnitude $0.09 \sqrt { } x \mathrm {~N}$ directed away from $O$ acts on $P$. An additional force of constant magnitude 0.3 N directed towards $O$ acts on $P$.\\
(i) Show that $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 0.45 \sqrt { } x - 1.5$.\\
(ii) Find the value of $x$ for which the acceleration of $P$ is zero.\\
(iii) Given that the minimum value of $v$ is positive, find the set of possible values for the speed of projection.\\
\hfill \mbox{\textit{CAIE M2 2019 Q6 [9]}}