| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - non-gravitational context |
| Difficulty | Challenging +1.2 This is a standard M2 variable force problem requiring Newton's second law with F=ma in the form v(dv/dx), integration of a sum including an inverse square term, and verification that the particle remains at rest. The setup is straightforward with clearly stated forces, and the techniques are standard for this topic, though it requires careful handling of multiple force components and checking equilibrium conditions. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.2a = -0.2g0.5 - 0.4x^2\), \(vdv/dx = -(5 + 2x^{-2})\) | M1, A1 | Uses Newton's Second law; AG [2] |
| (ii) \(\int vdv = -\int(5 + 2x^{-2})dx\), \(v^2 / 2 = -5x + 2/x(+ c)\), \(3^2/2 = -5 \times 0.5 + 2/0.5 + c\), \(x = 1\), Travels (\(= 1 - 0.5) = 0.5\)m, F towards O (0.4) less than maximum friction (\(= 1\)) | M1, A1, M1, A1, A1, M1, A1 | Separates variables and integrates; Hence \(c = 3\), or \([v^2/2]^1_{0.5} = [-5x + 2/x]^1_{0.5}\); From \(0 = -5x + 2/x = 3\); Compares \(0.5 \times 0.2g\) and \(0.4/1^2\) [7] |
**(i)** $0.2a = -0.2g0.5 - 0.4x^2$, $vdv/dx = -(5 + 2x^{-2})$ | M1, A1 | Uses Newton's Second law; AG [2]
**(ii)** $\int vdv = -\int(5 + 2x^{-2})dx$, $v^2 / 2 = -5x + 2/x(+ c)$, $3^2/2 = -5 \times 0.5 + 2/0.5 + c$, $x = 1$, Travels ($= 1 - 0.5) = 0.5$m, F towards O (0.4) less than maximum friction ($= 1$) | M1, A1, M1, A1, A1, M1, A1 | Separates variables and integrates; Hence $c = 3$, or $[v^2/2]^1_{0.5} = [-5x + 2/x]^1_{0.5}$; From $0 = -5x + 2/x = 3$; Compares $0.5 \times 0.2g$ and $0.4/1^2$ [7]
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\includegraphics[max width=\textwidth, alt={}, center]{9d377c95-09b8-4893-b29f-8517a5016e8b-3_151_949_1206_598}\\
$O$ and $A$ are fixed points on a horizontal surface, with $O A = 0.5 \mathrm {~m}$. A particle $P$ of mass 0.2 kg is projected horizontally with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from $A$ in the direction $O A$ and moves in a straight line (see diagram). At time $t \mathrm {~s}$ after projection, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its displacement from $O$ is $x \mathrm {~m}$. The coefficient of friction between the surface and $P$ is 0.5 , and a force of magnitude $\frac { 0.4 } { x ^ { 2 } } \mathrm {~N}$ acts on $P$ in the direction $P O$.\\
(i) Show that, while the particle is in motion, $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - \left( 5 + \frac { 2 } { x ^ { 2 } } \right)$.\\
(ii) Calculate the distance travelled by $P$ before it comes to rest, and show that $P$ does not subsequently move.
\hfill \mbox{\textit{CAIE M2 2011 Q6 [9]}}