| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Standard +0.8 This is a connected particles problem with velocity-dependent resistance requiring differential equations. Students must correctly apply Newton's second law to both particles simultaneously, handle friction, set up and solve a separable DE, then integrate to find distance. The multiple forces, coupled system, and calculus requirements make this moderately challenging, though the steps are systematic once the equation of motion is established. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03v Motion on rough surface: including inclined planes6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \([P = \pm F + 0.6g\sin25°]\) | M1 | For resolving forces in the direction of \(P\) |
| \(P_{\max} = F + 0.6g\sin25°\) or '\(P = F + 0.6g\sin25°\) when the particle is about to slide upwards' | A1 | |
| \(P_{\min} = -F + 0.6g\sin25°\) or '\(P = -F + 0.6g\sin25°\) when the particle is about to slide downwards' | A1 | |
| \(R = 0.6g\cos25°\) | B1 | |
| \([F = 0.36 \times 0.6g\cos25°]\) | M1 | For using \(F = \mu R\) |
| \([P_{\max} = 0.36 \times 0.6g\cos25° + 0.6g\sin25°,\) \(P_{\min} = -0.36 \times 0.6g\cos25° + 0.6g\sin25°]\) \(P_{\max} = 4.49, P_{\min} = 0.578\) (accept 0.58) | DM1 | For substituting for \(F\) to obtain values of \(P_{\max}\) and \(P_{\min}\) (Dependent on first M mark) |
| A1 | Dependent on first M mark | |
| M1 | For identifying range of value for equilibrium; AEF; Accept 0.58 instead of 0.578 and accept \(<\) instead of \(\leq\) | |
| Set of values is \(\{P; 0.578 \leq P \leq 4.49\}\) | A1 | 9 marks total |
| $[P = \pm F + 0.6g\sin25°]$ | M1 | For resolving forces in the direction of $P$ |
| $P_{\max} = F + 0.6g\sin25°$ or '$P = F + 0.6g\sin25°$ when the particle is about to slide upwards' | A1 | |
| $P_{\min} = -F + 0.6g\sin25°$ or '$P = -F + 0.6g\sin25°$ when the particle is about to slide downwards' | A1 | |
| $R = 0.6g\cos25°$ | B1 | |
| $[F = 0.36 \times 0.6g\cos25°]$ | M1 | For using $F = \mu R$ |
| $[P_{\max} = 0.36 \times 0.6g\cos25° + 0.6g\sin25°,$ $P_{\min} = -0.36 \times 0.6g\cos25° + 0.6g\sin25°]$ $P_{\max} = 4.49, P_{\min} = 0.578$ (accept 0.58) | DM1 | For substituting for $F$ to obtain values of $P_{\max}$ and $P_{\min}$ (Dependent on first M mark) |
| | A1 | Dependent on first M mark |
| | M1 | For identifying range of value for equilibrium; AEF; Accept 0.58 instead of 0.578 and accept $<$ instead of $\leq$ |
| Set of values is $\{P; 0.578 \leq P \leq 4.49\}$ | A1 | 9 marks total |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{e30ba526-db21-4904-96dc-c12a1f67c81a-4_238_725_258_712}
Two particles $P$ and $Q$, of masses 0.4 kg and 0.2 kg respectively, are attached to opposite ends of a light inextensible string. $P$ is placed on a horizontal table and the string passes over a small smooth pulley at the edge of the table. The string is taut and the part of the string attached to $Q$ is vertical (see diagram). The coefficient of friction between $P$ and the table is 0.5 . $Q$ is projected vertically downwards with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and at time $t \mathrm {~s}$ after the instant of projection the speed of the particles is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The motion of each particle is opposed by a resisting force of magnitude $0.9 v \mathrm {~N}$. The particle $P$ does not reach the pulley.\\
(i) Show that $\frac { \mathrm { d } v } { \mathrm {~d} t } = - 3 v$.\\
(ii) Find the value of $t$ when the particles have speed $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the distance that each particle has travelled in this time.
\hfill \mbox{\textit{CAIE M2 2012 Q6 [11]}}