Standard +0.3 This is a standard connected particles problem requiring resolution of forces on two inclined planes, friction at limiting equilibrium (F=μR), and solving simultaneous equations. While it involves multiple components and careful bookkeeping of forces, it follows a well-established method taught in M1 with no novel insight required—slightly easier than average due to its routine nature.
5
\includegraphics[max width=\textwidth, alt={}, center]{f1f33ef0-0d4d-4a4a-aadb-28de8dc0ea8d-08_483_840_258_649}
The diagram shows a particle \(A\), of mass 1.2 kg , which lies on a plane inclined at an angle of \(40 ^ { \circ }\) to the horizontal and a particle \(B\), of mass 1.6 kg , which lies on a plane inclined at an angle of \(50 ^ { \circ }\) to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley \(P\) fixed at the top of the planes. The parts \(A P\) and \(B P\) of the string are taut and parallel to lines of greatest slope of the respective planes. The two planes are rough, with the same coefficient of friction, \(\mu\), between the particles and the planes.
Find the value of \(\mu\) for which the system is in limiting equilibrium.
Resolving parallel to the slope at \(A\) or \(B\) to form an equation.
\*M1
Correct number of terms; allow sign errors; allow sin/cos mix.
\(1.6g\sin 50 - T - F_B = 0\)
A1
If using the same \(F\)s, then M1A1A0B1 max.
\(T - F_A - 1.2g\sin 40 = 0\)
A1
System equation (must be four different terms): \(1.6g\sin 50 - F_B - F_A - 1.2g\sin 40 = 0\) only scores M1A1A1. Any sign errors scores M1 only.
\(R_A = 1.2g\cos 40\) or \(R_B = 1.6g\cos 50\)
\*B1
Either correct. Must be explicitly linked to the correct contact (could be seen on a diagram), or as part of a resolving parallel to the slope equation(s) (so must be combined with \(\mu\)).
\(F_A = 1.2g\mu\cos 40\) or \(F_B = 1.6g\mu\cos 50\)
\*M1
Use of \(F = \mu R\) at either \(A\) or \(B\). Must be explicitly linked to the correct contact (could be seen on a diagram) or as part of a resolving parallel to the slope equation(s). Allow sin/cos mix error only.
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolving parallel to the slope at $A$ or $B$ to form an equation. | **\*M1** | Correct number of terms; allow sign errors; allow sin/cos mix. |
| $1.6g\sin 50 - T - F_B = 0$ | **A1** | If using the same $F$s, then M1A1A0B1 max. |
| $T - F_A - 1.2g\sin 40 = 0$ | **A1** | System equation (must be four different terms): $1.6g\sin 50 - F_B - F_A - 1.2g\sin 40 = 0$ only scores M1A1A1. Any sign errors scores M1 only. |
| $R_A = 1.2g\cos 40$ or $R_B = 1.6g\cos 50$ | **\*B1** | Either correct. Must be explicitly linked to the correct contact (could be seen on a diagram), or as part of a resolving parallel to the slope equation(s) (so must be combined with $\mu$). |
| $F_A = 1.2g\mu\cos 40$ or $F_B = 1.6g\mu\cos 50$ | **\*M1** | Use of $F = \mu R$ at either $A$ or $B$. Must be explicitly linked to the correct contact (could be seen on a diagram) or as part of a resolving parallel to the slope equation(s). Allow sin/cos mix error only. |
| $1.6g\sin 50 - 1.6g\mu\cos 50 = 1.2g\sin 40 + 1.2g\mu\cos 40$ | **DM1** | Eliminating $T$, $F_A$ and $F_B$ to form an equation in $\mu$ only. |
| $\mu = \dfrac{1.6g\sin 50 - 1.2g\sin 40}{1.2g\cos 40 + 1.6g\cos 50} \Rightarrow \mu = 0.233$ | **A1** | $0.23326119\ldots$ |
| | **7** | |
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5\\
\includegraphics[max width=\textwidth, alt={}, center]{f1f33ef0-0d4d-4a4a-aadb-28de8dc0ea8d-08_483_840_258_649}
The diagram shows a particle $A$, of mass 1.2 kg , which lies on a plane inclined at an angle of $40 ^ { \circ }$ to the horizontal and a particle $B$, of mass 1.6 kg , which lies on a plane inclined at an angle of $50 ^ { \circ }$ to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley $P$ fixed at the top of the planes. The parts $A P$ and $B P$ of the string are taut and parallel to lines of greatest slope of the respective planes. The two planes are rough, with the same coefficient of friction, $\mu$, between the particles and the planes.
Find the value of $\mu$ for which the system is in limiting equilibrium.\\
\hfill \mbox{\textit{CAIE M1 2023 Q5 [7]}}