Standard +0.3 This is a standard equilibrium problem with a smooth ring on a string requiring resolution of forces in two directions. The geometry is clearly specified (60° and 90° angles), and students need to apply the key principle that tension is equal throughout the string at a smooth ring, then resolve forces. While it requires careful diagram work and simultaneous equations, it follows a well-practiced method with no novel insight needed, making it slightly easier than average.
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\includegraphics[max width=\textwidth, alt={}, center]{2a680bda-4ba2-44eb-8592-95b4e1aed263-04_337_661_262_740}
A smooth ring \(R\) of mass 0.2 kg is threaded on a light string \(A R B\). The ends of the string are attached to fixed points \(A\) and \(B\) with \(A\) vertically above \(B\). The string is taut and angle \(A B R = 90 ^ { \circ }\). The angle between the part \(A R\) of the string and the vertical is \(60 ^ { \circ }\). The ring is held in equilibrium by a force of magnitude \(X \mathrm {~N}\), acting on the ring in a direction perpendicular to \(A R\) (see diagram).
Calculate the tension in the string and the value of \(X\).
Must use 0.2 substituted for \(m\) if just awarding M1 for vertical equation. Must have correct number of relevant terms. Allow sin/cos mix. Allow sign errors. Allow \(g\) missing.
\(X\sin 60 + T\sin 30 - 0.2g = 0\)
A1
OE. Correct vertical.
\(X\cos 60 - T - T\cos 30 = 0\)
A1
OE. Correct horizontal. If the two \(T\)s are different, they can get max M1A1A0M0A0, unless they subsequently state that the two \(T\)s are equal.
For attempt to solve for tension or \(X\)
M1
Must have correct number of relevant terms in both equations. Must get to '\(T=\)' or '\(X=\)'. Allow \(g\) missing. Can be implied by correct answers. If no working shown their values must follow from their equations.
\(X = 2\), tension in string \(= 0.536\) [N]
A1
Allow exact value of tension \(= 4 - 2\sqrt{3}\). Allow awrt 2.00 for \(X\).
Alternative method (resolving parallel and perpendicular to \(X\)):
Answer
Marks
Guidance
Answer
Mark
Guidance
For attempt to resolve in one direction, with 0.2 substituted for \(m\)
M1
Must have correct number of relevant terms. Allow sin/cos mix. Allow sign errors. Allow \(g\) missing.
\(X - 0.2g\cos 30 - T\cos 60 = 0\)
A1
OE. Correct parallel to \(X\).
\(T + T\cos 30 - 0.2g\cos 60 = 0\)
A1
OE. Correct perp to \(X\). If the two \(T\)s are different, they can get max M1A1A0M0A0 unless they subsequently state that the two \(T\)s are equal.
For attempt to solve for the tension or for \(X\)
M1
Must have correct number of relevant terms in both equations. Must get to '\(T=\)' or '\(X=\)'. Allow \(g\) missing. Can be implied by correct answers. If no working shown their values must follow from their equations.
\(X = 2\), Tension in string \(= 0.536\) [N] \([0.53589\ldots]\)
A1
Allow exact value of tension \(= 4 - 2\sqrt{3}\). Allow awrt 2.00 for \(X\).
5
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| For attempt to resolve in one direction | M1 | Must use 0.2 substituted for $m$ if just awarding M1 for vertical equation. Must have correct number of relevant terms. Allow sin/cos mix. Allow sign errors. Allow $g$ missing. |
| $X\sin 60 + T\sin 30 - 0.2g = 0$ | A1 | OE. Correct vertical. |
| $X\cos 60 - T - T\cos 30 = 0$ | A1 | OE. Correct horizontal. If the two $T$s are different, they can get max M1A1A0M0A0, unless they subsequently state that the two $T$s are equal. |
| For attempt to solve for tension or $X$ | M1 | Must have correct number of relevant terms in both equations. Must get to '$T=$' or '$X=$'. Allow $g$ missing. Can be implied by correct answers. If no working shown their values must follow from their equations. |
| $X = 2$, tension in string $= 0.536$ [N] | A1 | Allow exact value of tension $= 4 - 2\sqrt{3}$. Allow awrt 2.00 for $X$. |
**Alternative method (resolving parallel and perpendicular to $X$):**
| Answer | Mark | Guidance |
|--------|------|----------|
| For attempt to resolve in one direction, with 0.2 substituted for $m$ | M1 | Must have correct number of relevant terms. Allow sin/cos mix. Allow sign errors. Allow $g$ missing. |
| $X - 0.2g\cos 30 - T\cos 60 = 0$ | A1 | OE. Correct parallel to $X$. |
| $T + T\cos 30 - 0.2g\cos 60 = 0$ | A1 | OE. Correct perp to $X$. If the two $T$s are different, they can get max M1A1A0M0A0 unless they subsequently state that the two $T$s are equal. |
| For attempt to solve for the tension or for $X$ | M1 | Must have correct number of relevant terms in both equations. Must get to '$T=$' or '$X=$'. Allow $g$ missing. Can be implied by correct answers. If no working shown their values must follow from their equations. |
| $X = 2$, Tension in string $= 0.536$ [N] $[0.53589\ldots]$ | A1 | Allow exact value of tension $= 4 - 2\sqrt{3}$. Allow awrt 2.00 for $X$. |
| | **5** | |
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\includegraphics[max width=\textwidth, alt={}, center]{2a680bda-4ba2-44eb-8592-95b4e1aed263-04_337_661_262_740}
A smooth ring $R$ of mass 0.2 kg is threaded on a light string $A R B$. The ends of the string are attached to fixed points $A$ and $B$ with $A$ vertically above $B$. The string is taut and angle $A B R = 90 ^ { \circ }$. The angle between the part $A R$ of the string and the vertical is $60 ^ { \circ }$. The ring is held in equilibrium by a force of magnitude $X \mathrm {~N}$, acting on the ring in a direction perpendicular to $A R$ (see diagram).
Calculate the tension in the string and the value of $X$.\\
\hfill \mbox{\textit{CAIE M1 2023 Q3 [5]}}