| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Smooth ring on string |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem with three forces meeting at a point. Part (i) tests understanding of smooth rings (routine recall), part (ii) requires resolving forces in two directions to find tension (standard technique), and part (iii) uses the result to find mass. The geometry is straightforward with nice angles (30°, 45°), making this slightly easier than average for M1 equilibrium questions. |
| Spec | 3.03b Newton's first law: equilibrium3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks |
|---|---|
| \(R\) is smooth | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| For resolving forces horizontally | M1 | Requires 3 relevant terms and at least one force resolved |
| \(T + T\cos60° = 1.6\cos45°\) | A1 | |
| Tension is \(0.754\) N | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| For resolving forces vertically to obtain equation for \(m\) | M1 | Requires 3 relevant terms with both \(T\) and the \(1.6\) N force resolved |
| \(mg = T\sin60° + 1.6\sin45°\) | A1ft | ft sin/cos mix from (ii) |
| \(m = 0.182\) | A1 | Total: 3 |
| SR: \(m = T\sin60° + 1.6\sin45°\), \(m = 1.78\) | M1, B1 |
# Question 1:
## Part (i)
| $R$ is smooth | B1 | |
## Part (ii)
| For resolving forces horizontally | M1 | Requires 3 relevant terms and at least one force resolved |
| $T + T\cos60° = 1.6\cos45°$ | A1 | |
| Tension is $0.754$ N | A1 | **AG** | **Total: 3** |
## Part (iii)
| For resolving forces vertically to obtain equation for $m$ | M1 | Requires 3 relevant terms with both $T$ and the $1.6$ N force resolved |
| $mg = T\sin60° + 1.6\sin45°$ | A1ft | ft sin/cos mix from (ii) |
| $m = 0.182$ | A1 | **Total: 3** |
| SR: $m = T\sin60° + 1.6\sin45°$, $m = 1.78$ | M1, B1 | |
---1\\
\begin{tikzpicture}[scale=1.2, >=stealth]
% Fixed points
\coordinate (A) at (0,4);
\coordinate (B) at (0,0);
\coordinate (R) at (2,0);
% Dashed vertical line from A downward
\draw[dashed] (A) -- (0,0.5);
% String AR (at 30 to downward vertical from A)
\draw[thick] (A) -- (R);
% String BR (horizontal)
\draw[thick] (B) -- (R);
% 30 angle arc at A (between downward vertical and AR)
\draw (A) ++(0,-0.8) arc[start angle=-90, end angle=-90+30, radius=0.8];
\node at ($(A) + (0.25,-0.7)$) {$30°$};
% Force arrow at R: 1.6 N at 45 above horizontal
\draw[->, thick] (R) -- ++(45:1.8) node[above] {$1.6$ N};
% 45 angle arc at R (between horizontal and force)
\draw[dashed] (R) -- ++(1.5,0);
\draw (R) ++(0.6,0) arc[start angle=0, end angle=45, radius=0.6];
\node at ($(R) + (0.85,0.25)$) {$45°$};
% Labels
\node[above] at (A) {$A$};
\node[left] at (B) {$B$};
\node[below] at (R) {$m$ kg};
\node[above left] at (R) {$R$};
% Dots at points
\fill (A) circle (1.5pt);
\fill (B) circle (1.5pt);
\fill (R) circle (1.5pt);
\end{tikzpicture}
A light inextensible string has its ends attached to two fixed points $A$ and $B$. The point $A$ is vertically above $B$. A smooth ring $R$ of mass $m \mathrm {~kg}$ is threaded on the string and is pulled by a force of magnitude 1.6 N acting upwards at $45 ^ { \circ }$ to the horizontal. The section $A R$ of the string makes an angle of $30 ^ { \circ }$ with the downward vertical and the section $B R$ is horizontal (see diagram). The ring is in equilibrium with the string taut.\\
(i) Give a reason why the tension in the part $A R$ of the string is the same as that in the part $B R$.\\
(ii) Show that the tension in the string is 0.754 N , correct to 3 significant figures.\\
(iii) Find the value of $m$.
\hfill \mbox{\textit{OCR M1 2005 Q1 [7]}}