| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Ring or bead on wire/rod, equilibrium |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem with friction at limiting equilibrium. It requires resolving forces in two directions for ring A (part i), applying equilibrium at point P (part ii), and resolving for ring B (part iii). The trigonometric values are given, eliminating calculation complexity. While multi-part, each step follows routine mechanics procedures without requiring novel insight—slightly easier than average A-level. |
| Spec | 3.03n Equilibrium in 2D: particle under forces3.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| For resolving forces on \(A\) vertically (3 terms) | M1 | |
| \(F = 2 + 7\cos\alpha\) | A1 | |
| \(F = 3.96\) (may be implied) | A1 | |
| For resolving forces on \(A\) horizontally (2 terms) | M1 | |
| \(N = 7\sin\alpha\) | A1 | |
| \(N = 6.72\) (may be implied) | A1 | |
| \(3.96 = \mu\times6.72\) | M1 | For using \(F = \mu N\) |
| Coefficient is \(0.589\) or \(33/56\) | A1 | Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\cos\beta = 7\cos\alpha\) | M1 | For resolving forces at \(P\) vertically (2 terms) |
| \(T\cos\beta = 7\times0.28\ (= 1.96)\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\cos\beta - mg = 0\) | M1 | For resolving forces on \(B\) vertically (2 terms) |
| Mass is \(0.2\) kg | A1 | Total: 3 |
# Question 5:
## Part (i)
| For resolving forces on $A$ vertically (3 terms) | M1 | |
| $F = 2 + 7\cos\alpha$ | A1 | |
| $F = 3.96$ (may be implied) | A1 | |
| For resolving forces on $A$ horizontally (2 terms) | M1 | |
| $N = 7\sin\alpha$ | A1 | |
| $N = 6.72$ (may be implied) | A1 | |
| $3.96 = \mu\times6.72$ | M1 | For using $F = \mu N$ |
| Coefficient is $0.589$ or $33/56$ | A1 | **Total: 7** |
## Part (ii)
| $T\cos\beta = 7\cos\alpha$ | M1 | For resolving forces at $P$ vertically (2 terms) |
| $T\cos\beta = 7\times0.28\ (= 1.96)$ | A1 | **AG** | **Total: 2** |
## Part (iii)
| $T\cos\beta - mg = 0$ | M1 | For resolving forces on $B$ vertically (2 terms) |
| Mass is $0.2$ kg | A1 | **Total: 3** |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{99d30766-9c1b-43a8-986a-112b78b08146-3_697_579_1238_781}
Two small rings $A$ and $B$ are attached to opposite ends of a light inextensible string. The rings are threaded on a rough wire which is fixed vertically. $A$ is above $B$. A horizontal force is applied to a point $P$ of the string. Both parts $A P$ and $B P$ of the string are taut. The system is in equilibrium with angle $B A P = \alpha$ and angle $A B P = \beta$ (see diagram). The weight of $A$ is 2 N and the tensions in the parts $A P$ and $B P$ of the string are 7 N and $T \mathrm {~N}$ respectively. It is given that $\cos \alpha = 0.28$ and $\sin \alpha = 0.96$, and that $A$ is in limiting equilibrium.\\
(i) Find the coefficient of friction between the wire and the ring $A$.\\
(ii) By considering the forces acting at $P$, show that $T \cos \beta = 1.96$.\\
(iii) Given that there is no frictional force acting on $B$, find the mass of $B$.
\hfill \mbox{\textit{OCR M1 2005 Q5 [12]}}