| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ring on wire with string |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem involving resolving forces and moments. Part (i) uses basic trigonometry with a 3-4-5 triangle to find tensions. Part (ii) applies friction laws with vertical/horizontal force resolution. Part (iii) reverses the friction direction—straightforward but requires careful bookkeeping. Slightly above average due to the three-part structure and friction reversal, but all techniques are routine for M1. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For resolving forces at \(J\) horizontally or vertically | |
| \(0.8T_A + 0.6T_R = 5.6\) | A1 | Allow \(T_A\cos 36.9 + T_R\cos 53.1 = 5.6\) oe |
| \(0.6T_A = 0.8T_R\) | A1 | Allow \(T_A\sin 36.9 = T_R\sin 53.1\) oe |
| M1 | For solving the simultaneous equations for \(T_A\) and \(T_R\) | |
| Tension in \(AJ\) is \(4.48\) N and tension in \(RJ\) is \(3.36\) N | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{5.6}{\sin 90} = \dfrac{T_A}{\sin\alpha} = \dfrac{T_R}{\sin(270-\alpha)}\) m | M1 | For applying Lami's theorem to two of the three forces \(T_A\), \(T_R\), and \(5.6\) where \(\alpha\) is an obtuse angle |
| \(\dfrac{5.6}{\sin 90} = \dfrac{T_A}{0.8} = \dfrac{T_R}{0.6}\) m | A1, A1 | Allow \(\sin 126.9\) for \(0.8\) and \(\sin 143.1\) for \(0.6\) here |
| M1 | Solve for \(T_A\) and \(T_R\) | |
| \(T_A = 4.48\) and \(T_R = 3.36\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{5.6}{\sin 90} = \dfrac{T_A}{\sin\alpha} = \dfrac{T_R}{\sin(90-\alpha)}\) m | M1 | For applying triangle of forces to two of the three forces \(T_A\), \(T_R\), and \(5.6\) |
| \(\dfrac{5.6}{\sin 90} = \dfrac{T_A}{0.8} = \dfrac{T_R}{0.6}\) m | A1, A1 | Allow \(\sin 53.1\) for \(0.8\) and \(\sin 36.9\) for \(0.6\) here |
| M1 | Solve for \(T_A\) and \(T_R\) | |
| \(T_A = 4.48\) and \(T_R = 3.36\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.2g + F = T_R \times \cos 36.9\) | B1↑ | ft on \(T_R\) and \(36.9\) |
| \(N = T_R \times \sin 36.9\) | B1↑ | ft on \(T_R\) and \(36.9\) |
| \(\left[0.2g + \mu \times T_R \times 0.6 = T_R \times 0.8\right]\) | M1 | For using \(\mu = F \div N\) and obtaining an equation in \(\mu\) |
| \(\mu = 0.688 \div 2.016 = 0.341\) | A1 | [4] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left[0.2g + mg = \mu N + 0.8T_R\right]\) | M1 | For a four term equation from resolving forces acting on \(R\) vertically |
| \(0.2g + mg = 0.341 \times 2.016 + 3.36 \times 0.8\) | A1 | |
| \(m = 0.137\) or \(0.138\) | A1 | [3] |
# Question 7:
## Part (i) - Main Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces at $J$ horizontally **or** vertically |
| $0.8T_A + 0.6T_R = 5.6$ | A1 | Allow $T_A\cos 36.9 + T_R\cos 53.1 = 5.6$ oe |
| $0.6T_A = 0.8T_R$ | A1 | Allow $T_A\sin 36.9 = T_R\sin 53.1$ oe |
| | M1 | For solving the simultaneous equations for $T_A$ and $T_R$ |
| Tension in $AJ$ is $4.48$ N **and** tension in $RJ$ is $3.36$ N | A1 | **[5]** |
## Part (i) - First Alternative Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{5.6}{\sin 90} = \dfrac{T_A}{\sin\alpha} = \dfrac{T_R}{\sin(270-\alpha)}$ m | M1 | For applying Lami's theorem to two of the three forces $T_A$, $T_R$, and $5.6$ where $\alpha$ is an obtuse angle |
| $\dfrac{5.6}{\sin 90} = \dfrac{T_A}{0.8} = \dfrac{T_R}{0.6}$ m | A1, A1 | Allow $\sin 126.9$ for $0.8$ and $\sin 143.1$ for $0.6$ here |
| | M1 | Solve for $T_A$ **and** $T_R$ |
| $T_A = 4.48$ **and** $T_R = 3.36$ | A1 | **[5]** |
## Part (i) - Second Alternative Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{5.6}{\sin 90} = \dfrac{T_A}{\sin\alpha} = \dfrac{T_R}{\sin(90-\alpha)}$ m | M1 | For applying triangle of forces to two of the three forces $T_A$, $T_R$, and $5.6$ |
| $\dfrac{5.6}{\sin 90} = \dfrac{T_A}{0.8} = \dfrac{T_R}{0.6}$ m | A1, A1 | Allow $\sin 53.1$ for $0.8$ and $\sin 36.9$ for $0.6$ here |
| | M1 | Solve for $T_A$ **and** $T_R$ |
| $T_A = 4.48$ **and** $T_R = 3.36$ | A1 | **[5]** |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.2g + F = T_R \times \cos 36.9$ | B1↑ | ft on $T_R$ and $36.9$ |
| $N = T_R \times \sin 36.9$ | B1↑ | ft on $T_R$ and $36.9$ |
| $\left[0.2g + \mu \times T_R \times 0.6 = T_R \times 0.8\right]$ | M1 | For using $\mu = F \div N$ and obtaining an equation in $\mu$ |
| $\mu = 0.688 \div 2.016 = 0.341$ | A1 | **[4]** AG |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[0.2g + mg = \mu N + 0.8T_R\right]$ | M1 | For a four term equation from resolving forces acting on $R$ vertically |
| $0.2g + mg = 0.341 \times 2.016 + 3.36 \times 0.8$ | A1 | |
| $m = 0.137$ or $0.138$ | A1 | **[3]** |7\\
\includegraphics[max width=\textwidth, alt={}, center]{d5f48bef-2518-4abd-b3e1-5e48ce56cf62-4_657_618_255_760}
A small ring $R$ is attached to one end of a light inextensible string of length 70 cm . A fixed rough vertical wire passes through the ring. The other end of the string is attached to a point $A$ on the wire, vertically above $R$. A horizontal force of magnitude 5.6 N is applied to the point $J$ of the string 30 cm from $A$ and 40 cm from $R$. The system is in equilibrium with each of the parts $A J$ and $J R$ of the string taut and angle $A J R$ equal to $90 ^ { \circ }$ (see diagram).\\
(i) Find the tension in the part $A J$ of the string, and find the tension in the part $J R$ of the string.
The ring $R$ has mass 0.2 kg and is in limiting equilibrium, on the point of moving up the wire.\\
(ii) Show that the coefficient of friction between $R$ and the wire is 0.341 , correct to 3 significant figures.
A particle of mass $m \mathrm {~kg}$ is attached to $R$ and $R$ is now in limiting equilibrium, on the point of moving down the wire.\\
(iii) Given that the coefficient of friction is unchanged, find the value of $m$.
{www.cie.org.uk} after the live examination series.
}
\hfill \mbox{\textit{CAIE M1 2015 Q7 [12]}}