| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough horizontal surface, particle hanging |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with friction requiring Newton's second law applied to a two-particle system. Students must find acceleration using F=ma for both particles, then use constant acceleration equations. While it involves multiple steps and careful force analysis, it follows a well-practiced template with straightforward calculations and no novel problem-solving required. |
| Spec | 3.03l Newton's third law: extend to situations requiring force resolution3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([F = 0.7 \times 3,\ \text{WD} = 2.1 \times 0.9]\) | M1 | For using \(F = \mu R\) and \(\text{WD} = Fs\) |
| Work done is \(1.89 \text{ J}\) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Loss of PE \(= 3 \times 0.9 = 2.7 \text{ J}\) | B1 | 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([\text{KE gain} = 2.7 - 1.89]\) | M1 | For 'gain in KE \(=\) loss in PE \(-\) WD by friction' |
| Gain in KE \(= 0.81 \text{ J}\) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([T - 2.1 = 0.3a\) and \(3 - T = 0.3a \Rightarrow a = 1.5]\); \([v^2 = 2 \times 1.5 \times 0.9 = 2.7]\) | M1 | For applying Newton's 2nd law to both particles, finding \(a\) and using \(v^2 = 0 + 2as\), and attempting KE |
| \(\text{KE} = 0.5 \times (0.3 + 0.3) \times 2.7 = 0.81 \text{ J}\) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([\frac{1}{2}(0.3 + 0.3)v_{\text{at break}}^2 = 0.81]\) | M1 | For using \(\frac{1}{2}(m_A + m_B)v^2 =\) gain in KE |
| \(v_{\text{floor}}^2 = v_{\text{at break}}^2 + 2g \times 0.54\) | M1 | For using \(v^2 = u^2 + 2gs\) |
| Speed at the floor is \(3.67 \text{ ms}^{-1}\) | A1 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([0.3 \times g \times 0.54]\) or \([\frac{1}{2} \times 0.3 \times (v^2 - 2.7)]\) | M1 | For attempting PE loss or KE gain for falling particle only |
| \([1.62 = \frac{1}{2} \times 0.3 \times (v^2 - 2.7)]\) | M1 | For using PE loss \(=\) KE gain of this particle |
| Speed at floor \(= 3.67 \text{ ms}^{-1}\ (= 1.5\sqrt{6})\) | A1 | 3 marks total |
## Question 5:
### Part (i)(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $[F = 0.7 \times 3,\ \text{WD} = 2.1 \times 0.9]$ | M1 | For using $F = \mu R$ and $\text{WD} = Fs$ |
| Work done is $1.89 \text{ J}$ | A1 | 2 marks total |
### Part (i)(b):
| Working | Mark | Guidance |
|---------|------|----------|
| Loss of PE $= 3 \times 0.9 = 2.7 \text{ J}$ | B1 | 1 mark total |
### Part (i)(c):
| Working | Mark | Guidance |
|---------|------|----------|
| $[\text{KE gain} = 2.7 - 1.89]$ | M1 | For 'gain in KE $=$ loss in PE $-$ WD by friction' |
| Gain in KE $= 0.81 \text{ J}$ | A1 | 2 marks total |
*Alternative method for (i)(c):*
| Working | Mark | Guidance |
|---------|------|----------|
| $[T - 2.1 = 0.3a$ and $3 - T = 0.3a \Rightarrow a = 1.5]$; $[v^2 = 2 \times 1.5 \times 0.9 = 2.7]$ | M1 | For applying Newton's 2nd law to both particles, finding $a$ and using $v^2 = 0 + 2as$, and attempting KE |
| $\text{KE} = 0.5 \times (0.3 + 0.3) \times 2.7 = 0.81 \text{ J}$ | A1 | 2 marks total |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $[\frac{1}{2}(0.3 + 0.3)v_{\text{at break}}^2 = 0.81]$ | M1 | For using $\frac{1}{2}(m_A + m_B)v^2 =$ gain in KE |
| $v_{\text{floor}}^2 = v_{\text{at break}}^2 + 2g \times 0.54$ | M1 | For using $v^2 = u^2 + 2gs$ |
| Speed at the floor is $3.67 \text{ ms}^{-1}$ | A1 | 3 marks total |
*Alternative method for (ii) [using falling particle only]:*
| Working | Mark | Guidance |
|---------|------|----------|
| $[0.3 \times g \times 0.54]$ **or** $[\frac{1}{2} \times 0.3 \times (v^2 - 2.7)]$ | M1 | For attempting PE loss or KE gain for falling particle only |
| $[1.62 = \frac{1}{2} \times 0.3 \times (v^2 - 2.7)]$ | M1 | For using PE loss $=$ KE gain of this particle |
| Speed at floor $= 3.67 \text{ ms}^{-1}\ (= 1.5\sqrt{6})$ | A1 | 3 marks total |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{c7133fc4-9a14-43fd-b5ed-788da72291cd-3_289_567_1233_788}
Particles $A$ and $B$, each of mass 0.3 kg , are connected by a light inextensible string. The string passes over a small smooth pulley fixed at the edge of a rough horizontal surface. Particle $A$ hangs freely and particle $B$ is held at rest in contact with the surface (see diagram). The coefficient of friction between $B$ and the surface is 0.7 . Particle $B$ is released and moves on the surface without reaching the pulley.\\
(i) Find, for the first 0.9 m of $B$ 's motion,\\
\hfill \mbox{\textit{CAIE M1 2014 Q5 [8]}}