CAIE M1 2016 June — Question 5 7 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2016
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeParticle on rough incline, particle hanging
DifficultyStandard +0.3 This is a standard two-particle pulley system requiring resolution of forces on an incline, friction calculation, and Newton's second law applied to a connected system. While it involves multiple steps (finding sin α and cos α from tan α, resolving forces parallel to slope, applying F=ma to both particles), these are routine mechanics techniques with no novel insight required. Slightly easier than average due to straightforward setup and clear solution path.
Spec3.03k Connected particles: pulleys and equilibrium3.03r Friction: concept and vector form

5 \includegraphics[max width=\textwidth, alt={}, center]{099c81e0-a95a-4f98-801c-32d905ef7c7d-3_432_710_258_721} Two particles of masses 5 kg and 10 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The 5 kg particle is on a rough fixed slope which is at an angle of \(\alpha\) to the horizontal, where \(\tan \alpha = \frac { 3 } { 4 }\). The 10 kg particle hangs below the pulley (see diagram). The coefficient of friction between the slope and the 5 kg particle is \(\frac { 1 } { 2 }\). The particles are released from rest. Find the acceleration of the particles and the tension in the string.

Question 5:
AnswerMarks Guidance
AnswerMark Guidance
\(R = 5g\cos\alpha = 4g\); \(F = 0.5 \times 4g = 2g\)B1 For finding the normal reaction \(R\) acting on the 5 kg particle and using \(F = \mu R\)
M1For applying Newton's second law to one or both particles or to the system
\(T - 2g - 5g\sin\alpha = 5a \rightarrow T - 5g = 5a\)A1 System equation is \(10g - 5g\sin\alpha - 2g = 5g = 15a\)
\(10g - T = 10a\)A1
\([5g = 15a]\)M1 For eliminating \(T\) and solve for \(a\)
\(a = g/3 = 3.33\) ms\(^{-2}\)A1
\(T = 10g - 10(g/3) = 20g/3 = 66.7\) NB1 [7]
## Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 5g\cos\alpha = 4g$; $F = 0.5 \times 4g = 2g$ | B1 | For finding the normal reaction $R$ acting on the 5 kg particle and using $F = \mu R$ |
| | M1 | For applying Newton's second law to one or both particles or to the system |
| $T - 2g - 5g\sin\alpha = 5a \rightarrow T - 5g = 5a$ | A1 | System equation is $10g - 5g\sin\alpha - 2g = 5g = 15a$ |
| $10g - T = 10a$ | A1 | |
| $[5g = 15a]$ | M1 | For eliminating $T$ and solve for $a$ |
| $a = g/3 = 3.33$ ms$^{-2}$ | A1 | |
| $T = 10g - 10(g/3) = 20g/3 = 66.7$ N | B1 | [7] |

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\includegraphics[max width=\textwidth, alt={}, center]{099c81e0-a95a-4f98-801c-32d905ef7c7d-3_432_710_258_721}

Two particles of masses 5 kg and 10 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The 5 kg particle is on a rough fixed slope which is at an angle of $\alpha$ to the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$. The 10 kg particle hangs below the pulley (see diagram). The coefficient of friction between the slope and the 5 kg particle is $\frac { 1 } { 2 }$. The particles are released from rest. Find the acceleration of the particles and the tension in the string.

\hfill \mbox{\textit{CAIE M1 2016 Q5 [7]}}