| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Variable mass or unknown mass |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem requiring application of Newton's second law to both particles and use of constant acceleration equations. The setup is straightforward with given time and velocity to find acceleration, then solving simultaneous equations for mass and tension. Slightly above average difficulty due to the algebraic manipulation with variable mass m, but follows a well-practiced method with no conceptual surprises. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0.6 = 0 + 0.3a]\) | M1 | For using \(v = 0 + at\) |
| Acceleration is \(2\ \text{ms}^{-2}\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mg - T = 2m,\ T - (1-m)g = 2(1-m)\) | M1 | For applying Newton's 2nd law to A or to B |
| \([m = T/8 \Rightarrow T - (10 - 1.25T) = 2 - 0.25T]\) or \(T = 8m \Rightarrow 8m - (10 - 10m) = 2 - 2m\) | M1 | For eliminating or evaluating \(m\) |
| \(m = 0.6\) and \(T = 8m\) | A1 | |
| \(m = 0.6\) and tension is 4.8 N | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\{m + (1-m)\} \times 2 = \{m - (1-m)\} \times g]\) | M1 | For using \((m_A + m_B)a = (m_A - m_B)g\) |
| \(m = 0.6\) | A1 | |
| \([mg - T = 2m\) or \(T - (1-m)g = 2(1-m)]\) | M1 | For applying Newton's 2nd law to A or to B, substituting for \(m\) and solving for \(T\) |
| Tension is 4.8 N | A1 |
# Question 2:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.6 = 0 + 0.3a]$ | M1 | For using $v = 0 + at$ |
| Acceleration is $2\ \text{ms}^{-2}$ | A1 | **[2]** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg - T = 2m,\ T - (1-m)g = 2(1-m)$ | M1 | For applying Newton's 2nd law to A or to B |
| $[m = T/8 \Rightarrow T - (10 - 1.25T) = 2 - 0.25T]$ or $T = 8m \Rightarrow 8m - (10 - 10m) = 2 - 2m$ | M1 | For eliminating or evaluating $m$ |
| $m = 0.6$ and $T = 8m$ | A1 | |
| $m = 0.6$ and tension is 4.8 N | A1 | **[4]** |
### Alternative for part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\{m + (1-m)\} \times 2 = \{m - (1-m)\} \times g]$ | M1 | For using $(m_A + m_B)a = (m_A - m_B)g$ |
| $m = 0.6$ | A1 | |
| $[mg - T = 2m$ or $T - (1-m)g = 2(1-m)]$ | M1 | For applying Newton's 2nd law to A or to B, substituting for $m$ and solving for $T$ |
| Tension is 4.8 N | A1 | |
---2 Particles $A$ and $B$ of masses $m \mathrm {~kg}$ and $( 1 - m ) \mathrm { kg }$ respectively are attached to the ends of a light inextensible string which passes over a fixed smooth pulley. The system is released from rest with the straight parts of the string vertical. $A$ moves vertically downwards and 0.3 seconds later it has speed $0.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find\\
(i) the acceleration of $A$,\\
(ii) the value of $m$ and the tension in the string.
\hfill \mbox{\textit{CAIE M1 2012 Q2 [6]}}