| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | String breaks during motion |
| Difficulty | Standard +0.8 This is a multi-phase mechanics problem requiring analysis of motion before and after string breaks, involving friction, connected particles, and SUVAT equations across three distinct phases. It demands careful tracking of velocities between phases and understanding when friction changes direction, which is significantly more complex than standard single-phase pulley problems. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using Newton's 2nd law for both particles | |
| \(T - 0.2 \times 3 = 0.3a\) and \(7 - T = 0.7a\) | A1 | |
| Acceleration \(= 6.4\) ms\(^{-2}\) | A1 | |
| \([v = 0 + 6.4 \times 0.25]\) | M1 | For using \(v = 0 + at\) to find speed when string breaks |
| \(v = 1.6\) ms\(^{-1}\) | A1 | |
| \(\left[\text{Distance} = 0 + \frac{1}{2}6.4\times0.25^2\right]\) | M1 | For using \(s = ut + \frac{1}{2}at^2\) to find distance moved before break |
| Distance \(= 0.2\) m | A1 | |
| \([v^2 = 1.6^2 + 2g\times(0.5 - 0.2)]\) | M1 | For using \(v^2 = u^2 + 2gs\) to find speed when \(B\) hits floor |
| Speed is \(2.93\) ms\(^{-1}\) | A1 | |
| Total: 9 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using \(v^2 = u^2 + 2as\) | M1 | For finding distance travelled by \(A\) after break from \(v^2 = u^2 + 2as\) |
| Distance travelled after break \(= (0 - 1.6^2) \div (2 \times -2) = 0.64\) | A1 | For \(A\), \(F = 0.2 \times 3\) and so \(-0.2 \times 3 = 0.3a\) so \(a = -2\) |
| Total distance travelled \(= 0.2 + 0.64 = 0.84\) | B1 [3] | Distance \(= 0.84\) m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = 2.52\), \(F = 0.2 \times 3\); WD by \(T = 2.52 \times 0.2\); WD by \(F = 0.2 \times 3 \times d\) | B1 | For stating WD by \(T\) on \(A\) and WD by \(F\) |
| \([0.6d = 2.52 \times 0.2]\) | M1 | Using WD by \(F\) = WD by \(T\) (No change in KE or PE for \(A\)) |
| WD by \(T\) = WD by \(F \rightarrow d = 0.84\) | A1 [3] | Distance \(= 0.84\) m |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using Newton's 2nd law for both particles |
| $T - 0.2 \times 3 = 0.3a$ and $7 - T = 0.7a$ | A1 | |
| Acceleration $= 6.4$ ms$^{-2}$ | A1 | |
| $[v = 0 + 6.4 \times 0.25]$ | M1 | For using $v = 0 + at$ to find speed when string breaks |
| $v = 1.6$ ms$^{-1}$ | A1 | |
| $\left[\text{Distance} = 0 + \frac{1}{2}6.4\times0.25^2\right]$ | M1 | For using $s = ut + \frac{1}{2}at^2$ to find distance moved before break |
| Distance $= 0.2$ m | A1 | |
| $[v^2 = 1.6^2 + 2g\times(0.5 - 0.2)]$ | M1 | For using $v^2 = u^2 + 2gs$ to find speed when $B$ hits floor |
| Speed is $2.93$ ms$^{-1}$ | A1 | |
| **Total: 9 marks** | | |
## Question 7(ii):
**Main Method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $v^2 = u^2 + 2as$ | M1 | For finding distance travelled by $A$ after break from $v^2 = u^2 + 2as$ |
| Distance travelled after break $= (0 - 1.6^2) \div (2 \times -2) = 0.64$ | A1 | For $A$, $F = 0.2 \times 3$ and so $-0.2 \times 3 = 0.3a$ so $a = -2$ |
| Total distance travelled $= 0.2 + 0.64 = 0.84$ | B1 [3] | Distance $= 0.84$ m |
---
**Alternative Method for 7(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = 2.52$, $F = 0.2 \times 3$; WD by $T = 2.52 \times 0.2$; WD by $F = 0.2 \times 3 \times d$ | B1 | For stating WD by $T$ on $A$ and WD by $F$ |
| $[0.6d = 2.52 \times 0.2]$ | M1 | Using WD by $F$ = WD by $T$ (No change in KE or PE for $A$) |
| WD by $T$ = WD by $F \rightarrow d = 0.84$ | A1 [3] | Distance $= 0.84$ m |7\\
\includegraphics[max width=\textwidth, alt={}, center]{f4f2996b-5382-4b0d-9804-b5f5945946b3-3_376_1052_1171_548}
Particles $A$ and $B$, of masses 0.3 kg and 0.7 kg respectively, are attached to the ends of a light inextensible string. Particle $A$ is held at rest on a rough horizontal table with the string passing over a smooth pulley fixed at the edge of the table. The coefficient of friction between $A$ and the table is 0.2 . Particle $B$ hangs vertically below the pulley at a height of 0.5 m above the floor (see diagram). The system is released from rest and 0.25 s later the string breaks. A does not reach the pulley in the subsequent motion. Find\\
(i) the speed of $B$ immediately before it hits the floor,\\
(ii) the total distance travelled by $A$.
\hfill \mbox{\textit{CAIE M1 2015 Q7 [12]}}