| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| 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 two-particle pulley system problem requiring Newton's second law, kinematics, and friction. Part (i) involves straightforward application of F=ma and SUVAT equations. Part (ii) adds friction and requires analyzing motion in two phases (before and after B hits ground), but follows well-established procedures taught in M1. Slightly above average difficulty due to the two-phase analysis in part (ii), but still a routine mechanics question. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03r Friction: concept and vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([2.4g - T = 2.4aT = 1.6a]\) or the system equation \(2.4g = (1.6 + 2.4)a\) | M1 | For applying Newton's second law to one of the particles or to the combined system |
| M1 | For applying Newton's second law to a second particle if needed and/or solving for \(a\) | |
| \(a = 6 \text{ ms}^{-2}\) | A1 | |
| \(0.5 = \frac{1}{2} \times 6 \times t^2\) | M1 | For using \(s = ut + \frac{1}{2}at^2\) |
| \(t = 0.408\text{s}\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\text{PE loss} = 2.4 \times g \times 0.5 = 12\), \(\text{KE gain} = \frac{1}{2}(1.6 + 2.4)v^2 = 2v^2]\) | M1 | For attempting to find PE and KE as \(B\) reaches the ground |
| \([12 = 2v^2]\) | M1 | Using PE loss = KE gain |
| \(v^2 = 6 \rightarrow v = 2.45 \text{ ms}^{-1}\) | A1 | |
| \([0.5 = \frac{1}{2} \times (0 + 2.45) \times t]\) | M1 | Using \(s = \frac{1}{2}(u + v)t\) |
| \(t = 0.408\text{s}\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 1.6g = 16\) and \(F = \frac{3}{8}R = 6\) | B1 | |
| System: \([2.4g - 6 = (1.6 + 2.4)a]\) | M1 | For using Newton's second law for both particles or the system |
| \(2.4g - T = 2.4a\) and \(T - 6 = 1.6a\) | A1 | Both or system equation |
| \([a = 4.5]\) | M1 | For finding \(a\) and using \(v^2 = u^2 + 2as\) to find \(v\) as \(B\) reaches the ground |
| \(v = \sqrt{2 \times 4.5 \times 0.5} = \sqrt{4.5} = 2.12 \text{ ms}^{-1}\) | A1 | |
| \(-6 = 1.6a \rightarrow a = -3.75 \text{ ms}^{-2}\) | M1 | For finding the deceleration of \(A\) and using \(v^2 = u^2 + 2as\) to find \(s\) the total distance travelled by \(A\) |
| \(0 = 4.5 + 2 \times (-3.75) \times (s - 0.5)\) | ||
| \(s = 1.1\text{ m}\) | A1 | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 1.6g = 16\) and \(F = \frac{3}{8}R = 6\) | B1 | |
| \(\text{PE loss} = 2.4 \times g \times 0.5 [= 12]\), \(\text{KE gain} = \frac{1}{2} \times (1.6 + 2.4) \times v^2 [= 2v^2]\) | M1, A1 | For attempting PE loss and KE gain as \(B\) reaches the ground; for both PE and KE correct |
| M1 | For using PE loss = KE gain + WD against \(F\) | |
| \(12 = 2v^2 + 6 \times 0.5 \rightarrow v^2 = 4.5 \rightarrow v = 2.12\) | A1 | |
| Loss of KE = WD against \(F\): \([\frac{1}{2} \times 1.6 \times 4.5 = 6 \times (s - 0.5)]\) | M1 | For considering the motion of \(A\) after \(B\) reaches the ground to find \(s\) the total distance travelled |
| \(s = 1.1\text{ m}\) | A1 | 7 |
# Question 7:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[2.4g - T = 2.4aT = 1.6a]$ or the system equation $2.4g = (1.6 + 2.4)a$ | M1 | For applying Newton's second law to one of the particles or to the combined system |
| | M1 | For applying Newton's second law to a second particle if needed and/or solving for $a$ |
| $a = 6 \text{ ms}^{-2}$ | A1 | |
| $0.5 = \frac{1}{2} \times 6 \times t^2$ | M1 | For using $s = ut + \frac{1}{2}at^2$ |
| $t = 0.408\text{s}$ | A1 | 5 | Accept $t = \sqrt{6}/6$ |
### Alternative for 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{PE loss} = 2.4 \times g \times 0.5 = 12$, $\text{KE gain} = \frac{1}{2}(1.6 + 2.4)v^2 = 2v^2]$ | M1 | For attempting to find PE and KE as $B$ reaches the ground |
| $[12 = 2v^2]$ | M1 | Using PE loss = KE gain |
| $v^2 = 6 \rightarrow v = 2.45 \text{ ms}^{-1}$ | A1 | |
| $[0.5 = \frac{1}{2} \times (0 + 2.45) \times t]$ | M1 | Using $s = \frac{1}{2}(u + v)t$ |
| $t = 0.408\text{s}$ | A1 | 5 | Accept $t = \sqrt{6}/6$ |
---
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 1.6g = 16$ and $F = \frac{3}{8}R = 6$ | B1 | |
| System: $[2.4g - 6 = (1.6 + 2.4)a]$ | M1 | For using Newton's second law for both particles or the system |
| $2.4g - T = 2.4a$ and $T - 6 = 1.6a$ | A1 | Both or system equation |
| $[a = 4.5]$ | M1 | For finding $a$ and using $v^2 = u^2 + 2as$ to find $v$ as $B$ reaches the ground |
| $v = \sqrt{2 \times 4.5 \times 0.5} = \sqrt{4.5} = 2.12 \text{ ms}^{-1}$ | A1 | |
| $-6 = 1.6a \rightarrow a = -3.75 \text{ ms}^{-2}$ | M1 | For finding the deceleration of $A$ and using $v^2 = u^2 + 2as$ to find $s$ the total distance travelled by $A$ |
| $0 = 4.5 + 2 \times (-3.75) \times (s - 0.5)$ | | |
| $s = 1.1\text{ m}$ | A1 | 7 |
### First Alternative for 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 1.6g = 16$ and $F = \frac{3}{8}R = 6$ | B1 | |
| $\text{PE loss} = 2.4 \times g \times 0.5 [= 12]$, $\text{KE gain} = \frac{1}{2} \times (1.6 + 2.4) \times v^2 [= 2v^2]$ | M1, A1 | For attempting PE loss and KE gain as $B$ reaches the ground; for both PE and KE correct |
| | M1 | For using PE loss = KE gain + WD against $F$ |
| $12 = 2v^2 + 6 \times 0.5 \rightarrow v^2 = 4.5 \rightarrow v = 2.12$ | A1 | |
| Loss of KE = WD against $F$: $[\frac{1}{2} \times 1.6 \times 4.5 = 6 \times (s - 0.5)]$ | M1 | For considering the motion of $A$ after $B$ reaches the ground to find $s$ the total distance travelled |
| $s = 1.1\text{ m}$ | A1 | 7 |7\\
\includegraphics[max width=\textwidth, alt={}, center]{fd2fbf13-912c-46c5-a470-306b2269aa0b-3_378_1001_1672_573}
A particle $A$ of mass 1.6 kg rests on a horizontal table and is attached to one end of a light inextensible string. The string passes over a small smooth pulley $P$ fixed at the edge of the table. The other end of the string is attached to a particle $B$ of mass 2.4 kg which hangs freely below the pulley. The system is released from rest with the string taut and with $B$ at a height of 0.5 m above the ground, as shown in the diagram. In the subsequent motion $A$ does not reach $P$ before $B$ reaches the ground.\\
(i) Given that the table is smooth, find the time taken by $B$ to reach the ground.\\
(ii) Given instead that the table is rough and that the coefficient of friction between $A$ and the table is $\frac { 3 } { 8 }$, find the total distance travelled by $A$. You may assume that $A$ does not reach the pulley.
\hfill \mbox{\textit{CAIE M1 2016 Q7 [12]}}