| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Connected particles with pulley |
| Difficulty | Standard +0.3 This is a standard connected particles problem requiring work-energy principles. Part (i) involves straightforward calculation of work done by tension (which can be found from F=ma), and part (ii) requires energy conservation after B stops. The setup is familiar and the methods are routine for M1 students, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (applying Newton's 2nd law) | M1 | For applying Newton's 2nd law to A or B |
| \(T - 0.4g = 0.4a\) or \(1.6g - T = 1.6a\) | A1 | |
| \(1.6g - T = 1.6a\) or \(T - 0.4g = 0.4a\) or \(1.6g - 0.4g = (1.6 + 0.4)a\) | B1 | |
| \(T = 6.4\) | A1 | |
| Work done by tension is \(7.68\) J | B1ft | ft answer for total distance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (applying Newton's 2nd law) | M1 | For applying Newton's 2nd law to A or B |
| \(T - 0.4g = 0.4a\) or \(1.6g - T = 1.6a\) | A1 | |
| \(1.6g - T = 1.6a\) or \(T - 0.4g = 0.4a\) or \(1.6g - 0.4g = (1.6 + 0.4)a\) | B1 | |
| \(\text{WD by }T = \text{initial PE} - \text{final KE} = 1.6 \times g \times 1.2 - \frac{1}{2} \times 1.6 \times 14.4\) | M1 | For finding \(v^2\) and applying Work/Energy equation to B |
| \(\text{WD by }T = 19.2 - 11.52 = 7.68\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([1.6 \times 10 \times 1.2 = \frac{1}{2} \times 1.6 \times v^2 + 7.68]\) | M1 | For using PE loss = KE gain + WD by T to find \(v^2\) |
| \(v^2 = 14.4\) | A1 | |
| \(14.4 = 2 \times 10 \times h\), \(h = 0.72\), \(H = 2 \times 1.2 + h\) | M1 | For using PCE for A's motion after B reaches the ground or \(0 = u^2 - 2gh\) and \(H = 2 \times 1.2 + h\) |
| Greatest height is \(3.12\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v^2 = 2 \times 6 \times 1.2]\) | M1 | For using \(v^2 = 2as\) to find \(v^2\) |
| \(v^2 = 14.4\) | A1 | |
| \(14.4 = 2 \times 10 \times h\), \(h = 0.72\), \(H = 2 \times 1.2 + h\) | M1 | For using PCE for A's motion after B reaches the ground or \(0 = u^2 - 2gh\) and \(H = 2 \times 1.2 + h\) |
| Greatest height is \(3.12\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{WD by }T = \text{Increase in PE}\): \(7.68 = 0.4 \times g \times s\) | M1 | For applying WD by T to particle A's complete motion |
| \(s = 1.92\) | A1 | |
| \(H = 1.2 + s\) | M1 | For adding \(1.2\) to \(s\) |
| \(H = 1.2 + 1.92 = 3.12\) Height \(= 3.12\) m | A1 |
# Question 6 (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (applying Newton's 2nd law) | M1 | For applying Newton's 2nd law to A or B |
| $T - 0.4g = 0.4a$ or $1.6g - T = 1.6a$ | A1 | |
| $1.6g - T = 1.6a$ or $T - 0.4g = 0.4a$ or $1.6g - 0.4g = (1.6 + 0.4)a$ | B1 | |
| $T = 6.4$ | A1 | |
| Work done by tension is $7.68$ J | B1ft | ft answer for total distance |
**Alternative Mark Scheme for 6(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| (applying Newton's 2nd law) | M1 | For applying Newton's 2nd law to A or B |
| $T - 0.4g = 0.4a$ or $1.6g - T = 1.6a$ | A1 | |
| $1.6g - T = 1.6a$ or $T - 0.4g = 0.4a$ or $1.6g - 0.4g = (1.6 + 0.4)a$ | B1 | |
| $\text{WD by }T = \text{initial PE} - \text{final KE} = 1.6 \times g \times 1.2 - \frac{1}{2} \times 1.6 \times 14.4$ | M1 | For finding $v^2$ and applying Work/Energy equation to B |
| $\text{WD by }T = 19.2 - 11.52 = 7.68$ | A1 | |
---
# Question 6 (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[1.6 \times 10 \times 1.2 = \frac{1}{2} \times 1.6 \times v^2 + 7.68]$ | M1 | For using PE loss = KE gain + WD by T to find $v^2$ |
| $v^2 = 14.4$ | A1 | |
| $14.4 = 2 \times 10 \times h$, $h = 0.72$, $H = 2 \times 1.2 + h$ | M1 | For using PCE for A's motion after B reaches the ground or $0 = u^2 - 2gh$ and $H = 2 \times 1.2 + h$ |
| Greatest height is $3.12$ m | A1 | |
**First Alternative Marking Scheme for 6(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v^2 = 2 \times 6 \times 1.2]$ | M1 | For using $v^2 = 2as$ to find $v^2$ |
| $v^2 = 14.4$ | A1 | |
| $14.4 = 2 \times 10 \times h$, $h = 0.72$, $H = 2 \times 1.2 + h$ | M1 | For using PCE for A's motion after B reaches the ground or $0 = u^2 - 2gh$ and $H = 2 \times 1.2 + h$ |
| Greatest height is $3.12$ m | A1 | |
**Second Alternative Marking Scheme for 6(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{WD by }T = \text{Increase in PE}$: $7.68 = 0.4 \times g \times s$ | M1 | For applying WD by T to particle A's complete motion |
| $s = 1.92$ | A1 | |
| $H = 1.2 + s$ | M1 | For adding $1.2$ to $s$ |
| $H = 1.2 + 1.92 = 3.12$ Height $= 3.12$ m | A1 | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{79b90ef5-ef3a-4c59-b662-d0fbfba813ca-3_526_519_902_813}
Particles $A$ of mass 0.4 kg and $B$ of mass 1.6 kg are attached to the ends of a light inextensible string which passes over a fixed smooth pulley. $A$ is held at rest and $B$ hangs freely, with both straight parts of the string vertical and both particles at a height of 1.2 m above the floor (see diagram). $A$ is released and both particles start to move.\\
(i) Find the work done on $B$ by the tension in the string, as $B$ moves to the floor.
When particle $B$ reaches the floor it remains at rest. Particle $A$ continues to move upwards.\\
(ii) Find the greatest height above the floor reached by particle $A$.
\hfill \mbox{\textit{CAIE M1 2013 Q6 [9]}}