| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | String breaks during motion |
| Difficulty | Standard +0.3 This is a standard two-part pulley problem requiring Newton's second law to find tension (routine calculation with connected particles), followed by projectile motion after string breaks. The calculations are straightforward with clear setup, slightly above average due to the two-phase motion analysis but well within typical M1 scope. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For applying Newton's 2nd law to A or to B | |
| \(T - 0.3g = 0.3a\) or \(0.7g - T = 0.7a\) | A1 | |
| \(0.7g - T = 0.7a\) or \(T - 0.3g = 0.3a\) or \(0.7g - 0.3g = (0.7 + 0.3)a\) | B1 | |
| Tension is \(4.2\) N | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 4\) | B1 | May be scored in (i) |
| \(s_\text{taut} = \frac{1.6^2}{2 \times 4}\) \((= 0.32)\) | B1 | |
| \((0.52 + 0.32) = -1.6t + 5t^2\) | M1 | For using \(s = ut + \frac{1}{2}gt^2\) |
| \((t - 0.6)(5t + 1.4) = 0\) | M1 | For solving the resultant quadratic equation |
| Time taken is \(0.6\) s | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0^2 = 1.6^2 - 2gs_\text{up}\), \(t_\text{up} = \frac{2s_\text{up}}{(1.6+0)}\) \((= 0.16)\) | M1 | For using kinematic formulae to find \(t_\text{up}\) |
| \(0.52 + s_\text{taut} + s_\text{up} = 0 + \frac{1}{2}gt_\text{down}^2\) \((t_\text{down} = 0.44)\) | M1 | For using kinematic formulae to find \(t_\text{down}\) |
| Time taken \(= t_\text{up} + t_\text{down} = 0.6\) s | B1 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For applying Newton's 2nd law to A or to B |
| $T - 0.3g = 0.3a$ or $0.7g - T = 0.7a$ | A1 | |
| $0.7g - T = 0.7a$ or $T - 0.3g = 0.3a$ or $0.7g - 0.3g = (0.7 + 0.3)a$ | B1 | |
| Tension is $4.2$ N | A1 [4] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 4$ | B1 | May be scored in (i) |
| $s_\text{taut} = \frac{1.6^2}{2 \times 4}$ $(= 0.32)$ | B1 | |
| $(0.52 + 0.32) = -1.6t + 5t^2$ | M1 | For using $s = ut + \frac{1}{2}gt^2$ |
| $(t - 0.6)(5t + 1.4) = 0$ | M1 | For solving the resultant quadratic equation |
| Time taken is $0.6$ s | A1 [5] | |
**Alternative Marking Scheme (last three marks):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0^2 = 1.6^2 - 2gs_\text{up}$, $t_\text{up} = \frac{2s_\text{up}}{(1.6+0)}$ $(= 0.16)$ | M1 | For using kinematic formulae to find $t_\text{up}$ |
| $0.52 + s_\text{taut} + s_\text{up} = 0 + \frac{1}{2}gt_\text{down}^2$ $(t_\text{down} = 0.44)$ | M1 | For using kinematic formulae to find $t_\text{down}$ |
| Time taken $= t_\text{up} + t_\text{down} = 0.6$ s | B1 | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{3e58aa5a-3789-4aaf-8656-b5b98cd7f693-3_518_515_1436_815}
Particles $A$ and $B$, of masses 0.3 kg and 0.7 kg respectively, are attached to the ends of a light inextensible string. The string 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 0.52 m above the floor (see diagram). $A$ is released and both particles start to move.\\
(i) Find the tension in the string.
When both particles are moving with speed $1.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ the string breaks.\\
(ii) Find the time taken, from the instant that the string breaks, for $A$ to reach the floor.\\
$7 \quad$ A particle $P$ starts from rest at a point $O$ and moves in a straight line. $P$ has acceleration $0.6 t \mathrm {~m} \mathrm {~s} ^ { - 2 }$ at time $t$ seconds after leaving $O$, until $t = 10$.\\
(i) Find the velocity and displacement from $O$ of $P$ when $t = 10$.
After $t = 10 , P$ has acceleration $- 0.4 t \mathrm {~m} \mathrm {~s} ^ { - 2 }$ until it comes to rest at a point $A$.\\
(ii) Find the distance $O A$.
\hfill \mbox{\textit{CAIE M1 2013 Q6 [9]}}