| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Multi-part pulley system, subsequent motion |
| Difficulty | Standard +0.3 This is a standard connected particles problem requiring systematic application of Newton's second law, followed by projectile motion after the string breaks. While multi-part with several steps (finding k, tension, final speed, time, and sketching), each component uses routine mechanics techniques without requiring novel insight. The numerical values work out cleanly, and the problem structure is typical for M1 level. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[0.81 = 0 + \frac{1}{2}\times a\times 0.9^2\right]\) | M1 | For use of \(s = ut + \frac{1}{2}at^2\) |
| \(a = 2\) | A1 | |
| \(T - mg = ma\) or \(kmg - T = kma\) | M1 | Use of Newton's Second Law for \(A\) or \(B\) or use of \(a = \frac{(m_B - m_A)g}{(m_B+m_A)}\) |
| \(T - mg = ma\) and \(kmg - T = kma\) or \(\left[a = \frac{(km-m)g}{(km+m)}\right]\) | A1 | |
| \(a = \frac{(kg-g)}{(k+1)} = 2 \rightarrow k = \ldots\) | M1 | Solves to find \(k\) |
| \(k = 1.5\) | A1 | |
| \(T = 10m + 2m = 12m\) N | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Velocity of \(A\) when string breaks \(= 2\times0.9\) \((= 1.8 \text{ ms}^{-1}\) upwards\()\) | B1FT | For use of \(v = u+at\) ft \(a\) from (i) |
| \(v^2 = 1.8^2 + 2g\times1.62 \rightarrow v = \ldots\) | M1 | For use of suvat to find \(v_A\) at ground |
| Speed is \(5.97 \text{ ms}^{-1}\) | A1 | AG |
| Time taken \(= \frac{(1.8+5.97)}{g} = 0.777s\) \((0.7769\ldots)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Straight line from \((0,0)\) to \((0.9, 1.8)\) | B1 | |
| Straight line from \((0.9, 1.8)\) to approx. \((1.7, -6)\) | B1FT | FT \(0.9 + t\) from (ii) for \(1.7\) |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[0.81 = 0 + \frac{1}{2}\times a\times 0.9^2\right]$ | M1 | For use of $s = ut + \frac{1}{2}at^2$ |
| $a = 2$ | A1 | |
| $T - mg = ma$ **or** $kmg - T = kma$ | M1 | Use of Newton's Second Law for $A$ or $B$ or use of $a = \frac{(m_B - m_A)g}{(m_B+m_A)}$ |
| $T - mg = ma$ **and** $kmg - T = kma$ or $\left[a = \frac{(km-m)g}{(km+m)}\right]$ | A1 | |
| $a = \frac{(kg-g)}{(k+1)} = 2 \rightarrow k = \ldots$ | M1 | Solves to find $k$ |
| $k = 1.5$ | A1 | |
| $T = 10m + 2m = 12m$ N | B1 | AG |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Velocity of $A$ when string breaks $= 2\times0.9$ $(= 1.8 \text{ ms}^{-1}$ upwards$)$ | B1FT | For use of $v = u+at$ ft $a$ from (i) |
| $v^2 = 1.8^2 + 2g\times1.62 \rightarrow v = \ldots$ | M1 | For use of suvat to find $v_A$ at ground |
| Speed is $5.97 \text{ ms}^{-1}$ | A1 | AG |
| Time taken $= \frac{(1.8+5.97)}{g} = 0.777s$ $(0.7769\ldots)$ | B1 | |
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Straight line from $(0,0)$ to $(0.9, 1.8)$ | B1 | |
| Straight line from $(0.9, 1.8)$ to approx. $(1.7, -6)$ | B1FT | FT $0.9 + t$ from (ii) for $1.7$ |7\\
\includegraphics[max width=\textwidth, alt={}, center]{60a41d3b-62a0-40d9-a30d-0560903429af-12_565_511_260_817}
Two particles $A$ and $B$ have masses $m \mathrm {~kg}$ and $k m \mathrm {~kg}$ respectively, where $k > 1$. The particles are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley and the particles hang vertically below it. Both particles are at a height of 0.81 m above horizontal ground (see diagram). The system is released from rest and particle $B$ reaches the ground 0.9 s later. The particle $A$ does not reach the pulley in its subsequent motion.\\
(i) Find the value of $k$ and show that the tension in the string before $B$ reaches the ground is equal to $12 m \mathrm {~N}$.\\
At the instant when $B$ reaches the ground, the string breaks.\\
(ii) Show that the speed of $A$ when it reaches the ground is $5.97 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, correct to 3 significant figures, and find the time taken, after the string breaks, for $A$ to reach the ground.\\
(iii) Sketch a velocity-time graph for the motion of particle $A$ from the instant when the system is released until $A$ reaches the ground.
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M1 2019 Q7 [13]}}