| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Elastic string equilibrium |
| Difficulty | Standard +0.3 This appears to be a standard M1 elastic string equilibrium problem involving resolving forces and taking moments. The fragmented text suggests a typical setup with an elastic string, angle θ, and force P. Such problems follow routine procedures: resolve horizontally/vertically, apply Hooke's law, and use moment equilibrium—all standard M1 techniques requiring minimal problem-solving insight. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = (16 \div 65)g\) | B1 | |
| \([8^2 = 2(16 \div 65)gS]\) | M1 | For using \(v^2 = 2as\) to find \(S\) |
| \(S = 13\) | A1 | |
| \([v^2 = 2(16 \div 65)g \times 6.5\) or \(v^2 \div 8^2 = \frac{1}{2}]\) | M1 | For using \(v^2 = 2a(\frac{1}{2}S)\) or \(v^2 \propto s\) |
| Speed is \(5.66 \text{ ms}^{-1}\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\frac{1}{2}mv^2 = mgh\) and \(S = h \div \sin\alpha]\) | M1 | For using KE gain \(=\) PE loss |
| \(S = (8^2 \div 20) \div (16 \div 65)\) | A1 | Or AEF |
| \(S = 13\) | A1 | |
| \(\frac{1}{2}mv^2 = mg(\frac{1}{2} \times 13 \times (16/65))\) | M1 | Or AEF |
| Speed is \(5.66 \text{ ms}^{-1}\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([s = \frac{1}{2}a \times (64 \div 4a^2)\) or \(s \div 13 = (\frac{1}{2})^2]\) | M1 | For using \(8 = 0 + aT\) and \(s = \frac{1}{2}a(T/2)^2\) or \(s \propto t^2\) |
| Distance is \(3.25\text{ m}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For eliminating \(at^2\) from \(s = \frac{1}{2}at^2\) and \(13 = \frac{1}{2}a(2t)^2\) | |
| Distance is \(3.25\text{ m}\) | A1 [2] |
# Question 4:
## Part (i) — Main Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = (16 \div 65)g$ | B1 | |
| $[8^2 = 2(16 \div 65)gS]$ | M1 | For using $v^2 = 2as$ to find $S$ |
| $S = 13$ | A1 | |
| $[v^2 = 2(16 \div 65)g \times 6.5$ or $v^2 \div 8^2 = \frac{1}{2}]$ | M1 | For using $v^2 = 2a(\frac{1}{2}S)$ or $v^2 \propto s$ |
| Speed is $5.66 \text{ ms}^{-1}$ | A1 [5] | |
## Part (i) — Alternative Marking Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\frac{1}{2}mv^2 = mgh$ and $S = h \div \sin\alpha]$ | M1 | For using KE gain $=$ PE loss |
| $S = (8^2 \div 20) \div (16 \div 65)$ | A1 | Or AEF |
| $S = 13$ | A1 | |
| $\frac{1}{2}mv^2 = mg(\frac{1}{2} \times 13 \times (16/65))$ | M1 | Or AEF |
| Speed is $5.66 \text{ ms}^{-1}$ | A1 [5] | |
## Part (ii) — Main Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[s = \frac{1}{2}a \times (64 \div 4a^2)$ or $s \div 13 = (\frac{1}{2})^2]$ | M1 | For using $8 = 0 + aT$ and $s = \frac{1}{2}a(T/2)^2$ or $s \propto t^2$ |
| Distance is $3.25\text{ m}$ | A1 [2] | |
## Part (ii) — Alternative Marking Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For eliminating $at^2$ from $s = \frac{1}{2}at^2$ and $13 = \frac{1}{2}a(2t)^2$ |
| Distance is $3.25\text{ m}$ | A1 [2] | |
---4\\
$A$
B
$A \quad B$\\
$\begin{array} { l l } B & \\ A & B \end{array}$\\
$P B$
$$P \theta$$
$\theta$\\
P\\
$\theta$\\
L\\
$P$\\
$P$\\
(i)(i)\\
P
$\theta$\\
$P \quad \underline { \theta }$\\
(ii) -\\
$\underline { \theta }$\\
$5 \theta$
$$\begin{gathered}
\\
\theta
\end{gathered} \quad P$$
(i) $P$\\
(ii)
\hfill \mbox{\textit{CAIE M1 2013 Q4}}