## Question 5:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_2\cos\theta = m_2g$ | M1 | 1.1a – Resolving $T_2$ vertically and balancing forces on $R$; do not allow extra forces present. In this solution $\theta$ is the angle between $RP$ and $RA$; sin may be seen instead if $\theta$ is measured horizontally. |
| $T_2 = \frac{m_2 \times 9.8}{0.8} = 12.25m_2$ | A1 | 1.1 – Allow use of g, e.g. $\frac{5}{4}gm_2$; do not allow incomplete expressions e.g. $\frac{m_2 g}{\sin 53.13}$ |
| **[2]** | | |

### Part (b)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_2\cos\theta + m_1g = T_1\cos\theta$ | M1 | 3.1b – Vertical forces on $P$; 3 terms including resolving of $T_1$; allow sign error. Or $T_1\cos\theta = m_1g + m_2g$ (equation for the system as a whole) |
| $T_1 = T_2 + \frac{9.8m_1}{0.8} =$ | A1 | 2.1 – **AG** Dividing by $\cos\theta$ (= 0.8), substituting their $T_2$ and rearranging. Allow 12.25 instead of $\frac{49}{4}$. At least one intermediate step must be seen. |
| **[2]** | | |

### Part (b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_1\sin\theta + T_2\sin\theta = m_1 a$ | M1 | 3.1b – NII horizontally for $P$; 3 terms including resolving of tensions; allow sign error. Could see $a$ or $0.6\omega^2$ or $\frac{v^2}{0.6}$ or $\omega^2 r$ or $\frac{v^2}{r}$ |
| $12.25(m_1+m_2)\times 0.6 + 12.25m_2\times 0.6 = m_1\times 0.6\omega^2$ | M1 | 1.1 – Substituting for $T_1$, their $T_2$, $\sin\theta$ and $a$. $\sin\theta = 0.6$; must be $a = 0.6\omega^2$ |
| $\omega^2 = \frac{7.35m_1 + 14.7m_2}{0.6m_1} = \frac{49(m_1+2m_2)}{4m_1}$ | A1 | 2.1 – **AG** Must see an intermediate step |
| **[3]** | | |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $m_1 \gg m_2 \Rightarrow \frac{2m_2}{m_1} \approx 0$ or $\frac{49m_2}{4m_1} \approx 0$ | M1 | 1.1 – Allow argument such as if $m_1 \gg m_2$ then $m_1 + 2m_2 \approx m_1$. Do not allow the assumption that $m_2 = 0$ |
| $\omega \approx \sqrt{\frac{49m}{4m}} = 3.5$ | A1 | 1.1 – **AG** $m$ may be missing. SC1 for result following argument that $m_2$ is negligible (by comparison with $m_1$) without justification, or using trial values of $m_1$ and $m_2$ with $m_1 \gg m_2$. If using trial values, $m_1$ must be at least $70 \times m_2$ to give $\omega = 3.5$ to 1dp. |
| **[2]** | | |

### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = r\omega = 0.6\sqrt{\frac{49\times2.5+98\times2.8}{4\times2.5}}$ | M1 | 1.2 – Use of $v = r\omega$ with values for $m_1$ and $m_2$. ($v = 3.78$, $v^2 = 14.2884$). NB $\omega = 6.3$ |
| Final energy $= 2.5\times g\times 1$ | B1 | 1.1 – Assuming zero PE level at 2m below $A$; other values possible. (24.5) |
| Initial KE $= \frac{1}{2}\times 2.5\times 0.6^2\times\frac{49\times2.5+98\times2.8}{4\times2.5}$ | M1 | 1.1 – Do not allow use of $\omega = 3.5$. (17.8605) |
| Initial PE $= 2.5\times g\times 1.2 + 2.8\times g\times 0.4$ | M1 | 1.1 – oe with different zero PE level. (40.376) |
| Energy loss $= 17.8605 + 40.376 - 24.5 = 33.7365$ | A1 | 3.2a – awrt 33.7 |
| **Alternate method:** | | |
| $v = r\omega = 0.6\sqrt{\frac{49\times2.5+98\times2.8}{4\times2.5}}$ | M1 | Use of $v = r\omega$ with values for $m_1$ and $m_2$. ($v=3.78$, $v^2=14.2884$). NB $\omega=6.3$ |
| Initial KE $= \frac{1}{2}\times2.5\times0.6^2\times\frac{49\times2.5+98\times2.8}{4\times2.5}$ | M1 | (17.8605) |
| $\Delta PE$ for $m_1 = \pm2.5\times9.8\times(0.8-1)$ | M1 | ($\pm4.9$) |
| $\Delta PE$ for $m_2 = \pm2.8\times9.8(1.6-2)$ | M1 | ($\pm10.976$). Or $-\Delta PE = 2.5\times9.8\times0.2 + 2.8\times9.8\times0.4$ ($\pm15.876$) |
| Energy loss $= 17.8605 + 4.9 + 10.976$ | A1 | awrt 33.7. Or $15.876 + 17.8605$ |
| **[5]** | | |

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