# Question 7:

## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For $P$: $T_{S_1} = 1.5 \times 5\omega^2\ (= 7.5\omega^2)$ | M1 | NII for particle $P$ using $a = r\omega^2$. $T_{S_1}$ and $T_{S_2}$ must be in terms of $\omega$. Condone use of $m$ and $r$. SC1 only for omitting the element of $m$, or for using a specific value of $\omega$ to get the result that $T_{S_1} = T_{S_1}$. |
| For $Q$: $T_{S_2}\sin\theta = 1.5r\omega^2$ | M1 | Resolving tension for $Q$ and using NII in the horizontal with $a = r\omega^2$. Allow sin/cos confusion. Allow use of specific value of $\theta$. Allow use of $r\sin\theta$ or $r\cos\theta$ |
| $r = 5\sin\theta \Rightarrow T_{S_2} = \frac{1.5 \times 5\sin\theta \times \omega^2}{\sin\theta} = 7.5\omega^2$ | A1 | AG. Two identical expressions clearly seen. A0 if a specific value of $\theta$ has been used |
| $\therefore T_{S_1} = T_{S_2}$ | | |

## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P$: $\frac{1}{2} \times 1.5v^2 = 39.2$, so $v^2 = \frac{784}{15}$ | M1 | Using kinetic energy is 39.2 to find $v$ or $v^2$. $v^2 = 52.666...$, $v = \frac{28\sqrt{15}}{15} = 7.229...$ (allow 7.22) |
| $t_P = \frac{2\pi}{\omega_P} = \frac{2\pi \times 5}{v} = \frac{5\sqrt{15}\pi}{14}$ | A1 | Use of "$\omega = \frac{2\pi}{t}$" and finding the time using "$v = r\omega$" for $P$. NB $t_P = 4.3454...$, $\omega_P \approx 1.45$. Allow 4.33–4.35. Allow unsimplified. Penalise inexact value only at the end |
| $Q$: $\omega_Q = \frac{v}{5\sin\theta}\ \left(= \frac{28\sqrt{15}}{75}\right)$ | B1 | Use of "$v = r\omega$" for $Q$ with correct radius. Or $t_Q = \frac{2\pi \times 5\sin\theta}{v}$ (may be seen later). Or $\omega_Q = \omega_P\sin\theta$. B0 for assuming specific value of $\theta$ (and for subsequent marks). May be seen later as $\omega_Q = \frac{v}{4} = \frac{7}{\sqrt{15}}$ or $t_Q = \frac{4\sqrt{15}\pi}{14}$ |
| $Q$: $T_{S_2}\cos\theta = 1.5g$ | M1 | Resolving the tension vertically and balancing with weight (condone missing $g$ here). $4g = 39.2$. Condone use of $m$ |

## Question (Circular Motion/Conical Pendulum):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $Q: T_{S_2} \sin\theta = 1.5a = \dfrac{1.5v^2}{5\sin\theta}$ | **M1** | Resolving tension horizontally and using NII and $a = \dfrac{v^2}{r}$ with correct radius. Or: $T_{S_2} = 7.5\left(\dfrac{v}{5\sin\theta}\right)^2$. Could see use of $v^2 = \dfrac{784}{15}$ here so $T_{S_2}\sin\theta = 1.5a = \dfrac{1176}{75\sin\theta}$. Condone use of $m$ and $r$. |
| $\dfrac{\sin^2\theta}{\cos\theta} = \dfrac{\left(\frac{1.5v^2}{5}\right)}{1.5g} = \dfrac{v^2}{5g} = \dfrac{784}{15\times 5g} = \dfrac{16}{15}$ | **M1** | Finding an equation in $\theta$, using $v^2 = \dfrac{784}{15}$ and substituting $\sin^2\theta = 1 - \cos^2\theta$ to get an equation in $\cos\theta$ only. |
| $15(1-\cos^2\theta) = 16\cos\theta$ | | |
| $(\Rightarrow 15c^2 + 16c - 15 = 0)$ | | |
| $((5c-3)(3c+5) = 0 \Rightarrow) \cos\theta = 3/5$ since $\cos\theta$ cannot be $-5/3$ | **A1** | Allow $\cos\theta = 3/5$ or $\sin\theta = 4/5$ to appear without working. |
| $\Delta t = \dfrac{5\sqrt{15}\pi}{14} - \dfrac{4\sqrt{15}\pi}{14} = \dfrac{\sqrt{15}}{14}\pi$, so difference in time periods is $\dfrac{\sqrt{15}}{14}\pi$ (s) oe | | Use of $"\omega = \dfrac{2\pi}{t}"$ and finding the time difference. (For reference: $0.86909\ldots$). Or $t_Q = \dfrac{2\pi r}{v} = \dfrac{2\pi \times 5\sin\theta}{v} = \dfrac{2\pi\times 5\times\frac{4}{5}}{\sqrt{\frac{784}{15}}} = \dfrac{4\sqrt{15}\pi}{14}$ |
| **[7]** | | |