## Question 7(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{20e}{1.8} = \frac{15(1.5-e)}{0.9}$ | M1, A1 | |
| $2e = 4.5 - 3e \Rightarrow e = \frac{4.5}{5} = 0.9$ | A1 | |
| $AO = 2.7\text{ m}$ | A1cso (4) | |
| **ALT:** $\frac{20(AO-1.8)}{1.8} = \frac{15(4.2-AO-0.9)}{0.9}$ | M1A1A1 | With $AO$ as unknown |
| $AO = 2.7\text{ m}$ | A1cso | |

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## Question 7(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{15(0.6-x)}{0.9} - \frac{20(0.9+x)}{1.8} = m\ddot{x}$ or $\frac{20(0.9-x)}{1.8} - \frac{15(0.6+x)}{0.9} = m\ddot{x}$ | M1, A1, A1 | |
| $-\frac{50}{1.8m}x = \ddot{x}$ | M1 | |
| $(m>0)$ $\therefore$ SHM | A1cso (5) | |

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## Question 7(c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $-\frac{50}{1.8\times10}x = \ddot{x}$ | | |
| $\omega = \sqrt{\frac{50}{18}} = \frac{5}{3}$ | B1ft | |
| $a = 0.2\text{ m}$ | B1 | |
| $v_{\max} = a\omega = 0.2\times\frac{5}{3}$ | M1 | |
| **(i)** $J = 10\times0.2\times\frac{5}{3} = \frac{10}{3}$ (= 3.3... = 3.3 or better), $x=-0.1$ | M1, A1 | |
| **(ii)** $-0.1 = 0.2\sin\left(\frac{5}{3}t\right)$ | M1 | |
| $t = \frac{3}{5}\sin^{-1}(-0.5) = \frac{3}{5}\times\frac{7\pi}{6} = 2.199\ldots\text{ s}$ (accept 2.2 or better inc $\frac{7\pi}{10}$) | M1, A1 (8) | |

# Question (a):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Attempt equation equating 2 tensions using Hooke's Law, two extensions must add to 1.5 | M1 | Tensions to be obtained using Hooke's Law |
| Correct equation | A1 | |
| Correct extension for either string | A1 | |
| Correct completion to length $AO$ | A1cso | |

# Question (b):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Form equation of motion with difference of 2 tensions with different variable extensions. Tensions must be of form $k\dfrac{\lambda x}{l}$. Acceleration can be $\ddot{x}$ or $a$ | M1 | NB: $x$ can be measured in either direction |
| Deduct one per error | A1 A1 | Award A1A0 for one deduction. Allow if $a$ used instead of $\ddot{x}$ **provided** direction of $a$ is same as that of $\ddot{x}$ |
| Attempt to simplify equation to standard form for SHM. Acceleration must be $\ddot{x}$. Equation must simplify to $\ddot{x} = \pm\omega^2 x$ | M1 | |
| Correct equation with conclusion | A1cso | |

# Question (c):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Correct value of $\omega$, seen explicitly or used. Follow through from equation of form $\ddot{x}$ or $a = \pm\omega^2 x$ | B1ft | If equation left in form $m\ddot{x}$ or $ma = \ldots$ in (b) but RHS divided by $m$ to obtain $\omega$, this mark available (but no extra marks for (b)) |
| Correct amplitude, seen explicitly or used | B1 | |
| Attempt maximum speed with their $a$ and $\omega$ | M1 | |
| Use impulse-momentum equation with their max speed to obtain value for $J$ | (i)M1 | |
| Correct value for $J$ | A1 | |
| Use $x = a\sin\omega t$ or $x = a\cos\omega t$ with $x = \pm 0.1$, their $\omega$ and $a$ | (ii)M1 | |
| Solve equation and use correct method to find required time. Must use radians | M1 | |
| Correct time 2.2 or better | A1 | |