## Question 7(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $F = ma$ when string stretched | M1 | Must have $mg$ – tension term (involving $39.2m$, $0.8$ and $x$) $= m\ddot{x}$; allow if sign errors; $x$ could be length or ext of string, or from eq$^\text{m}$ pos. |
| $mg - \frac{39.2m(x-0.8)}{0.8} = m\ddot{x}$ | | |
| $\ddot{x} = -49(x-1)$ | A1 | |
| Show $x = 1$ is centre of SHM or that $x = 1$ is equilibrium position | B1 | And state about $x = 1$; Convincingly |
| **[3]** | | |

## Question 7(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| By energy | M1 | Must be PE term and EE term; Allow for missing '2', wrong '$g$' or inconsistent lengths |
| $mg(0.8+e) = \frac{39.2me^2}{2 \times 0.8}$ | A1 | Or $mgh = \frac{39.2m(h-0.8)^2}{2\times0.8}$ and $h = 0.8 + e$, $2.5e^2 - e - 0.8 = 0$ |
| $e = 0.8$ satisfies this equation AG | A1 | Or by solving quadratic in $e$; Convincingly; Allow full credit if done correctly from $v^2 = \omega^2(a^2 - x^2)$; Allow integration of $v\frac{\text{d}v}{\text{d}x} = g - 49x$ |
| **[3]** | | |

## Question 7:

### Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| For SHM, $\omega = 7$ | B1 | To be awarded if seen in (i) or (iv) or seen or used here |
| $a = 0.6$ | B1 | or seen or used here |
| Correct use of appropriate SHM distance equation | M1 | $-0.2 = 0.6\cos(7t)$ or $-0.2 = 0.6\sin(7t)$ — Allow $+0.2$, allow their $a$ and $\omega$ |
| $t = 0.272(9476)$ from bottom ($x = 1.6$) to $x = 0.8$ | A1 | Could be $0.0485 + 0.224$ |
| $t = 0.404(061)$ from $O$ to $x = 0.8$ | B1 | Or $\dfrac{2\sqrt{2}}{7}$ — May be seen first |
| Time to reach lowest point $= 0.677$ s | A1ft | ($\text{`}0.273\text{'} + \text{`}0.404\text{'}$) |
| **[6]** | | |

### Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $v = -a\omega\sin\omega t$ or $a\omega\cos\omega t$ | M1 | Must ft from their $x$ equation in (iii), or shown here — Allow use of their $a$ and $\omega$, sign error |
| $v = -0.6 \times 7\sin 7t$ | A1 | or $0.6 \times 7\cos 7t$ |
| Use of $t = 0.8 - 0.677 = 0.123$ after bottom point | B1ft | Or use of $t = 0.3475$ in 'cos' version — Must be between 0 and 0.8 |
| $v = 3.19 \quad (3.185677\ldots)$ | A1 | $(-)3.187$ — Do not allow if direction stated to be down |
| **[4]** | | |