# Question 8:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| At equilibrium: $0.5g = \frac{25e}{1.25}$, $e = \frac{0.5\times10\times1.25}{25} = \frac{1}{4}$ | B1 | Correct only. Award if $mg = \frac{\lambda e}{l}$ used in equation of motion |
| For taut string, equation of motion with $x$ from equilibrium: $\frac{25(e+x)}{1.25} - 0.5g = -0.5\ddot{x}$ | M1, A1ft | Need all terms, dimensionally correct. Allow with their $e \neq 0$. Condone sign errors. Correct unsimplified equation |
| $\ddot{x} = -40x$, hence SHM | A1* | Reach given conclusion from correct working |
| Periodic time: $T = \frac{2\pi}{\sqrt{40}} = \frac{\pi}{\sqrt{10}}$ | M1, A1* | Use $T = \frac{2\pi}{\omega}$ from $\ddot{x} = -\omega^2 x$. Obtain given answer from correct working |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Max KE $= 2.5 = \frac{1}{2}\times\frac{1}{2}\times\max v^2 \Rightarrow \max v^2 = 10$ | B1 | Use KE to find max $v$ or max $v^2$ |
| Max speed $= a\omega$: $\sqrt{10} = a\sqrt{40}$ | M1 | Use model to find amplitude |
| $AB = 1.25 + \frac{1}{4} + \frac{1}{2} = 2\ (\text{m})$ | A1* | Obtain given answer from correct working |

## Part (b) alt:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Energy: $\frac{25e^2}{2.5} + 2.5 + 0.5ga = \frac{25(e+a)^2}{2.5}$ | B1 | Using correct $\lambda$ and $l$ and $e$ or their $e$ |
| Solve for $a$ | M1 | Requires energy equation with all right terms |
| $AB = 1.25 + \frac{1}{4} + \frac{1}{2} = 2\ (\text{m})$ | A1* | Obtain given answer from correct working |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 0.5$, $x = 0.5\cos\sqrt{40}t$ | B1 | Correct equation for SHM seen or implied |
| $-0.25 = 0.5\cos\sqrt{40}t \Rightarrow t = 0.3311\ldots$ | M1 | Find time until string goes slack. If working from $x = 0.5\sin\sqrt{40}t$ need $\frac{T}{4} + \frac{1}{\sqrt{40}}\sin^{-1}\frac{1}{2}$ |
| $v^2 = 40\left(0.5^2 - 0.25^2\right) = \frac{15}{2}$ | M1 | Use model to find $v$ or $v^2$ at instant string goes slack ($v = 2.738\ldots$). Using SHM formula or conservation of energy |
| Total time $= 2\times0.3311\ldots + \frac{2\times\sqrt{7.5}}{10}$ | DM1 | Complete method to find total time until return to $B$. Requires preceding M marks |
| $= 1.2\ (\text{s})$ or better | A1 | |

## Mark Scheme Content

**Question (Projectile time calculation):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{\dfrac{15}{2}} = -\sqrt{\dfrac{15}{2}} + gt$ or a combination of $v^2 = u^2 + 2as$ and $s = ut + \dfrac{1}{2}at^2$ | M1 | If they use suvat to find the time as a projectile it must be a complete method |
| $= 1.2$ (s) or better | A1 | Condone an answer to $> 2$ s.f. Not scored if they have used $9.8$ |