## Question 5(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $[s_P \text{ up}=] \pm\left(2t - \frac{1}{2}gt^2\right)$ OR $[s_Q \text{ down}=] \pm\frac{1}{2}gt^2$ | M1 | For use of $s = ut + \frac{1}{2}at^2$ at least once with $a = \pm g$ and $u = 0$ or $u = \pm 2$. Seen anywhere |
| $2t - \frac{1}{2}gt^2 + \frac{1}{2}gt^2 = 2 - 1 \Rightarrow t = \frac{1}{2}$ | M1 | Use $s_P + s_Q = \pm(2-1)$ OR $\pm 1$ ONLY with $s_P$ and $s_Q$ of the correct form |
| $v_P = 2 - \frac{1}{2}g = -3\ \text{ms}^{-1}$ so speed $= 3\ \text{ms}^{-1}$ | A1 | Must be positive |
| $v_Q = -\frac{1}{2}g = -5\ \text{ms}^{-1}$ so speed $= 5\ \text{ms}^{-1}$ | A1 | Must be positive. If A0A0, allow SCB1 if both are negative |

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## Question 5(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $2m \times (3[v_P]) + m \times (5[v_Q]) = 2mv + m \times 3.5$ | *M1 | Use of conservation of momentum; 4 non-zero terms; using their $\pm 3\ \text{ms}^{-1}$ and their $\pm 5\ \text{ms}^{-1}$; allow sign errors and $m$ missing ONLY. Do not allow with $v_p = \pm 2$ or $v_Q = 0$. Do not allow made up values for 3 and 5. Allow LHS to have only one term but only if getting $v_P = 0$ in (a) from $s_P = s_Q$ |
| $v = 3.75$ | A1 | Allow $v = -3.75$ from correct work |
| At $t = \frac{1}{2}$: $s_p = \pm\left(2 \times \frac{1}{2} - \frac{1}{2}g\left(\frac{1}{2}\right)^2\right) = \mp\frac{1}{4}$ OR $s_p = \pm\left(-3 \times \frac{1}{2} + \frac{1}{2}g\left(\frac{1}{2}\right)^2\right) = \mp\frac{1}{4}$ OR $s_p = \pm\left(\frac{(-3)^2 - 2^2}{2g}\right) = \pm\frac{1}{4}$ OR At $t=\frac{1}{2}$: $s_Q = \pm\frac{1}{2}g\left(\frac{1}{2}\right)^2 = \pm\frac{5}{4}$ OR $s_Q = \pm\left(5 \times \frac{1}{2} - \frac{1}{2}g\left(\frac{1}{2}\right)^2\right) = \pm\frac{5}{4}$ OR $s_Q = \pm\left(\frac{5^2 - 0^2}{2g}\right) = \pm\frac{5}{4}$; so height above ground $= \frac{3}{4}$ | *B1 | This may be seen in part (a), but do not award the mark until stated/used in part (b) |

## Question 5(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $v^2 = (their\ 3.75)^2 + 2 \times (\pm g) \times \left(\pm \frac{3}{4}\right)$ | **DM1** | Use of $v^2 = u^2 + 2as$ using their $v$ and $a = \pm g$; their $s$ is either $\pm\left[1+\left(-\frac{1}{4}\right)\right]$ or $\pm\left[2-\frac{5}{4}\right]$ OE. Dependent on previous M and B marks. |
| OR $0^2 = (their\ 3.75)^2 - 2\times(\pm g)\times s \Rightarrow s = \frac{45}{64}$ AND $v^2 = \left[0^2+\right]2\times(\pm g)\times\left[\left(\frac{3}{4}\right)+\frac{45}{64}\right]$ | | |
| $v = 5.39\ \text{ms}^{-1}$ | **A1** | Or $\frac{\sqrt{465}}{4}$; AWRT 5.39; 5.390964... |

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