## Question 3:

### Part 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments equation (M1A0 for a moments inequality) | M1 | Part (a) is a 'Show that...' so equations need to be given in full to earn A marks |
| e.g. $M(A)$: $mga\cos\theta = 2Sa\sin\theta$ **OR** $M(B)$: $mga\cos\theta + 2Fa\sin\theta = 2Ra\cos\theta$ **OR** $M(C)$: $F\times2a\sin\theta = mga\cos\theta$ **OR** $M(D)$: $2Ra\cos\theta = mga\cos\theta + 2Sa\sin\theta$ **OR** $M(G)$: $Ra\cos\theta = Fa\sin\theta + Sa\sin\theta$ | A1 | Correct equation |
| $(\uparrow)\ R = mg$ **OR** $(\leftrightarrow)\ F = S$ | B1 | |
| Use equations and $F \leq \mu R$ to give inequality in $\mu$ and $\theta$ only (allow DM1 for use of $F = \mu R$ to give an equation in $\mu$ and $\theta$ only) | DM1 | |
| $\mu \geq \frac{1}{2}\cot\theta$ | A1* | Correct answer correctly obtained |

### Part 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments equation | M1 | |
| e.g. $M(A)$: $mga\cos\theta = 2Na\sin\theta$ **OR** $M(B)$: $mga\cos\theta + 2kmga\sin\theta = 2Ra\cos\theta + \frac{1}{2}mg\cdot2a\sin\theta$ **OR** $M(D)$: $2Ra\cos\theta = mga\cos\theta + N\cdot2a\sin\theta$ **OR** $M(G)$: $kmga\sin\theta + Na\sin\theta = \frac{1}{2}mga\sin\theta + Ra\cos\theta$ | A1 | Correct equation |

## Question 3b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $mga\cos\theta + \frac{1}{2}mg \cdot 2a\sin\theta = kmg \cdot 2a\sin\theta$ | M1A1B1 | Any moments equation with correct terms, condone sign errors |
| $1 + \frac{5}{4} = \frac{5k}{2}$ | M1 | Dependent on M1, using equations with trig substituted, to solve for $k$ numerically |
| $k = 0.9$ | A1 | cao |

---

## Question 4a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $s = ut + \frac{1}{2}at^2$ vertically to give equation in $t$ only | M1 | Complete method, correct no. of terms, condone sign errors and sin/cos confusion |
| $-70 = 65\sin\alpha \times t - \frac{1}{2} \times g \times t^2$ | A1, M(A)1 | A1: correct equation with at most one error; M(A)1: correct equation in $t$ only |
| $t = 7$ (s) | A1 | cao $(g = 9.8, 7.1$ or $7.11)$ |

---

## Question 4b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal velocity component at $A = 65\cos\alpha$ (60) | B1 | Seen, including on a diagram |
| Complete method to find vertical velocity component at $A$ | M1 | Condone sign errors and sin/cos confusion |
| $65\sin\alpha - g \times 7$ **OR** $\sqrt{(-25)^2 + 2g \times 70}$ (45) | A1ft | Correct expression; accept negative, follow their $t$ |
| $\sqrt{60^2 + (-45)^2}$ | M1 | Sub for trig and use Pythagoras |
| 75, accept 80 (m s$^{-1}$) | A1 | cao $(g = 9.8$ or $9.81$, $75$ or $74.8)$ |

---

## Question 4c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. approximate value of $g$ used; dimensions/spin of stone; $g = 10$ instead of $9.8$; $g$ assumed constant; wind; shape of stone | B1 | B0 if incorrect extras |

---

## Question 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate **v** wrt $t$ | M1 | Both powers decreasing by 1 (M0 if vectors disappear but allow recovery) |
| $\frac{3}{2}t^{-\frac{1}{3}}\mathbf{i} - 2\mathbf{j}$ | A1 | cao |

---

## Question 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3t^{\frac{1}{2}} = 2t$ | M1 | Complete method using **v** to obtain equation in $t$ only, allow sign error |
| Solve for $t$ | DM1 | Dependent on M1, solve for $t$ |
| $t = \frac{9}{4}$ | A1 | cao |

---

## Question 5c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrate **v** wrt $t$ | M1 | Both powers increasing by 1 (M0 if vectors disappear but allow recovery) |
| $\mathbf{r} = 2t^{\frac{3}{2}}\mathbf{i} - t^2\mathbf{j} (+\mathbf{C})$ | A1 | Correct expression without **C** |
| $t=1, \mathbf{r} = -\mathbf{j} \Rightarrow \mathbf{C} = -2\mathbf{i}$ so $\mathbf{r} = 2t^{\frac{3}{2}}\mathbf{i} - t^2\mathbf{j} - 2\mathbf{i}$ | A1 | cao |

---

## Question 5d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{(3t^{\frac{1}{2}})^2 + (2t)^2} = 10$ or $(3t^{\frac{1}{2}})^2 + (2t)^2 = 10^2$ | M1 | Use of Pythagoras on **v** and 10 to set up equation in $t$ |
| $9t + 4t^2 = 100$ | M(A)1 | Correct 3 term quadratic in $t$ |
| $t = 4$ | A1 | cao |
| $\mathbf{r} = 14\mathbf{i} - 16\mathbf{j}$ | M1 | Substitute their numerical $t$ into their **r** |
| $\sqrt{14^2 + (-16)^2}$ | M1 | Use of Pythagoras to find magnitude of their **r** |
| $\sqrt{452}\ (2\sqrt{113})$ (m) | A1 | cso |