# Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| M(A): $2aT = mga\cos\theta$ $\left(T = \frac{1}{2}mg\cos\theta\right)$ | M1A1 | First equation. Need all terms. Condone sign errors and sin/cos confusion |
| M(B): $mga\cos\theta + Fr\times 2a\sin\theta = R\times 2a\cos\theta$ | M1A1 | Second equation. Need all terms. Condone sign errors and sin/cos confusion |
| Resolve $\leftrightarrow$: $Fr = T\sin\theta\left(= \frac{1}{2}mg\cos\theta\sin\theta\right)$ | M1A1 | Third equation. Need all terms. Condone sign errors and sin/cos confusion |
| Use $Fr = \mu R$: $\mu R = T\sin\theta$ | B1 | Condone correct inequality |
| $R = mg - \frac{1}{2}mg\cos\theta\cos\theta$ and $\mu R = \frac{1}{2}mg\cos\theta\sin\theta$ | DM1 | Eliminate $T$ and $R$. Dependent on first 3 M marks |
| $\mu = \dfrac{\frac{1}{2}mg\cos\theta\sin\theta}{mg - \frac{1}{2}mg\cos\theta\cos\theta}$ | DM1 | Solve for $\mu$. Dependent on previous M |
| $\mu = \dfrac{\cos\theta\sin\theta}{2 - \cos^2\theta}$ | A1 | Obtain **given answer** from correct working. Must explain if inequality becomes equality |
| **(10 marks total)** | | |

**Alternative 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $B$: $mga\cos\theta + Fr\times 2a\sin\theta = R\times 2a\cos\theta$ | M1 A1 | Correct unsimplified |
| Resolving parallel to rod: $Fr\cos\theta + R\sin\theta = mg\sin\theta$ | M2 A2 | -1 each error |
| Use $Fr = \mu R$: $mg\cos\theta + \mu R\times 2\sin\theta = R\times 2\cos\theta$ and $\mu R\cos\theta + R\sin\theta = mg\sin\theta$ | B1 | |
| $\frac{mg\sin\theta}{mg\cos\theta} = \frac{\mu R\cos\theta + R\sin\theta}{2R\cos\theta - 2\mu R\sin\theta}$ | DM1 | Eliminate $T$ and $R$ |
| $2\cos\theta\sin\theta - 2\mu\sin^2\theta = \mu\cos^2\theta + \cos\theta\sin\theta$ | DM1 | Solve for $\mu$ |
| $\mu = \dfrac{\sin\theta\cos\theta}{\cos^2\theta + 2\sin^2\theta} = \dfrac{\sin\theta\cos\theta}{2-\cos^2\theta}$ | A1 | Obtain **given answer** from correct working |

# Question Alt 2 (Friction/Statics):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan(\theta + \alpha) = \dfrac{\tan\theta + \tan\alpha}{1 - \tan\theta\tan\alpha}$ | M1A1 | |
| $\tan\theta = \dfrac{a}{2a\tan\alpha} \Rightarrow \tan\alpha = \dfrac{1}{2\tan\theta}$ | M1 | |
| $\tan(\theta+\alpha) = \dfrac{\tan\theta + \dfrac{1}{2\tan\theta}}{1 - \tan\theta \times \dfrac{1}{2\tan\theta}} = 2\left(\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{2\sin\theta}\right)$ | M1A1 A1 | |
| $F = \mu R \Rightarrow \mu = \dfrac{1}{\tan(\theta+\alpha)}$ | B1 DM1 | |
| $= \dfrac{1}{2}\left(\dfrac{2\sin\theta\cos\theta}{2\sin^2\theta+\cos^2\theta}\right) = \dfrac{\cos\theta\sin\theta}{2-\cos^2\theta}$ | DM1 A1 | Obtain **given answer** from correct working |
| | (10) | |

---

# Question 5a:

| Answer/Working | Marks | Guidance |
|---|---|---|
| CLM: $3m\times 2u - 2m\times u = 3mv + 2mw$, i.e. $(4u = 3v+2w)$ | M1A1 | |
| Impact law: $w - v = \dfrac{1}{3}(2u+u) = u$ | M1A1 | |
| Solve simultaneously: $3w-3v=3u$, $2w+3v=4u$; $5w=7u$, $w=\dfrac{7}{5}u$ | DM1 A1 | Dependent on both previous M marks. **Must see working - Given Answer** |
| $v = \dfrac{2}{5}u$ | A1 | Or equivalent. Must be positive |
| | (7) | |

---

# Question 5b:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Speed of B after collision with wall: $\dfrac{1}{2}\times\dfrac{7}{5}u = \dfrac{7}{10}u$ | B1 | Accept +/- |
| Total time for either particle | B1 | |
| Equate the time travelled for each particle | M1 | |
| $\dfrac{x}{\frac{7}{5}u}+\dfrac{y}{\frac{7}{10}u}=\dfrac{x-y}{\frac{2}{5}u}$ | A1 | Correct unsimplified |
| $\dfrac{5x}{7u}+\dfrac{10y}{7u}=\dfrac{5x}{2u}-\dfrac{5y}{2u}$, $\quad 10x+20y=35x-35y$ | DM1 | Dependent on previous M1 |
| $55y=25x$, $\quad y=\dfrac{5}{11}x$ | A1 | Or equivalent. $0.45x$ or better |
| | (6) | |

---

# Question 6a:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Differentiate **v**: $\mathbf{a}=(4-6t)\mathbf{i}+(-8+2t)\mathbf{j}$ | M1A1 | Anywhere in (a) |
| Use $\mathbf{F}=m\mathbf{a}$ and substitute $t=3$: $\mathbf{F}=0.5((4-6\times3)\mathbf{i}+(-8+2\times3)\mathbf{j})=-7\mathbf{i}-\mathbf{j}$ | DM1 | Dependent on first M1 |
| Use Pythagoras' theorem | DM1 | Dependent on first M1. NB could use Pythagoras then use $\mathbf{F}=m\mathbf{a}$: 1st M1=1st step, 2nd M1=2nd step |
| $|\mathbf{F}|=\sqrt{49+1}=\sqrt{50}\left(=5\sqrt{2}=7.07...\right)$ | A1 | 7.1 or better |
| For **v**, **i** component = **j** component: $(4t-3t^2)=(-40-8t+t^2)$ | M1 | With no incorrect equations in $t$ seen |
| Solve for $t$: $4t^2-12t-40=0 \Rightarrow t^2-3t-10=0$; $(t-5)(t+2)=0$, $t=5$ | DM1 A1 | Dependent on previous M. Must see method if solving incorrect quadratic. Only - could be implied by later rejection of $-2$ |
| $\mathbf{a}=(4-30)\mathbf{i}+(-8+10)\mathbf{j}=-26\mathbf{i}+2\mathbf{j}$ (ms$^{-2}$) | A1 | Only |
| | (9) | |

---

# Question 6b:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Integrate **v**: $\mathbf{r}=\left(2t^2-t^3(+p)\right)\mathbf{i}+\left(-40t-4t^2+\dfrac{1}{3}t^3(+q)\right)\mathbf{j}$ | M1 A2 | $-1$ each error |
| $\mathbf{r}_1=\mathbf{i}-43\dfrac{2}{3}\mathbf{j}$, $\mathbf{r}_2=-93\dfrac{1}{3}\mathbf{j}$; $\overrightarrow{AB}=\mathbf{r}_2-\mathbf{r}_1$ | DM1 | Use limits in definite integral or evaluate constant of integration. Dependent on previous M1. $\left(\dfrac{131}{3}, \dfrac{280}{3}\right)$ |
| $\overrightarrow{AB}=-\mathbf{i}-49\dfrac{2}{3}\mathbf{j}\left(=-\mathbf{i}-\dfrac{149}{3}\mathbf{j}\right)$ | A1 | 49.7 or better |
| | (5) | |
| | [14] | |

## Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal distance: $x = u\cos\alpha \cdot t$ | B1 | $\frac{1}{\sqrt{5}}ut$ |
| Vertical distance: $y = u\sin\alpha \cdot t - \frac{1}{2}gt^2$ | M1A1 | $\frac{2}{\sqrt{5}}ut - \frac{1}{2}gt^2$; condone sign errors and sin/cos confusion |
| $y = u\sin\alpha \times \frac{x}{u\cos\alpha} - \frac{g}{2}\times\left(\frac{x}{u\cos\alpha}\right)^2 = x\tan\alpha - \frac{gx^2}{2u^2}\times\frac{1}{\cos^2\alpha} = 2x - \frac{gx^2}{2u^2}\times\frac{1}{1/5}$ | DM1 | Substitute for $t$ and $\alpha$; dependent on previous M1 |
| $= 2x - \frac{5g}{2u^2}x^2$ | A1 | Obtain **given answer** from exact working |
| **(5)** | | |

---

## Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=36, y=0$: $0 = 2 - \frac{5g}{2u^2}\times 36$ | M1 | Use given equation or a complete method using suvat to find $u$ |
| $u^2 = \frac{5g\times 36}{4}$, $\quad u = 21 \text{ (m s}^{-1})$ | A1 | Accept $\sqrt{45g}$ |
| **(2)** | | |

---

## Question 7c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Min speed $= u\cos\alpha$ | M1 | $u\cos\alpha = 21\times\frac{1}{\sqrt{5}} = 9.39 \text{ (m s}^{-1})$; consistent with their B1 in (a) |
| Minimum KE: $\frac{1}{2}\times 0.3\times(u\cos\alpha)^2 = \frac{0.3}{2}\left(\frac{21}{\sqrt{5}}\right)^2 = 13.2 \text{ (13) (J)}$ | DM1, A1 | Dependent on previous M1 |
| **(3)** | | |

**Alternative 7c:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Max height when $\frac{dy}{dx}=0$, $\quad x = \frac{2u^2}{5g}\ (=18)$ | M1 | Or from $\frac{1}{2}\times 36$ (symmetry) |
| Conservation of energy: $\frac{1}{2}mu^2 - mgh = \frac{1}{2}mv^2$ | M1 | |
| $= \frac{1}{2}\times 0.3\times 21^2 - 0.3\times g\times\frac{2\times 21^2}{5g} = 13.2 \text{ (J)}$ | A1 | |
| **(3)** | | |

---

## Question 7d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of trajectory at $B = -\frac{1}{2}$ | B1 | Accept $\frac{-1}{\tan\alpha}$ |
| Differentiate and equate: $\frac{dy}{dx} = 2 - \frac{5g}{u^2}x = -\frac{1}{2}$ | M1A1 | (their $u$) |
| Solve for $x$: $-\frac{1}{2} = 2 - \frac{5g}{u^2}x$, $\quad \frac{5}{2} = \frac{5gx}{u^2}$ | DM1 | Dependent on previous M1 |
| $x = \frac{21^2}{2g} = 22.5 \text{ (23) (m)}$ | A1 | |
| **(5)** | | |

**Alternative 7d (alt1):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of trajectory at $B = -\frac{1}{2}$ | B1 | Accept $\frac{-1}{\tan\alpha}$ |
| Use components of velocity: $-\frac{1}{2} = \frac{u\sin\alpha - gt}{u\cos\alpha}$ | M1 | |
| $t = \frac{u\sin\alpha + \frac{1}{2}u\cos\alpha}{g}\left(= \frac{105}{2g\sqrt{5}}\right)$ | A1 | $(t = 2.40)$ |
| Horizontal distance: $u\cos\alpha \cdot t = 22.5 \text{ (23) (m)}$ | DM1, A1 | Dependent on previous M1 |
| **(5)** | | |

**Alternative 7d (alt2):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of trajectory at $B = -\frac{1}{2}$ | B1 | Can be implied by downward velocity $\frac{21\sqrt{5}}{2}$ or $\frac{u\cos\alpha}{\tan\alpha}$ |
| $v_y = -\frac{1}{2}\times 21\times\frac{1}{\sqrt{5}}$, $\quad -\frac{21}{2\sqrt{5}} = 21\times\frac{2}{\sqrt{5}} - gt$ | M1 | Use suvat to find $t$ |
| $t = \frac{\frac{5}{2}\times\frac{21}{\sqrt{5}}}{g}\ (= 2.39)$ | A1 | |
| Horizontal distance: $u\cos\alpha \cdot t = 22.5 \text{ (23) (m)}$ | DM1, A1 | Dependent on previous M1 |
| **(5)** | | |

**Alternative 7d (alt3):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}u\cos\alpha \\ u\sin\alpha\end{pmatrix}\cdot\begin{pmatrix}u\cos\alpha \\ u\sin\alpha - gt\end{pmatrix} = 0$ | B1 | Scalar product $= 0$ |
| $u^2(\cos^2\alpha + \sin^2\alpha) - u\sin\alpha \cdot gt = 0 \Rightarrow u = \sin\alpha \cdot gt$ | M1A1 | Must have $-gt$ in second vector; solve for $t$ |
| Horizontal distance: $u\cos\alpha \cdot t = u\cos\alpha\times\frac{u}{g\sin\alpha} = \frac{21^2}{2g} = 22.5$ | M1A1 | |
| **(5)** | **[15]** | |