## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Driving force $F = \dfrac{3500}{V}$ | B1 | Use of $P = Fv$ |
| Equation of motion: $F - 20V + 480g\sin\theta = 0$ | M1 | Need all terms. Dimensionally correct. Condone sign errors and sin/cos confusion |
| $\dfrac{3500}{V} - 20V + 40g = 0$ | A1 | Correct unsimplified equation in $V$ |
| $20V^2 - 392V - 3500 = 0$ | M1 | Form a 3 term quadratic equation $(=0)$ in $V$ |
| $V = 26.3$ (26) | A1 | 3 sf or 2 sf. Not $\dfrac{49 + 22\sqrt{14}}{5}$ (follows use of 9.8) |

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## Question 2a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\mathbf{a} = \dfrac{d\mathbf{v}}{dt}$ | M1 | Differentiate – at least 3 powers going down by 1 |
| $\mathbf{a} = (10t - 3t^2)\mathbf{i} + (6t^2 - 8)\mathbf{j}$ | A1 | |
| $\mathbf{F} = 1.5 \times ((20-12)\mathbf{i} + (24-8)\mathbf{j})$ | DM1 | Substitute $t=2$ and use $\mathbf{F} = m\mathbf{a}$. Dependent on preceding M1 |
| $= 12\mathbf{i} + 24\mathbf{j}$ | A1 | Ignore magnitude of $\mathbf{F}$ if found |

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## Question 2b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $5t^2 - t^3 = 0 \Rightarrow t = 5$ | B1 | Not moving when $t=0$ so no need to mention $t=0$ |
| Use of $\mathbf{r} = \int \mathbf{v}\, dt$ | M1 | Integrate to find $\mathbf{r}$ – at least 3 powers going up by 1 |
| $\mathbf{r} = \left(\dfrac{5}{3}t^3 - \dfrac{1}{4}t^4\right)\mathbf{i} + \left(\dfrac{1}{2}t^4 - 4t^2\right)\mathbf{j}$ | A1 | Condone if no constant of integration seen (since $t=0$, $\mathbf{r} = \mathbf{0}$) |
| $\mathbf{r} = \left(\dfrac{625}{12}\right)\mathbf{i} + \left(\dfrac{425}{2}\right)\mathbf{j}$ | A1 | Final answer $52\mathbf{i} + 210\mathbf{j}$ or better $(52.08\dot{3}\mathbf{i} + 212.5\mathbf{j})$ |

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## Question 3a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratio: square 36, triangle 8, circle $\pi$, total $T = 28-\pi$; distances from $AD$: square $3a$, triangle $\frac{7}{3}a$, circle $4a$, total $d$ | B1 | Mass ratio |
| Distances from $AD$ or parallel axis | B1 | |
| $M(AD)$: $36\times 3a - 8\times\dfrac{7}{3}a - \pi\times 4a = (28-\pi)d$ | M1 | Moments equation. Need all terms and dimensionally correct. Condone sign errors |
| $36\times 3a - 8\times\dfrac{7}{3}a - \pi\times 4a = (28-\pi)d$ | A1 | Correct unsimplified equation for their parallel axis |
| $d = \dfrac{324 - 56 - 12\pi}{3(28-\pi)}a = \dfrac{4(67-3\pi)}{3(28-\pi)}a$ * | A1* | Obtain given answer from correct working. Distance from $BC$ is $\dfrac{(236-6\pi)a}{3(28-\pi)}$. Allow 4/5 if seen. |

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## Question 3b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(A)$: $W\times\dfrac{4(67-3\pi)}{3(28-\pi)}a = kW\times 6a$ or resolve vertically and use $M(G)$ where $G$ is centre of mass. $T_A + kW = W$; $T_A\left(\dfrac{4(67-3\pi)}{3(28-\pi)}a\right) = kW\left(6a - \dfrac{4(67-3\pi)}{3(28-\pi)}a\right)$ | M1 | Complete method to form equation in $k$ and $W$ only. Dimensionally correct but condone use of incorrect distances |
| Correct unsimplified equation | A1 | NB $\dfrac{4(67-3\pi)}{3(28-\pi)}a = 3.088a$ |
| $k = 0.51$ | A1 | Q asks for 2dp |

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