# Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $12500 = 5F$ | B1 | Use of $P = Fv$ |
| $F + 900g\sin\theta - 570 = 900a$ | M1 | Use of $F = ma$ parallel to slope. All 4 terms required. Condone sign errors and sin/cos confusion. 12500 in place of F is M0 – dimensionally incorrect |
| | A2 | Correct unsimplified equation. -1 each error |
| $a = 2.47 \quad (2.5)$ | A1 | Working with positive direction up the slope acceptable for first 4 marks, but final answer must be positive |

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# Question 2a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Ratio of masses $1:3:4$ | B1 | Correct ratios for their division. Also common: 3 equal triangles, 6 equal triangles, a rectangle and two equal triangles |
| Centre of mass of triangle $\frac{2}{3}a$ from $O$ | B1 | Correct centres for triangles in their division consistent with their axis |
| Moments about horizontal axis through $O$: $4\times0 - 1\times\frac{2}{3}a = 3d$ | M1 | Condone $4\times0$ not seen. Terms must be of correct form. Condone use of moments about a parallel axis. Signs must be consistent with their axis and distances. Watch out for people who add a triangle to the square |
| $\left(d = -\frac{2}{9}a\right)$ Distance $= \frac{2}{9}a$ | A1 | Reach **Given answer** with no errors seen. Their answer must be positive |

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# Question 2b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{2}akM = \frac{2}{9}a\cos 45° M$ | M1 | Moments about $A$. Lengths must be resolved as necessary (need use of trig) |
| | A2 | -1 each error |
| $k = \frac{1}{9}$ | A1 | |

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# Alt 2b (Method 1 – components):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Take $A$ as origin and axes along $AD$ and $AB$ to find components of distance of c of m from $A$ | M1 | Could choose a different origin |
| $\bar{x} = \frac{a(1+2k)}{1+k}$ | A1 | |
| $\bar{y} = \frac{11a}{9(1+k)}$ | A1 | |
| $\bar{x} = \bar{y} \Rightarrow \frac{11}{9} = 1+2k \Rightarrow k = \frac{1}{9}$ | A1 | CWO. This mark is not available if they have assumed that the c of m of the system is at $O$ |

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# Alt 2b (Method 2 – ratios):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2}{9}a\cos 45° = \frac{1}{9}\cdot\sqrt{2}a = \frac{1}{9}OD$ | M1A1 | Using ratios |
| $kM \times OD = M \times \frac{1}{9}OD$ | A1 | |
| $\Rightarrow k = \frac{1}{9}$ | A1 | |

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# Question 3a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.75\mathbf{v} = 6\mathbf{i} + 6\mathbf{j} + 0.75\times4\mathbf{i} \quad (= 9\mathbf{i}+6\mathbf{j})$ | M1 | Impulse momentum equation. Must be considering $\pm m(\mathbf{v}-\mathbf{u})$ |
| | A1 | Correct unsimplified |
| $\mathbf{v} = 12\mathbf{i} + 8\mathbf{j}$ | A1 | Award in (a) if seen or if (a) completed correctly. Award in (b) if (a) incomplete and this mark not yet awarded and correct **v** seen for first time in (b) |
| $\theta = \tan^{-1}\!\left(\frac{2}{3}\right)$ or $\theta = \cos^{-1}\!\left(\frac{1+13-8}{2\sqrt{13}}\right)$, or equivalent | M1 | Correct trig to find the required angle |
| $33.7°$ or $0.588$ radians | A1 | Accept $34°$ or better. Must be the final answer |

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# Alt 3a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}6\\6\end{pmatrix} = 0.75\begin{pmatrix}v\cos\theta\\v\sin\theta\end{pmatrix} - 0.75\begin{pmatrix}4\\0\end{pmatrix}$ | M1A1 | |
| $\Rightarrow 0.75\times v\cos\theta = 9, \quad 0.75\times v\sin\theta = 6$ | A1 | |
| $\Rightarrow \tan\theta = \frac{2}{3}$ | M1 | |
| $33.7°$ or $0.588$ radians | A1 | |

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# Question 3b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Change in KE $= \frac{1}{2}\times\frac{3}{4}(144+64) - \frac{1}{2}\times\frac{3}{4}(16)$ | M1 | Finding a difference between KE terms. Must use $\frac{1}{2}mv^2$ in both terms (all of **v**, not just one component) |
| | A1ft | Follow their **v**. Allow $\pm$ |
| $= 72$ (J) | A1 | CAO |