## Question 4(a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Moments about $AC$ | M1 | All terms. Dimensionally correct. Condone sign errors |
| $18\times\dfrac{3a}{2} - 2\pi\times\dfrac{8a}{3\pi} + 2\pi\left(3a + \dfrac{8a}{3\pi}\right) = 18\bar{y}$ | A1 | Unsimplified equation with at most one error |
| NB: valid to use $18\times\dfrac{3a}{2} - 2\pi\times d + 2\pi(3a+d) = 18\bar{y}$ for $d\neq 0$ without stating value for $d$. Use of $d=0 \Rightarrow$ M0 | | |
| $\left(27a + 6\pi a = 18\bar{y}\right)$ | A1 | Correct unsimplified equation. The same incorrect distance used twice in place of $\frac{8a}{3\pi}$ is one error. The same incorrect area for the semicircle used twice is one error. |
| $27a + 6\pi a = 18\bar{y} \Rightarrow \bar{y} = \dfrac{9+2\pi}{6}a$ | A1 (4) | Obtain **given answer** from sufficient correct exact working. Must see a separate conclusion for $\bar{y}$ |

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## Question 4(b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $M\bar{x} + kM\times 6a = (1+k)M\bar{x}_T$ | M1 | |
| $3a + 6ak = (1+k)\bar{x}_T$ o.e. | A1 | |
| $M\bar{y} = (1+k)M\bar{y}_T$ | M1 | |
| $(1+k)\bar{y}_T = \dfrac{9+2\pi}{6}a$ | A1 | e.g. $\bar{y}_T = \dfrac{2(3+6k)a}{3(1+k)}$ |
| $\tan\phi = \dfrac{3}{2} = \dfrac{\bar{x}_T}{\bar{y}_T} \Rightarrow \dfrac{3}{2} = \dfrac{6(3a+6ak)}{(9+2\pi)a}$ | DM1 | Form equation in $k$ and solve for $k$. Dependent on the previous 2 M marks |
| $\Rightarrow k = \dfrac{\pi}{12} - \dfrac{1}{8}$ or equivalent | A1 (6) | $k = 0.137\ \ (0.14)$ or better |

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## Question 4(b) alt:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Distance of original c of m from vertical through $A$ | M1 | |
| $\left(\dfrac{9+2\pi}{6}a - 2a\right)\times\sin\phi \left(= \dfrac{\sqrt{13}(2\pi-3)a}{26}\right)$ | A1 | Or equivalent |
| Distance of additional particle from vertical through $A$ | M1 | |
| $6a\times\cos\phi\left(= \dfrac{12a}{\sqrt{13}}\right)$ | A1 | Or equivalent |
| $mg\times\dfrac{\sqrt{13}(2\pi-3)a}{26} = kmg\times\dfrac{12a}{\sqrt{13}}$ | DM1 | Moments about $A$. Dependent on the 2 previous M marks |
| $k = 0.137\ \ (0.14)$ | A1 (6) | 0.14 or better |

**Total: [10]**

## Question 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{a} = \frac{d\mathbf{v}}{dt}$ : $\mathbf{a} = 6t\mathbf{i} + 2\mathbf{j}$ | M1 | Powers going down |
| $t = 0 \Rightarrow \mathbf{a} = 2\mathbf{j}$ $(\text{ms}^{-2})$ | A1 (2) | Must see vector answer but ISW if go on to state the magnitude |

## Question 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $11(2t-4) = (3t^2-4)$ | M1 | Use of velocity parallel to $11\mathbf{i}+\mathbf{j}$; 11 must be on the correct side |
| $3t^2 - 22t + 40 = 0 \Rightarrow \left(t = \frac{10}{3}\right) t = 4$ | M1 | |
| $\mathbf{v} = 44\mathbf{i} + 4\mathbf{j}$, speed $= \sqrt{44^2 + 4^2}$ | DM1 | Select the larger root (dependent on previous 2 M1 marks and on 2 positive roots) and use Pythagoras. Condone if they find both speeds |
| $= 4\sqrt{122}$ $(\text{m s}^{-1})$ | A1 (4) | Any equivalent simplified surd form $(\sqrt{1952})$ ISW. 44.18... implies M1 if correct surd form not seen. Both values for speed given is A0 |

## Question 5c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{r} = \int \mathbf{v}\, dt$ | M1 | Powers going up |
| $\mathbf{r} = (t^3 - 4t)\mathbf{i} + (t^2 - 4t)\mathbf{j}$ | A1 | If a constant of integration is introduced, they must conclude it is equal to the zero vector |
| Set $\mathbf{r} = \mathbf{0}$ and solve for $t$ | M1 | Consider both components |
| $t^3 - 4t = 0 \Rightarrow t = 0, 2, (-2)$ | | |
| $t^2 - 4t = 0 \Rightarrow t = 0, 4$ | | |
| The only common value is $t = 0$, so does not return to $O$ | A1* (4) | Or equivalent clear explanation of given result. Condone if they ignore $t=0$. Do not need to see the roots. But do need to see the factorised form for each component if using this method |

## Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically: $R + N\cos\alpha = W$ | M1 | |
| Correct unsimplified equation | A1 | |
| Take moments about $A$: $7aN = 4a\cos\alpha \times W$ | M1 | |
| Correct unsimplified equation | A1 | |
| Obtain equation in $R$, $W$ and $\alpha$ | DM1 | Solve for $R$ in terms of $W$. Dependent on the 2 preceding M marks |
| $N = W \times \frac{4}{7}\cos\alpha \Rightarrow R = W - \frac{4}{7}W\cos^2\alpha = W\left(1 - \frac{4}{7}\cos^2\alpha\right)$ | A1* (6) | Obtain given answer from correct working |
| Alternative: $R\sin\alpha + F\cos\alpha = W\sin\alpha$ | M1 | Parallel to the rod; equation in $R$. All terms needed. Condone sin/cos confusion and sign errors |
| $N + R\cos\alpha = W\cos\alpha + F\sin\alpha$ | A1 | Correct unsimplified equation |
| $W.3a\cos\alpha + F.7a\sin\alpha = R.7a\cos\alpha$ | M1 | Sufficient additional equations to solve for $R$ in terms of $W$. Dimensionally correct. All terms needed. Condone sin/cos confusion and sign errors |
| | A1 | Correct unsimplified equation |
| | DM1 | |
| | A1* (6) | |

## Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = W\left(1 - \frac{4}{7} \times \frac{9}{10}\right) = \frac{17W}{35}$ | B1 | Seen or implied |
| Resolve horizontally: $F = N\sin\alpha = \frac{4}{7} \times \frac{3}{\sqrt{10}} \times \frac{1}{\sqrt{10}}W$ | M1, A1 | Obtain equation in $F$. Correct unsimplified equation in $F$ and $W$ (trig. substituted) $(0.171W)$ |
| $\left(= \frac{6}{35}W\right)$ | | |
| Use of $F \leq \mu R$ | M1 | Correct method to find the critical value. Condone with any symbol |
| $\Rightarrow \mu \geq \frac{6}{17}$ | A1 (5) | 0.35 or better $(0.3529...)$ from correct working. Final answer. Do not ISW |

## Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of conservation of energy: $\frac{1}{2}m \times 25^2 = \frac{1}{2}m \times 15^2 + mgh$ | M1, A1 | Need energy equation with all 3 terms. Must be dimensionally correct. Condone sign errors. Correct unsimplified equation. NB: sine/cosine confusion is not condoned in projectile questions |
| $\Rightarrow h = 20$ or $20.4$ (m) | A1 (3) | Max 3 sf. Not $\frac{1000}{49}$ nor $\frac{200}{g}$ |

## Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertical distance: $20.4 = 25\sin\alpha \times 3 - 4.5 \times 9.8$ | M1 | Use of suvat to find $\alpha$ |
| $\alpha = 59°$ or $59.3°$ | A1ft, A1 (3) | Correct unsimplified equation in their $h$. 0.554 rads. Max 3 sf From CWO |

## Question 7c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal component of speed is constant: $25\cos\alpha = 15\cos\beta$ | M1 | Or horizontal distance travelled |
| $\beta = 32°$ or $31.8°$ | A1ft, A1 (3) | Correct unsimplified in $\alpha$ or their $\alpha$. 0.554 rads. Max 3 sf From CWO |
| **Alternative:** Vertical distance: $20.4 = -15\sin\beta \times 3 + 4.5 \times 9.8$ | M1 | Use of suvat to find $\beta$; using $s = vt - \frac{1}{2}gt^2$. Correct unsimplified equation in their $h$ |
| $\beta = 32°$ or $31.8°$ | A1ft, A1 (3) | 0.554 rads. Max 3 sf From CWO |

## Question 7d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Min speed = horizontal component $= 25\cos\alpha$ $(= 15\cos\beta)$ $= 13$ or $12.8$ $(\text{m s}^{-1})$ | M1, A1 (2) | Follow their angle. Must show working if using incorrect angle. Max 3 sf From CWO |

## Question 7e:

| Answer/Working | Mark | Guidance |
|---|---|---|
| By considering vertical component of speed at $B$: $15\sin 31.8° - gT = -15\sin 31.8°$ | M1 | Complete method using suvat to find $T$ |
| $T = 1.6$ or $1.61$ (s) | A1ft, A1 (3) | Correct unsimplified equation in $T$ - follow their angles. Max 3 sf From CWO |