# Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| 1st equation e.g. change in KE | M1 | Dimensionally correct. Must be subtracting but condone sign error. |
| $\frac{1}{2} \times 0.5\left(x^2 + y^2 - (5^2 + 3^2)\right) = 22$ | A1 | Correct unsimplified equation seen or implied (may have used impulse-momentum first) |
| $\left(x^2 + y^2 = 122\right) \quad \left(1^2 + (2\lambda+3)^2 = 122\right)$ | | |
| 2nd equation e.g. Impulse-momentum | M1 | Dimensionally correct. Must be subtracting but condone sign error. |
| $0.5(x\mathbf{i} + y\mathbf{j}) - 0.5(5\mathbf{i} + 3\mathbf{j}) = (-2\mathbf{i} + \lambda\mathbf{j})$ | A1 | Correct unsimplified equation |
| $\left((x-5)\mathbf{i} + (y-3)\mathbf{j} = -4\mathbf{i} + 2\lambda\mathbf{j}\right)$ | | |
| Form quadratic in $\lambda$: $1^2 + (3+2\lambda)^2 = 122$ | DM1 | Dependent on the 2 preceding M marks |
| Solve for 2 values: $4\lambda^2 + 12\lambda - 112 = 0$ or $(3+2\lambda)^2 = 121$ | DM1 | Dependent on preceding M1 |
| $\lambda = 4$ or $\lambda = -7$ | A1 | Correct only and no errors seen (watch out for $x = -1$ used) |

**Alternative:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Form quadratic in $y$: $1 + y^2 = 122 \quad (y^2 = 121)$ | DM1 | Dependent on the 2 preceding M marks |
| Solve for 2 values of $y$ and use to obtain 2 values of $\lambda$ | DM1 | Dependent on preceding M1 |
| $\lambda = 4$ or $\lambda = -7$ | A1 | |

**Total: 7 marks**

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# Question 3a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Rectangle: area $48a^2$, distance from $AE$: $4a$ | B1 | Mass ratio correct |
| Triangle: area $18a^2$, distance from $AE$: $8a - 2a = 6a$ | B1 | Distances from $AE$ (or parallel axis) correct |
| Lamina: area $30a^2$ | | |
| Moments equation $M(AE)$ | M1 | Allow use of parallel axis. Equation should include $a$ but condone if mass ratio does not include factor of $a^2$. Dimensionally correct. |
| $48a^2 \times 4a - 18a^2 \times 6a = 30a^2\bar{x}$ | A1 | Correct unsimplified equation for their axis. Accept as part of vector equation. |
| $\bar{x} = \frac{84}{30}a = \frac{14}{5}a$ * | A1* | Obtain given answer from correct working (including correct use of $a$) |

*Note: If moments taken about $BD$: $d = 5.2a$. Allow B1B1M1A1A0 if they get this far.*

**Total: 5 marks**

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# Question 3b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Find trig ratio of a relevant angle | M1 | Correct use of trig. |
| $\tan\theta° = \frac{3a}{2.8a}$ | A1 | Correct equation for the required angle. (DO NOT ISW: If they obtain 47 and then use $90 - 47 = 43$ they score M1A0A0) |
| $\theta = 47$ | A1 | The question asks for a whole number of degrees. 0.82 radians scores M1A1A0 |

**Total: 3 marks (Question 3 total: 8 marks)**

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# Question 4a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $t = 2$ and $3t^2 + 2t = t^3 + kt$: $(12 + 4 = 8 + 2k)$ | M1 | Allow verification. |
| $k = 4$ * | A1* | Obtain given answer from correct working. Verification requires a clear conclusion. |

**Total: 2 marks**

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# Question 4b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\mathbf{a} = \frac{d\mathbf{v}}{dt}$ | M1 | Differentiate the vector $\mathbf{v}$. Majority of powers going down. |
| $\mathbf{a} = (6t+2)\mathbf{i} + (3t^2+4)\mathbf{j}$ | A1 | Correct only |
| Use $|\mathbf{F}| = m|\mathbf{a}|$ | DM1 | Correct use of Pythagoras and N2L. Dependent on preceding M1. |
| $|\mathbf{F}| = 1.5 \times \sqrt{14^2 + 16^2} = 3\sqrt{113}$ | A1 | Or $\frac{3}{2}\sqrt{452}$ or 32 or better (31.89…) |

**Total: 4 marks**

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# Question 4c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\mathbf{r} = \int \mathbf{v}\, dt$ | M1 | Majority of powers going up |
| $\mathbf{r} = \left(t^3 + t^2 (+A)\right)\mathbf{i} + \left(\frac{1}{4}t^4 + \frac{4}{2}t^2 (+B)\right)\mathbf{j}$ | A1 | Allow without constant of integration |
| Correct use of $\mathbf{r} = 3\mathbf{i} + 4\mathbf{j}$ when $t = 0$ to find $\mathbf{r}$ when $t = 2$ | DM1 | $\mathbf{r} = (t^3+t^2+3)\mathbf{i} + \left(\frac{1}{4}t^4 + \frac{4}{2}t^2+4\right)\mathbf{j}$. Dependent on preceding M1. Use of $\mathbf{r} = -3\mathbf{i} - 4\mathbf{j}$ is M0. |
| $\mathbf{r} = 15\mathbf{i} + 16\mathbf{j}$ | A1 | Correct answer only. Accept column vector. |

**Total: 4 marks (Question 4 total: 10 marks)**

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# Question 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $F_{\max} = \mu R$: $F_{\max} = \frac{2}{7} \times 1.5g\cos\theta$ | M1 | (3.87…) Condone trig confusion. Trig substitution not required. Allow M1 if there is a clear statement for $F_{\max}$ "correct" and then used in a calculation including the gain in GPE. |
| Use $WD = 2.5F_{\max}$ | M1 | Trig substitution not required. M0 if they have included the gain in GPE. If method for $F$ is incorrect but involves $\mu$ to obtain $F$ and they use the "work done" formula correctly allow M0M1. |
| $WD = 9.69 \quad (9.7) \text{ (J)}$ | A1 | 3 sf or 2 sf, not $\frac{126}{13}$ |

**Total: 3 marks**

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# Question 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Work-energy equation | M1 | Need all terms and dimensionally correct. Condone sign errors and sin/cos confusion. |
| $\frac{1}{2} \times 1.5U^2 = WD + 1.5 \times 9.8 \times 2.5 \times \sin\theta$ | A1ft | Unsimplified equation with at most one error. |
| | A1ft | Correct unsimplified equation. Follow their $WD$ against friction. |
| $U = 5.64 \quad (5.6)$ | A1 | 3 sf or 2 sf |

*Note: If answer to (a) included the GPE then it must be used for total work done here to score M1.*

**Total: 4 marks**

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# Question 5c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Work-energy equation for $A$ to $A$ | M1 | Need all terms and dimensionally correct. |
| $\frac{1}{2} \times 1.5v^2 = \frac{1}{2} \times 1.5U^2 - 2WD$ | A1ft | Correct unsimplified equation. Follow their $WD$ against friction and their $U$. |
| $v = 2.43 \; (2.4) \; (\text{ms}^{-1})$ | A1 | 3 sf or 2 sf |

**Alternative (B to A):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Work-energy equation for $B$ to $A$ | M1 | Need all terms and dimensionally correct. |
| $\frac{1}{2} \times 1.5v^2 = 1.5 \times 9.8 \times 2.5 \times \sin\theta - WD$ | A1ft | Correct unsimplified equation. Follow their $WD$. |
| $v = 2.43 \; (2.4) \; (\text{ms}^{-1})$ | A1 | 3 sf or 2 sf |

**Total: 3 marks (Question 5 total: 10 marks)**

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# Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $A$: $M(A)$ | M1 | Or equivalent method to form equation in $W$ only. Must be dimensionally correct and contain all relevant terms. Condone sin/cos confusion and sign errors. |
| $50 \times 3\cos30° + W \times 6\cos30° = 60\sqrt{3} \times 4\sin30°$ | A1 | Unsimplified equation with at most one error. |
| | A1 | Correct unsimplified equation |
| $W = 15$ * | A1* | Correct answer only |

**Total: 4 marks**

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# Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| First equation e.g. Resolve vertically: $(\pm)V + 50 + 15 = T\cos30° \quad (V = 25)$ | M1, A1 | Or resolve parallel to pole. Or: $P + 50\cos60° + 15\cos60° = 60\sqrt{3} \times \frac{\sqrt{3}}{2}$ |
| Second equation e.g. Resolve horizontally: $(\pm)H = T\cos60° \left(= 30\sqrt{3} = 51.96...\right)$ | M1, A1 | Or resolve perpendicular to pole. Or: $50\cos30° + 15\cos30° = 60\sqrt{3}\cos60° + Q$ |
| $|R| = \sqrt{25^2 + \left(30\sqrt{3}\right)^2}$ | DM1 | Dependent on 2 preceding M marks. $\left(\sqrt{57.5^2 + 3 \times 6.25}\right)$ |
| $= 5\sqrt{133} \; (57.662...) \text{ (N)}$ | A1 | 58 N or better |

**Alternative:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Form vector triangle for vertical forces, thrust and resultant. Correct triangle. | M1, A1 | |
| Use cosine rule: $R^2 = T^2 + (50+W)^2 - 2T(50+W)\cos30°$ | M1 | |
| $R^2 = \left(60\sqrt{3}\right)^2 + (65)^2 - 2 \times 60\sqrt{3} \times 65\cos30°$ | A1 | Correct unsimplified equation |
| Substitute and solve: | DM1 | |
| $|R| = 5\sqrt{133} \; (57.662...) \text{ (N)}$ | A1 | 58 N or better |

*Note: Full marks available using $\pm V, \pm H, \pm P, \pm Q$*

**Total: 6 marks (Question 6 total: 10 marks)**

## Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use CLM | M1 | Need all terms and dimensionally correct. Condone sign errors. Might see them using equal (and opposite) impulses. |
| $6mu - 3kmu = 3mu + kmv \quad ((3-3k)u = kv)$ | A1 | Correct unsimplified equation |
| $\Rightarrow v = \frac{(3-3k)}{k}u$ | A1* | Obtain **given answer** from full and correct working |

## Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of Impulse = change in momentum | M1 | Must be subtracting. Can be for either particle. |
| $\|I_Q\| = \|I_P\| = \|3mu - 3m.2u\| = 3mu$ or $\|kmv-(-3mku)\| = \left\|km.\frac{3-3k}{k}u + 3mku\right\| = 3mu$ | A1 | Correct only. (Do not need to state that $\|I_Q\|=\|I_P\|$ if find $\|I_P\|$) |

## Question 7c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use impact law | M1 | Seen or implied. If stated in (a) must be used here. Must be used correctly but condone sign errors |
| $\frac{v-u}{5u} = e$ or $\frac{3-3k}{k}u - u = 5ue$ | A1 | Correct unsimplified equation |
| Use $v > u$ or $e > 0$ to form an inequality in $k$ | M1 | Could use $e \ldots 0$ followed by $v \neq u$ |
| Use $e < 1$ to form an inequality in $k$ | M1 | |
| $\frac{3-3k}{k} > 1$ and $3-3k < 6k \Rightarrow \frac{1}{3} < k < \frac{3}{4}$ | A1 | Correct answer only |

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## Question 8a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation for horizontal distance | M1 | Complete method using suvat. Condone sine/cosine confusion |
| $x = u\cos\alpha \, t$ | A1 | Correct only |
| Equation for vertical distance | M1 | Complete method using suvat. Condone sine/cosine confusion and sign error |
| $y = u\sin\alpha \, t - \frac{1}{2}gt^2$ | A1 | Correct only |
| $t = \frac{x}{u\cos\alpha} \Rightarrow y = u\sin\alpha \cdot \frac{x}{u\cos\alpha} - \frac{g}{2}\left(\frac{x}{u\cos\alpha}\right)^2$ | DM1 | Substitute for $t$ to obtain $y$ in terms of $x$ and $\alpha$. Dependent on the 2 preceding M marks |
| $\Rightarrow y = x\tan\alpha - \frac{gx^2}{2u^2}(1+\tan^2\alpha)$ | A1* | Obtain **given answer** from full and correct working. Need some evidence for the final step. $\frac{1}{\cos^2\alpha} = 1+\tan^2\alpha$ is not sufficient. |

## Question 8b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Conservation of energy | M1 | Method specified in the question. Need all terms and dimensionally correct. Condone sign errors |
| $\frac{1}{2}m \times 25^2 = \frac{1}{2}mU^2 + mg \times 20$ | A1 | Correct unsimplified equation |
| $U = 15.3 \;\;(15)$ | A1 | 3 sf or 2 sf only |

## Question 8c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use part (a) or work from first principles to form an equation in $\tan\theta$ | M1 | $\left(-20 = 30\tan\theta - \frac{9.8\times 900}{2U^2}(1+\tan^2\theta)\right)$ |
| Obtain $18.9\tan^2\theta - 30\tan\theta - 1.07 = 0$ $\left(\frac{4410}{233}\tan^2\theta - 30\tan\theta - \frac{250}{233} = 0\right)$ | A1ft | Or 3 term equivalent. Follow their $U$. Can be implied by a correct final answer |
| $\Rightarrow \theta = 58.3°$ or $58°$ | A1 | 3 sf or 2 sf only |