| Impulse-momentum equation (2D, dimensionally correct, condone subtraction in wrong order) | M1 | Must be working in 2 dimensions. Dimensionally correct. Condone subtraction in wrong order |
| $\begin{pmatrix} 6\cos 50° \\ 6\sin 50° \end{pmatrix} = 0.8v - 0.8\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ | A1 | Correct unsimplified equation |
| $v = \begin{pmatrix} 4 + 7.5\cos 50° \\ 7.5\sin 50° \end{pmatrix}$ | A1 | $\begin{pmatrix} 8.82 \\ 5.74 \end{pmatrix}$ |
| Pythagoras: $\|v\| = \sqrt{(4 + 7.5\cos 50°)^2 + (7.5\sin 50°)^2}$ | M1 | Must have 2 components |
| $\|v\| = \sqrt{8.82^2 + 5.75^2} = 10.5 \text{ (m s}^{-1}\text{)}$ | A1 | Accept 10.52 and 10.53 |

**Total: [5]**

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# Question 1 Alternative:

| Impulse-momentum equation (dimensionally correct, condone subtraction in wrong order) | M1 | Dimensionally correct. Condone subtraction in wrong order |
| $\begin{pmatrix} 6 \\ 0 \end{pmatrix} = 0.8\begin{pmatrix} v\cos\theta - 4\cos 50° \\ v\sin\theta - 4\sin 50° \end{pmatrix}$ | A1 | Working parallel and perpendicular to the impulse |
| $\begin{pmatrix} v\cos\theta \\ v\sin\theta \end{pmatrix} = \begin{pmatrix} 7.5 + 4\cos 50° \\ 4\sin 50° \end{pmatrix}$ | A1 | $\begin{pmatrix} 10.071... \\ 3.064... \end{pmatrix}$ |
| Use of Pythagoras | M1 | |
| $\|v\| = 10.5 \text{ (m s}^{-1}\text{)}$ | A1 | |

**Total: [5]**

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# Question 1 Alternative 2:

Momentum (or velocity) triangle with sides labeled $0.8v$, $6$, $50°$ angle, and base $3.2$

**Total: [5]**

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# Question 1 (continued):

| Cosine rule: $(0.8v)^2 = 3.2^2 + 6^2 - 2 \times 3.2 \times 6\cos 130°$ | M1 | |
| Solve for $v$ | M1 | |
| $v = 10.5 \text{ (m s}^{-1}\text{)}$ | A1 | |

**Total: [5]**

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# Question 2a:

| Any two of: KE change / PE change / work done against resistance | B1, B1 | Correct unsimplified expression required |
| Work done: $\frac{1}{2} \times 1200(8^2 - 5^2) + 1200g \times 90\sin\alpha + 250 \times 90$ | M1 | All terms required. Dimensionally correct. Condone sign errors and sin/cos confusion. |
| $(23400) = (70560) = (22500)$ $(=116460) = 116000\text{ (J)} = (120000)$ | A1, A1 | Correct unsimplified equation. Max 3 sf |

**Total: (5)**

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# Question 2a Alternative:

| Use $\text{suvat to obtain } a = \frac{13}{60}\text{ (m s}^{-2}\text{)}$ | B1 | Accept correct equation in $a$ e.g. $8^2 = 5^2 + 2 \times a \times 90$ |
| Use $F = ma$ to obtain net force $= 260\text{ (N)}$ | B1 | Accept $1200a$ |
| Work done: $= 90(260 + 250 + 1200g\sin\alpha)$ | M1 | All terms required. Dimensionally correct. Condone sign errors and sin/cos confusion. |
| $(=116460) = 116000\text{ (J)} = (120000)$ | A1, A1 | Correct unsimplified equation. Max 3 sf |

**Total: (5)**

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# Question 2b:

| Equation of motion: $F + 1200g\sin\alpha - 250 = 1200a$ | M1 | All terms required. Condone sign errors and sin/cos confusion |
| Use of $F = \frac{P}{v}$: $F = \frac{8000}{6}$ | M1 | Independent |
| $a = \frac{1867...}{1200} - 1.56\text{ (m s}^{-2}\text{)} \text{ (1.6)}$ | A1 | Max 3 sf |

**Total: (4) [9]**

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# Question 3a:

| Use of $v = \frac{dr}{dt}$ | M1 | Differentiate – powers going down |
| $v = (16 - 9t^2)i + (3t^2 - 2t)j$ | A1 | |
| i component of velocity = 0: $16 - 9t^2 = 0 \Rightarrow t = \frac{4}{3}$ | DM1 | Solve for $t$ and find $v$ or $\|v\|$. Dependent on previous M1 |
| $v = \left(3 \times \frac{16}{9} - 2 \times \frac{4}{3}\right)j = -\frac{8}{3}j\quad (2.67j)$ | A1 | Answer must be a vector. ISW |

**Total: (5)**

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# Question 3b:

| Use of $a = \frac{dv}{dt}$ | M1 | Differentiate – powers going down |
| $a = (-18t)i + (6t - 2)j\quad (= -72i + 22j)$ | A1ft | Follow their $v$ |
| Use of Pythagoras' theorem: $\|a\| = \sqrt{72^2 + 22^2}$ | M1 | |
| $\|a\| = \sqrt{5668} = 75.3\text{ (m s}^{-2}\text{)} \text{ (75)}$ | A1 | Or better. From correct work |

**Total: (4) [9]**

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# Question 4a:

| Velocity at $T$: $\rightarrow 12\cos 30° = u_n (= u\cos\theta°)$ | M1 | |
| $(u\cos\theta° = 6\sqrt{3} = 10.39...)$ | A1 | Correct unsimplified equation for horizontal component of $u$ |
| $\uparrow -12\sin 30° = u_v - 2g (= u\sin\theta° - 2 \times 9.8)$ | M1 | |
| $(u\sin\theta° = 13.6)$ | A1 | Correct unsimplified equation for vertical component of $u$ |
| $\tan\theta° = \frac{13.6}{6\sqrt{3}}$ | DM1 | Solve equations for $u$ or $\theta$. Dependant on both preceding M marks |
| $\theta = 52.6\text{ (53)}$ | A1 | One correct (max 3 s.f.) |
| $u = 17.1\text{ (17)}$ | A1 | Both correct (max 3 s.f.) |

**Total: (7)**

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# Question 4b:

| Vertical distance: $h = -12\sin 30° \times 2 + \frac{1}{2} \times 9.8 \times 2^2$ | M1 | Complete method using $\text{suvat}$ to find $h$. |
| $(or\quad h = 17.1\sin 52.6° \times 2 - \frac{1}{2} \times 9.8 \times 2^2)$ | A1 | Or equivalent correct unsimplified equation in $h$ |
| $(or\quad 6^2 = (u\sin\theta)^2 - 2gh)$ | A1 | |
| $h = 7.6\text{ (7.60)}$ | A1 | |

**Total: (3)**

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# Question 4b Alternative:

| Using energy: $\frac{1}{2}mu^2 - \frac{1}{2}m \times 2^2 = mgh$ | M1A1 | |
| $h = 7.6\text{ (7.60)}$ | A1 | |

**Total: (3)**

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# Question 4c:

| Double the time from max ht to $T$: $-12\sin 30° = -gt$ | M1 | |
| Time above $T$: $2t = 2 \times \frac{12\sin 30}{g}$ | A1 | |
| $= 1.22\text{ (1.2) (s)}$ | A1 | |

**Total: (3)**

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# Question 4c Alternative:

| Vertical component of speed equal magnitude and opposite sign: $-12\sin 30° = 12\sin 30° - gt$ | M1 | |
| $\frac{24\sin 30°}{g}$ | A1 | |
| $t = 1.22$ | A1 | |

**Total: (3)**

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# Question 4c Alternative 2:

| Equation for vertical distance and solve for values of $t$: $7.6 = u\sin\theta° \times t - \frac{1}{2}gt^2, \quad 4.9t^2 - 13.6t + 7.6 = 0$ | M1 | |
| $t_2 - t_1 = \frac{\sqrt{13.6^2 - 4 \times 4.9 \times 7.6}}{4.9}$ | A1 | $2 - \frac{38}{49}(2 - 0.7785)$ |
| $t = 1.22$ | A1 | From correct work only |

**Total: (3)**

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# Question 4c (continued):

| For other alternatives: | | |
| Complete strategy | M1 | |
| Correct equation in $t$ | A1 | |
| $t = 1.22$ | A1 | |

**Total: [13]**

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# Question 5a:

| | area | From $AB$ | From $AE$ |
|---|---|---|---|
| rectangle | $3ka^2$ | $\frac{1}{2}ka$ | $\frac{3}{2}a$ |
| Triangle - | $\frac{9}{2}a^2$ | $ka - 3a + 2a$ | $a$ |
| Triangle + | $\frac{9}{2}a^2$ | $ka - 3a + a$ | $2a$ |
| $L$ | $3ka^2$ | $\bar{x}$ | $\bar{y}$ |

| mass ratio: $3k : \frac{9}{2} : \frac{9}{2} : 3k$ | B1 | |
| Horizontal distances: | B1 | From $AB$ or from a parallel axis. Need all terms and dimensionally correct |
| Moments about $AB$: $3k \times \frac{ka}{2} - \frac{2}{2} \times (k-1)a + \frac{9}{2} \times (k-2)a = 3k\bar{x}$ | M1 | Need all terms and dimensionally correct. Accept on vector form |
| $\frac{3k^2}{2}a - \frac{9}{2}a = 3k\bar{x}, \quad \bar{x} = \frac{(k^2-3)}{2k}a$ | A1 | Correct unsimplified equation in $\bar{x}$ |

**Total: (5)**

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# Question 5b:

| Vertical distances | B1 | From $AE$ or from a parallel axis. Need all terms and dimensionally correct |
| Moments about $AE$: $3k \times \frac{3a}{2} - \frac{9}{2} \times a + \frac{9}{2} \times 2a = 3k\bar{y}$ | M1 | |
| $\frac{9ka}{2} - \frac{9a}{2} = 3k\bar{y}, \quad \bar{y} = \frac{3(k+1)a}{2k}$ | A1 | Correct unsimplified equation in $\bar{y}$. Distance from $BD = \frac{3a(k-1)}{2k}$ |
| | A1 | Or equivalent |

**Total: (4)**

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# Question 5a-5c (Alternative):

If there is a triangle "missing", the total marks available for the question are: B0B1M1A0A0 B1M1M1A0 M1M1M1A0 (7/13)

See over for alternative working

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# Question 5a-5c (continued):

| Alternative working for parts (a) and (b) | | |
|---|---|---|
| | mass | From $AB$ | From $AE$ |
| rectangle | $3(k-3)a^2$ | $\frac{1}{2}(k-3)a$ | $\frac{3}{2}a$ |
| Triangle | $2 \times \frac{9}{2}a^2$ | $(k-3)a + a$ | $2a$ |
| $L$ | $3ka^2$ | $\bar{x}$ | $\bar{y}$ |

| Moments about $AB$: $3(k-3) \times \frac{(k-3)a}{2} + 9 \times (k-2)a = 3k\bar{x}$ | | |
| Moments about $AE$: $3(k-3) \times \frac{3a}{2} + 9 \times 2a = 3k\bar{y}$ | | |

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# Question 5c:

| $\bar{x} = \bar{y}: (k^2 - 3) = 3(k+1)$ for their values | M1 | |
| Simplify to 3 term quadratic in $k$: $k^2 - 3k - 6 = 0$ | DM1 | Dependent on preceding M1 |
| Solve for $k$: $k = \frac{3 \pm \sqrt{9+24}}{2}$ | DM1 | Dependent on preceding M1 |
| $k = 4.37$ only | A1 | The Q asks for 3 sf |

**Total: (4) [13]**

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# Question 6a:

| Moments about $A$: | M1 | Need all terms and dimensionally correct |
| $kmg \times 0.5a\sin 60° + 8mg \times a\sin 60° = T\sin 30° \times 2a$ | A1 | Unsimplified equation. -1 each error |
| | A1 | $\cos 60°$ for $\sin 60°$ twice counts as one error |
| $T = g\sin 60°\left(\frac{km}{2} + 8m\right) - \frac{\sqrt{3}}{4}(16+k)mg \quad \text{Given Answer}$ | A1 | Obtain given answer from correct working |

**Total: (4)**

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# Question 6b:

| Resolving: $\rightarrow T\cos 60° = H$ | M1 | Condone sin/cos confusion |
| $\uparrow V + T\cos 60° = 8mg + kmg$ | M1 | Condone sin/cos confusion & sign errors |
| | A1 | Both equations correct unsimplified |
| | | Allow M1M1A1 for alternative equations that are sufficient to solve for $k$ |
| Use $F = \mu R$ with their $V$ and $H$ | M1 | Dependent on having expressions for $V$ and $H$ |
| $\left(V = \mu H \Rightarrow (8+k)mg - T\cos 30° = \frac{2}{3}\sqrt{3} \times T\cos 60°\right)$ | | |
| Substitute for $T$ and solve for $k$: $(8+k) - \frac{3}{8}(16+k) = \frac{\sqrt{3}\sqrt{3}}{3}\times 4(16+k)$ | DM1 | Dependent on 3 preceding M marks |
| $2 + \frac{5}{8}k = 4 + \frac{1}{8}k, \quad \frac{3}{8}k = 2, \quad k = \frac{16}{3}\text{ (or 5.33)}$ | A1 | |

**Total: (6) [10]**

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# Question 7a:

| Impact law: $\frac{3}{2}u + v = e(3u - u)(= 2eu)$ | M1 | Used the right way round |
| $\left(v = 2eu - \frac{3}{2}u\right)$ | A1 | Correct unsimplified equation |
| $v > 0 \Rightarrow 2e > \frac{3}{2}$ | M1 | Form and solve correct inequality for their $v$ |
| $(1 \geq)e > \frac{3}{4}$ | A1 | Accept $1 > e > \frac{3}{4}$ and $e > \frac{3}{4}$ |

**Total: (4)**

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# Question 7a Alternative:

| Impact law: $\frac{3}{2}u + v = e(3u - u)(= 2eu)$ | M1 | Used the right way round |
| $\left(v = 2eu - \frac{3}{2}u\right)$ | A1 | Correct unsimplified equation |
| CLM $\Rightarrow v = \frac{u}{k}(1-3k) > 0 \Rightarrow k < \frac{1}{3}$ | M1 | Use CLM to form inequality in $k$ and substitute into impact equation |
| $e = \frac{1}{2k} - \frac{3}{4} - \frac{3}{4} - \frac{3}{4} < e(\leq 1)$ | A1 | |

**Total: (4)**

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# Question 7b:

| $e = \frac{7}{8} \Rightarrow v = \frac{7}{4}u - \frac{3}{2}u = \frac{1}{4}u$ | B1 | |
| CLM: $3kmu + 2mu = 2m \times \frac{3}{2}u - kmu$ | M1 | Need all terms and dimensionally consistent. If only seen in (a) it must be used in (b) to score. |
| $\left(3k + 2 = 3 - \frac{1}{4}k\right)$ | A1 | Correct unsimplified equation |
| $k = \frac{4}{13}$ | A1 | |
| KE lost: $\frac{1}{2} \times \frac{4}{13}m\left(9u^2 - \frac{u^2}{16}\right)$ | M1 | Accept in terms of $k$ e.g. $\frac{1}{2}km(9u^2 - \frac{1}{16}u^2)$ |
| $= \frac{2}{13} \times \frac{143}{16}u^2 = \frac{11}{8}mu^2\text{ *Given answer*}$ | A1 | Obtain given answer from correct working. Fully correct substitution seen |

**Total: (6)**

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# Question 7c:

| Time for $Q$ to reach wall: $\frac{2d}{3u}$ | B1 | |
| Speed of $Q$ after collision with wall: $\frac{1}{3} \times \frac{3}{2}u = \frac{1}{2}u$ | B1 | |
| $P$ has moved: $\frac{u}{4} \times \frac{2d}{3u} = \frac{d}{6}$ | B1 | |
| Gap: $d + \frac{d}{6} - \frac{7d}{6}$ closing at $\frac{1}{2}u - \frac{1}{4}u = \frac{1}{4}u$ | M1 | |
| takes: $\frac{7d}{6} \div \frac{u}{4} = \frac{14d}{3u}$ | M1 | Terms dimensionally correct |
| Total time: $\frac{14d}{3u} + \frac{2d}{3u} = \frac{16d}{3u}$ | A1 | |

**Total: (6)**

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# Question 7c Alternative:

| Time for $Q$ to reach wall: $\frac{2d}{3u}$ | B1 | |
| Speed of $Q$ after collision with wall: $\frac{1}{3} \times \frac{3}{2}u = \frac{1}{2}u$ | B1 | |
| Total time for $Q$: $\frac{2d}{3u} + \frac{2x}{u}$ | B1 | |
| Equal times: $\frac{2d}{3u} + \frac{2x}{u} = \frac{4(x-d)}{u}$ | M1 | Terms dimensionally correct. Condone a sign error |
| Solve for $x$: $2d + 6x = 12x - 12d, \quad x = \frac{7d}{3}$ | M1 | |
| Time $= \frac{u}{4} \times \frac{3}{u} = \frac{3u}{3u}$ | A1 | |

**Total: (6) [16]**