Since this question is about the magnitude of the impulse, condone subtraction in the "wrong order" throughout.

| $m\mathbf{v} - m\mathbf{u} = 0.6(2c\mathbf{i} - c\mathbf{j} - c\mathbf{i} - 2c\mathbf{j})$ | M1 | Impulse = change in momentum. Marking the RHS only |
| $= 0.6(c\mathbf{i} - 3c\mathbf{j})$ | A1 | |
| Magnitude $= 0.6\sqrt{c^2 + 9c^2}$ | DM1 | Correct use of Pythagoras' theorem on $m\mathbf{v} - m\mathbf{u}$ or $\mathbf{v} - \mathbf{u}$. Marking the RHS only. Dependent on the previous M1 |
| $= 0.6\sqrt{10c} = (-0.6\sqrt{10c^2})$ | A1 | Accept $\sqrt{10}c$ for change in velocity |
| The next two marks are not available to a candidate who has equated a scalar to a vector. | | |
| $2\sqrt{10} = 0.6\sqrt{10}c$ | DM1 | Equate & solve for $c$. Dependent on the previous M1 |
| $c = \frac{10}{3}$ | A1 | Accept 3.3 or better |
| | (6) | |

## Question 3 alt

| $m\mathbf{v} - m\mathbf{u} = 0.6(2c\mathbf{i} - c\mathbf{j} - c\mathbf{i} - 2c\mathbf{j})$ | M1 | change in momentum |
| $= 0.6(c\mathbf{i} - 3c\mathbf{j})$ | A1 | |
| Square of magnitude | DM1 | |
| $= 0.36(10c^2)$ | A1 | |
| The next two marks are not available to a candidate who has equated a scalar to a vector. | | |
| $40 = 0.36(c^2 + 9c^2)$ | DM1 | Equate & solve for $c$ |
| $c = \frac{10}{3}$ | A1 | |
| | (6) | |

## Question 3 alt (impulse momentum equation)

| $\begin{pmatrix} 2\sqrt{10} \cos \theta \\ 2\sqrt{10} \sin \theta \end{pmatrix} = 0.6 \begin{pmatrix} 2c - c \\ -c - 2c \end{pmatrix}$ | M1 | Impulse momentum equation |
| $= 0.6c \begin{pmatrix} 1 \\ -3 \end{pmatrix}$ | A1 | Correct equation |
| $2\sqrt{10} \cos \theta = 0.6c$ | DM1 | Compare coefficients and form equation for $\theta$ |
| $2\sqrt{10} \sin \theta = -3 \times 0.6c$ | | |
| $\tan \theta = -3 \Rightarrow \cos \theta = (\pm) \frac{1}{\sqrt{10}}$ | A1 | $\cos \theta$ or $\sin \theta$ correct |
| $2\sqrt{10} \cos \theta = 0.6c$ | DM1 | |
| $\Rightarrow c = \frac{10}{3}$ | A1 | |

## Question 3 alt (impulse momentum triangle)

| Impulse momentum triangle. Units used for the vectors must be dimensionally correct | M1 | |
| Sides of magnitude $\sqrt{5c}, \sqrt{5c}, \frac{10\sqrt{10}}{3}$ or $\frac{3\sqrt{5c}}{5}, \frac{3\sqrt{5c}}{5}, 2\sqrt{10}$ | A1 | |
| $\mathbf{u} \cdot \mathbf{v} = \begin{pmatrix} c \\ 2c \end{pmatrix} \cdot \begin{pmatrix} 2c \\ -c \end{pmatrix}$ | DM1 | Use of scalar product |
| $= 2c^2 - 2c^2 = 0 \therefore$ at $90°$ | A1 | ...to show velocities perpendicular. Use of Pythagoras' theorem in a right angled triangle |
| $(0.6 \times \sqrt{5c})^2 + (0.6 \times \sqrt{5c})^2 = (2\sqrt{10})^2$ | DM1 | |
| $\frac{18c^2}{5} = 40, c = \frac{10}{3}$ | A1 | |
| | (6) | |

## Question 4a

| | Triangle | sector |
|---|---|---|
| mass | 4.5 | $4\pi$ (= 12.56..) |
| c of m from O | $\frac{2}{3}, \frac{3\sqrt{2}}{2}$ (= 1.41...) | $\frac{16\sqrt{2}}{3\pi}$ (= 2.40....) |

| B1 | Mass ratios |
| B1 | Distances |
| | | Distances from $AD$ are $-\frac{1}{3} \times \frac{3\sqrt{2}}{2}$ and $\frac{16\sqrt{2}}{3\pi} - \frac{3\sqrt{2}}{2}$ (= 0.280) |

| $4\pi \times \frac{16\sqrt{2}}{3\pi} - 4.5\sqrt{2} = \left(\frac{101\sqrt{2}}{6}\right) = (4\pi - 4.5)d$ | M1 | Moments about an axis through O and parallel to DA. Terms must be dimensionally correct. Shapes combined correctly. |
| | A1 | Correct unsimplified equation |
| $d = \frac{101\sqrt{2}}{6(4\pi - 4.5)} = 2.951.....$ | | (distance from O) |
| Distance from $DA = 2.951... - \frac{3\sqrt{2}}{2} = 0.830 \text{ (0.83) m}$ | A1 | Accept $\frac{101\sqrt{2}}{6(4\pi - 4.5)} - \frac{3\sqrt{2}}{2}$ |
| | (5) | |

## Question 4a alt

| | Triangle | sector |
|---|---|---|
| mass | 4.5 | $4\pi (= 12.56..)$ |
| c of m from both axes $OC, OB$ | 1 | $\frac{16\sqrt{2}}{3\pi} \times \frac{1}{\sqrt{2}} = \frac{16}{3\pi}$ |

| $4\pi \times \frac{16}{3\pi} - 4.5 \times 1 = (4\pi - 4.5)\bar{x}$ | M1 | Moments about an axis through O. Terms must be dimensionally correct. Condone sign error(s) |
| $(\bar{x} = \bar{y} = \frac{101}{6(4\pi - 4.5)})$ | A1 | Correct unsimplified equation |
| $d = \frac{101\sqrt{2}}{6(4\pi - 4.5)}$ | | Distance from O |
| Distance from $DA = 2.951... - \frac{3\sqrt{2}}{2} = 0.830 \text{ (0.83) m}$ | A1 | |
| | (5) | |

## Question 4b

| $\tan \theta = \frac{\text{their } 0.830}{2.12}$ or $\tan \phi = \frac{2.12}{\text{their } 0.830}$ | M1 | Use of tan to find a relevant angle: |
| $21.4°$ or $68.6°$ | A1 | |
| Angle between DC and downward vertical = $135° -$ their $\theta$ | M1 | Correct method for the required angle |
| $= 114°$ | A1 | The Q asks for the angle to the nearest degree. |
| | (4) | |

## Question 4b alt

| $GD^2 = OD^2 + OG^2 - 2OD \cdot OG \cos 45$ | M1 | Complete method to find angle $ODG$ |
| $(GD = 2.28)$ | | |
| $\sin 45 = \frac{\sin \theta}{DG}$ and $\frac{\sin \theta}{OG}$ | | |
| $\Rightarrow \theta = 66.4°$ | A1 | |
| Required angle = $180° - 66.4° = 114°$ | M1 A1 | Correct method for the required angle. The Q asks for the angle to the nearest degree. |
| | (4) [9] | |

## Question 5a

| M(A): $d \cos \theta \times 5g = 4P$ | M1 | Terms must be dimensionally correct. Condone trig confusion |
| Resolving horizontally: $P \sin \theta = F$ | B1 | |
| Resolving vertically: $P \cos \theta + R = 5g$ | M1 | Requires all 3 terms. Condone trig confusion and sign errors |
| | A1 | Correct equation |
| | DM1 | Substitute for $P$ to find $R$ or $F$. Dependent on both previous M marks. One force correct. Accept equivalent forms e.g. $R = \frac{20g - 5gd + 20g \tan^2 \theta}{4(1 + \tan^2 \theta)}$ |
| $R = 5g - \frac{5gd \cos^2 \theta}{4}$ | A1 | One force correct |
| $F = \frac{5gd \cos \theta \sin \theta}{4}$ | A1 | Both forces correct. Accept equivalent forms e.g. $F = \frac{5gd \tan \theta}{4 \sec^2 \theta}$ |
| | (8) | |

## Question 5a alt

| M(B): | M1 | Needs all three terms. Terms must be dimensionally correct. Condone trig confusion |
| $5g \cos \theta \times (4 - d) + F \sin \theta \times 4 = R \cos \theta \times 4$ | | |
| Resolve parallel to the rod: | M1 | Requires all 3 terms. Condone trig confusion and sign errors |
| $5g \sin \theta = R \sin \theta + F \cos \theta$ | B1 | At most one error |
| | A1 | Correct equation |
| $\Rightarrow R = 5g - \frac{F \cos \theta}{\sin \theta}$ | | |
| $5g \cos \theta \times (4 - d) + F \sin \theta \times 4 = 4 \cos \theta \left(5g - \frac{F \cos \theta}{\sin \theta}\right)$ | DM1 | Eliminate one variable to find $F$ or $R$. Dependent on both previous M marks |
| $4F \left(\sin \theta + \frac{\cos^2 \theta}{\sin \theta}\right) = 20g \cos \theta - 20g \cos \theta + 5gd \cos \theta$ | | |
| $= 20g \cos \theta - 20g \cos \theta + 5gd \cos \theta$ | | |
| $F = \frac{5gd \cos \theta \sin \theta}{4}$ | A1 | One force correct |
| $R = 5g - \frac{5gd \cos^2 \theta}{4}$ | A1 | Both forces correct |

## Question 5b

| $\mu = \frac{\frac{5gd \cos \theta \sin \theta}{4}}{5g - \frac{5gd \cos^2 \theta}{4}}$ | M1 | Use of $F = \mu R$ |
| $\frac{1}{2}\left(5g - \frac{5gd \cos^2 \theta}{4}\right) = \frac{5gd \cos \theta \sin \theta}{4}$ | A1 | $(4 - d \cos^2 \theta = 2d \cos \theta \sin \theta)$ |
| $4 \times 169 - 120d + 144d$ | M1 | Use $\tan \theta = \frac{5}{12}$ and solve for $d$ |
| $d = \frac{169}{66}$ | A1 | (= 2.6 m or better) |
| | (4) | |

## Question 5b alt

| $F = 5gd \times \frac{12}{13} \times \frac{5}{13} \times \frac{1}{4} = \left(\frac{75gd}{169}\right)$ | M1 | Use $\tan \theta = \frac{5}{12}$ |
| $R = 5g - \frac{5gd}{4} \times \frac{144}{169}$ | A1 | Both unsimplified expressions |
| $75gd = \frac{1}{2}(5 \times 169 g - 180gd)$ | M1 | Use of $F = \mu R$ and solve for $d$ |
| $150gd + 180gd = 845g, d = \frac{169}{66}$ | A1 | (= 2.6 m or better) |
| | (4) | |

## Question 5b alt t

| $R = 5g - \frac{12}{13}P, F = \frac{5}{13}P$ | M1 | Substitute trig in their equations from resolving. |
| $\frac{5}{13}P = \frac{1}{2}\left(5g - \frac{12}{13}P\right)$ | M1 | use $F = \mu R$ and solve for $d$ |
| $\Rightarrow P = \frac{65}{22}g$ | A1 | |
| $d = \frac{4P}{5g \cos \theta} = \frac{169}{66}$ | A1 | |
| | [12] | |

## Question 6a

| Horizontal motion: $x = 3t$ | B1 | |
| Vertical motion: $y = 4t - \frac{g}{2}t^2$ | M1 | Correct use of suvat. Condone sign error(s) |
| | A1 | |
| $\left(y = 4 \times \frac{x}{3} - \frac{g}{2} \times \frac{x^2}{9}\right), \lambda = -\left(\frac{4 \times}{3} - \frac{gx^2}{18}\right)$ | M1 | Use $y = -x$ and form an equation in one variable |
| $\frac{7\lambda}{3} - \frac{g\lambda^2}{18}$ | M1 | solve for $\lambda$ |
| $\lambda = \frac{42}{g}$ or 4.3 (4.29) | A1 | (6) | Not $\frac{30}{7}$ |

## Question 6a alt

| Horizontal motion: $x = 3t$ | B1 | |
| Vertical motion: $y = 4t - \frac{g}{2}t^2$ | M1 | Correct use of suvat. Condone sign error(s) |
| | A1 | |
| $\Rightarrow -3t = 4t - \frac{1}{2}gt^2, \left(t = \frac{14}{g}\right)$ | M1 | Use $y = -x$ and form an equation in one variable |
| $\lambda = 3t$ | M1 | Solve for $\lambda$ |
| $\lambda = 4.3$ (4.29) | A1 (6) | |

## Question 6b

| At A: $v \to 3 \text{ (m s}^{-1})$ | B1 | |
| $v \uparrow \quad 4 - g \times \frac{14}{g}$ | M1 | Complete method using suvat to find $v \uparrow$ with their $t$ or $\lambda$ |
| | | $= -10 \text{ (m s}^{-1})$ | A1 | Accept +10 with direction confirmed by diagram |
| Speed $= \sqrt{(\text{their } 10)^2 + (3)^2}$ | DM1 | Dependent on the first M1 in (b) |
| $= \sqrt{109 \text{ (m s}^{-1})$ | A1 | (10.4) Allow for $v \uparrow = 10$ |
| $\tan^{-1}\left(\frac{\text{their } 10}{3}\right)$ or $\tan^{-1}\left(\frac{3}{\text{their } 10}\right)$ | DM1 | Use trig to find a relevant angle. Dependent on the first M1 in (b) |
| Direction $= 73.3°$ below the horizontal | A1 | (1.28 radians) Accept direction $3\mathbf{i} - 10\mathbf{j}$. Do not accept a bearing |
| | (7) | |

## Question 6b alt

| Loss in GPE: $mg \lambda A = 42m$ | B1 | |
| Gain in KE: $\frac{1}{2}mv^2 - \frac{1}{2}m \times 25$ | M1 | Terms must be dimensionally correct. Condone sign error. |
| Solve for v: $42 = \frac{1}{2}v^2 - \frac{25}{2}$ | M1 | |
| $v = \sqrt{109}$ | A1 | |
| $v \cos \theta = 3$ | M1 | Use trig. to find a relevant angle |
| $\theta = 73.3°$ below the horizontal | A1 (7) | Accept correct angle marked correctly on a diagram. |
| | [13] | |

## Question 7a

| CLM: $6mu = 2mv + 3mw$ | M1 | Requires all 3 terms. Must be dimensionally correct. Condone sign error(s) |
| $(6u = 2v + 3w)$ | A1 | This equation defines their directions |
| Impact: $w - v = \frac{3}{4} \times 3u \left(= \frac{9}{4}u\right)$ | M1 | Must be used with $e$ on the correct side |
| | A1 | Penalise inconsistent directions here |
| $6u = 2w - \frac{9}{2}u + 3w$ | DM1 | Solve simultaneous equations for $v$ or $w$. Dependent on the 2 previous M marks |
| $w = \frac{21}{10}u = v_B$ | A1 | One correct |
| $v = w - u = \left(\frac{21}{10} - \frac{9}{4}\right)u = -\frac{3}{20}u, v_A = -\frac{3}{20}u$ | A1 | Both correct |
| | (7) | |

## Question 7b

| Speed of B after hitting wall = $\frac{21}{10}ue$ | M1(B1) | $e \times$ their $w$ |
| Impulse = $\frac{27}{4}mu = 3m\left(\frac{21}{10}u + \frac{21}{10}ue\right)$ | M1 | for their $w$. Must be trying to use the correct equation with $3m$. |
| $\frac{9}{4} \times \frac{10}{10}(1 + e), e = \frac{1}{14}$ | A1 | (3) |

## Question 7b alt

| Impulse = $\frac{27}{4}mu = 3m\left(\frac{21}{10} + V\right), \left(V = \frac{3u}{20}\right)$ | M1(B1) | Use impulse to find $V$. Must be trying to use the correct equation with $3m$. |
| $\frac{21u}{10} = e \times$ their $w$. | M1 | $V = e \times$ their $w$. |
| $e = \frac{1}{14}$ | A1 | (3) |

## Question 7c

| Speed of B after second impact = $\frac{1}{14} \times \frac{21}{10}u = \frac{3}{20}u$ | B1 ft | Compare two relevant speeds. (ft on their $V$ or their $e \times$ their $w$) |
| Same velocity (and A has a head start), so no collision. | B1 | From correct work only |
| | (2) [12] | |