**Part (i) and (ii) together**
For marking: 1st equation in one unknown M1A1, 2nd equation in one unknown M1A1, 1st value A1, 2nd value A1

| Impulse - momentum equation for $P$ | M1 | Must be trying to subtract. Terms dimensionally consistent. |
| $5mu = 3m(v_p - u)$ | A1 | Correct unsimplified equation |
| $v_p = \frac{2u}{3}$ | A1 | Final answer positive. Condone unexplained sign change |
| Impulse momentum equation for $Q$ | M1 | Must be trying to subtract. Terms dimensionally consistent. |
| $5mu = m(v_Q - -3u)$ | A1 | Correct unsimplified equation |
| $v_Q = 2u$ | A1 | |

**Part (iii) alt**
Use of CLM | M1 | Need all terms and dimensionally consistent. Condone sign errors. |
| $3mu - 3mu = -3m\frac{2u}{3} + mv_Q$ or $3mu - 3mu = 3mv_p + 2mu$ | A1 | Correct unsimplified equation |
| $v_Q = 2u$ | A1 | Final answer positive. Condone unexplained sign change |
| | [6] | |

## Question 2a

**Part i**
| Moments equation | M1 | Use moments to form an equation in $R_C$ and/or $R_D$. All terms required. Dimensionally correct. Condone sign errors. |
| $M(D): (60g \times 0.6) + (20g \times 1.6) = R_c \times 2$ | | |
| $M(C): (60g \times 1.4) + (20g \times 0.4) = R_D \times 2$ | | |
| $M(A): 2 \times 20g + 3 \times 60g = 1.6R_c + 3.6R_D$ | | |
| $M(B): 0.4R_D + 2.4R_c = 60g \times 1 + 20g \times 2$ | | |
| $R_c = 34g$ | A1 | Correct unsimplified equation |
| | A1 | 333 (333.2) is an accuracy error |

**Part ii**
| Resolve vertically | M1 | Or form a moments equation in $R_D$ |
| $(\uparrow) R_c + R_D = 80g$ | A1 | Correct unsimplified equation |
| $R_D = 46g$ | A1 | 451 (450.8) is an accuracy error (penalise once only if $g$ substituted in both answers and correct versions not seen) |
| | (6) | |

## Question 2b

| Set $R_D = 0$ and use moments to form equation in a relevant distance (One unknown only) | M1 | Complete method for a relevant distance. Dimensionally correct equation. Using their answers from (a) is M0 |
| $M(C), (20g \times 0.4) = (60g \times x)$ where $x =$ distance from C when beam tilts | A1 | Correct unsimplified equation for a relevant distance |
| $\left(x = \frac{2}{15}\right)$ | | |
| Use their distance to find the distance walked | DM1 | Dependent on the previous M1 |
| Distance $= 1.4 + \frac{2}{15} - \frac{23}{15} = 1.53 \text{ m}$ | A1 | |
| | (4) | |
| | [10] | |

## Question 3a

| | B1 shape | Correct shape graph for cyclist |
| | B1 figs | 4 marked |
| | B1 shape | Motorcyclist graph in relatively correct position. Must start at $t = 4$ and must continue beyond point of intersection of the graphs. $T + 4$ marked |
| | B1 figs | (4) |
| | | Treat two separate graphs as two attempts and award the marks for the better attempt |

## Question 3b

| $\frac{1}{2}T.4T = \left(\frac{T + T + 4}{2}\right)8$ | M1 | Equate distances to form equation in $T$ |
| | A1 | One distance correct |
| | A1 | Both distances correct |
| $T^2 - 4T - 8 = 0$ | A1 | Simplify to 3 term quadratic |
| $T = 2 \pm \sqrt{12}$ | M1 | Solve a 3 term quadratic for $T$ |
| $T = 5.5$ | A1 | Q asks for answer to 1 dp. Must reject negative solution if seen. |
| | (6) | |
| | [10] | |

## SC1

| B1B1 | |
| B1B0 | |
| | $16 + 8(T - 4) = \frac{1}{2} \times 4(T - 4)^2$ | M1A1A1 |
| $T^2 - 12T + 24 = 0$ (or equivalent) | A1 |
| $T = 6 + 2\sqrt{3} = 9.5$ | M1A0 |
| (marking the $T$ as a misread) | |
| | B1B1 |
| B0B0 | |
| | $16 + 8(T - 4) = \frac{1}{2} \times 4T^2$ | M1A1A1 |
| $27T^2 - 8T + 16 = 0$ | A0M0A0 |
| | (completely changed the question but some evidence of correct thinking) |

## SC2

## Question 4a

| Resolve perpendicular to the surface | M1 | Condone sin/cos confusion |
| $R = 2g \cos \alpha$ (15.68) | A1 | Correct resolution |
| $F = \frac{1}{4}R = \frac{2g}{5} = 3.9 \text{ N or } 3.92 \text{ N}$ | A1 | Max 3 sf for decimal answer |
| | (3) | |

## Question 4b

| $-2g \sin \alpha - F = 2a$ | M1 | Equation of motion parallel to the plane. Require all terms and dimensionally correct. Condone sign errors and sin/cos confusion |
| | A1ft | Correct unsimplified equation in $F$ (or their $F$) |
| $\frac{-4g}{5} = a$ | A1 | Or $-7.84 \text{ (ms}^{-2}\text{)}$ Accept +/- |
| $0^2 = 6^2 - \frac{8g}{5} s$ | DM1 | Complete method using $suvat$ and $a \neq g$ to find $s$. Dependent on the previous M1 |
| $s = \frac{45}{2g} = 2.3 \text{ m or } 2.30 \text{ m}$ | A1 | Max 3 sf |
| | (5) | |

## Question 4c

| $2g \sin \alpha - F = 2a'$ | M1 | Equation for motion down the plane to find new acceleration. Require all terms and dimensionally correct. Condone sign errors and sin/cos confusion |
| | A1ft | Correct unsimplified equation in $F$ (or their $F$) |
| $\frac{2g}{5} = a'$ | A1 | Or $3.92 \text{ (ms}^{-2}\text{)}$ |
| $v^2 = \frac{4g}{5} \times \frac{45}{2g} = 18 \Rightarrow$ | DM1 | Complete method using $suvat$, $a' \neq g$ and $a' \neq a$, to find $v$. Dependent on the previous M1 |
| $v = \sqrt{18} = 4.2 \text{ m s}^{-1}\text{(or better)}$ | A1 | $g$ cancels. Condone 4.25 (from using rounded values). |
| | (5) | |
| | [13] | |

## Question 5a

| Correct equation for $v_p$ or find displacement | M1 | Use of $\mathbf{r}_p = \mathbf{r}_0 + v_pt$ to find $v$. Allow for $\lambda(-\mathbf{i} - 5\mathbf{j})$ |
| $\mathbf{v}_p = 3(6\mathbf{i} - (7\mathbf{i} + 5\mathbf{j})) = -3\mathbf{i} - 15\mathbf{j}$ | A1 | |
| $\sqrt{(-3)^2 + (-15)^2}$ | M1 | Use of Pythagoras to find magnitude of their $\mathbf{v}$ |
| $= \sqrt{234} = 15.3 \left(\text{km h}^{-1}\right)$ (or better) | A1 | CSO $\left(3\sqrt{26}\right)$ A0 if it comes from $3\mathbf{i} + 15\mathbf{j}$ |
| | | NB Could score the M marks in reverse order - find displacement in 20 minutes and then multiply by 3 |
| | (4) | |

## Question 5b

| Use of $\mathbf{r}_p = \mathbf{r}_0 + v_pt : \mathbf{r}_p = 7\mathbf{i} + 5\mathbf{j} + t(-3\mathbf{i} - 15\mathbf{j})$ | M1 | For their $\mathbf{v}_p$ |
| $\Rightarrow \mathbf{r}_p = (7 - 3t)\mathbf{i} + (5 - 15t)\mathbf{j}$ | A1 | Obtain given answer from correct working |
| | (2) | |

## Question 5c

| $\frac{(7 - 3t)}{(5 - 15t)} = \frac{16}{5}$ | M1 | Use given answer and direction to form equation in $t$ |
| $35 - 15t = 80 - 240t$ | A1 | Correct unsimplified equation |
| $t = 0.2$ | DM1 | Solve for $t$. Dependent on the previous M1 |
| | A1 | |
| | (4) | |

## Question 5d

| $P$ and $Q$ in the same place at the same time | M1 | Equate i or j components of position vectors and solve for $t$ |
| $\Rightarrow 7 - 3t = 5 + 2t$ or $5 - 15t = -3 + 5t$ | A1 | Either |
| $t = 0.4$ | A1 | |
| Check that the same value of $t$ gives equal values for the other component | DM1 | Dependent on the previous M mark |
| $\mathbf{r} = (5.8\mathbf{i} - \mathbf{j}) \text{ km}$ | A1 | Must be a vector |
| | (5) | |
| | [15] | |

## Question 6a

| For the trailer: | M1 | Complete method to form an equation in $T$. e.g. equation of motion for the trailer. Need all 3 terms. Condone sign errors. |
| $-100 - T = 600 \times (-4)$ | A1 | Correct unsimplified equation. Allow with $\pm T$ |
| $T = 2300 \text{ N}$ | A1 | Must be positive |
| | (3) | |

## Question 6b

| For the car and trailer: | M1 | Complete method to solve for $M$. Equation of motion for the car + trailer. Need all terms. Condone sign errors. |
| $6500 + 100 + 200 = 4(M + 600)$ | A1 | Correct unsimplified equation |
| $M = 1100$ | A1 | |
| | | Allow M1A1 if a correct equation is seen in (a) and used in (b) |

## Question 6balt

| For the car: | M1 | Equation of motion for the car. Need all terms. Condone sign errors. |
| $6500 + 200 - T = 4M$ | A1 | Correct unsimplified equation in $T$ or their $T$ |
| $M = 1100$ | A1 | |
| | (3) | |

## Question 6c

| $s = vt - \frac{1}{2}at^2$ | M1 | Complete method using $suvat$ to find $t$. Clear use of $s = ut + \frac{1}{2}at^2$ with $u = 0, a = 4$ is M0. e.g. $40.5 = -2t^2$ from no working is M0A0 |
| $40.5 = \frac{1}{2}.4.t^2$ | A1 | Correct unsimplified equation |
| $t = 4.5 \text{ s}$ | A1 | |
| | (3) | |
| | [9] | |

## Question 7a

| $\sin \alpha = \frac{3}{5}$ or $\cos \alpha = \frac{4}{5}$ | B1 | Correct trig ratios for $\alpha$ seen or implied. Watch out - it could be up beside the diagram |
| At $B, (\uparrow)$ | M1 | Complete method to form equation in $T_{AB}$ |
| $\Rightarrow T_{AB} \sin \alpha = 3g$ | A1 | Correct unsimplified equation |
| $T_{AB} = 5g = 49 \text{ N}$ | A1 | |
| | (4) | |

## Question 7b

| At $B, (\rightarrow)$ | M1 | Complete method to form equation in $T_{BC}$ |
| $\Rightarrow T_{AB} \cos \alpha = T_{BC}$ | A1 | Correct unsimplified equation. Allow with their $T_{AB}$ |
| $T_{BC} = 4g = 39 \text{ or } 39.2 \text{ N}$ | A1 | |
| | (3) | |

## Question 7c

| Resolve at $C$: | M1 | Resolve to form equation in $T_{CD}$. There is a lot of confusion over the labelling of the tensions. Allow if a value is used correctly, whatever it is called. |
| At $C, (\rightarrow) T_{CD} \cos \beta = T_{BC}$ | A1 | One correct equation in $T_{CD}$. Could be whole system equations e.g. $T_{AB} \cos \alpha = T_{CD} \cos \beta$ or $T_{AB} \sin \alpha + T_{CD} \sin \beta = (3 + M) g$ |
| At $C, (\uparrow) T_{CD} \sin \beta = Mg$ | A1 | Two correct equations in $T_{CD}$ $(=101.92)$ |
| $\tan \beta = \frac{Mg}{T_{BC}}$ | DM1 | Dependent on previous M1. Use $\tan \beta$ and solve for $M$ |
| $Mg = 4g \times \frac{12}{5} \Rightarrow M = 9.6$ | A1 | |
| | (5) | |
| | [12] | |