| $76 = 4u + \frac{1}{2}a \cdot 4^2$ or $76 = \frac{1}{2}(u + u + 4u) \times 4$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ for $t=4, s=76$ and $u \neq 0$ (use of $u=0$ is M0) |
| $(38 = 2u + 4a)$ | A1 | Correctly substituted equation |
| $295 = 10u + \frac{1}{2}a \cdot 10^2$ or $295 = \frac{1}{2}(u + u + 10a) \times 10$ or $295 = (u+10a) \times 10 - \frac{1}{2}a \times 100$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ for $t=10, s=295$ or $s = u't + \frac{1}{2}at^2$ for $t=6, s=219, u' \neq u$ |
| $(59 = 2u + 10a)$ or $219 = (19 + 2a) \times 6 + \frac{1}{2}a \times 6^2$ or $219 = (38-u) \times 6 + \frac{1}{2}a \times 6^2$ or $219 = (u+4a) \times 6 + \frac{1}{2}a \times 6^2$ or $219 = \frac{1}{2}(u + 4a + u + 10) \times 6$ or $219 = (u + 10a) \times 6 - \frac{1}{2}a \times 36$ | A1 | Correctly substituted equation |
| | DM1 | Solve simultaneous for $u$ or for $a$. This marks is not available if they have assumed a value for $u$ or $a$ in the preceding work - it is dependent on the first 2 M marks |
| $u = 12$ | A1 | |
| $a = 3.5$ | A1 | |
| | [7] | |

# Question 1 Alt:

| $t = 2, v_2 = \frac{76}{4} = 19$ and $t = 7, v_7 = \frac{219}{6} = 36.5$ | M1 A1 | Find the speed at $t=2, t=7$. Both values correct. Averages with no links to times is M0 |
| $36.5 = 19 + 5a \Rightarrow a = 3.5$ | M1 A1 | Use of $v = u + 5a$ with their $u,v$. Correct $a$ |
| $19 = u + 2a$ | DM1 A1 | Complete method for finding $u$. Correct equation in $u$ |
| $u = 19 - 7 = 12$ | A1 | |

# Question 2(a):

| $mu - 2kmu = -\frac{1}{2}mu + kmu$ or $m(\frac{1}{2}u + u) = -km(-u - 2u)$ | M1 | Use of CLM. OR Equal and opposite impulses. Need all 4 terms dimensionally correct. Masses and speeds must be paired correctly. Condone sign errors. Condone factor of $g$ throughout. Unsimplified equation with at most one error |
| | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| $k = \frac{1}{2}$ | A1 | From correct working only |
| | (4) | |

# Question 2(b):

| For $P: I = \pm m(\frac{1}{2}u \pm -u)$ or For $Q: I = \pm km(u \pm -2u)$ | M1 | Impulse on $P$ or impulse on $Q$. Mass must be used with the correct speeds. e.g. $km \times -\frac{1}{2}u$ is M0. If working on $Q$, allow equation using their $k$. Terms must be dimensionally correct. Use of $g$ is M0 |
| $\frac{3mu}{2}$ | A1 | Only. From correct working only |
| | (2) | |
| | [6] | |

# Question 3(a):

| $7^2 = 2 \times 9.8h$ | M1 | Use of $v^2 = u^2 + 2as$ with $u = 0, v = 7$ or alternative complete method to find $h$ |
| $h = 2.5$ | A1 | Condone $h = -2.5$ in the working but the final answer must be positive |
| | (2) | |

# Question 3(b):

| $9 \times 7 = 10.5u$ | M1 | Use CLM to find the speed of the blocks after the impact. Condone additional factor of $g$ throughout. |
| $u = 6$ | A1 | |
| $0^2 = 6^2 - 2a \times 0.12$ | M1 | Use of $v^2 = u^2 + 2as$ with $u = 6, v = 0$. Allow for their $u$ and $v = 0$. Allow for $u = 7, v = 0$. Accept alternative suvat method to form an equation in $a$. Condone use of 12 for 0.12 |
| | A1 | Correctly substituted equation in $a$ with $u = 6, s = 0.12$ (implied by $a = 150$) |
| $(\downarrow) 10.5g - R = 10.5 \times (-a)$ | M1 | Use of $F = ma$ with their $a \neq \pm g$. Must have all 3 terms and 10.5. Condone sign error(s). |
| $(\downarrow) 10.5g - R = 10.5 \times (-150)$ | A1 | Unsimplified equation with $a$ substituted and at most one error (their $a$ with the wrong sign is 1 error) |
| | A1 | Correct unsimplified equation with $a$ substituted |
| $R = 1680$ or $1700$ | A1 | |
| | (8) | |
| **Alternative for the last 6 marks:** | | |
| $\frac{1}{2} \times 10.5 \times 6^2 + 10.5 \times 9.8 \times 0.12 = R \times 0.12$ | M2 | Energy equation (needs all three terms) |
| | A3 | -1 each error. A1A1A0 for 1 error, A1A0A0 for 2 errors |
| $R = 1680$ or $1700$ | A1 | |
| | [10] | |

# Question 4(a):

| $M(A)$ $(30g \times 2) + (50g \times 4) = 0.6S$ | M1 | Moments equation. Requires all terms and dimensionally correct. Condone sign errors. Allow M1 if $g$ missing |
| $M(C)$ $(0.6 \times R) = (1.4 \times 30g) + (3.4 \times 50g)$ or $M(G)$ $(2 \times R) = (1.4 \times S) + (2 \times 50g)$ or $M(B)$ $(4 \times R) + (2 \times 30g) = (3.4 \times S)$ | A1 | Correct unsimplified equation |
| $(\uparrow) R + 30g + 50g = S$ or $(R + 784 = S)$ | M1 | Resolve vertically. Requires all 4 terms. Condone sign errors |
| | A1 | Correct equation (with $R$ or their $R$) |
| **NB: The second M1A1 can also be earned for a second moments equation** | | |
| $R = 3460$ or $3500$ or $\frac{1060g}{3}$ (N) Not 353.3g | A1 | One force correct |
| $S = 4250$ or $4200$ or $\frac{1300g}{3}$ (N) Not 433.3g | A1 | Both forces correct. If both forces are given as decimal multiples of $g$ or mark this as an accuracy penalty A0A1 |
| | (6) | |

# Question 4(b):

| $M(C)$ $(30g \times 1.4) + (Mg \times 3.4) = 0.6 \times 5000$ | M1 | Use $R = 5000$ and complete method to form an equation in $M$ or weight. Needs all terms present and dimensionally correct. Condone sign errors. Accept inequality. Use of $R$ and $S$ from (a) is M0 |
| | A1 | Correct equation in $M$ (not weight) (implied by $M = 77.68$) |
| $M = 77$ kg | A1 | 77.7 is A0 even is the penalty for over-specified answers has already been applied |
| | (3) | |

# Question 4(c):

| The weight of the diver acts at a point. | B1 | Accept "the mass of the diver is at a point". |
| | (1) | |
| | [10] | |

# Question 5(a):

| $(2\mathbf{i} - 3\mathbf{j}) + (p\mathbf{i} + q\mathbf{j}) = (p + 2)\mathbf{i} + (q - 3)\mathbf{j}$ | M1 | Resultant force = $\mathbf{F}_1 + \mathbf{F}_2$ in the form $a\mathbf{i} + b\mathbf{j}$ |
| $\frac{p + 2}{q - 3} = \frac{1}{2}$ or $\frac{p + 2}{q - 3} = 2n$ for $n \neq 1$ | M1 | Use parallel vector to form a scalar equation in $p$ and $q$. |
| | A1 | Correct equation (accept any equivalent form) |
| $4 + 2p = -3 + q$ | DM1 | Dependent on no errors seen in comparing the vectors. Rearrange to obtain given answer. At least one stage of working between the fraction and the given answer |
| $2p - q + 7 = 0$ | A1 | **Given Answer** |
| | (5) | |

# Question 5(b):

| $q = 11 \Rightarrow p = 2$ | B1 | |

# Question 5(b) continued:

| $\mathbf{R} = 4\mathbf{i} + 8\mathbf{j}$ | M1 | $(2 + p)\mathbf{i} + 8\mathbf{j}$ for their $p$ |
| $4\mathbf{i} + 8\mathbf{j} = 2\mathbf{a}$ $(\mathbf{a} = 2\mathbf{i} + 4\mathbf{j})$ | M1 | Use of $\mathbf{F} = m\mathbf{a}$ |
| $|\mathbf{a}| = \sqrt{2^2 + 4^2}$ | DM1 | Correct method for $|\mathbf{a}|$. Dependent on the preceding M1 |
| $= \sqrt{20} = 4.5$ or $4.47$ or better (m s$^{-2}$) | A1 | $2\sqrt{5}$ |
| | (5) | |
| **Alternative for the last two M marks:** | | |
| $|\mathbf{F}| = \sqrt{16 + 64} \left(= \sqrt{80}\right)$ | M1 | Correct method for $|\mathbf{F}|$ |
| $\sqrt{80} = 2 \times |\mathbf{a}|$ | DM1 | Use of $|\mathbf{F}| = m|\mathbf{a}|$. Dependent on the preceding M1 |
| | [10] | |

# Question 6(a):

| $v = u + at \Rightarrow 14 = 3.5a$ | M1 | Use of $suvat$ to form an equation in $a$ |
| $a = 4$ | A1(2) | |

# Question 6(b):

| | B1 | Graph for A or B |
| | B1 | Second graph correct and both graphs extending beyond the point of intersection |
| | B1 | Values 3.5, 14, $T$ shown on axes, with $T$ not at the point of intersection. Accept labels with delineators. |
| | (3) | NB 2 separate diagrams scores max B1B0B1 |

# Question 6(c):

| $\frac{1}{2}T.3T$, $\frac{(T + T - 3.5)}{2} \cdot 14$ | M1 | Find distance for A or B in terms of $T$ only. Correct area formulae: must see $\frac{1}{2}$ in area formula and be adding in trapezium |
| | A1 | One distance correct |
| | A1 | Both distances correct |
| $\frac{1}{2}T.3T = \frac{(T + T - 3.5)}{2} \cdot 14$ or $\frac{1}{2}T.3T = \frac{1}{2} \times 4 \times 3.5^2 + 14(T - 3.5)$ | M1 | Equate distances and simplify to a 3 term quadratic in $T$ in the form $aT^2 + bT + c = 0$ |
| $3T^2 - 28T + 49 = 0$ | A1 | Correct quadratic |
| $(3T - 7)(T - 7) = 0$ | M1 | Solve 3 term quadratic for $T$ |
| $T = \frac{7}{3}$ or $7$ | A1 | Correct solution(s) - can be implied if only ever see $T = 7$ from correct work |
| but $T > 3.5$, $T = 7$ | A1(8) | |

# Question 6(d):

| $73.5$ m | B1(1) | From correct work only. B0 if extra answers. |

# Question 6(e):

| | B1 | (A) Condone missing 4 |
| | B1 | (B) Condone graph going beyond $T = 7$. Must go beyond 3.5. Condone no 3. |
| | B1(3) | (A) Condone graph going beyond $T = 7$. Must go beyond 3.5. B0 if see a solid vertical line. Sometimes very difficult to see. If you think it is there, give the mark. Condone separate diagrams. |
| | [17] | See next page |

# Alternative for (c) for candidates with a sketch like this:

| $\frac{1}{2} \times 3 \times (T + 3.5) = \frac{1}{2} \times 4 \times 3.5^2 + 14T$ | M1 | Use diagram to find area |
| | A1 | One distance correct |
| | A1 | Both distances correct |
| $12T^2 - 28T - 49 = 0$ | M1 | Simplify to a 3 term quadratic in $T$ |
| | A1 | Correct quadratic |
| $(2T - 7)(6T + 7) = 0$ | M1 | Complete method to solve for the $T$ in the equation |
| $T = \frac{7}{2}$ or $\frac{-7}{6}$ | A1 | Correct solution(s) - can be implied if only ever see Total = 7 |
| Total time = 7 | A1(8) | |

# Question 7(a):

| $F = 0.25R$ | B1 | |
| $\sin \alpha = \frac{3}{5}$ or $\cos \alpha = \frac{4}{5}$ and $\sin \beta = \frac{4}{5}$ or $\cos \beta = \frac{3}{5}$ | B1 | Use of correct trig ratios for $\alpha$ or $\beta$ |
| $R = 4g \cos \alpha$ $(31.36)$ | M1 | Normal reaction on $P$. Condone trig confusion (using $\alpha$) |
| | A1 | Correct equation |
| $T + F = 4g \sin \alpha$ | M1 | Equation of motion for $P$. Requires all 3 terms. Condone consistent trig confusion. Condone an acceleration not equated to $0$: $T + F - 4g \sin \alpha = 4a$ |
| $(T + 7.84 = 23.52)$ $(T = 15.68)$ | A1 | Correct equation |
| $T = mg \sin \beta$ | M1 | Equation of motion for $Q$. Condone trig confusion. Condone an acceleration not equated to 0: $T - mg \sin \beta = -ma$ |
| $(T = 7.84m)$ | A1 | Correct equation |
| Solve for $m$ | DM1 | Dependent on the 3 preceding M marks. Not available if their equations used $a \neq 0$ |
| $m = 2$ | A1 | |
| **NB Condone a whole system equation** $4g \sin \alpha - F = mg \sin \beta$ **followed by** $m = 2$ **for 6/6. M2 for an equation with all 3 terms. Condon trig confusion. Condon an acceleration $\neq 0$. A2 (-1 each error) for a correct equation;** | | |
| | (10) | |

# Question 7(b):

| $F = \sqrt{T^2 + T^2}$ or $2T \cos 45°$ or $\frac{T}{\cos 45}$ | M1 | Complete method for finding $F$ in terms of $T$. Accept $\sqrt{(R_y)^2 + (R_x)^2}$ |
| | A1 | Correct expression in $T$ |
| | DM1 | Substitute their $T$ into a correct expression. Dependent on the previous M mark |
| $F = \sqrt{2} \times \frac{8g}{5} = 22$ or $22.2$ (N) | A1 | Watch out - resolving vertically is not a correct method and gives 21.9 N. |
| | (4) | |

# Question 7(c):

| Along the angle bisector at the pulley | B1(1) | Or equivalent - accept angle + arrow shown on diagram. (8.1° to downward vertical). Do not accept a bearing |
| | [15] | |