# Question 5:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Moments about **vertical** axis ($AF$): | M1 | Requires all terms and dimensionally correct but condone $g$ missing |
| $\frac{Mg}{2} \times \frac{1}{2}a + \frac{Mg}{2} \times 1.5a + 3akMg = Mg(1+k)\bar{x}$ | A2 | $-1$ each error. Accept with $M$ and/or $g$ not seen. |
| $\left(\bar{x} = \frac{1+3k}{1+k}a\right)$ | | |
| Moments about **horizontal** axis ($AB$ or $FE$): | M1 | Requires all terms and dimensionally correct but condone $g$ missing |
| $\frac{Mg}{2} \times 1.5a + \frac{Mg}{2} \times 3.5a + 4akMg = Mg(1+k)\bar{y}$ | A2 | $-1$ each error. Accept with $M$ and/or $g$ not seen. Do not penalise repeated errors. |
| $\left(\bar{y} = \frac{2.5+4k}{1+k}a\right)$ | | |
| Use of $\tan\theta$ with their distances from $AF$ & $AB$ | M1 | Must be considering the whole system. Allow for inverted ratio. |
| $\tan\theta = \frac{M + 3kM}{2.5M + 4kM}\left(= \frac{4}{7}\right)$ | A1 | or exact equivalent |
| Equate their $\tan\theta$ to $\frac{4}{7}$ and solve for $k$: $7M + 21kM = 10M + 16kM$ | M1 | |
| $k = \frac{3}{5}$ | A1 | CSO |

# Question (Centre of Mass / Tilting Problem):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\tan\theta$ with their distances for finding $d_1$ or $d_2$ | M1 | |
| $\left(\frac{5}{2}a\right)\tan\theta - a \left(=\frac{3}{7}a\right)$ | A1 | Correct for their centre of mass; obtain length of a side in a triangle containing $d_1$ |
| $d_1 = \left(\frac{3}{7}a\right)\cos\theta$ | A1 | Correct for their centre of mass |
| Use of $\tan\theta$ to find second distance: $3a - 4a\tan\theta = \frac{5}{7}a$ | M1 | |
| $d_2 = \frac{5}{7}a\cos\theta$ | A1 | Correct for their centre of mass |
| Moments about $A$: $Md_1 = kMd_2$ | M1 | |
| $\frac{3}{7}a\cos\theta = k \times \frac{5}{7}a\cos\theta \Rightarrow k = \frac{3}{5}$ | A1 | |
| **Total** | **(10)** | Candidate may start by finding centre of mass at $\left(a, \frac{3}{2}a\right)$ relative to $F$; M1A2 scored |

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# Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Taking moments about $A$ | M1 | Requires all terms; condone trig confusion and sign errors |
| $bF = 3mga\cos\theta + mg \times 2a\cos\theta$ | A2 | -1 each error |
| $bF = 5mga\cos\theta \quad F = \frac{5mga}{b}\cos\theta$ | A1 | **Given answer** |
| **Total** | **(4)** | |

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# Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Component of $\mathbf{R}$ parallel to $AB$: $R\cos(\phi-\theta)$ | M1 | Requires all terms; condone trig confusion |
| $= 3mg\sin\theta + mg\sin\theta = 4mg\sin\theta$ | A1 | Correct unsimplified |
| Component of $\mathbf{R}$ perpendicular to $AB$ | M1 | Requires all terms; condone consistent trig confusion and sign errors |
| $R\sin(\phi-\theta) + F = 4mg\cos\theta$ | A1 | Correct unsimplified |
| $R\sin(\phi-\theta) = 4mg\cos\theta - \frac{5mga}{b}\cos\theta$ | A1 | Correct with $F$ substituted |
| **Total** | **(5)** | ISW for incorrect work after correct components seen |

Alternatives via $M(B)$: $2aR\sin(\phi-\theta) + 3mga\cos\theta = F(2a-b)$

Via $M(C)$: $bR\sin(\phi-\theta) + (2a-b)mg\cos\theta = 3mg(b-a)\cos\theta$

**Special Case (misread of directions in 6b):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $X = F\sin\theta = \frac{5mga}{b}\cos\theta\sin\theta$ | M1 | Allow with $F$; requires all terms |
| | A1 | $F$ substituted |
| $Y = 4mg - F\cos\theta = 4mg - \frac{5mga}{b}\cos^2\theta$ | M1 | Allow with $F$; requires all terms |
| | A1 | Correct unsimplified |
| | A1 | Correct substituted |

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# Question 6c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $R\sin(\phi-\theta) > 0$ | M1 | |
| $4 > \frac{5a}{b}$, $\quad (2a\geq)b > \frac{5}{4}a$ | A1 | $2a$ not required; CSO |
| **Total** | **(2)** | |

**Alternative (from SC misread):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| For $\varphi > \theta$, $\tan\phi > \tan\theta$ | | |
| $\tan\varphi = \frac{Y}{X} = \dfrac{4 - \frac{5a}{b}\cos^2\theta}{\frac{5a}{b}\cos\theta\sin\theta} > \tan\theta$ | M1 | |
| $4 - \frac{5a}{b}\cos^2\theta > \frac{5a}{b}\sin^2\theta$ | | |
| $4 > \frac{5a}{b}(\cos^2\theta + \sin^2\theta) \Rightarrow b > \frac{5}{4}a$ | A1 | CSO |

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# Question 7a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equate horizontal components of speeds | M1 | |
| $u\cos\theta^\circ = 6\cos 45^\circ \left(= 3\sqrt{2}\right)$ (4.24...) | A1 | Correct unsimplified |
| Use suvat for vertical speeds: $u\sin\theta^\circ - 2g = -6\sin 45^\circ$ | M1 | Condone sign errors |
| $u\sin\theta = 2g - 3\sqrt{2}$ | A1 | Correct unsimplified |
| Divide to find $\tan\theta$: $\tan\theta = \dfrac{2g - 6\sin 45}{6\cos 45}$ | DM1 | Dependent on previous 2 Ms; follow their components |
| $\left(= \dfrac{2g - 3\sqrt{2}}{3\sqrt{2}} = 3.61..\right) \Rightarrow \theta = 74.6\ (75)$ | A1 | $(u = 15.93...)$ |
| **Total** | **(6)** | |

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# Question 7b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| At max height, speed $= u\cos\theta \left(= 3\sqrt{2}\ \text{ms}^{-1}\right)$ | B1 | |
| $KE = \frac{1}{2} \times 0.7 \times \left(3\sqrt{2}\right)^2$ (J) | M1 | Correct for their $v$ at the top, $v\neq 0$ |
| $= 6.3$ (J) | A1 | Accept awrt 6.30; CSO |
| **Total** | **(3)** | |

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# Question 7c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| When P is moving upwards at $6\ \text{ms}^{-1}$ | M1 | Use suvat to find first time $v=6$ |
| $u\sin\theta - gt = 3\sqrt{2}$ | A1 | |
| $2g - 3\sqrt{2} - gt = 3\sqrt{2}$ | M1 | Solve for $t$ |
| $t = \dfrac{2g - 6\sqrt{2}}{g} = 1.13...$ | A1 | Sensitive to premature approximation; allow 1.14 |
| $T = 2 - 1.13... = 0.87$ | A1 | CAO accept awrt 0.87 |
| **Total** | **(5)** | |

**Alternative 7c (Method 1):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $6\sin 45 = 0 + gt$ | M1A1 | Find time from top to $A$ |
| $T = 2t = \dfrac{12\sqrt{2}/2}{g} = 0.87$ | M1, A1, A1 | Correct strategy; correct unsimplified |

**Alternative 7c (Method 2):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u\sin\theta = gt$ (their $u,\theta$) | M1 | Time to top |
| $t = 1.567...$ | A1 | |
| $T = 2(2 - 1.567...)$ | M1A1 | |
| $= 0.87$ | A1 | |

**Alternative 7c (Method 3):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertical speed at $A = -(\text{vertical speed at }B) = \sqrt{36-(3\sqrt{2})^2} = 3\sqrt{2}$ | M1, A1 | Or use the $45°$ angle |
| Use $v = u + at$ for $A\to B$: $-3\sqrt{2} = 3\sqrt{2} - gT$ | M1, A1 | Correct use for their values |
| $T = 0.87$ | A1 | |

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# Question Alt 7d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v^2 = \left(3\sqrt{2}\right)^2 + \left(u\sin\theta - gt\right)^2 \leq 36$ | M1 | Form expression for $v^2$; inequality not needed at this stage |
| | A1 | Correct inequality for $v^2$ |
| $-\sqrt{18} \leq u\sin\theta - gt \leq \sqrt{18}$ | M1 | |
| $\dfrac{u\sin\theta - \sqrt{18}}{g} \leq t \leq \dfrac{u\sin\theta + \sqrt{18}}{g}$ | A1 | |
| $T = \dfrac{u\sin\theta + \sqrt{18}}{g} - \dfrac{u\sin\theta - \sqrt{18}}{g} = \dfrac{2\sqrt{18}}{g} = 0.866$ | A1 | |
| **Total** | **(5)** | |