# Question 4:

## Part 4a(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $D$: $20g \times 2 + 8g \times 4.5 = R_C \times 4.5$ **OR** Resolve: $R_C + R_D = 28g$ | M1A1 | |
| $R_C = \dfrac{152}{9}g\ (= 166 \text{ or } 170)$ | A1 | |

## Part 4a(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $C$: $20g \times 2.5 = R_D \times 4.5$ **OR** Resolve: $R_C + R_D = 28g$ | M1A1 | |
| $R_D = \dfrac{100}{9}g\ (= 109 \text{ or } 110)$ | A1 | Remember to only penalise overaccuracy after use of $g$, once per whole question |

## Part 4b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $A$: $R \times 1.5 + 2R \times 6 = 20g \times 4 + 8g \times x$ | M1A1 | |
| Resolve: $3R = 28g$, $\left(R = \dfrac{28}{3}g\ (= 91.5)\right)$ | M1A1 | |
| Substitute for $R$ and solve for $x$: $\dfrac{27}{2} \times \dfrac{28}{3}g = 80g + 8g \times x$ | M1 | |
| $126 = 80 + 8x,\quad 8x = 46,\quad x = 5.75$ (m) | A1 | |

## Part 4c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| The weight of the package acts at point $C$ (or $E$) | B1 | |

## Question 4a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Any moments equation or vertical resolution | M1 | First moments equation (even if contains both reactions) |
| Correct equation | A1 | First correct equation |
| $R_C = \frac{152}{9}g$ (= 166 or 170) | A1 | Second A1 for correct value |
| Another moments equation or vertical resolution | M1 | Second moments equation (even if contains both reactions) |
| Correct equation | A1 | Third correct equation |
| $R_D = \frac{100}{9}g$ (= 109 or 110) | A1 | Fourth A1 for correct value |

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## Question 4b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments equation in $R$ and $x$ only ($x$ may not be $AE$) | M1 | First M1 |
| Correct equation e.g. $R \times 1.5 + 2R \times 6 = 20g \times 4 + 8g \times x$ | A1 | First correct equation |
| Another moments equation in $R$ and $x$ only, or vert resolution in $R$ only | M1 | Second M1 |
| Correct equation | A1 | Second correct equation |
| Solving for $AE$ | M1 | Third M1 |
| $AE = 5.75$ m | A1 | Must be EXACT |

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## Question 4c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass or weight of package acts at point $C$ (or $E$) | | |

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## Question 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt at difference in momenta for $Q$ | M1 | Clear attempt at *difference* in momenta for $Q$ only; M0 if mass omitted, g included, or clearly adding; in terms of $k$, $m$ and $u$ only |
| $= \pm km(2u - -u)$ | A1 | First A1 for $\pm km(2u--u)$ |
| Magnitude $= 3kmu$ | A1 | Second A1 for $3kmu$ |

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## Question 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Case 1: $P$ continues same direction: $4m \times 3u - km \times u = 4m \times 2u + km \times 2u$ **OR** $-3kmu = 4m(2u-3u)$ | M1A1 | First M1 for equation in $k$, $m$, $u$ only; dim. correct with correct no. of terms; condone sign errors. First A1 for correct equation |
| $k = \frac{4}{3}$, 1.33 (3 SF or better) | A1 | Second A1 for correct value of $k$ |
| Case 2: $P$ changes direction: $4m \times 3u - km \times u = -4m \times 2u + km \times 2u$ **OR** $3kmu = 4m(2u--3u)$ | M1 | Second M1: must clearly have $P$ moving in opposite direction; in $k$, $m$, $u$ only; condone sign errors |
| $k = \frac{20}{3}$, 6.67 (3 SF or better) | A1 | Third A1 for other correct value of $k$ |

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## Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 4g\cos 30$ | B1 | First B1 |
| $F = 0.3R$ seen | B1 | Second B1 (could just be on diagram) |
| $F = ma$ parallel to plane: $4a = 4g\sin 30 - F$ | M1A1 | M1 equation of motion with usual rules, condone sign errors; A1 correct equation ($F$ not substituted) |
| $4a = 4g\sin 30 - 0.3 \times 4g\cos 30$ | A1 | Second A1: correct equation in $a$ only, without trig ratios substituted |
| $v = \sqrt{(10a)}$ using $v^2 = u^2 + 2as$ | M1 | Complete method for finding $v$ (must have found an $a$ value) |
| $v = 4.9$ or $4.85$ m s$^{-1}$ | A1 | Third A1 |

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## Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Perpendicular to plane: $R = 4g\cos 30 + H\cos 60$ | M1A1 | First M1 resolution with usual rules, condone sign errors; A1 correct equation |
| Parallel to plane: $H\cos 30 = F + 4g\sin 30$ | M1A1 | Second M1 another resolution; A1 correct equation |
| Use $F = 0.3R$ | M1 | Third M1 (must appear in equation; M0 if using $R$ from part **a**) |
| Solve for $H$: $H = \dfrac{g(1.2\cos 30 + 4\sin 30)}{\cos 30 - 0.3\cos 60}$ | DM1 | Fourth M1 **dependent** on first, second and third M1s; eliminating $F$ and $R$, solving for $H$ |
| $H = 42$ or $41.6$ | A1 | Third A1 |

**Alternative (6b alt):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically: $R\cos 30 = 4g + F\cos 60$ | M1A1 | |
| Resolve horizontally: $H = R\cos 60 + F\cos 30$ | M1A1 | |
| Use $F = 0.3R$ | M1 | |
| Solve for $H$ | DM1 | |
| $H = 42$ or $41.6$ | A1 | |