## Question 3:

### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| 25 N | B1 | Condone no units. Do not accept −25 N |
| **[1]** | | |

### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $50\cos 25$ | M1 | Attempt to resolve 50 N. Accept $s \leftrightarrow c$. No extra forces |
| $= 45.31538\ldots$ so 45.3 N (3 s.f.) | A1 | cao but accept −45.3 |
| **[2]** | | |

### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolving vertically: $R + 50\sin 25 - 8 \times 9.8 = 0$ | M1 | All relevant forces with resolution of 50 N. No extras. Accept $s \leftrightarrow c$ |
| $R = 57.26908\ldots$ so 57.3 N (3 s.f.) | A1 | All correct |
| | A1 | |
| **[3]** | | |

### Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Newton's 2nd Law in direction DC: $50\cos 25 - 20 = 18a$ | M1 | Newton's 2nd Law with $m=18$. Accept $F=mga$. Attempt at resolving 50 N. Allow 20 N omitted and $s \leftrightarrow c$. No extra forces |
| $a = 1.4064105\ldots$ so 1.41 m s$^{-2}$ (3 s.f.) | A1 | Allow only sign error and $s \leftrightarrow c$ |
| | A1 | cao |
| **[3]** | | |

### Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolution of weight down the slope | B1 | $mg\sin 5°$ where $m = 8$ or 10 or 18, wherever first seen |
| **Either:** Newton's 2nd Law down slope overall: $18 \times 9.8 \times \sin 5° - 20 = 18a$ | M1 | $F=ma$. Must have 20 N and $m=18$. Allow weight not resolved and use of mass. Accept $s \leftrightarrow c$ and sign errors |
| $a = -0.2569\ldots$ | A1 | cao |
| Newton's 2nd Law down slope, force in rod can be taken as tension or thrust. Taking it as tension $T$ gives: | M1 | $F=ma$. Must consider motion of either C or D and include: component of weight, resistance and $T$. No extra forces. Condone sign errors and $s \leftrightarrow c$. Do not condone inconsistent value of mass |
| For D: $10 \times 9.8 \times \sin 5° - 15 - T = 10a$ | | |
| (For C: $8 \times 9.8 \times \sin 5° - 5 + T = 8a$) | F1 | FT only applies to $a$, and only if direction is consistent. '$+T$' if $T$ taken as a thrust; '$-T$' if $T$ taken as a thrust |
| $T = -3.888\ldots = -3.89$ N (3 s.f.) | A1 | If $T$ taken as thrust, then $T = +3.89$ |
| The force is a thrust | A1 | Dependent on $T$ correct |

**Or:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Newton's 2nd Law down slope, force in rod taken as tension $T$: | M1 | $F=ma$. Must consider motion of C and include: component of weight, resistance and $T$. No extra forces. Condone sign errors and $s \leftrightarrow c$. Do not condone inconsistent value of mass |
| | M1 | $F=ma$. Must consider motion of D and include: component of weight, resistance and $T$. No extra forces. Condone sign errors and $s \leftrightarrow c$. Do not condone inconsistent value of mass |
| For C: $8 \times 9.8 \times \sin 5° - 5 + T = 8a$ | A1 | Award for either equation for C or equation for D correct. '$-T$' if $T$ taken as thrust |
| For D: $10 \times 9.8 \times \sin 5° - 15 - T = 10a$ | | |
| $a = -0.2569\ldots$, $T = -3.888\ldots = -3.89$ N (3 s.f.) | A1 | First of $a$ and $T$ found is correct. If $T$ taken as thrust, then $T = +3.89$ |
| | F1 | The second of $a$ and $T$ found is FT |
| The force is a thrust | A1 | Dependent on $T$ correct |

**Then:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| After 2 s: $v = 3 + 2 \times a$ | M1 | Allow sign of $a$ not followed. FT their value of $a$. Allow change to correct sign of $a$ at this stage |
| $v = 2.4860303\ldots$ so 2.49 m s$^{-1}$ (3 s.f.) | F1 | FT from magnitude of **their** $a$ but must be consistent with its direction |
| **[9]** | | |
| **Total [18]** | | |