**(i)**

$25 \text{ N}$ | B1 | Condone no units. Do not accept $-25 \text{ N}$.

| | 1 |

**(ii)**

$50 \cos 25$ $= 45.31538\ldots$ so $45.3 \text{ N}$ (3 s.f.) | M1 | Attempt to resolve 50 N. Accept $s \leftrightarrow c$. No extra forces.

| A1 | cao but accept $= -45.3$.

| | 2 |

**(iii)**

Resolving vertically | M1 | All relevant forces with resolution of 50 N. No extras. Accept $s \leftrightarrow c$.

$R + 50 \sin 25 - 8 \times 9.8 = 0$ | A1 | All correct.

$R = 57.26908\ldots$ so $57.3 \text{ N}$ (3 s.f.) | A1 |

| | 3 |

**(iv)**

Newton's 2nd Law in direction DC | M1 | Newton's 2nd Law with $m = 18$. Accept $F = ma$. Attempt at resolving 50 N. Allow 20 N omitted and $s \leftrightarrow c$. No extra forces.

$50\cos 25 - 20 = 18a$ | A1 | Allow only sign error and $s \leftrightarrow c$.

$a = 1.4064105\ldots$ so $1.41 \text{ m s}^{-2}$ (3 s.f.) | A1 | cao

| | 3 |

**Q8 continued**

**(v)**

Resolution of weight down the slope | B1 | $mg\sin 5°$ where $m = 8$ or 10 or 18, whichever first seen

**either**

Newton's 2nd Law down slope overall | 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 (including inconsistency between the 15 N and the 5 N).

$18 \times 9.8 \times \sin 5 - 20 = 18a$ | A1 | cao

Newton's 2nd Law down slope. Force in rod can be taken as tension or thrust. Taking it as tension | M1 | $F = ma$. Must consider the 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.

$T$ gives | | |

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

$T = -3.888\ldots = -3.89 \text{ N}$ (3 s.f.) | A1 | If $T$ taken as thrust, then $T = +3.89$. Dependent on $T$ correct

The force is a thrust | A1 |

**or**

Newton's 2nd Law down slope. Force in rod can be taken as tension or thrust. Taking it as tension | M1 | $F = ma$. Must consider the 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.

$T$ gives | | |

For C: $8 \times 9.8 \times \sin 5 - 5 + T = 8a$ | | |

$a = -0.2569\ldots$, $T = -3.888\ldots = -3.89 \text{ N}$ (3s.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. Award for either the equation for C or the equation for D correct. '-T' if $T$ taken as a thrust. '+T' if $T$ taken as a thrust

| A1 | Dependent on $T$ correct

**then**

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 $2.49 \text{ m s}^{-1}$ (3 s.f.) | F1 | FT from magnitude of their $a$ but must be consistent with its direction.

| | 9 |

| | 18 |