# Question 8:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| N2L **i** direction: $150 = 250a$, $a = 0.6$ so $0.6 \text{ m s}^{-2}$ | M1 | Use of N2L. Allow $F = mga$. |
| | A1 | Accept no reference to direction |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $150 \text{ N}$ | B1 | |
| $-$**i** direction | B1 | Allow correct description or arrow. [Accept '$-150$ in **i** direction' for B1 B1] |

## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| For force only in direction perp to **i**: $300\sin 40 = 450\sin\theta$ | M1 | Resolution of both terms attempted. Allow $\sin \leftrightarrow \cos$ if in both terms. Allow 250 or $250g$ present. |
| | B1 | $300\sin40$ or $450\sin\theta$ |
| $\theta = 25.37300...$ so $25.4°$ (3 s.f.) | A1 | Accept $\pm$. Accept answer rounding to 25.5. Allow SC1 if seen in this part. |
| In **i** direction: $300\cos 40 + 150 + 450\cos\theta$ | M1 | Proper resolution attempted of 450 **and** 300. Allow $\sin \leftrightarrow \cos$ if in both terms. Accept use of **their** $\theta$ or just $\theta$. |
| | A1 | Either resolution correct. Accept **their** $\theta$ or just $\theta$. Accept sin/cos consistent with use for cpt perpendicular to **i**. |
| $786.4017...$ so $786$**i** N (3 s.f.) | A1 | Accept no reference to direction cao. Allow SC1 WW |

## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $s = ut + 0.5at^2$: $1 = 0.5a \times 2^2$, $a = 0.5$ | M1 | Appropriate (sequence of) suvat. [WW M0 A0] |
| | A1 | |
| Using N2L in **i** direction: $786.4017... - F = 250\times0.5$ | M1 | Use of $F = ma$ with **their** 786.4 and **their** $a$. No extra forces. Allow sign errors. |
| $661.4017...$ so $661 \text{ N}$ (3 s.f.) | A1 | All correct using **their** 786.4 and $a$ |
| | E1 | Use of N2L clearly shown. (Accept 0.5 used WW) |

## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using N2L in **i** direction **either**: $125 - 200 = 250a_1$ **or** (starting again): $786.4017... - (200 + 661.4017...) = 250a_1$ | M1 | Use of $F = ma$ with **their** values. Allow 1 force missing. |
| so $a_1 = -0.3$ | F1 | FT only **their** 786... and **their** 661 |
| Using $v^2 = u^2 + 2a_1 s$: $v^2 = 1.8^2 + 2\times(-0.3)\times1.65$ | M1 | Appropriate (sequence of) suvat with $u \neq 0$. Must be 'new' $a$ obtained by using N2L. |
| $v = 1.5$ so $1.5 \text{ m s}^{-1}$ | F1 | Only FT use of $\pm$ **their** $a_1$ |
| | A1 | cao |