## Question 7:

### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T - 0.3g\sin30 = 0.3a$ *OR* $0.4g\sin30 - T = 0.4a$ | B1 | Either correct N2L for one particle; Putting $a=0.7$ into correct equation for a single particle and working out $T$ correctly gets B1M0A0M1A1 |
| $0.4g\sin30 - 0.3g\sin30 = 0.7a$ | M1 | Allow combined approach as "method", must be components of weight, allow $mg(\cos/\sin)30$; Consult TL if this is done for both particles |
| $a = 0.7$ m s$^{-2}$ AG | A1 | May use the other equation |
| $T = 0.3g\sin30 + 0.3\times0.7$ | M1 | Allow $0.3g(\cos/\sin)30$. Accept cv(0.7) |
| $T = 1.68$ N | A1 | |

### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = 1.8 \times 0.7$ | M1 | Accept cv(0.7) |
| $V = 1.26$ m s$^{-1}$ | A1 | |

### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Dec $= 1.26/1.4$ | M1 | Accept $1.8\times0.7/1.4$; cv(1.26) |
| Dec $= 0.9$ m s$^{-2}$ | A1 | Or $a = +/-0.9$ |

### Part (iv)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T - 0.3g\sin30 = -0.3\times0.9$ | M1 | N2L, 2 forces including cmpt of weight cv(0.9) but signs must be consistent with the direction of motion; Allow $mg(\cos/\sin)30$ |
| | A1ft | |
| $T = 1.2$ | A1 | |

### Part (iv)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-0.4\times0.9 = 0.4g\sin30 - T - F_r$ | M1 | N2L, 3 forces including cmpt of weight cv(0.9) and cv(1.2) but signs must be consistent with the direction of motion; Allow $mg(\cos/\sin)30$ |
| $-0.4\times0.9 = 0.4g\sin30 - 1.2 - F_r$ | A1ft | |
| $F_r = 1.12$ N | A1 | |

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