# Question 7:

## Part (i)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.45a = 0.45g - 2.52$ | M1 | N2L for R. 2 vertical forces. Accept $+/-0.45a = 0.45g +/- 2.52$ |
| $a = 4.2 \text{ ms}^{-2}$ | A1 **[2]** | Accept $-4.2$ |

## Part (i)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.05 \times 4.2 = 0.05g + 2.52 - T$ | M1 | N2L for Q, 3 vertical forces, $0.05 \times 4.2 = 0.05g +/- 2.52 +/- T$; accn not 9.8; $0.5g$ is TWO vertical forces $(0.45g + 0.05g)$ not MR |
| $T = 0.05 \times 9.8 + 2.52 - 0.05 \times 4.2$ | A1 ft | ft cv$(a(\mathbf{i}))$. Any equivalent form of equation |
| $T = 2.8 \text{ N}$ | A1 **[3]** | |
| ACCEPT COMBINED Q AND R METHOD: $(0.45+0.05) \times 4.2 = 0.45g + 0.05g +/-T$ giving $T = 2.8\text{ N}$ | M1, A1ft, A1 | |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\pm 4.2m = T - mg$ OR $\pm 4.2 = (0.05g + 0.45g - mg)/(0.05 + 0.45 + m)$ | M1 | N2L for P, difference of 2 vertical forces, accn cv$(a(\mathbf{i}))$; $\pm\text{cv}(a(\mathbf{i})) = (\text{wt } P + \text{wt } Q - \text{wt } R)/\text{sum of masses}$ |
| $4.2m = 2.8 - mg$ OR $9.8m + 4.2m = 2.8$ | A1 ft | ft cv$(T(\mathbf{ib}))$. Any equivalent form of equation with cv$(a(\mathbf{i}))$ |
| $m = 0.2$ | A1 **[3]** | |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| **BEFORE R STRIKES SURFACE** | | |
| $v = 4.2 \times 0.5$ | M1* | Find speed when R hits surface, using $a(\mathbf{i})$ |
| $v = 2.1$ | A1 | |
| $s = 2.1^2/(2 \times 4.2) = 4.2 \times 0.5^2/2$ | M1 | Distance R falls (0.525 m). Accept $+/-4.2 \times 0.5^2/2$ |
| **AFTER R STRIKES SURFACE** | | |
| $+/-0.2a = T - 0.2g$ OR $+/-0.05a = 0.05g - T$ | M1 | N2L for either P (with cv$(m)$) or Q |
| $+/-0.2a = T - 0.2g$ AND $+/-0.05a = 0.05g - T$ | A1 | Correct equations for both P and Q. OR combination $0.05g(-T+T) - 0.2g = +/-(0.2a + 0.05a)$ |
| $a = +/-5.88$ | A1 | |
| $S = 2.1^2/(2 \times 5.88)$ | D*M1 | Distance P rises after R hits ground (0.375), $a$ not $a(\mathbf{i})$ or 9.8 |
| **TOTAL JOURNEY** Distance $= (0.375 + 0.525) = 0.9\text{ m}$ | A1 **[8]** | |

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