## Question 8:

### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4v^2/0.6 = 6,\ v = 3$ | M1, A1 | Attempt at use of N2L with $a = v^2/r$ or $a = r\omega^2$; allow verification. **AG** If $+/-3$ then $-3$ must be rejected for A1. |

### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3(0.4) = 0.4v_A + 0.5v_B$ | *M1, A1 | Attempt at use of conservation of linear momentum |
| $v_B - v_A = 3(0.35)$ | *M1, A1 | Attempt at use of restitution equation, must be correct way round. Must be consistent with the directions used for CoLM |
| | dep*M1 | Solving simultaneous equations |
| $v_A = 0.75 \text{ m s}^{-1}$ and $v_B = 1.8 \text{ m s}^{-1}$ | A1 | Both values positive as final answer. |

### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Y = 0.4g\ (= 3.92)$ | B1 | Resolve vertically |
| $(4.9)^2 - (0.4g)^2\ (= 2.94^2)$ | B1 | Use of Pythagoras |
| $0.4(0.6)\omega^2 = \text{cv}(2.94)$ | M1 | Resolve horizontally, $\text{cv}(2.94)$ from a legitimate attempt at 2.94 |
| $\omega = 3.5$ | A1 | Exact |

**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4.9\cos\theta = 0.4g$ | B1 | Resolve vertically, $\theta$ angle with vertical |
| $\cos\theta = 0.8$ or $\theta = 36.9°$ | B1 | |
| $0.4(0.6)\omega^2 = 4.9\sin\theta$ | M1 | Resolve horizontally, $\theta$ substituted and from a legitimate attempt |
| $\omega = 3.5$ | A1 | Exact |

**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Y = 0.4g\ (= 3.92)$ | B1 | Resolve vertically |
| $X = 0.4(0.6)\omega^2\ (= 0.24\omega^2)$ | B1 | Resolve horizontally |
| $(0.4g)^2 + (0.4(0.6)\omega^2)^2 = 4.9^2$ | M1 | Attempt at use of Pythagoras, from a legitimate attempt at $X$ |
| $\omega = 3.5$ | A1 | Exact |