| Answer | Marks | Guidance |
|--------|-------|----------|
| COM from vertex $\times 3r + 3r$ | M1 | For COM centre of mass |
| $\frac{1}{3}r + 3r$ | A1 | AGCorrect answer with convincing working |

## Question 2(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| In equilibrium, $2mg = \frac{2mg}{d}$ | M1 | Use Hooke's law |
| $x = \frac{d}{12}$ | A1 | |

## Question 2(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Let the distance below A be $\frac{1}{2}x(d-a)$ | M1 | Loss on gravitational potential energy (EPE) of spring |
| $= \frac{1}{2} \times 2mg \times (d-a)$ | M1 | |
| No change in kinetic energy (KE) so | M1 | |
| $2mg \times d - \frac{1}{2}(d-a)^2 = a$ | A1 | Attempt to solve |

## Question 3(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \frac{1}{2g} - k^2$ | M1 | Use Newton's 2nd law |
| $e^2 = (1-e^2)$ | A1 | AG |
| $v = c = \frac{1}{2}mg$ | A1 | |

## Question 3(b)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = 31.62...$ | B1 | |
| When $v = \frac{1}{2} \times 31.62... = 2 \times \frac{4}{m}$ | M1 | |
| distance $= 14.4$ m | A1 | |

## Question 3(b)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Speed of A perpendicular to lines of centres unaligned in expression for speed | M1 | |
| $= \sqrt{\frac{\sin 30^{\circ} + \sqrt{3}}{6 \sqrt{(1-2a)}}}$ | A1 | Difference in KEs with $e = \frac{1}{3}$ |
| $= mg^2$ | A1 | Correct answer |

## Question 4(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Speeds after collision: $A_1 = \frac{m+m_1}{2m} + m_1$ | M1 | Momentum: $2pm + m_1 = minus 30^{\circ}$ |
| Restitution: $v - v = euros 30^{\circ}$ | A1 | |
| $v\sin 30^{\circ} = \frac{\sqrt{3}}{6}$ | M1 | |
| $e^2 = \frac{1 - e^2}{3}$ | A1 | Correct answer |

## Question 4(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Loss in kinetic energy (KE): $= \frac{1}{m^2} - \frac{2m}{(6 \times \frac{2^2}{m}) \times \frac{2^2}{m}}$ | M1 | |
| $= \frac{1}{m^2}$ | A1 | Correct answer |

## Question 5(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Tosoff $mg$ | B1 | Vertically |
| Fushoff $= m \times 0.60^{\circ}$ | M1A1 | Horizontally |
| Divide: $o^2 = 2g$ | A1 | |
| Loading to gives $2cos$ single gas slack tension is zero | M1 | When string goes slack |
| $mg cos \theta = \frac{m g}{m^2}$ | A1 | |

## Question 5(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Energy from 60° with downward vertical to point when string goes slack: | M1 | |
| $\frac{1}{m}\left(\frac{2g}{2}\right)^2 - mg(cos 60^{\circ} + cos\theta)$ | M1 | |
| When single goes slack: tension is zero | A1 | |
| $mg cos \theta = \frac{1}{3}$ | M1A1 | |

## Question 6(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $-v = \frac{cos v t}{1}$ | B1 | Both needed |
| () = timist.1 g^2 | M1 | |
| Eliminate $v$ using $\frac{cos v \times \text{cons}7}{a}$ | A1 | |
| For greatest height, using vertical motion: $H = \frac{2g}{3B}$ | M1 | |
| $H = \frac{2g}{3B}$ | A1 | |

## Question 6(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| using umax $= 2$ | M1 | |
| Where $single goes slack$ tension is zero | A1 | |
| $mg cos \theta = \frac{3 H + \frac{3 H}{2} = 0$ | M1 | |
| $d = \frac{3 H}{3 H - 3 H}$ | A1 | |
| CAO | A1 | |