## Question 2(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| N2L: $F - 11.2 = 8 \times 2$ | M1 | Use of N2L (allow $F = mga$). Allow 11.2 omitted; no extra forces |
| All correct | A1 | |
| $F = 27.2$ N | A1 | Cao |

**Total: [3]**

---

## Question 2(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| The string is inextensible | E1 | Allow 'light inextensible' but not other irrelevant reasons given as well (e.g. smooth pulley) |

**Total: [1]**

---

## Question 2(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Force diagram | B1 | One diagram with all forces present; no extras; correct arrows and labels; accept use of words |
| Both diagrams correct with a common label | B1 | |

**Total: [2]**

---

## Question 2(iv):

**Method (1):**

| Answer | Mark | Guidance |
|--------|------|----------|
| For either box or sphere, $F = ma$ | M1 | Allow omitted force and sign errors but not extra forces. Need correct mass. Allow use of mass not weight |
| Box N2L: $105 - T - 11.2 = 8a$ | A1 | Correct and in any form |
| Sphere N2L $\uparrow$: $T - 58.8 = 6a$ | A1 | Correct and in any form. [box and sphere equations with consistent signs] |
| Adding $35 = 14a$ | M1 | Eliminating 1 variable from 2 equations in 2 variables |
| $a = 2.5$ m s$^{-2}$ | E1 | |
| Substitute $a = 2.5$ giving $T = 58.8 + 15$ | M1 | Attempt to substitute in either box or sphere equation |
| $T = 73.8$ N | A1 | |

**Method (2):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $105 - 11.2 - 58.8 = 14a$ | M1 | For box and sphere, $F = ma$. Must be correct mass. Allow use of mass not weight |
| $a = 2.5$ | A1 | |
| Method made clear | E1 | |
| For **either** box **or** sphere, $F = ma$ | M1 | Allow omitted force and sign errors but not extra forces. Need correct mass |
| Box N2L: $105 - T - 11.2 = 8a$ **or** Sphere N2L $\uparrow$: $T - 58.8 = 6a$ | A1 | Correct and in any form |
| Substitute $a = 2.5$ in either equation | M1 | Attempt to substitute in either box or sphere equation |
| $T = 73.8$ N | A1 | [If AG used in either equation award M1 A1 for that equation as above and M1 A1 for finding $T$. For full marks, both values must be shown to satisfy the second equation.] |

**Total: [7]**

---

## Question 2(v)(A):

| Answer | Mark | Guidance |
|--------|------|----------|
| $g$ downwards | B1 | Accept $\pm g$, $\pm 9.8$, $\pm 10$, $\pm 9.81$ |

**Total: [1]**

---

## Question 2(v)(B):

| Answer | Mark | Guidance |
|--------|------|----------|
| Taking $\uparrow$ +ve, $s = -1.8$, $u = 3$ and $a = -9.8$, so $-1.8 = 3T - 4.9T^2$ | M1 | Some attempt to use $s = ut + 0.5at^2$ with $a = \pm 9.8$ etc. $s = \pm 1.8$ and $u = \pm 3$. Award for $a = g$ even if answer to (A) wrong |
| $4.9T^2 - 3T - 1.8 = 0$ | E1 | Clearly shown. No need to show +ve required |

**Total: [2]**

---

## Question 2(v)(C):

| Answer | Mark | Guidance |
|--------|------|----------|
| Time to reach $3$ m s$^{-1}$ given by $3 = 0 + 2.5t$ so $t = 1.2$ | B1 | |
| Remaining time is root of quadratic | M1 | Quadratic solved and +ve root added to time to break |
| Time is $0.98513\ldots$ s | B1 | Allow $0.98$. [Award for answer seen WW] |
| Total $2.1851\ldots$ so $2.19$ s (3 s.f.) | A1 | Cao |

**Total: [4]**

---

*With the 11.2 N resistance acting to the right:*

| Answer | Mark | Guidance |
|--------|------|----------|
| **(i)** $F + 11.2 = 8 \times 2$ so $F = 4.8$ | — | The same scheme as above |
| **(iii)** | — | The 11.2 N force may be in either direction, otherwise the same scheme |
| **(iv)** Method (1): Box N2L $\rightarrow$ $105 - T + 11.2 = 8a$; sphere as before. Method (2): $105 + 11.2 - 58.8 = 14a$; giving $a = 4.1$ and $T = 83.4$ | — | Allow 2.5 substituted in box equation to give $T = 96.2$. If the sign convention gives as positive the direction of the sphere descending, $a = -4.1$. Allow substituting $a = 2.5$ in the equations to give $T = 43.8$ (sphere) or $136.2$ (box) |
| **(v)** | — | In (C) allow use of $a = 4.1$ to give time to break as $0.73117\ldots$ s and total time as $1.716\ldots$ s |

**Overall Total: [20]**