At max height

$0 = 30^2 - 2 \times 10 \times h$

$\therefore h = 45 \text{ m}$ | M1 A1 A1 | Use of an appropriate 'suvat' equation. Correct equation. Correct outcome. Allow $g = 9.8$, giving $h = 45.918$

On return to ground level

$-30 = 30 - 10 \times t$

$\therefore t = 6 \text{ sec}$ | M1 A1 | Correct use of a second appropriate 'suvat' equation. Allow any valid method, e.g. (time to max ht) $\times 2$. Correct outcome. Allow $g = 9.8$, giving $t = 6.122$

## Question 8(i):

$F_1 + F_2 + F_3 + F_4 = 0$

$\therefore (5i - 8j) + (-3i - 4j) + (6i + 6j) + F_4 = 0$

$\therefore F_4 = (-8i + 6j)$ | M1 A1 | Sum of 4 forces set equal to 0 o.e. c.a.o.

## Question 8(ii):

$|F_4| = \sqrt{(-8)^2 + 6^2} = 10 \text{ N}$ | M1 A1 | Use of Pythagoras. ft (i).

$\theta = \text{inv} \tan\left(\frac{6}{-8}\right) = 143(13...)°$ | M1 A1 | Correct use of inverse tan (or cos or sin). ft (i), but not c's magnitude. Must have a clear reference direction. Allow sketch as evidence if convincing.

## Question 9(i):

Diagram showing weight, normal contact force and friction, and no others | B1 |

## Question 9(ii):

Resolve perpendicular to slope:

$N = mg \cos \theta$

N2L used & resolve down slope:

$ma = mg \sin \theta - F$ | B1 M1 | Equation of motion with 3 terms, and at least 2 correct. Condone consistent sin/cos error.

Friction law:

$F = \mu N$

$\therefore ma = mg \sin \theta - \mu mg \cos \theta$ | B1 M1 | Limiting friction only. Attempt to eliminate $N$ and $F$, and cancel $m$.

$\therefore a = g(\sin \theta - \mu \cos \theta)$ | A1 | c.a.o.

## Question 9(iii):

If $\mu > \tan \theta$ then the particle will not move. | B1 |

## Question 10(i):

$v = \int(12 - 6t)dt = 12t - 3t^2 (+c)$ | M1 A1 | Set up integral for $v$. Correct integration. Condone omission of "c".

$v = 0$ when $t = 0 \therefore c = 0$

When $t = 4, v = 48 - 48 = 0 \text{ ms}^{-1}$ | M1 A1 | "c" dealt with explicitly. c.a.o. from correctly integrated $a$. Accept correct answer obtained from a definite integral.

## Question 10(ii):

$x = \int(12t - 3t^2)dt = (6t^2 - t^3)|^4_0 = (96 - 64) - (0) = 32 \text{ m}$ | M1 A1 A1 | Correct integral of c's $v$, including limits (which may appear or be dealt with later). Correct integration. ft c's $v$. c.a.o. following use of limits or explicit treatment of "c".

## Question 10(iii):

When $x = 0, 6t^2 - t^3 = 0$ | M1 | Equation for $x = 0$. ft c's expression for $x$ in (ii) only if obtained by integration. Condone omission of consideration of "c" (=0).

$t \neq 0 \therefore t = 6 \text{ sec}$

$\therefore v = 72 - 108 = -36 \text{ ms}^{-1}$ | A1 A1 | Solved and non-zero solution chosen. c.a.o.

## Question 11(i):

N2L & Resolve vertically for either particle

$0.3a = 0.3g - T$

$0.2a = T - 0.2g$

$\therefore 0.5a = 0.1g = 1$

$\therefore a = 2 \text{ ms}^{-2}$

$\therefore T = 0.3 \times 10 - 0.3 \times 2 = 2.4 \text{ N}$ | M1 A1 A1 M1 A1 A1 | Accept use of $g = 9.8$ throughout. Eliminate either $T$ or $a$. Correct value for one. c.a.o. Correct value for the other. c.a.o.

**ALTERNATIVE:**

N2L for whole system

$(0.3 + 0.2)a = 0.3g - 0.2g$

$\therefore a = 2 \text{ ms}^{-2}$

N2L & Resolve vertically for either particle

$0.3a = 0.3g - T$ or $0.2a = T - 0.2g$

$\therefore T = 2.4 \text{ N}$ | M1 A1 A1 M1 A1 A1 | Allow 1 error. All correct. c.a.o. All correct. ft c's $a$. c.a.o.

## Question 11(ii):

$v^2 = 0^2 + 2 \times 2 \times 2.25 = 9$

$\therefore v = 3 \text{ ms}^{-1}$ | M1 A1 | Use of appropriate 'suvat' equation. ft c's $a$.

## Question 11(iii):

$I = (0.3 \times 3) - (0) = 0.9 \text{ Ns}$ | M1 A1 | Use of Impulse = change in momentum. ft c's $v$, including units. Allow −0.9 and/or kgms$^{-1}$.

## Question 11(iv):

$0.9 = P \times 0.005$

$\therefore P = 180 \text{ N}$ | M1 A1 | Use of Impulse = force $\times$ time, o.e. ft c's $I$. Allow −180.