# Question 1:

## Part 1a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of CLM: $2m \times 2u - 3mu = -2m \times 0.5u + 3mw$ | M1 | Use of CLM, correct no. of terms, condone consistent extra $g$'s, sign errors, cancelled $m$'s, to produce an equation in $(m)$, $u$ and $w$ only. **OR:** Use of two impulse-momentum equations with $I$ eliminated to produce equation in $(m)$, $u$ and $w$ only. |
| **OR:** $(I=)\ 2m(0.5u - -2u) = 3m(w - -u)$ | A1 | Correct equation in $(m)$, $u$ and $w$ only. |
| $w = \dfrac{2u}{3}$, accept $0.67u$ or better | A1 | cao. Must see $w = ku$ |
| **(3 marks)** | | |

## Part 1b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of NEL | M1 | Must be right way up with 'correct' terms but condone sign errors. Allow without the $u$'s |
| $e = \dfrac{0.5u + \dfrac{2u}{3}}{2u + u}$ | A1ft | Correct unsimplified expression, ft on their $w$ |
| $\dfrac{7}{18}$, accept $0.39$ or better | A1 | cao |
| **(3 marks)** | | |

## Part 1c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of impulse-momentum principle for $A$ or $B$ | M1 | Correct structure but condone sign errors (M0 if $g$ included or $m$'s missing). Can score if $m$ used for mass with a 'correct' pair of velocities. |
| $\pm 2m(0.5u - -2u)$ **OR** $\pm 3m\!\left(\dfrac{2u}{3} - -u\right)$ | A1 | Correct equation |
| $5mu$ | A1 | cao |
| **(3 marks)** | | |

**Total: 9 marks**

## Question 2a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $F = \frac{84000}{12}$ | M1 | Allow use of 84 |
| Equation of motion horizontally | M1 | Correct no. of terms, condone sign errors. Allow if they use 84 or 84000 as the driving force |
| $\frac{84000}{12} - 490 \times 12 = 5000a$ | A1 | Correct equation |
| $\frac{28}{125}$ or $0.224$ or $0.22$ ($\text{m s}^{-2}$) | A1 | Accept 0.22 |

## Question 2b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $D = \frac{84000}{V}$ | M1 | Allow use of 84 |
| Equation of motion parallel to road: $D - 490V - 5000g\sin\alpha = 0$ | M1 | Equation in $V$ only with correct no. of terms, condone sign errors and sin/cos confusion and omitted $g$ with $D$ in terms of $V$ |
| $\frac{84000}{V} - 490V - 5000g\sin\alpha = 0$ | A1 | Correct equation in $V$ only |
| $V = 10$ only | A1 | cao |

## Question 3a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use the principle of conservation of mechanical energy and model | M1 | Correct number of terms, dimensionally correct, condone sign errors. M0 if they use $v^2 = u^2 + 2as$ |
| $\frac{1}{2}mW^2 - \frac{1}{2}m\left(\sqrt{2gh}\right)^2 = mgh$ | A1 | Correct equation with at most one error |
| | A1 | Correct equation |
| $W = \sqrt{4gh} = 2\sqrt{gh}$ | A1 | Either (need $W =$ ) |

## Question 3b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = mg\cos\theta$ | M1 | Condone sin/cos confusion and allow $\cos\left(\frac{4}{5}\right)$ etc |
| $F = \frac{1}{3}R$ | M1 | $F = \frac{1}{3}R$ |
| $F = \frac{4}{15}mg$ | A1 | Accept $0.27\ mg$ or better |

## Question 3c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use the work-energy principle and the model | M1 | Correct number of terms, dimensionally correct, condone sign errors and sin/cos confusion and allow $\cos\left(\frac{4}{5}\right)$ etc |
| $A$ to $B$: $mgd\sin\theta - \frac{1}{2}m\left(\sqrt{2gh}\right)^2 = \frac{4}{15}mgd$ | A1ft | Correct equation with at most one error |
| or $mg(h + d\sin\theta) - mgh - \frac{1}{2}m\left(\sqrt{2gh}\right)^2 = \frac{4}{15}mgd$ | A1ft | Correct equation ft on their answer to (b) (and (a) if they use $A$ to the ground) |
| **OR** $A$ to ground: $mg(h + d\sin\theta) - \frac{1}{2}m\left(\sqrt{4gh}\right)^2 = \frac{4}{15}mgd$ | | |
| Solve for $d$ in terms of $h$ or $h$ in terms of $d$ | M1 | Solve for $d$, must have at least 3 terms, with two of them in $d$ |
| $d = 3h$ | A1 | cao. N.B. No marks available if they don't use work-energy |

## Question 4a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of CLM **OR** NEL | M1 | CLM: Correct no. of terms, condone consistent extra $g$'s, sign errors, cancelled $m$'s. **OR** NEL: $e$ on the correct side but condone sign errors |
| $mu = -m\frac{u}{5}(4e-1) + 4mv_Q$ **OR** $v_Q + \frac{u}{5}(4e-1) = eu$ | A1 | Correct equation |
| $v_Q = \frac{u}{5}(e+1)$ | A1 | Cao. Accept any equivalent two term expression |

## Question 4b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_P = \pm\frac{fu}{5}(4e-1)$ | B1 | Seen or implied |
| $= \pm\frac{2fu}{5}$ | B1 | Seen or implied |
| $\text{KE Loss} = \frac{1}{2}m\left(\frac{2u}{5}\right)^2 - \frac{1}{2}m\left(\frac{2fu}{5}\right)^2$ | M1 | Allow negative of this and without $e$ being substituted. Allow anything of the form $\pm\left(\frac{1}{2}m(v_p)^2 - \frac{1}{2}m(fv_p)^2\right)$, provided that $v_p$ has come from an attempt to put $e = \frac{3}{4}$ in the given expression |
| $= \frac{2mu^2}{25}(1-f^2)$ | A1 | Cao. Accept any equivalent two term expression, isw |

## Question 4c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v_Q = \frac{7u}{20}$ | M1 | For attempt to put $e = \frac{3}{4}$ in their $v_Q$ expression to give a multiple of $u$, seen or implied at some stage |
| $\frac{7u}{20} < \frac{2fu}{5}$ | M1 | Correct inequality for their speeds (which could involve $e$), provided it's dimensionally correct. Not available if $Q$ is moving towards the wall |
| $\frac{7}{8} < f$ | A1 | $\frac{7}{8} < f$ oe |
| $,\ 1$ | B1 | For $f \leqslant 1$. N.B. All marks available if they go straight to $\frac{7}{20} < \frac{2f}{5}$ |

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