# Question 1:

## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3 \times 4 + 21 \times 2 = 4U$ | M1 | Use of PCLM and $I = Ft$ |
| $4U = 54$ so $U = 13.5$, speed is $13.5 \text{ m s}^{-1}$ | A1 | |
| **OR** $21 = 4a : a = 5.25$ and $v = 3 + 2 \times 5.25$, speed is $13.5 \text{ m s}^{-1}$ | M1, A1 | Use of $F = ma$ and suvat |

## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $V$ be speed of S in direction PQ; $54 - 2 \times 3 = (4+2)V$ | M1 | PCLM for coalescence |
| $6V = 48$ so $V = 8$, velocity is $8 \text{ m s}^{-1}$ in direction PQ | E1 | Answer given. Accept no reference to direction |

## Part (a)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6 \times 8 + 4u = 6 \times 5 + 4v$ | M1 | Use of PCLM. Allow any sign convention. All masses and speeds must be correct |
| $v - u = 4.5$ | A1 | Any form |
| $\dfrac{v-5}{u-8} = -\dfrac{1}{4}$ | M1 | Use of NEL correct way up; allow sign errors |
| $4v + u = 28$ | A1 | Any form, signs consistent with PCLM eqn |
| $u = 2$ so $2 \text{ m s}^{-1}$ in direction SR | A1 | cao **NOTE that a sign error in NEL leads to $u = -2$; this gets A0** |
| $v = 6.5$ so $6.5 \text{ m s}^{-1}$ in direction SR | A1 | cao. Withhold only 1 of final A marks if directions not clear. Directions can be inferred from a CLEAR diagram |

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# Question 1(b):

## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 0.2 \times v^2 - \frac{1}{2} \times 0.2 \times 5^2 = 0.2 \times 10 \times 10$ | M1 | Use of WE or suvat, must use distance of 10; allow $g = 9.8$ |
| $v^2 = 225$ and $v = 15$ | A1 | Answer not required ($v = 14.9$ if $g = 9.8$) |
| $\cos\alpha = \dfrac{4}{5}$, $\sin\alpha = \dfrac{3}{5}$ | B1 | Use of either expression or use of $36.9°$ |
| Parallel to plane: $15\cos\alpha = 13\cos\beta$ | M1 | Attempt to conserve velocity component parallel to plane. Allow use of 5 instead of 15 |
| $\cos\beta = \dfrac{12}{13}$ and $\beta = 22.61°$ so $22.6°$ | A1 | ($\beta = 23.8°$ if $g = 9.8$) |
| Perpendicular to plane: $13\sin\beta = e \times 15\sin\alpha$ | M1 | Attempt to use NEL perpendicular to plane. Allow use of 5 instead of 15, or use $\tan\beta = e\tan\alpha$ |
| $\sin\beta = \dfrac{5}{13}$, o.e. find $\tan\beta = \dfrac{5}{12}$ | A1 | |
| $13 \times \dfrac{5}{13} = 15 \times \dfrac{3}{5} \times e$ and $e = \dfrac{5}{9}$ | A1 | cao. Accept $0.56$ ($e = 0.589$ if $g = 9.8$) |

**OR (alternative method):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| First three marks as above | M1A1B1 | |
| Parallel to plane: $u_x = 15\cos\alpha (= 12)$ and $v_x = u_x (= 12)$ | M1 | Attempt to conserve velocity component parallel to plane. Allow use of 5 instead of 15 |
| $\cos\beta = \dfrac{v_x}{v} = \dfrac{12}{13}$, $\beta = 22.6°$ | A1 | |
| Perpendicular to plane: $u_y = 15\sin\alpha (= 9)$ and $v_y = eu_y (= 9e)$ | M1 | Attempt to use NEL perpendicular to plane. Allow use of 5 instead of 15 |
| $v_x^2 + v_y^2 = 13^2$ | A1 | Use of Pythagoras for velocities after collision in attempt to find $e$ |
| $12^2 + (9e)^2 = 13^2$ so $e^2 = \dfrac{25}{81}$, $e = \dfrac{5}{9}$ | A1 | |

## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Impulse is perp to plane with mod $0.2(13\sin\beta - (-15\sin\alpha)) = 0.2(5-(-9))$ | M1 | For use of $I = m(v-u)$ perp to plane; $0.2(5-9)$ gets M1A0 |
| $= 2.8 \text{ N s}$ | A1 | cao |

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