$mu + 2m \times (-2u) = mv_A + 2mv_B$ (oe) | M1, A1 | Conservation of momentum attempted; Completely correct equation (any form)

$v_B - v_A = e(u - (-2u))$ | M1, A1 | oe; restitution equation attempted; Completely correct equation (any form)

$v_A = -u(1 + 2e)$ | M1, A1 | Solving for $v_A$ and $v_B$

$v_B = u(1-e)$ | A1 |

KE of A is $\frac{1}{2}mu^2(1+2e)^2$ | Bift [1] |

KE of B is $\frac{1}{2} \times 2mu^2(1-e)^2$ | Bift [1] |

$\frac{1}{2}mu^2(1+2e)^2 = 8 \times \frac{1}{2} \times 2mu^2(1-e)^2$ | M1 | Equation for e using ratio of $\text{KE}_A : \text{KE}_B$

$4e^2 - 12e + 5 = 0$ or $1 + 2e = \pm4(1-e)$ | M1 | Forming and solving 3-term quadratic | oe

$e = \frac{5}{2} > 1$ rejected | A1 | Must see explicit rejection for this mark

$e = \frac{1}{2}$ | A1 [12] |

**Alternative solution**

$mu + 2m \times (-2u) = mv_A + 2mv_B$ | M1A1 | As in solution above

$v_A + 2v_B = -3u$ | A1 | soi

$\frac{1}{2}mv_A^2 = 8 \times \frac{1}{2} \times 2mv_B^2$ | M1 | Use of given KE ratio

$v_A = \pm4v_B$ | A1 | Correct equation (unsimplified); Must have ± at this stage

$v_A = 4v_B \Rightarrow v_B = -\frac{3}{4}u$ and $v_A = -2u$ | M1 | Using either of the two possible cases

so $-\frac{1}{4}u - (-2u) = e(u - (-2u))$ | M1 | oe; restitution equation attempted

$e = \frac{1}{2}$ | A1 |

$v_A = -4v_B \Rightarrow v_B = \frac{3}{5}u$ and $v_A = -6u$ | M1 | Using either of the two possible cases

so $\frac{3}{5}u - (-6u) = e(u - (-2u))$ | M1 | oe

$e = \frac{2}{5} > 1$ rejected | A1 [12] | Or argument that $v_B > v_A$ is not physically possible

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