A smooth sphere \(A\) of mass \(m\) is projected with speed \(u\) along a smooth horizontal surface and strikes a stationary smooth sphere \(B\) of equal radius but of mass \(M\). The direction of motion of \(A\) before the impact makes an acute angle \(\theta\) with the line of centres at the moment of contact. After the impact, the direction of motion of \(A\) is perpendicular to the initial direction of motion of \(A\). The coefficient of restitution between the two spheres is \(e\). Given that \(Me \geq m\), prove that $$\tan^2 \theta = \frac{Me - m}{m + M}.$$ [8]

Taking $v_1$ and $v_2$ as the components of velocity of $A$ and $B$ respectively along the line of centres after impact:

NEL: $-eu\cos\theta = v_1 - v_2 \Rightarrow -Meu\cos\theta = Mv_1 - Mv_2$ | M1, A1 |

CLM: $mu\cos\theta = mv_1 + Mv_2$ | M1, A1 |

Adding: $u\cos\theta(m - Me) = (m + M)v_1$ | M1 |

$\therefore v_1 = u\cos\theta\frac{(Me-m)}{m+M}$ | A1 |

For perpendicularity: $\frac{-v_1}{u\sin\theta} = \tan\theta$ | B1 |

Hence: $\frac{u}{\tan\theta}\left(\frac{Me-m}{m+M}\right) = \tan\theta \Rightarrow \tan^2\theta = \frac{Me-m}{m+M}$ | A1, (AG) | **8**

A smooth sphere $A$ of mass $m$ is projected with speed $u$ along a smooth horizontal surface and strikes a stationary smooth sphere $B$ of equal radius but of mass $M$. The direction of motion of $A$ before the impact makes an acute angle $\theta$ with the line of centres at the moment of contact. After the impact, the direction of motion of $A$ is perpendicular to the initial direction of motion of $A$. The coefficient of restitution between the two spheres is $e$. Given that $Me \geq m$, prove that
$$\tan^2 \theta = \frac{Me - m}{m + M}.$$ [8]

\hfill \mbox{\textit{Pre-U Pre-U 9795/2  Q1 [8]}}