1. **Problem statement:** We need to find (d) the time between the two collisions of balls P and Q, and (e) the ratio $d_1 : d_2$ given the setup and previous results. 2. **Given data and known results:** - Masses: $Q = 2m$, $P = m$ - Initial speed of $P$ before first collision: $\frac{13}{12}u$ - After first collision, $Q$ moves towards $C$ with speed $\frac{3}{5}u$ - Coefficient of restitution between balls: $e$ (to be used from part b) - Coefficient of restitution between ball and wall: $\frac{1}{2}$ - $\tan \alpha = \frac{5}{12}$ - Distance from $B$ to wall $W$ is $d_2$, and from $B$ to $C$ along $BC$ is $d_1$ - After collision, velocity components of $P$ parallel and perpendicular to $CB$ are $\frac{1}{5}u$ and $\frac{5}{12}u$ respectively. 3. **Step (d): Find time between two collisions of balls** - After the first collision, ball $Q$ moves towards $C$ with speed $\frac{3}{5}u$ along $BC$. - Ball $Q$ travels distance $d_1$ to reach $C$. - Time for $Q$ to reach $C$ is: $$t_Q = \frac{d_1}{\frac{3}{5}u} = \frac{5 d_1}{3 u}$$ - Ball $P$ after collision has velocity components: - Parallel to $CB$: $v_{P,\parallel} = \frac{1}{5}u$ - Perpendicular to $CB$: $v_{P,\perp} = \frac{5}{12}u$ - The perpendicular component $v_{P,\perp}$ is towards the wall $W$ at distance $d_2$. - Ball $P$ hits wall $W$ and rebounds with coefficient of restitution $\frac{1}{2}$. - Time for $P$ to reach wall $W$: $$t_{P,\text{to wall}} = \frac{d_2}{\frac{5}{12}u} = \frac{12 d_2}{5 u}$$ - After collision with wall, perpendicular velocity component becomes: $$v'_{P,\perp} = e_w v_{P,\perp} = \frac{1}{2} \times \frac{5}{12}u = \frac{5}{24}u$$ - Time for $P$ to return from wall to line $BC$: $$t_{P,\text{return}} = \frac{d_2}{v'_{P,\perp}} = \frac{d_2}{\frac{5}{24}u} = \frac{24 d_2}{5 u}$$ - Total time for $P$ to return to line $BC$: $$t_P = t_{P,\text{to wall}} + t_{P,\text{return}} = \frac{12 d_2}{5 u} + \frac{24 d_2}{5 u} = \frac{36 d_2}{5 u}$$ - During this time, $P$ moves parallel to $CB$ at speed $\frac{1}{5}u$: $$\text{distance along } CB = v_{P,\parallel} \times t_P = \frac{1}{5}u \times \frac{36 d_2}{5 u} = \frac{36 d_2}{25}$$ - The second collision occurs when $P$ catches $Q$ again. - $Q$ moves distance $d_1$ to $C$ in time $t_Q$, then rebounds with coefficient of restitution $\frac{1}{2}$, so speed after rebound: $$v'_{Q} = \frac{1}{2} \times \frac{3}{5}u = \frac{3}{10}u$$ - $Q$ returns from $C$ to $B$ in time: $$t_{Q,\text{return}} = \frac{d_1}{v'_{Q}} = \frac{d_1}{\frac{3}{10}u} = \frac{10 d_1}{3 u}$$ - Total time between collisions is sum of $Q$'s travel to $C$ and back plus $P$'s travel to wall and back: $$t_{\text{between collisions}} = t_Q + t_{Q,\text{return}} = \frac{5 d_1}{3 u} + \frac{10 d_1}{3 u} = \frac{15 d_1}{3 u} = \frac{15 d_1}{u}$$ **Hence, time between two collisions is** $$\boxed{\frac{15 d_1}{u}}$$ 4. **Step (e): Find ratio $d_1 : d_2$** - From part (c), it was shown that when $Q$ reaches $C$, $P$ is at distance $\frac{4}{3} d_1$ from wall $W$. - The distance of $P$ from wall $W$ at that time is: $$\text{distance} = d_2 - \text{perpendicular displacement of } P$$ - Perpendicular velocity of $P$ after collision is $\frac{5}{12}u$. - Time for $Q$ to reach $C$ is $t_Q = \frac{5 d_1}{3 u}$. - Perpendicular displacement of $P$ in time $t_Q$: $$s = v_{P,\perp} \times t_Q = \frac{5}{12}u \times \frac{5 d_1}{3 u} = \frac{25 d_1}{36}$$ - Distance of $P$ from wall $W$ when $Q$ reaches $C$: $$d_2 - s = \frac{4}{3} d_1$$ - Substitute $s$: $$d_2 - \frac{25 d_1}{36} = \frac{4}{3} d_1$$ - Rearranged: $$d_2 = \frac{4}{3} d_1 + \frac{25 d_1}{36} = \frac{48 d_1}{36} + \frac{25 d_1}{36} = \frac{73 d_1}{36}$$ - Therefore, ratio: $$d_1 : d_2 = 1 : \frac{73}{36} = 36 : 73$$ **Hence, the ratio is** $$\boxed{d_1 : d_2 = 36 : 73}$$