Solution 20
1. Analysis of Motion
The ball bounces elastically. Let the component of velocity perpendicular to the incline be $v_{\perp}$ and parallel be $v_{\parallel}$. Since the impact is elastic, $v_{\perp}$ is reversed with the same magnitude. The acceleration component $g_{\perp} = g \cos\theta$ acts downwards. The time of flight $T$ between bounces is determined solely by $v_{\perp}$ and $g_{\perp}$: $$ T = \frac{2 v_{\perp}}{g_{\perp}} $$ Since $v_{\perp}$ is restored after every elastic collision, the time of flight $T$ is constant for all bounces.
The ball bounces elastically. Let the component of velocity perpendicular to the incline be $v_{\perp}$ and parallel be $v_{\parallel}$. Since the impact is elastic, $v_{\perp}$ is reversed with the same magnitude. The acceleration component $g_{\perp} = g \cos\theta$ acts downwards. The time of flight $T$ between bounces is determined solely by $v_{\perp}$ and $g_{\perp}$: $$ T = \frac{2 v_{\perp}}{g_{\perp}} $$ Since $v_{\perp}$ is restored after every elastic collision, the time of flight $T$ is constant for all bounces.
2. Parallel Motion
The acceleration parallel to the incline is $g_{\parallel} = g \sin\theta$. Let $u_1$ be the parallel velocity at the start of the first bounce (just after impact). Distance $R_{12} = u_1 T + \frac{1}{2} g_{\parallel} T^2$. At the end of the first bounce, the parallel velocity is: $$ v_1 = u_1 + g_{\parallel} T $$ This becomes the initial velocity for the second bounce, $u_2 = v_1$. Distance $R_{23} = u_2 T + \frac{1}{2} g_{\parallel} T^2 = (u_1 + g_{\parallel} T) T + \frac{1}{2} g_{\parallel} T^2$. $$ R_{23} = u_1 T + \frac{3}{2} g_{\parallel} T^2 $$
The acceleration parallel to the incline is $g_{\parallel} = g \sin\theta$. Let $u_1$ be the parallel velocity at the start of the first bounce (just after impact). Distance $R_{12} = u_1 T + \frac{1}{2} g_{\parallel} T^2$. At the end of the first bounce, the parallel velocity is: $$ v_1 = u_1 + g_{\parallel} T $$ This becomes the initial velocity for the second bounce, $u_2 = v_1$. Distance $R_{23} = u_2 T + \frac{1}{2} g_{\parallel} T^2 = (u_1 + g_{\parallel} T) T + \frac{1}{2} g_{\parallel} T^2$. $$ R_{23} = u_1 T + \frac{3}{2} g_{\parallel} T^2 $$
3. Specific Condition (Dropped Ball)
Let’s calculate $R_{12}$: Substitute $T = \frac{2 v \cos\theta}{g \cos\theta} = \frac{2v}{g}$. $$ R_{12} = (v \sin\theta) \frac{2v}{g} + \frac{1}{2} (g \sin\theta) \frac{4v^2}{g^2} $$ $$ R_{12} = \frac{2v^2}{g}\sin\theta + \frac{2v^2}{g}\sin\theta = \frac{4v^2}{g}\sin\theta $$ Now $R_{23}$: $$ R_{23} = R_{12} + g_{\parallel} T^2 = \frac{4v^2}{g}\sin\theta + (g \sin\theta)\frac{4v^2}{g^2} $$ $$ R_{23} = \frac{4v^2}{g}\sin\theta + \frac{4v^2}{g}\sin\theta = \frac{8v^2}{g}\sin\theta $$
Let’s calculate $R_{12}$: Substitute $T = \frac{2 v \cos\theta}{g \cos\theta} = \frac{2v}{g}$. $$ R_{12} = (v \sin\theta) \frac{2v}{g} + \frac{1}{2} (g \sin\theta) \frac{4v^2}{g^2} $$ $$ R_{12} = \frac{2v^2}{g}\sin\theta + \frac{2v^2}{g}\sin\theta = \frac{4v^2}{g}\sin\theta $$ Now $R_{23}$: $$ R_{23} = R_{12} + g_{\parallel} T^2 = \frac{4v^2}{g}\sin\theta + (g \sin\theta)\frac{4v^2}{g^2} $$ $$ R_{23} = \frac{4v^2}{g}\sin\theta + \frac{4v^2}{g}\sin\theta = \frac{8v^2}{g}\sin\theta $$
Result:
The ratio of the distances is: $$ \frac{R_{12}}{R_{23}} = \frac{4}{8} = \frac{1}{2} $$ Answer: **1 : 2**
The ratio of the distances is: $$ \frac{R_{12}}{R_{23}} = \frac{4}{8} = \frac{1}{2} $$ Answer: **1 : 2**