The story of a ball and a slope SJPO question

By using the formula for kinetic energy and gravitational potential energy, it can be seen that the kinetic energy is directly proportional to the height that the ball would rise:

$\frac { 1 }{ 2 } m{ v }^{ 2 }=mgh\\ KE\quad \alpha \quad h\\ KE\quad \alpha \quad { v }^{ 2 }$

So since the speed decreases to $\frac { 4 }{ 5 }$ after every collision, the kinetic energy and height decreases to ${ \frac { 4 }{ 5 } }^{ 2 }$ after every collision.

Since the ball is going to bounce until it stops, it is assumed to bounce infinity times, and the distance it travels is thus as:

$2\sum _{ n=1 }^{ \infty }{ { \left[ \frac { 4 }{ 5 } \right] }^{ 2n } } +1$

Let $\sum _{ n=1 }^{ \infty }{ { \left[ \frac { 4 }{ 5 } \right] }^{ 2n } }$ be $S$

$\sum _{ n=1 }^{ \infty }{ { \left[ \frac { 4 }{ 5 } \right] }^{ 2n } } -\sum _{ n=2 }^{ \infty }{ { \left[ \frac { 4 }{ 5 } \right] }^{ 2n } } ={ \left[ \frac { 4 }{ 5 } \right] }^{ 2 }=S-S{ \left[ \frac { 4 }{ 5 } \right] }^{ 2 }$

Therefore, $S=\frac { 16 }{ 9 }$ which makes $L\left[ 2\sum _{ n=1 }^{ \infty }{ { \left[ \frac { 4 }{ 5 } \right] }^{ 2n } } +1 \right] =\frac { 41 }{ 9 } L$

$a+b=41+9=\boxed { 50 }$