Conservation of energy for a series of elastic collisions

[Andrew1235]

Homework Statement

A link to the problem: https://i.stack.imgur.com/tDhbm.png

A link to the solution: https://i.stack.imgur.com/h5s1g.png

Relevant Equations

Conservation of energy

The speed of the block after the nth collision is

$$ V_n=(2e)^n*v_0 $$

By conservation of energy the block travels a distance $$V_n^2/(2ug)$$ on the nth bounce. So the total distance is

$$ d=1/(2ug)∗(v_0^2+(2ev_0)^2...) $$

$$ d=1/(2ug)∗(v_0^2/(1−4e^2)) $$

$$ d=1/(2ug)∗(v_0^2∗M^2/(M^2−4m^2)) $$


Can someone explain why this is incorrect?

 

[TSny]

I don't believe the following is correct (other than for n = 1).

The speed of the block after the nth collision is

$$ V_n=(2e)^n*v_0 $$

Can you explain how you got this?

 

[haruspex]

The speed of the block after the nth collision is
$$ V_n=(2e)^n*v_0 $$

According to 5.155 and 5.156, it should be
$$ V_n=2e(1-2e)^{n-1}*v_0 $$