COM CYU 12

System Analysis:

We have an infinite array of blocks. Let the mass of the first (heaviest) block be $m_1 = m$. The mass of the second block is $m_2 = \eta m$, the third is $m_3 = \eta^2 m$, and so on, where $\eta = 0.9$. The distance between adjacent blocks is $l$.

Step 1: Velocity after collision

Consider the elastic collision between the $n$-th block (mass $m_n$) moving with velocity $v_n$ and the stationary $(n+1)$-th block (mass $m_{n+1} = \eta m_n$).

Using the formula for velocity of the second body after an elastic collision:

$$ v_{n+1} = \frac{2m_n}{m_n + m_{n+1}} v_n $$

Substituting $m_{n+1} = \eta m_n$:

$$ v_{n+1} = \frac{2m_n}{m_n + \eta m_n} v_n = \left( \frac{2}{1+\eta} \right) v_n $$

This shows the velocities form a Geometric Progression (GP) with common ratio $r = \frac{2}{1+\eta}$.

The velocity of the $n$-th block is: $v_n = u \cdot r^{n-1}$.

Step 2: Time Calculation

The time taken for the $n$-th block to travel the distance $l$ to hit the $(n+1)$-th block is:

$$ t_n = \frac{l}{v_n} = \frac{l}{u \cdot r^{n-1}} = \frac{l}{u} \left( \frac{1}{r} \right)^{n-1} $$

The term $\frac{1}{r} = \frac{1+\eta}{2}$. Since $\eta = 0.9$, $\frac{1+\eta}{2} = \frac{1.9}{2} = 0.95$, which is less than 1. Thus, the series converges.

Step 3: Total Time Summation

The total time $T$ is the sum of time intervals for all collisions:

$$ T = \sum_{n=1}^{\infty} t_n = \frac{l}{u} \sum_{n=1}^{\infty} \left( \frac{1+\eta}{2} \right)^{n-1} $$

This is an infinite geometric series with first term $a = \frac{l}{u}$ and common ratio $R = \frac{1+\eta}{2}$.

$$ T = \frac{a}{1 – R} = \frac{l/u}{1 – \frac{1+\eta}{2}} $$ $$ T = \frac{l/u}{\frac{2 – (1+\eta)}{2}} = \frac{2l}{u(1-\eta)} $$