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Large number of identical blocks are placed on a surface with adjacent blocks $$d$$ apart. The left most block is imparted a velocity $$V$$ to the right. On striking its neighbour it comes to rest and the struck block takes off at $$0.8V$$. This process repeats. What is the time interval between the third and fourth hits? Assume that the motion between the hits is uniform. Neglect the duration of collision.
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Correct Answer: 0
After each hit the moving block's speed is multiplied by 0.8. After the first hit the second block moves at $$0.8V$$, after the second hit the third block moves at $$0.64V$$, and after the third hit the fourth block moves at $$(0.8)^3V = 0.512V$$. The fourth block covers the gap $$d$$ before the fourth hit, so the required interval is $$\frac{d}{0.512V} = \frac{125d}{64V} \approx 1.95\frac{d}{V}$$.
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