question_answer

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 neighbor it comes to rest and the struck block takes off at 0.8 V. This process repeats. What is the time interval between the third and the fourth hits? Assume that the motion between the hits is uniform. Neglect the duration of collision.

Answer:

\[{{t}_{1}}=\frac{d}{v},{{t}_{2}}=\frac{d}{{{v}_{1}}}\]where\[{{V}_{1}}=0.8v\,\,{{t}_{3}}=\frac{d}{{{v}_{2}}}\] where \[{{V}_{2}}=0.8{{V}_{1}}\] \[{{t}_{4}}=\frac{d}{{{v}_{3}}}\]where\[{{v}_{3}}=0.8{{v}_{2}},\] we get the required by answer by subtracting\[{{t}_{3}}\]from\[{{t}_{4}}.\]