Question

Two blocks \(A\) and \(B,\) each of mass \(2~\text{kg}\) are connected to two identical springs of spring constant \(800~Nm^{-1}\) and are placed on a smooth horizontal surface as shown in the figure. Initially the springs are relaxed. Now, the block \(A\) is slightly displaced to left and block \(B\) to right by the same distance and then released simultaneously. If the resulting collision is elastic, the time period of oscillation of the system is . The value of \(n\) is \(\dfrac{\pi}{5 n} s.\) The value of \(n\) is

Answer 1

(i)
(ii)
(iii)
(iv)
(v)
Since collision is elastic, total mechanical energy of the system is conserved. Initially, $A$ and $B$ are displaced, then released and collide with each other, exchange their velocities, and once again $A$ and $B$ will be in initial state of zero speed. Thereafter, complete motion is getting repeated. For each block, total time period of oscillation is
$T=2\pi \sqrt{\frac{M}{R}}=2\pi \sqrt{\frac{2}{800}}=\frac{2\pi }{20}=\frac{\pi }{10}s$.
$\therefore$ Time taken for the system to go from figure (ii) to Figure (v) is $t=\frac{T}{2}=\frac{\pi }{20}s$