Question

Two blocks A(3kg) and B(2kg) resting on a smooth horizontal surface is connected by a spring of stiffness 480N/m. Initially the spring is undefined and a velocity of 2m/s is imparted to A along the line of the spring away from B .The max extension in meters of the spring subsequent motion is.

Answer 1

Initially, only block-A is moving with velocity $2\frac{m}{s}$.

Hence initial momentum $=3\times 2=6kg\frac{-m}{s}..................\left(1\right)$

At time of maximum compression of spring, let the common velocity be v m/s

Hence final momentum $=(3+2)\times v=5\times vkg\frac{-m}{s}................\left(2\right)$

By conservation of momentum, we equate (1) and (2) to get common velocity $v=1.2\frac{m}{s}$

Initial kinetic energy $=\left(\frac{1}{2}\right)\times 3\times 2\times 2=6Joules................\left(3\right)$

Final kinetic energy $=\left(\frac{1}{2}\right)\times 5\times 1.2\times 1.2=3.6Joules..............\left(4\right)$

difference in kinetic energy is stored as potential energy in spring,

i.e., $\left(\frac{1}{2}\right)k{x}^2=\left(\frac{1}{2}\right)\times 480\times x^2=6-3.6=2.4Joules..........\left(5\right)$

from $\left(5\right)$, we get the maximum displacement, $x=0.1m$