Question

Two blocks $A\left(3kg\right)$ and $B\left(2kg\right)$ resting on a smooth horizontal surface is connected by a spring of stiffness $480N/m$. Initially the spring is undeformed and a velocity of $2m/s$ is imparted to $A$ along the line of the spring away form $B$. The maximum extension in meters of the spring during subsequent motion is

• A. $\frac{1}{10}$
• B. $\frac{1}{2\sqrt{10}}$
• C. $\frac{1}{2\sqrt{15}}$
• D. $0.15$

Answer 1

The correct option is A $\frac{1}{10}$

Initially only block $A$ is moving with velocity $2m/s.$

Hence initial momentum $=3\times 2=6kgm/s$ ..................$\left(1\right)$

At time of maximum compression of spring, let the common velocity be $vm/s$

Hence final momentum $=\left(3+2\right)\times v=5\times vkgm/s$ ................$\left(2\right)$

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

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

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

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

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

from $\left(5\right),$ we get the maximum displacement, $x=0.1m=\frac{1}{10}m$

Hence,

option $\left(A\right)$ is correct answer.