Question

In the figure shown, two identical blocks A and B of mass $m=1/2kg$ each are placed on the two opposite edges of a table. A light spring of stiffness $k={\pi }^2$ N/m and having natural length equal to the separation between the blocks, is placed between the blocks. Neither block is attached to the spring. The blocks are displaced towards each other by equal distance $x=2cm$ and then released at time $t=0$.
Calculate the magnitude of the net impulse (in SI unit) on the system of the blocks and spring from $t=0$ to $t=1.05s$. Ignore any friction, and take $g=10m/s^2$ and assume that the blocks do not reach the ground below the table, before 1.05 second.

• A. 8 N/s
• B. 0.2 N/s
• C. 0.1 N/s
• D. 0.4/3 N/s

Answer 1

The correct option is A 8 N/s

From work energy theorem,

$\frac{1}{2}K{x}^2=\frac{1}{2}m{v}^2$

$v=\sqrt{\frac{k}{m}}x$

$=\sqrt{\frac{\pi }{1}}2\times 2$

$=4\pi cm/s$

Net impulse = Total change in momentum $=8N/s$