Question

A massless spring of constant 1000 $N{m}^{-1}$ is compressed a distance of 20 cm between discs of 8 kg and 2 kg, spring is not attached to discs. The system is given an initial velocity $3m{s}^{-1}$ perpendicular to length of spring as shown in the figure. What is ground frame velocity of $2kg$ block (in $m{s}^{-1}$) when spring regains its natural length.

Answer 1

$\text{Solution}:$

Let $v_1$ and $v_2$ are the velocities along the spring of A and B when spring regains it natural length then

$2v_1=8v_2$…….(i)

$\frac{1}{2}2v_1^2+\frac{1}{28}{v}_2^2=\frac{1}{21000}(\frac{20}{100}{)}^2$…………(ii)

Form (i) and (ii),

$v_1=4m{s}^{-1},v_2=1m{s}^{-1}$

Ground frame velocity of A

$\sqrt{v^2+v_1^2}=\sqrt{3^2+4^2}=5m{s}^{-1}$

$\text{Hence correct answer is 5}$