Q.
A block C of mass $m$is moving with velocity $v_{0}$and collides elastically with block $A$of mass $m$ and connected to another block $B$of mass $2m$ through spring constant $k$.What is $k$ if $x_{0}$ is compression of spring when velocity of $A\ and\ B$ is same ?
A) $\frac{mv_{0}^{2}}{X_{0}^{2}}$
B) $\frac{mv_{0}^{2}}{2x_{0}^{2}}$
C) $\frac{3}{2}\frac{mv_{0}^{2}}{x_{0}^{2}}$
D) $\frac{2}{3}\frac{mv_{0}^{2}}{x_{0}^{2}}$
Explanation
(d) Using conservation of linear momentum, we have
$mv_{0} = mv + 2mv$
Or $v = \frac{v_{0}}{3}$
Using conservation of energy, we have
$\frac{1}{2}mv_{0}^{2} = \frac{1}{2}kx_{0}^{2} + \frac{1}{2}(3m)v^{2}$
Where $x_{0}$=compression in the spring ,
$\therefore mv_{0}^{2} = kx_{0}^{2} + (3m)\ \frac{v_{0}^{2}}{9}$
Or $kx_{0}^{2} = \ mv_{0}^{2} - \frac{mv_{0}^{2}}{3}$
Or $kx_{0}^{2} = \frac{2mv_{0}^{2}}{3}$
$\therefore k = \frac{2mv_{0}^{2}}{3x_{0}^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.