question_answer

A block C of mass m is moving with velocity \[{{\text{v}}_{0}}\] 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) \[\frac{m{{\text{v}}_{0}}^{2}}{{{x}_{0}}^{2}}\]                 

B) \[\frac{m{{\text{v}}_{0}}^{2}}{2{{x}_{0}}^{2}}\]

C) \[\frac{3}{2}\frac{m{{\text{v}}_{0}}^{2}}{{{x}_{0}}^{2}}\]

D)                    \[\frac{2}{3}\frac{m{{\text{v}}_{0}}^{2}}{{{x}_{0}}^{2}}\]

Correct Answer: D

Solution :

When C strikes A \[\frac{1}{2}m{{v}_{0}}^{2}=\frac{1}{2}mv{{'}^{2}}+\frac{1}{2}k{{x}_{0}}^{2}\]  (v=? = velocity of A)             \[k{{x}_{0}}^{2}=m({{v}_{0}}^{2}-v{{'}^{2}})\]                ..?..(i)             \[\frac{1}{2}2mv{{'}^{2}}=\frac{1}{2}k{{x}_{0}}^{2}\]        (When A and B Block attains K.E.) \[\therefore \,\,\,\frac{1}{2}2mv{{'}^{2}}=\frac{1}{2}k{{x}_{0}}^{2}\]                        ......... (ii) From (i) and (ii), \[k{{x}_{0}}^{2}=m{{v}_{0}}^{2}-mv{{'}^{2}}=m{{v}_{0}}^{2}-\frac{k}{2}{{x}_{0}}^{2}\] \[\Rightarrow \,\,\,k{{x}_{0}}^{2}+\frac{k}{2}{{x}_{0}}^{2}=m{{v}_{0}}^{2}\] \[\frac{3}{2}k{{x}_{0}}^{2}=m{{v}_{0}}^{2}\] \[\therefore \,\,k=\frac{2}{3}m\frac{{{v}_{0}}^{2}}{{{x}_{0}}^{2}}\]