question_answer

A spring is compressed between blocks of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] placed on a smooth horizontal surface as shown in the figure. When the blocks are release, they have initial velocity of \[{{v}_{1}}\] and \[{{v}_{2}}\] in the direction shown. The blocks travel distances \[{{x}_{1}}\] and \[{{x}_{2}}\] before coming to rest. The ratio of \[\frac{{{x}_{1}}}{{{x}_{2}}}\] is

A) \[\frac{{{m}_{1}}}{{{m}_{2}}}\]

B) \[\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\]

C) \[\frac{{{m}_{2}}}{{{m}_{1}}}\]

D) \[\sqrt{\frac{{{m}_{2}}}{{{m}_{1}}}}\]

Correct Answer: C

Solution :

[c] From linear momentum conservation, \[{{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}}\]. As they move spring exerts equal and opposite forces on \[{{m}_{1}}\] and \[{{m}_{2}}\]. \[\Rightarrow \]   \[{{m}_{1}}\frac{d{{x}_{1}}}{dt}={{m}_{2}}\frac{d{{x}_{2}}}{dt}\] \[\Rightarrow \]   \[{{m}_{1}}\int{\,d{{x}_{1}}={{m}_{2}}\int{\,d{{x}_{2}}}}\] \[\Rightarrow \]   \[{{m}_{1}}{{x}_{1}}={{m}_{2}}{{x}_{2}}\] \[\Rightarrow \]   \[\frac{{{x}_{1}}}{{{x}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\]