question_answer

Two masses \[{{m}_{1}}\]and \[{{m}_{2}}\left( {{m}_{1}}>{{m}_{2}} \right)\]are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of centre of mass is [J&K CET 2005]            

A)                         \[{{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}g\]            

B)                           \[\frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}g\]          

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

D)                         Zero

Correct Answer: A

Solution :

                Acceleration of each mass \[=a=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\ g\]             Now acceleration of centre of mass of the system \[{{A}_{cm}}=\frac{{{m}_{1}}\overrightarrow{{{a}_{1}}}+{{m}_{1}}\overrightarrow{{{a}_{2}}}}{{{m}_{1}}+{{m}_{2}}}\]             As both masses move with same acceleration but in opposite direction so \[\overrightarrow{{{a}_{1}}}=-\overrightarrow{{{a}_{2}}}\] = a (let)             \[\therefore \ \ {{A}_{cm}}=\frac{{{m}_{1}}a-{{m}_{2}}a}{{{m}_{1}}+{{m}_{2}}}\]             \[=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\times \left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\times g\]             \[={{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}\times g\]