question_answer

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

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 :

The free body diagram showing the various forces acting on the pulley mass are as follows: Equating the vertical forces, we have \[{{m}_{1}}g-T={{m}_{1}}a\] ?..(i) \[T-{{m}_{2}}g={{m}_{2}}a\] ?..(ii) From Eqs. (i) and (ii), we get \[a=\frac{{{m}_{1}}g-{{m}_{2}}g}{{{m}_{1}}+{{m}_{2}}}\] ?..(iii) The acceleration of centre of mass is \[{{a}_{CM}}=\frac{{{m}_{1}}a-{{m}_{2}}a}{{{m}_{1}}+{{m}_{2}}}\] Putting the value of a from Eq. .(iii), we get \[{{a}_{CM}}=\frac{{{({{m}_{1}}-{{m}_{2}})}^{2}}}{{{({{m}_{1}}+{{m}_{2}})}^{2}}}g\]