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 frictioniess 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) \[T-{{m}_{2}}g={{m}_{a}}a\]          ...(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                 \[{{a}_{CM}}=\frac{{{({{m}_{1}}-{{m}_{2}})}^{2}}}{{{({{m}_{1}}+{{m}_{2}})}^{2}}}g\]