Laws Of Motion Question 167

Question: 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 [J&K CET 2005]

Options:

A) $ {{( \frac{m _{1}-m _{2}}{m _{1}+m _{2}} )}^{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

Show Answer

Answer:

Correct Answer: A

Solution:

Acceleration of each mass $ =a=( \frac{m _{1}-m _{2}}{m _{1}+m _{2}} )\ 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}} $

$ =( \frac{m _{1}-m _{2}}{m _{1}+m _{2}} )\times ( \frac{m _{1}-m _{2}}{m _{1}+m _{2}} )\times g $

$ ={{( \frac{m _{1}-m _{2}}{m _{1}+m _{2}} )}^{2}}\times g $



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