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 $
BETA
