JEE Main · PhysicsHard

$Two blocks of masses m_{1} and m_{2} are connected by a light string passing over a pulley. If m_{1} > m_{2}, the acceleration of the center of mass of the system is:$

A. $\frac{( m _{1} - m _{2} ) ^{2} g}{( m _{1} + m _{2} ) ^{2}}$
B. $\frac{( m _{1} + m _{2} ) g}{( m _{1} - m _{2} )}$
C. $\frac{( m _{1} - m _{2} ) g}{( m _{1} + m _{2} )}$
D. $0$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{( m _{1} - m _{2} ) ^{2} g}{( m _{1} + m _{2} ) ^{2}}$

Solution

$The acceleration of each block is a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} .$
$The acceleration of the center of mass is a_{c m} = \frac{m _{1} a _{1} + m _{2} a _{2}}{m _{1} + m _{2}} .$
$Since a_{1} = a and a_{2} = - a, a_{c m} = \frac{( m _{1} - m _{2} ) a}{m _{1} + m _{2}} .$
$Substituting a, we get a_{c m} = \frac{( m _{1} - m _{2} ) ^{2} g}{( m _{1} + m _{2} ) ^{2}} .$
$Hence the answer is (A).$

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