Q38 / 105 · NTA Abhyas (JEE Main) 2020 · Physics · Center of Mass Momentum and Collision

Two masses $m$ and $M (> m)$ are joined by a light string passing over a smooth light pulley. The centre of mass of the system moves with an acceleration

A
$g (\frac{M - m}{M + m})$ downward
Correct answer
B
$g {(\frac{M - m}{M + m})}^{2}$ downward
C
$g {(\frac{M - m}{M + m})}^{2}$ upward if M < m
D
$z e r o$

Solution

a_{c o m} = \frac{m_{1} \bar{a_{1}} + m_{2} \bar{a_{2}}}{m_{1} + m_{2}}

a_{c o m} = \frac{1}{m + M} [\frac{M (M - m)}{m + M} - \frac{m (M - m)}{m + M}] g

= \frac{1}{m + M} [\frac{{(M - m)}^{2}}{m + M}] g = {(\frac{M - m}{M + m})}^{2} g

Hence, the correct answer is option: 'B'.

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