In the arrangement shown in the figure- - m - A -2kg and - m - B -1kg- String is light and inextensible- Find the acceleration of centre of mass of both the blocks -Neglect friction everywhere-g-3 downwardsg-9 downwardsg-9 upwardsg-3 upwards

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Toppr Admin · Accepted Answer

As mA-gt-mB so acceleration of A is downward and B is upward- For downward assume acceleration is positive and for upward acceleration negative- i-e- aA-a and aB-x2212-aHere- -mA-mB-a-mA-x2212-mB-g-2-1-a-2-x2212-1-g-x21D2-a-g3thus- acm-mAaA-mBaBma-mB-2-g-3-1-x2212-g-3-2-1-g-9 downward- -xA0-as it is positive-

In the arrangement shown in the figure, mA=2kg and mB=1kg. String is light and inextensible. Find the acceleration of centre of mass of both the blocks :Neglect friction everywhere.g/3 downwardsg/9 downwardsg/9 upwardsg/3 upwards

As mA>mB so acceleration of A is downward and B is upward. For downward assume acceleration is positive and for upward acceleration negative. i.e, aA=a and aB=−aHere, (mA+mB)a=(mA−mB)g(2+1)a=(2−1)g⇒a=g3thus, acm=mAaA+mBaBma+mB=2(g/3)+1(−g/3)2+1=g/9 downward. (as it is positive)

In the arrangement shown in the figure, $m_{A} = 2 k g$ and $m_{B} = 1 k g$ . String is light and inextensible. Find the acceleration of centre of mass of both the blocks :
Neglect friction everywhere.

$g / 3$ downwards
$g / 9$ downwards
$g / 9$ upwards
$g / 3$ upwards