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Two blocks of masses \mathrm{10kg} and \mathrm{30kg} are placed on the same straight line with coordinates \mathrm{\left ( 0,0 \right )cm} and \mathrm{\left ( x,0 \right )cm} respectively. The block of \mathrm{10kg} is moved on the same line through a distance of \mathrm{6cm} towards the other block. The distance through which the block of \mathrm{30kg} must be moved to keep the position of centre of mass of the system unchanged is :

Option: 1

\mathrm{4cm} towards the \mathrm{10kg} block


Option: 2

\mathrm{2cm} away from the \mathrm{10kg} block


Option: 3

\mathrm{2cm} towards the \mathrm{10kg} block


Option: 4

\mathrm{4cm} away from the \mathrm{10kg} block


Answers (1)

best_answer

If the centre of mass does not change then

\mathrm{\Delta x_{c m}=\frac{m_{1} \Delta x_{1}+m_{2} \Delta x_{2}}{m_{1}+m_{2}}=0}

\mathrm{m_{1} \Delta x_{1}=-m_{2} \Delta x_{2}}

A negative sign implies both blocks will move in the opposite direction

\mathrm{10 \times 6 =30 \times \Delta x_{2}} \\

\mathrm{\Delta x_{2} =2 \mathrm{~cm}}   towards left

Hence the correct answer is option 3.

Posted by

sudhir.kumar

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