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Standard XII

Physics

Displacement of COM:Application

A spring is c...

Question

A spring is compressed between two blocks of masses $m_{1}$ and $m_{2}$ placed on a horizontal frictionless surface. When the blocks are released, the blocks travel distance $x_{1}$ and $x_{2}$ respectively before coming to rest. The ratio $x_{1} / x_{2}$ is:

m_{1} / m_{2}

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m_{2} / m_{1}

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\sqrt{\frac{m_{1}}{m_{2}}}

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\sqrt{\frac{m_{2}}{m_{1}}}

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Solution

The correct option is D $m_{2} / m_{1}$
Initial momentum of the system is zero i.e. $P_{i} = 0$
Let the velocity acquire by masses $m_{1}$ and $m_{2}$ just after they are released be $v_{1}$ and $v_{2}$ .
Final momentum of the system $P_{f} = m_{1} v_{1} - m_{2} v_{2}$
Using conservation of momentum : $P_{i} = P_{f}$
$∴$ $0 = m_{1} v_{1} - m_{2} v_{2}$
$⟹$ $m_{1} v_{1} = m_{2} v_{2}$ ( $v = \frac{d x}{d t})$
$⟹ m_{1} x_{1} = m_{2} x_{2}$
$⟹$ $\frac{x_{1}}{x_{2}} = \frac{m_{2}}{m_{1}}$

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A spring is compressed between two blocks of masses m1 and m2 placed on a horizontal frictionless surface. When the blocks are released, the blocks travel distance x1 and x2 respectively before coming to rest. The ratio x1/x2 is:

A spring is compressed between two blocks of masses $m_{1}$ and $m_{2}$ placed on a horizontal frictionless surface. When the blocks are released, the blocks travel distance $x_{1}$ and $x_{2}$ respectively before coming to rest. The ratio $x_{1} / x_{2}$ is: