A spring is compressed between two blocks of masses m_1 and m_2 p… - Vidyadip

MCQ

A spring is compressed between two blocks of masses $m_1$ and $m_2$ placed on a horizontal frictionless surface as shown in the figure. When the blocks arc released, they have initial velocity of $v_1$ and $v_2$ as shown. The blocks travel distances $x_1$ and $x_2$ respectively before coming to rest. The ratio $\left( {\frac{{{x_1}}}{{{x_2}}}} \right)$ is

✓
$\frac{{{m_2}}}{{{m_1}}}$
B
$\frac{{{m_1}}}{{{m_2}}}$
C
$\sqrt {\frac{{{m_2}}}{{{m_1}}}} $
D
$\sqrt {\frac{{{m_1}}}{{{m_2}}}} $

✓

Answer

Correct option: A.

$\frac{{{m_2}}}{{{m_1}}}$

a
Initial momentum of the system is zero i.e. $\mathrm{P}_{\mathrm{i}}=0$

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 $\mathrm{P}_{\mathrm{f}}=\mathrm{m}_{1} \mathrm{v}_{1}-\mathrm{m}_{2} \mathrm{v}_{2}$

Using conservation of momentum: $P_{i}=P_{f}$

$\therefore 0=m_{1} v_{1}-m_{2} v_{2}$

$\Rightarrow m_{1} v_{1}=m_{2} v_{2}$ $\left(v=\frac{d x}{d t}\right)$

$\Rightarrow m_{1} x_{1}=m_{2} x_{2}$

$\Rightarrow \frac{x_{1}}{x_{2}}=\frac{m_{2}}{m_{1}}$

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