A spring is compressed between two toy carts of masses $m_{1}$ and $m_{2}$ . When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If $V_{1}$ and $V_{2}$ are the velocities of the toy carts and there is no friction between the toy carts and the ground, then:

\frac{V_{1}}{V_{1}} = \frac{m_{1}}{m_{2}}

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\frac{V_{1}}{V_{2}} = \frac{m_{2}}{m_{1}}

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\frac{V_{1}}{V_{2}} = - \frac{m_{2}}{m_{1}}

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\frac{V_{1}}{V_{2}} = - \frac{m_{1}}{m_{2}}

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Solution

The correct option is C $\frac{V_{1}}{V_{2}} = - \frac{m_{2}}{m_{1}}$
Here, $m_{1} \frac{d V_{1}}{d t} = - m_{2} \frac{d V_{2}}{d t}$
or $m_{1} a_{1} = m_{2} a_{2} . . . (1)$
now, $V_{1} = a_{1} t$ and $V_{2} = a_{2} t$ ( by using $v = u + a t$ )
from (1), $m_{1} \frac{V_{1}}{t} = - m_{2} \frac{V_{2}}{t}$
$∴ \frac{V_{1}}{V_{2}} = - \frac{m_{2}}{m_{1}}$

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