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A spring is compressed between two toy carts of mass $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$. 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 $\mathrm{v}_{2}$ are the velocities of the toy carts and there is no friction between the toy carts and the ground, then :
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The correct answer is:
$\mathrm{v}_{1} / \mathrm{v}_{2}=-\mathrm{m}_{2} / \mathrm{m}_{1}$
Applying law of conservation of linear momentum
$\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=0$
$\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=-\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$ or $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=-\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$
$\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=0$
$\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=-\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}$ or $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=-\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$
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