Question

show that there is always loss of kinetic energy during inelastic collision​

Answer 1

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From conservation of momentum

${m}_{1 }{v }_{1}$

$= ({m}_{1} +{m }_{2} ){v}-{2}→{v}_{2}$

$= \frac{{m}_{1}}{{m}_{1} +{m}_{2}}×{ v }_{1}$

The ratio of kinetic energies before & after collision is

$\frac{{KE}_{f}}{{KE}_{i}}$

$= \frac{\frac{1}{2}×({m}_{1}+{m}_{2}) × (\frac{{m}_{1}}{{m}_{1} +{m}_{2}}×{ v }_{1})²}{\frac{1}{2}×{m}_{1}{v²}_{1}}$

$= \frac{{m}_{1}}{{m}_{1} +{m}_{2}}{×}{ v }_{1}$

The fraction of kinetic energy lost is

$\frac{{KE}_{i} – {KE}_{f}}{{KE}_{i}}$

$= \frac{1 –( \frac{{m}_{1}}{{m}_{1} +{m}_{2}})×{ v }_{1}}{{KE}_{i}} × {KE}_{i}$

$= \frac{{m}_{2}}{{m}_{1} +{m}_{2}}×{ v }_{1}$

Hence energy always loss in inelastic collision.

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