Question

In inelastic collision the loss of kinetic energy is equal to

Answer 1

Answer:

An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction.

In collisions of macroscopic bodies, some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect, and the bodies are deformed.

The molecules of a gas or liquid rarely experience perfectly elastic collisions because kinetic energy is exchanged between the molecules' translational motion and their internal degrees of freedom with each collision. At any one instant, half the collisions are – to a varying extent – inelastic (the pair possesses less kinetic energy after the collision than before), and half could be described as “super-elastic” (possessing more kinetic energy after the collision than before). Averaged across an entire sample, molecular collisions are elastic.[citation needed]

Although inelastic collisions do not conserve kinetic energy, they do obey conservation of momentum.[1] Simple ballistic pendulum problems obey the conservation of kinetic energy only when the block swings to its largest angle.

Answer 2

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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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