Question

show that the amount of kinetic energy always decreases in perfectly inelastic collision in One dimension

Answer 1

For explanation , we take two bodies of mass m₁ and m₂ are moving with speed u₁ and u₂ ,after perfectly inelastic collision bodies collapse and formed a single body as shown in attachment , speed of single body is v and mass of it is (m₁ + m₂ )

Here it is clear that there is no external force act on bodies , so linear momentum of system of bodies will be conserved .
e.g., initial momentum = final momentum
m₁u₁ + m₂u₂ = (m₁ + m₂)v
v = (m₁u₁ + m₂u₂)/(m₁ + m₂) ------(1)

Now, use energy conservation theorem,
Change in kinetic energy = Kf - ki
final kinetic energy , Kf = 1/2 (m₁ + m₂) [(m₁u₁ + m₂u₂)/(m₁ + m₂)]² = 1/2(m₁u₁ + m₂u₂)²/(m₁ + m₂)
initial kinetic energy , ki = 1/2m₁u₁² + 1/2m₂u₂²
Now, change in kinetic energy , ∆K = 1/2 (m₁u₁ + m₂u₂)²/(m₁ + m₂) - 1/2m₁u₁² - 1/2m₂u₂²
= 1/{2(m₁ + m₂)} [m₁²u₁² + m₂²u₂² + 2m₁m₂u₁u₂ - m₁²u₁² - m₁m₂u₁² - m₂²u₂²- m₁m₂u₂²]
= 1/2 {m₁m₂/(m₁ + m₂)}[ 2u₁u₂ - u₁² - u₂² ]
= -1/2 {(m₁m₂/(m₁ + m₂)} [u₁ - u₂ ]² ⇒ negative

Here you can see that change in kinetic energy is negative , hence it is clear that amount of kinetic energy always decreases in perfectly inelastic collision in one dimension .

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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